test.jsonl

{"id": "Rudin|exercise_1_1b", "formal_statement": "theorem exercise_1_1b\n(x : \u211d)\n(y : \u211a)\n(h : y \u2260 0)\n: ( irrational x ) -> irrational ( x * y ) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $r$ is rational $(r \\neq 0)$ and $x$ is irrational, prove that $rx$ is irrational.", "nl_proof": "\\begin{proof}\n\n    If $r x$ were rational, then $x=\\frac{r x}{r}$ would also be rational.\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_4", "formal_statement": "theorem exercise_1_4\n(\u03b1 : Type*) [partial_order \u03b1]\n(s : set \u03b1)\n(x y : \u03b1)\n(h\u2080 : set.nonempty s)\n(h\u2081 : x \u2208 lower_bounds s)\n(h\u2082 : y \u2208 upper_bounds s)\n: x \u2264 y :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $E$ be a nonempty subset of an ordered set; suppose $\\alpha$ is a lower bound of $E$ and $\\beta$ is an upper bound of $E$. Prove that $\\alpha \\leq \\beta$.", "nl_proof": "\\begin{proof}\n\nSince $E$ is nonempty, there exists $x \\in E$. Then by definition of lower and upper bounds we have $\\alpha \\leq x \\leq \\beta$, and hence by property $i i$ in the definition of an ordering, we have $\\alpha<\\beta$ unless $\\alpha=x=\\beta$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_8", "formal_statement": "theorem exercise_1_8 : \u00ac \u2203 (r : \u2102 \u2192 \u2102 \u2192 Prop), is_linear_order \u2102 r :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that no order can be defined in the complex field that turns it into an ordered field.", "nl_proof": "\\begin{proof}\n\n    By Part (a) of Proposition $1.18$, either $i$ or $-i$ must be positive. Hence $-1=i^2=(-i)^2$ must be positive. But then $1=(-1)^2$, must also be positive, and this contradicts Part $(a)$ of Proposition 1.18, since 1 and $-1$ cannot both be positive.\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_12", "formal_statement": "theorem exercise_1_12 (n : \u2115) (f : \u2115 \u2192 \u2102) : \n  abs (\u2211 i in finset.range n, f i) \u2264 \u2211 i in finset.range n, abs (f i) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $z_1, \\ldots, z_n$ are complex, prove that $|z_1 + z_2 + \\ldots + z_n| \\leq |z_1| + |z_2| + \\cdots + |z_n|$.", "nl_proof": "\\begin{proof}\n\n    We can apply the case $n=2$ and induction on $n$ to get\n\n$$\n\n\\begin{aligned}\n\n\\left|z_1+z_2+\\cdots z_n\\right| &=\\left|\\left(z_1+z_2+\\cdots+z_{n-1}\\right)+z_n\\right| \\\\\n\n& \\leq\\left|z_1+z_2+\\cdots+z_{n-1}\\right|+\\left|z_n\\right| \\\\\n\n& \\leq\\left|z_1\\right|+\\left|z_2\\right|+\\cdots+\\left|z_{n-1}\\right|+\\left|z_n\\right|\n\n\\end{aligned}\n\n$$\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_14", "formal_statement": "theorem exercise_1_14\n  (z : \u2102) (h : abs z = 1)\n  : (abs (1 + z)) ^ 2 + (abs (1 - z)) ^ 2 = 4 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $z$ is a complex number such that $|z|=1$, that is, such that $z \\bar{z}=1$, compute $|1+z|^{2}+|1-z|^{2}$.", "nl_proof": "\\begin{proof}\n\n    $|1+z|^2=(1+z)(1+\\bar{z})=1+\\bar{z}+z+z \\bar{z}=2+z+\\bar{z}$. Similarly $|1-z|^2=(1-z)(1-\\bar{z})=1-z-\\bar{z}+z \\bar{z}=2-z-\\bar{z}$. Hence\n\n$$\n\n|1+z|^2+|1-z|^2=4 \\text {. }\n\n$$\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_17", "formal_statement": "theorem exercise_1_17\n  (n : \u2115)\n  (x y : euclidean_space \u211d (fin n)) -- R^n\n  : \u2016x + y\u2016^2 + \u2016x - y\u2016^2 = 2*\u2016x\u2016^2 + 2*\u2016y\u2016^2 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that $|\\mathbf{x}+\\mathbf{y}|^{2}+|\\mathbf{x}-\\mathbf{y}|^{2}=2|\\mathbf{x}|^{2}+2|\\mathbf{y}|^{2}$ if $\\mathbf{x} \\in R^{k}$ and $\\mathbf{y} \\in R^{k}$.", "nl_proof": "\\begin{proof}\n\n    The proof is a routine computation, using the relation\n\n$$\n\n|x \\pm y|^2=(x \\pm y) \\cdot(x \\pm y)=|x|^2 \\pm 2 x \\cdot y+|y|^2 .\n\n$$\n\nIf $\\mathrm{x}$ and $\\mathrm{y}$ are the sides of a parallelogram, then $\\mathrm{x}+\\mathrm{y}$ and $\\mathbf{x}-\\mathrm{y}$ are its diagonals. Hence this result says that the sum of the squares on the diagonals of a parallelogram equals the sum of the squares on the sides.\n\n\\end{proof}"}
{"id": "Rudin|exercise_1_18b", "formal_statement": "theorem exercise_1_18b\n  : \u00ac \u2200 (x : \u211d), \u2203 (y : \u211d), y \u2260 0 \u2227 x * y = 0 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $k = 1$ and $\\mathbf{x} \\in R^{k}$, prove that there does not exist $\\mathbf{y} \\in R^{k}$ such that $\\mathbf{y} \\neq 0$ but $\\mathbf{x} \\cdot \\mathbf{y}=0$", "nl_proof": "\\begin{proof}\n\n    Not true when $k=1$, since the product of two nonzero real numbers is nonzero.\n\n\\end{proof}"}
{"id": "Rudin|exercise_2_19a", "formal_statement": "theorem exercise_2_19a {X : Type*} [metric_space X]\n  (A B : set X) (hA : is_closed A) (hB : is_closed B) (hAB : disjoint A B) :\n  separated_nhds A B :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $A$ and $B$ are disjoint closed sets in some metric space $X$, prove that they are separated.", "nl_proof": "\\begin{proof}\n\n    We are given that $A \\cap B=\\varnothing$. Since $A$ and $B$ are closed, this means $A \\cap \\bar{B}=\\varnothing=\\bar{A} \\cap B$, which says that $A$ and $B$ are separated.\n\n\\end{proof}"}
{"id": "Rudin|exercise_2_25", "formal_statement": "theorem exercise_2_25 {K : Type*} [metric_space K] [compact_space K] :\n  \u2203 (B : set (set K)), set.countable B \u2227 is_topological_basis B :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that every compact metric space $K$ has a countable base.", "nl_proof": "\\begin{proof}\n\n    $K$ can be covered by a finite union of neighborhoods of radius $1 / n$, and this shows that this implies that $K$ is separable.\n\n\n\nIt is not entirely obvious that a metric space with a countable base is separable. To prove this, let $\\left\\{V_n\\right\\}_{n=1}^{\\infty}$ be a countable base, and let $x_n \\in V_n$. The points $V_n$ must be dense in $X$. For if $G$ is any non-empty open set, then $G$ contains $V_n$ for some $n$, and hence $x_n \\in G$. (Thus for a metric space, having a countable base and being separable are equivalent.)\n\n\\end{proof}"}
{"id": "Rudin|exercise_2_27b", "formal_statement": "theorem exercise_2_27b (k : \u2115) (E P : set (euclidean_space \u211d (fin k)))\n  (hE : E.nonempty \u2227 \u00ac set.countable E)\n  (hP : P = {x | \u2200 U \u2208 \ud835\udcdd x, (P \u2229 E).nonempty \u2227 \u00ac set.countable (P \u2229 E)}) :\n  set.countable (E \\ P) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $E\\subset\\mathbb{R}^k$ is uncountable, and let $P$ be the set of condensation points of $E$. Prove that at most countably many points of $E$ are not in $P$.", "nl_proof": "\\begin{proof}\n\n    If $x \\in W^c$, and $O$ is any neighborhood of $x$, then $x \\in V_n \\subseteq O$ for some n. Since $x \\notin W, V_n \\cap E$ is uncountable. Hence $O$ contains uncountably many points of $E$, and so $x$ is a condensation point of $E$. Thus $x \\in P$, i.e., $W^c \\subseteq P$.\n\nConversely if $x \\in W$, then $x \\in V_n$ for some $V_n$ such that $V_n \\cap E$ is countable. Hence $x$ has a neighborhood (any neighborhood contained in $V_n$ ) containing at most a countable set of points of $E$, and so $x \\notin P$, i.e., $W \\subseteq P^c$. Hence $P=W^c$.\n\nIt is clear that $P$ is closed (since its complement $W$ is open), so that we need only show that $P \\subseteq P^{\\prime}$. Hence suppose $x \\in P$, and $O$ is any neighborhood of $x$. (By definition of $P$ this means $O \\cap E$ is uncountable.) We need to show that there is a point $y \\in P \\cap(O \\backslash\\{x\\})$. If this is not the case, i.e., if every point $y$ in $O \\backslash\\{x\\}$ is in $P^c$, then for each such point $y$ there is a set $V_n$ containing $y$ such that $V_n \\cap E$ is at most countable. That would mean that $y \\in W$, i.e., that $O \\backslash\\{x\\}$ is contained in $W$. It would follow that $O \\cap E \\subseteq\\{x\\} \\cup(W \\cap E)$, and so $O \\cap E$ contains at most a countable set of points, contrary to the hypothesis that $x \\in P$. Hence $O$ contains a point of $P$ different from $x$, and so $P \\subseteq P^{\\prime}$. Thus $P$ is perfect.\n\n\\end{proof}"}
{"id": "Rudin|exercise_2_29", "formal_statement": "theorem exercise_2_29 (U : set \u211d) (hU : is_open U) :\n  \u2203 (f : \u2115 \u2192 set \u211d), (\u2200 n, \u2203 a b : \u211d, f n = {x | a < x \u2227 x < b}) \u2227 (\u2200 n, f n \u2286 U) \u2227\n  (\u2200 n m, n \u2260 m \u2192 f n \u2229 f m = \u2205) \u2227\n  U = \u22c3 n, f n :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that every open set in $\\mathbb{R}$ is the union of an at most countable collection of disjoint segments.", "nl_proof": "\\begin{proof}\n\n    Let $O$ be open. For each pair of points $x \\in O, y \\in O$, we define an equivalence relation $x \\sim y$ by saying $x \\sim y$ if and only if $[\\min (x, y), \\max (x, y)] \\subset$ 0 . This is an equivalence relation, since $x \\sim x([x, x] \\subset O$ if $x \\in O)$; if $x \\sim y$, then $y \\sim x$ (since $\\min (x, y)=\\min (y, x)$ and $\\max (x, y)=\\max (y, x))$; and if $x \\sim y$ and $y \\sim z$, then $x \\sim z([\\min (x, z), \\max (x, z)] \\subseteq[\\min (x, y), \\max (x, y)] \\cup$ $[\\min (y, z), \\max (y, z)] \\subseteq O)$. In fact it is easy to prove that\n\n$$\n\n\\min (x, z) \\geq \\min (\\min (x, y), \\min (y, z))\n\n$$\n\nand\n\n$$\n\n\\max (x, z) \\leq \\max (\\max (x, y), \\max (y, z))\n\n$$\n\nIt follows that $O$ can be written as a disjoint union of pairwise disjoint equivalence classes. We claim that each equivalence class is an open interval.\n\n\n\nTo show this, for each $x \\in O$; let $A=\\{z:[z, x] \\subseteq O\\}$ and $B=\\{z:[x, z] \\subseteq$ $O\\}$, and let $a=\\inf A, b=\\sup B$. We claim that $(a, b) \\subset O$. Indeed if $a<z<b$, there exists $c \\in A$ with $c<z$ and $d \\in B$ with $d>z$. Then $z \\in[c, x] \\cup[x, d] \\subseteq O$. We now claim that $(a, b)$ is the equivalence class containing $x$. It is clear that each element of $(a, b)$ is equivalent to $x$ by the way in which $a$ and $b$ were chosen. We need to show that if $z \\notin(a, b)$, then $z$ is not equivalent to $x$. Suppose that $z<a$. If $z$ were equivalent to $x$, then $[z, x]$ would be contained in $O$, and so we would have $z \\in A$. Hence $a$ would not be a lower bound for $A$. Similarly if $z>b$ and $z \\sim x$, then $b$ could not be an upper bound for $B$.\n\n\n\nWe have now established that $O$ is a union of pairwise disjoint open intervals. Such a union must be at most countable, since each open interval contains a rational number not in any other interval.\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_2a", "formal_statement": "theorem exercise_3_2a\n  : tendsto (\u03bb (n : \u211d), (sqrt (n^2 + n) - n)) at_top (\ud835\udcdd (1/2)) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that $\\lim_{n \\rightarrow \\infty}\\sqrt{n^2 + n} -n = 1/2$.", "nl_proof": "\\begin{proof}\n\n    Multiplying and dividing by $\\sqrt{n^2+n}+n$ yields\n\n$$\n\n\\sqrt{n^2+n}-n=\\frac{n}{\\sqrt{n^2+n}+n}=\\frac{1}{\\sqrt{1+\\frac{1}{n}}+1} .\n\n$$\n\nIt follows that the limit is $\\frac{1}{2}$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_5", "formal_statement": "theorem exercise_3_5 -- TODO fix\n  (a b : \u2115 \u2192 \u211d)\n  (h : limsup a + limsup b \u2260 0) :\n  limsup (\u03bb n, a n + b n) \u2264 limsup a + limsup b :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "For any two real sequences $\\left\\{a_{n}\\right\\},\\left\\{b_{n}\\right\\}$, prove that $\\limsup _{n \\rightarrow \\infty}\\left(a_{n}+b_{n}\\right) \\leq \\limsup _{n \\rightarrow \\infty} a_{n}+\\limsup _{n \\rightarrow \\infty} b_{n},$ provided the sum on the right is not of the form $\\infty-\\infty$.", "nl_proof": "\\begin{proof}\n\n    Since the case when $\\limsup _{n \\rightarrow \\infty} a_n=+\\infty$ and $\\limsup _{n \\rightarrow \\infty} b_n=-\\infty$ has been excluded from consideration, we note that the inequality is obvious if $\\limsup _{n \\rightarrow \\infty} a_n=+\\infty$. Hence we shall assume that $\\left\\{a_n\\right\\}$ is bounded above.\n\n\n\nLet $\\left\\{n_k\\right\\}$ be a subsequence of the positive integers such that $\\lim _{k \\rightarrow \\infty}\\left(a_{n_k}+\\right.$ $\\left.b_{n_k}\\right)=\\limsup _{n \\rightarrow \\infty}\\left(a_n+b_n\\right)$. Then choose a subsequence of the positive integers $\\left\\{k_m\\right\\}$ such that\n\n$$\n\n\\lim _{m \\rightarrow \\infty} a_{n_{k_m}}=\\limsup _{k \\rightarrow \\infty} a_{n_k} .\n\n$$\n\nThe subsequence $a_{n_{k_m}}+b_{n_{k_m}}$ still converges to the same limit as $a_{n_k}+b_{n_k}$, i.e., to $\\limsup _{n \\rightarrow \\infty}\\left(a_n+b_n\\right)$. Hence, since $a_{n_k}$ is bounded above (so that $\\limsup _{k \\rightarrow \\infty} a_{n_k}$ is finite), it follows that $b_{n_{k_m}}$ converges to the difference\n\n$$\n\n\\lim _{m \\rightarrow \\infty} b_{n_{k_m}}=\\lim _{m \\rightarrow \\infty}\\left(a_{n_{k_m}}+b_{n_{k_m}}\\right)-\\lim _{m \\rightarrow \\infty} a_{n_{k_m}} .\n\n$$\n\nThus we have proved that there exist subsequences $\\left\\{a_{n_{k_m}}\\right\\}$ and $\\left\\{b_{n_{k_m}}\\right\\}$ which converge to limits $a$ and $b$ respectively such that $a+b=\\limsup _{n \\rightarrow \\infty}\\left(a_n+b_n^*\\right)$. Since $a$ is the limit of a subsequence of $\\left\\{a_n\\right\\}$ and $b$ is the limit of a subsequence of $\\left\\{b_n\\right\\}$, it follows that $a \\leq \\limsup _{n \\rightarrow \\infty} a_n$ and $b \\leq \\limsup _{n \\rightarrow \\infty} b_n$, from which the desired inequality follows.\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_7", "formal_statement": "theorem exercise_3_7\n  (a : \u2115 \u2192 \u211d)\n  (h : \u2203 y, (tendsto (\u03bb n, (\u2211 i in (finset.range n), a i)) at_top (\ud835\udcdd y))) :\n  \u2203 y, tendsto (\u03bb n, (\u2211 i in (finset.range n), sqrt (a i) / n)) at_top (\ud835\udcdd y) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that the convergence of $\\Sigma a_{n}$ implies the convergence of $\\sum \\frac{\\sqrt{a_{n}}}{n}$ if $a_n\\geq 0$.", "nl_proof": "\\begin{proof}\n\n    Since $\\left(\\sqrt{a_n}-\\frac{1}{n}\\right)^2 \\geq 0$, it follows that\n\n$$\n\n\\frac{\\sqrt{a_n}}{n} \\leq \\frac{1}{2}\\left(a_n^2+\\frac{1}{n^2}\\right) .\n\n$$\n\nNow $\\Sigma a_n^2$ converges by comparison with $\\Sigma a_n$ (since $\\Sigma a_n$ converges, we have $a_n<1$ for large $n$, and hence $\\left.a_n^2<a_n\\right)$. Since $\\Sigma \\frac{1}{n^2}$ also converges ($p$ series, $p=2$ ), it follows that $\\Sigma \\frac{\\sqrt{a_n}}{n}$ converges.\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_13", "formal_statement": "theorem exercise_3_13\n  (a b : \u2115 \u2192 \u211d)\n  (ha : \u2203 y, (tendsto (\u03bb n, (\u2211 i in (finset.range n), |a i|)) at_top (\ud835\udcdd y)))\n  (hb : \u2203 y, (tendsto (\u03bb n, (\u2211 i in (finset.range n), |b i|)) at_top (\ud835\udcdd y))) :\n  \u2203 y, (tendsto (\u03bb n, (\u2211 i in (finset.range n),\n  \u03bb i, (\u2211 j in finset.range (i + 1), a j * b (i - j)))) at_top (\ud835\udcdd y)) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that the Cauchy product of two absolutely convergent series converges absolutely.", "nl_proof": "\\begin{proof}\n\n    Since both the hypothesis and conclusion refer to absolute convergence, we may assume both series consist of nonnegative terms. We let $S_n=\\sum_{k=0}^n a_n, T_n=\\sum_{k=0}^n b_n$, and $U_n=\\sum_{k=0}^n \\sum_{l=0}^k a_l b_{k-l}$. We need to show that $U_n$ remains bounded, given that $S_n$ and $T_n$ are bounded. To do this we make the convention that $a_{-1}=T_{-1}=0$, in order to save ourselves from having to separate off the first and last terms when we sum by parts. We then have\n\n$$\n\n\\begin{aligned}\n\nU_n &=\\sum_{k=0}^n \\sum_{l=0}^k a_l b_{k-l} \\\\\n\n&=\\sum_{k=0}^n \\sum_{l=0}^k a_l\\left(T_{k-l}-T_{k-l-1}\\right) \\\\\n\n&=\\sum_{k=0}^n \\sum_{j=0}^k a_{k-j}\\left(T_j-T_{j-1}\\right) \\\\\n\n&=\\sum_{k=0}^n \\sum_{j=0}^k\\left(a_{k-j}-a_{k-j-1}\\right) T_j \\\\\n\n&=\\sum_{j=0}^n \\sum_{k=j}^n\\left(a_{k-j}-a_{k-j-1}\\right) T_j\n\n&=\\sum_{j=0}^n a_{n-j} T_j \\\\\n\n&\\leq T \\sum_{m=0}^n a_m \\\\\n\n&=T S_n \\\\\n\n&\\leq S T .\n\n\\end{aligned}\n\n$$\n\nThus $U_n$ is bounded, and hence approaches a finite limit.\n\n\n\n\\end{proof}"}
{"id": "Rudin|exercise_3_21", "formal_statement": "theorem exercise_3_21\n  {X : Type*} [metric_space X] [complete_space X]\n  (E : \u2115 \u2192 set X)\n  (hE : \u2200 n, E n \u2283 E (n + 1))\n  (hE' : tendsto (\u03bb n, metric.diam (E n)) at_top (\ud835\udcdd 0)) :\n  \u2203 a, set.Inter E = {a} :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "If $\\left\\{E_{n}\\right\\}$ is a sequence of closed nonempty and bounded sets in a complete metric space $X$, if $E_{n} \\supset E_{n+1}$, and if $\\lim _{n \\rightarrow \\infty} \\operatorname{diam} E_{n}=0,$ then $\\bigcap_{1}^{\\infty} E_{n}$ consists of exactly one point.", "nl_proof": "\\begin{proof}\n\n    Choose $x_n \\in E_n$. (We use the axiom of choice here.) The sequence $\\left\\{x_n\\right\\}$ is a Cauchy sequence, since the diameter of $E_n$ tends to zero as $n$ tends to infinity and $E_n$ contains $E_{n+1}$. Since the metric space $X$ is complete, the sequence $x_n$ converges to a point $x$, which must belong to $E_n$ for all $n$, since $E_n$ is closed and contains $x_m$ for all $m \\geq n$. There cannot be a second point $y$ in all of the $E_n$, since for any point $y \\neq x$ the diameter of $E_n$ is less $\\operatorname{than} d(x, y)$ for large $n$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_1a", "formal_statement": "theorem exercise_4_1a\n  : \u2203 (f : \u211d \u2192 \u211d), (\u2200 (x : \u211d), tendsto (\u03bb y, f(x + y) - f(x - y)) (\ud835\udcdd 0) (\ud835\udcdd 0)) \u2227 \u00ac continuous f :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f$ is a real function defined on $\\mathbb{R}$ which satisfies $\\lim_{h \\rightarrow 0} f(x + h) - f(x - h) = 0$ for every $x \\in \\mathbb{R}$. Show that $f$ does not need to be continuous.", "nl_proof": "\\begin{proof}\n\n    $$\n\nf(x)= \\begin{cases}1 & \\text { if } x \\text { is an integer } \\\\ 0 & \\text { if } x \\text { is not an integer. }\\end{cases}\n\n$$\n\n(If $x$ is an integer, then $f(x+h)-f(x-h) \\equiv 0$ for all $h$; while if $x$ is not an integer, $f(x+h)-f(x-h)=0$ for $|h|<\\min (x-[x], 1+[x]-x)$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_3", "formal_statement": "theorem exercise_4_3\n  {\u03b1 : Type} [metric_space \u03b1]\n  (f : \u03b1 \u2192 \u211d) (h : continuous f) (z : set \u03b1) (g : z = f\u207b\u00b9' {0})\n  : is_closed z :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $f$ be a continuous real function on a metric space $X$. Let $Z(f)$ (the zero set of $f$ ) be the set of all $p \\in X$ at which $f(p)=0$. Prove that $Z(f)$ is closed.", "nl_proof": "\\begin{proof}\n\n    $Z(f)=f^{-1}(\\{0\\})$, which is the inverse image of a closed set. Hence $Z(f)$ is closed.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_4b", "formal_statement": "theorem exercise_4_4b\n  {\u03b1 : Type} [metric_space \u03b1]\n  {\u03b2 : Type} [metric_space \u03b2]\n  (f g : \u03b1 \u2192 \u03b2)\n  (s : set \u03b1)\n  (h\u2081 : continuous f)\n  (h\u2082 : continuous g)\n  (h\u2083 : dense s)\n  (h\u2084 : \u2200 x \u2208 s, f x = g x)\n  : f = g :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that if $g(p) = f(p)$ for all $p \\in P$ then $g(p) = f(p)$ for all $p \\in X$.", "nl_proof": "\\begin{proof}\n\n    The function $\\varphi: X \\rightarrow R^1$ given by\n\n$$\n\n\\varphi(p)=d_Y(f(p), g(p))\n\n$$\n\nis continuous, since\n\n$$\n\n\\left|d_Y(f(p), g(p))-d_Y(f(q), g(q))\\right| \\leq d_Y(f(p), f(q))+d_Y(g(p), g(q))\n\n$$\n\n(This inequality follows from the triangle inequality, since\n\n$$\n\nd_Y(f(p), g(p)) \\leq d_Y(f(p), f(q))+d_Y(f(q), g(q))+d_Y(g(q), g(p)),\n\n$$\n\nand the same inequality holds with $p$ and $q$ interchanged. The absolute value $\\left|d_Y(f(p), g(p))-d_Y(f(q), g(q))\\right|$ must be either $d_Y(f(p), g(p))-d_Y(f(q), g(q))$ or $d_Y(f(q), g(q))-d_Y(f(p), g(p))$, and the triangle inequality shows that both of these numbers are at most $d_Y(f(p), f(q))+d_Y(g(p), g(q))$.)\n\nBy the previous problem, the zero set of $\\varphi$ is closed. But by definition\n\n$$\n\nZ(\\varphi)=\\{p: f(p)=g(p)\\} .\n\n$$\n\nHence the set of $p$ for which $f(p)=g(p)$ is closed. Since by hypothesis it is dense, it must be $X$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_5b", "formal_statement": "theorem exercise_4_5b\n  : \u2203 (E : set \u211d) (f : \u211d \u2192 \u211d), (continuous_on f E) \u2227\n  (\u00ac \u2203 (g : \u211d \u2192 \u211d), continuous g \u2227 \u2200 x \u2208 E, f x = g x) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Show that there exist a set $E \\subset \\mathbb{R}$ and a real continuous function $f$ defined on $E$, such that there does not exist a continuous real function $g$ on $\\mathbb{R}$ such that $g(x)=f(x)$ for all $x \\in E$.", "nl_proof": "\\begin{proof}\n\n    Let $E:=(0,1)$, and define $f(x):=1 / x$ for all $x \\in E$. If $f$ has a continuous extension $g$ to $\\mathbb{R}$, then $g$ is continuous on $[-1,1]$, and therefore bounded by the intermediate value theorem. However, $f$ is not bounded in any neighborhood of $x=0$, so therefore $g$ is not bounded either, a contradiction.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_8a", "formal_statement": "theorem exercise_4_8a\n  (E : set \u211d) (f : \u211d \u2192 \u211d) (hf : uniform_continuous_on f E)\n  (hE : metric.bounded E) : metric.bounded (set.image f E) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $f$ be a real uniformly continuous function on the bounded set $E$ in $R^{1}$. Prove that $f$ is bounded on $E$.", "nl_proof": "\\begin{proof}\n\n    Let $a=\\inf E$ and $b=\\sup E$, and let $\\delta>0$ be such that $|f(x)-f(y)|<1$ if $x, y \\in E$ and $|x-y|<\\delta$. Now choose a positive integer $N$ larger than $(b-a) / \\delta$, and consider the $N$ intervals $I_k=\\left[a+\\frac{k-1}{b-a}, a+\\frac{k}{b-a}\\right], k=1,2, \\ldots, N$. For each $k$ such that $I_k \\cap E \\neq \\varnothing$ let $x_k \\in E \\cap I_k$. Then let $M=1+\\max \\left\\{\\left|f\\left(x_k\\right)\\right|\\right\\}$. If $x \\in E$, we have $\\left|x-x_k\\right|<\\delta$ for some $k$, and hence $|f(x)|<M$.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_11a", "formal_statement": "theorem exercise_4_11a\n  {X : Type*} [metric_space X]\n  {Y : Type*} [metric_space Y]\n  (f : X \u2192 Y) (hf : uniform_continuous f)\n  (x : \u2115 \u2192 X) (hx : cauchy_seq x) :\n  cauchy_seq (\u03bb n, f (x n)) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$ and prove that $\\left\\{f\\left(x_{n}\\right)\\right\\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\\{x_n\\}$ in $X$.", "nl_proof": "\\begin{proof}\n\n    Suppose $\\left\\{x_n\\right\\}$ is a Cauchy sequence in $X$. Let $\\varepsilon>0$ be given. Let $\\delta>0$ be such that $d_Y(f(x), f(u))<\\varepsilon$ if $d_X(x, u)<\\delta$. Then choose $N$ so that $d_X\\left(x_n, x_m\\right)<\\delta$ if $n, m>N$. Obviously $d_Y\\left(f\\left(x_n\\right), f\\left(x_m\\right)\\right)<\\varepsilon$ if $m, n>N$, showing that $\\left\\{f\\left(x_n\\right)\\right\\}$ is a Cauchy sequence.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_15", "formal_statement": "theorem exercise_4_15 {f : \u211d \u2192 \u211d}\n  (hf : continuous f) (hof : is_open_map f) :\n  monotone f :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Prove that every continuous open mapping of $R^{1}$ into $R^{1}$ is monotonic.", "nl_proof": "\\begin{proof}\n\n    Suppose $f$ is continuous and not monotonic, say there exist points $a<b<c$ with $f(a)<f(b)$, and $f(c)<f(b)$. Then the maximum value of $f$ on the closed interval $[a, c]$ is assumed at a point $u$ in the open interval $(a, c)$. If there is also a point $v$ in the open interval $(a, c)$ where $f$ assumes its minimum value on $[a, c]$, then $f(a, c)=[f(v), f(u)]$. If no such point $v$ exists, then $f(a, c)=(d, f(u)]$, where $d=\\min (f(a), f(c))$. In either case, the image of $(a, c)$ is not open.\n\n\\end{proof}"}
{"id": "Rudin|exercise_4_21a", "formal_statement": "theorem exercise_4_21a {X : Type*} [metric_space X]\n  (K F : set X) (hK : is_compact K) (hF : is_closed F) (hKF : disjoint K F) :\n  \u2203 (\u03b4 : \u211d), \u03b4 > 0 \u2227 \u2200 (p q : X), p \u2208 K \u2192 q \u2208 F \u2192 dist p q \u2265 \u03b4 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $K$ and $F$ are disjoint sets in a metric space $X, K$ is compact, $F$ is closed. Prove that there exists $\\delta>0$ such that $d(p, q)>\\delta$ if $p \\in K, q \\in F$.", "nl_proof": "\\begin{proof}\n\nFollowing the hint, we observe that $\\rho_F(x)$ must attain its minimum value on $K$, i.e., there is some point $r \\in K$ such that\n\n$$\n\n\\rho_F(r)=\\min _{q \\in K} \\rho_F(q) .\n\n$$\n\nSince $F$ is closed and $r \\notin F$, it follows from Exercise $4.20$ that $\\rho_F(r)>0$. Let $\\delta$ be any positive number smaller than $\\rho_F(r)$. Then for any $p \\in F, q \\in K$, we have\n\n$$\n\nd(p, q) \\geq \\rho_F(q) \\geq \\rho_F(r)>\\delta .\n\n$$\n\nThis proves the positive assertion.\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_1", "formal_statement": "theorem exercise_5_1\n  {f : \u211d \u2192 \u211d} (hf : \u2200 x y : \u211d, | (f x - f y) | \u2264 (x - y) ^ 2) :\n  \u2203 c, f = \u03bb x, c :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Let $f$ be defined for all real $x$, and suppose that $|f(x)-f(y)| \\leq (x-y)^{2}$ for all real $x$ and $y$. Prove that $f$ is constant.", "nl_proof": "\\begin{proof}\n\n    Dividing by $x-y$, and letting $x \\rightarrow y$, we find that $f^{\\prime}(y)=0$ for all $y$. Hence $f$ is constant.\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_3", "formal_statement": "theorem exercise_5_3 {g : \u211d \u2192 \u211d} (hg : continuous g)\n  (hg' : \u2203 M : \u211d, \u2200 x : \u211d, | deriv g x | \u2264 M) :\n  \u2203 N, \u2200 \u03b5 > 0, \u03b5 < N \u2192 function.injective (\u03bb x : \u211d, x + \u03b5 * g x) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $g$ is a real function on $R^{1}$, with bounded derivative (say $\\left|g^{\\prime}\\right| \\leq M$ ). Fix $\\varepsilon>0$, and define $f(x)=x+\\varepsilon g(x)$. Prove that $f$ is one-to-one if $\\varepsilon$ is small enough.", "nl_proof": "\\begin{proof}\n\n    If $0<\\varepsilon<\\frac{1}{M}$, we certainly have\n\n$$\n\nf^{\\prime}(x) \\geq 1-\\varepsilon M>0,\n\n$$\n\nand this implies that $f(x)$ is one-to-one, by the preceding problem.\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_5", "formal_statement": "theorem exercise_5_5\n  {f : \u211d \u2192 \u211d}\n  (hfd : differentiable \u211d f)\n  (hf : tendsto (deriv f) at_top (\ud835\udcdd 0)) :\n  tendsto (\u03bb x, f (x + 1) - f x) at_top at_top :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f$ is defined and differentiable for every $x>0$, and $f^{\\prime}(x) \\rightarrow 0$ as $x \\rightarrow+\\infty$. Put $g(x)=f(x+1)-f(x)$. Prove that $g(x) \\rightarrow 0$ as $x \\rightarrow+\\infty$.", "nl_proof": "\\begin{proof}\n\n    Let $\\varepsilon>0$. Choose $x_0$ such that $\\left|f^{\\prime}(x)\\right|<\\varepsilon$ if $x>x_0$. Then for any $x \\geq x_0$ there exists $x_1 \\in(x, x+1)$ such that\n\n$$\n\nf(x+1)-f(x)=f^{\\prime}\\left(x_1\\right) .\n\n$$\n\nSince $\\left|f^{\\prime}\\left(x_1\\right)\\right|<\\varepsilon$, it follows that $|f(x+1)-f(x)|<\\varepsilon$, as required.\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_7", "formal_statement": "theorem exercise_5_7\n  {f g : \u211d \u2192 \u211d} {x : \u211d}\n  (hf' : differentiable_at \u211d f 0)\n  (hg' : differentiable_at \u211d g 0)\n  (hg'_ne_0 : deriv g 0 \u2260 0)\n  (f0 : f 0 = 0) (g0 : g 0 = 0) :\n  tendsto (\u03bb x, f x / g x) (\ud835\udcdd x) (\ud835\udcdd (deriv f x / deriv g x)) :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f^{\\prime}(x), g^{\\prime}(x)$ exist, $g^{\\prime}(x) \\neq 0$, and $f(x)=g(x)=0$. Prove that $\\lim _{t \\rightarrow x} \\frac{f(t)}{g(t)}=\\frac{f^{\\prime}(x)}{g^{\\prime}(x)}.$", "nl_proof": "\\begin{proof}\n\n    Since $f(x)=g(x)=0$, we have\n\n$$\n\n\\begin{aligned}\n\n\\lim _{t \\rightarrow x} \\frac{f(t)}{g(t)} &=\\lim _{t \\rightarrow x} \\frac{\\frac{f(t)-f(x)}{t-x}}{\\frac{g(t)-g(x)}{t-x}} \\\\\n\n&=\\frac{\\lim _{t \\rightarrow x} \\frac{f(t)-f(x)}{t-x}}{\\lim _{t \\rightarrow x} \\frac{g(t)-g(x)}{t-x}} \\\\\n\n&=\\frac{f^{\\prime}(x)}{g^{\\prime}(x)}\n\n\\end{aligned}\n\n$$\n\n\\end{proof}"}
{"id": "Rudin|exercise_5_17", "formal_statement": "theorem exercise_5_17\n  {f : \u211d \u2192 \u211d}\n  (hf' : differentiable_on \u211d f (set.Icc (-1) 1))\n  (hf'' : differentiable_on \u211d (deriv f) (set.Icc 1 1))\n  (hf''' : differentiable_on \u211d (deriv (deriv f)) (set.Icc 1 1))\n  (hf0 : f (-1) = 0)\n  (hf1 : f 0 = 0)\n  (hf2 : f 1 = 1)\n  (hf3 : deriv f 0 = 0) :\n  \u2203 x, x \u2208 set.Ioo (-1 : \u211d) 1 \u2227 deriv (deriv (deriv f)) x \u2265 3 :=", "src_header": "import .common\n\nopen real complex\nopen topological_space\nopen filter\nopen_locale real \nopen_locale topology\nopen_locale big_operators\nopen_locale complex_conjugate\nopen_locale filter\n\n\nnoncomputable theory\n\n", "nl_statement": "Suppose $f$ is a real, three times differentiable function on $[-1,1]$, such that $f(-1)=0, \\quad f(0)=0, \\quad f(1)=1, \\quad f^{\\prime}(0)=0 .$ Prove that $f^{(3)}(x) \\geq 3$ for some $x \\in(-1,1)$.", "nl_proof": "\\begin{proof}\n\n    Following the hint, we observe that Theorem $5.15$ (Taylor's formula with remainder) implies that\n\n$$\n\n\\begin{aligned}\n\nf(1) &=f(0)+f^{\\prime}(0)+\\frac{1}{2} f^{\\prime \\prime}(0)+\\frac{1}{6} f^{(3)}(s) \\\\\n\nf(-1) &=f(0)-f^{\\prime}(0)+\\frac{1}{2} f^{\\prime \\prime}(0)-\\frac{1}{6} f^{(3)}(t)\n\n\\end{aligned}\n\n$$\n\nfor some $s \\in(0,1), t \\in(-1,0)$. By subtracting the second equation from the first and using the given values of $f(1), f(-1)$, and $f^{\\prime}(0)$, we obtain\n\n$$\n\n1=\\frac{1}{6}\\left(f^{(3)}(s)+f^{(3)}(t)\\right),\n\n$$\n\nwhich is the desired result. Note that we made no use of the hypothesis $f(0)=0$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_3b", "formal_statement": "theorem exercise_13_3b : \u00ac \u2200 X : Type, \u2200s : set (set X),\n  (\u2200 t : set X, t \u2208 s \u2192 (set.infinite t\u1d9c \u2228 t = \u2205 \u2228 t = \u22a4)) \u2192 \n  (set.infinite (\u22c3\u2080 s)\u1d9c \u2228 (\u22c3\u2080 s) = \u2205 \u2228 (\u22c3\u2080 s) = \u22a4) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that the collection $$\\mathcal{T}_\\infty = \\{U | X - U \\text{ is infinite or empty or all of X}\\}$$ does not need to be a topology on the set $X$.", "nl_proof": "\\begin{proof}\n\n    Let $X=\\mathbb{R}, U_1=(-\\infty, 0)$ and $U_2=(0, \\infty)$. Then $U_1$ and $U_2$ are in $\\mathcal{T}_{\\infty}$ but $U_1 \\cup U_2=\\mathbb{R} \\backslash\\{0\\}$ is not.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_4a2", "formal_statement": "theorem exercise_13_4a2 :\n  \u2203 (X I : Type*) (T : I \u2192 set (set X)),\n  (\u2200 i, is_topology X (T i)) \u2227 \u00ac  is_topology X (\u22c2 i : I, T i) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "If $\\mathcal{T}_\\alpha$ is a family of topologies on $X$, show that $\\bigcup \\mathcal{T}_\\alpha$ does not need to be a topology on $X$.", "nl_proof": "\\begin{proof}\n\n    On the other hand, the union $\\bigcup_\\alpha \\mathcal{T}_\\alpha$ is in general not a topology on $X$. For instance, let $X=\\{a, b, c\\}$. Then $\\mathcal{T}_1=\\{\\emptyset, X,\\{a\\}\\}$ and $\\mathcal{T}_2=\\{\\emptyset, X,\\{b\\}\\}$ are topologies on $X$ but $\\mathcal{T}_1 \\cup \\mathcal{T}_2=$ $\\{\\emptyset, X,\\{a\\},\\{b\\}\\}$ is not.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_4b2", "formal_statement": "theorem exercise_13_4b2 (X I : Type*) (T : I \u2192 set (set X)) (h : \u2200 i, is_topology X (T i)) :\n  \u2203! T', is_topology X T' \u2227 (\u2200 i, T' \u2286 T i) \u2227\n  \u2200 T'', is_topology X T'' \u2192 (\u2200 i, T'' \u2286 T i) \u2192 T' \u2286 T'' :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $\\mathcal{T}_\\alpha$ be a family of topologies on $X$. Show that there is a unique largest topology on $X$ contained in all the collections $\\mathcal{T}_\\alpha$.", "nl_proof": "\\begin{proof}\n\n    Now we prove that there exists a unique largest topology contained in all $\\mathcal{T}_\\alpha$. Uniqueness of such topology is clear. Consider $\\mathcal{T}=\\bigcap_\\alpha \\mathcal{T}_\\alpha$. We already know that $\\mathcal{T}$ is a topology by, and clearly $\\mathcal{T} \\subset \\mathcal{T}_\\alpha$ for all $\\alpha$. If $\\mathcal{O}$ is another topology contained in all $\\mathcal{T}_\\alpha$, it must be contained in their intersection, so $\\mathcal{O} \\subset \\mathcal{T}$. I follows that $\\mathcal{T}$ is the unique largest topology contained in all $\\mathcal{T}_\\alpha$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_5b", "formal_statement": "theorem exercise_13_5b {X : Type*}\n  [t : topological_space X] (A : set (set X)) (hA : t = generate_from A) :\n  generate_from A = generate_from (sInter {T | is_topology X T \u2227 A \u2286 T}) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $\\mathcal{A}$ is a subbasis for a topology on $X$, then the topology generated by $\\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\\mathcal{A}$.", "nl_proof": "\\begin{proof}\n\n    If we now considered $\\mathcal{A}$ as a subbasis, then the elements of $\\mathcal{T}$ are union of finite intersections of elements of $\\mathcal{A}$. The inclusion $\\mathcal{O} \\subset \\mathcal{T}$ is again clear and $\\mathcal{T} \\subset \\mathcal{O}$ holds since every union of finite intersections of elements of $\\mathcal{A}$ belongs to $\\mathcal{O}$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_13_8a", "formal_statement": "theorem exercise_13_8a :\n  topological_space.is_topological_basis {S : set \u211d | \u2203 a b : \u211a, a < b \u2227 S = Ioo a b} :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that the collection $\\{(a,b) \\mid a < b, a \\text{ and } b \\text{ rational}\\}$ is a basis that generates the standard topology on $\\mathbb{R}$.", "nl_proof": "\\begin{proof}\n\n    Exercise 13.8. (a) First note that $\\mathcal{B}$ is a basis for a topology on $\\mathbb{R}$. This follows from the fact that the union of its elements is all of $\\mathbb{R}$ and the intersection of two elements of $\\mathcal{B}$ is either empty or another element of $\\mathcal{B}$. Let $\\mathcal{T}$ be the standard topology on $\\mathbb{R}$. Clearly the topology generated by $\\mathcal{B}$ is coarser than $\\mathcal{T}$. Let $U \\in \\mathcal{T}$ and $x \\in U$. Then $U$ contains an open interval with centre $x$. Since the rationals are dense in $\\mathbb{R}$ with the standard topology, there exists $q \\in \\mathbb{Q}$ such that $x \\in(x-q, x+q) \\subset U$. This proves that $\\mathcal{T}$ is coarser than the topology generated by $\\mathcal{B}$. We conclude that $\\mathcal{B}$ generates the standard topology on $\\mathbb{R}$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_16_1", "formal_statement": "theorem exercise_16_1 {X : Type*} [topological_space X]\n  (Y : set X)\n  (A : set Y) :\n  \u2200 U : set A, is_open U \u2194 is_open (subtype.val '' U) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$.", "nl_proof": "\\begin{proof}\n\n    Exercise 16.1. Let $\\mathcal{T}$ be the topology $A$ inherits as a subspace of $Y$, and $\\mathcal{O}$ be the topology it inherits as a subspace of $X$. A (standard) basis element for $\\mathcal{T}$ has the form $U \\cap A$ where $U$ is open in $Y$, so is of the form $(Y \\cap V) \\cap A=V \\cap A$ where $V$ is open in $X$. Therefore every basis element for $\\mathcal{T}$ is also a basis element for $\\mathcal{O}$. Conversely, a (standard) basis element for $\\mathcal{O}$ have the form $W \\cap A=W \\cap Y \\cap A$ where $W$ is open in $X$. Since $W \\cap Y$ is open in $Y$, this is a basis element for $\\mathcal{T}$, so every basis element for $\\mathcal{O}$ is a basis element for $\\mathcal{T}$. It follows that $\\mathcal{T}=\\mathcal{O}$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_16_6", "formal_statement": "theorem exercise_16_6\n  (S : set (set (\u211d \u00d7 \u211d)))\n  (hS : \u2200 s, s \u2208 S \u2192 \u2203 a b c d, (rational a \u2227 rational b \u2227 rational c \u2227 rational d\n  \u2227 s = {x | \u2203 x\u2081 x\u2082, x = (x\u2081, x\u2082) \u2227 a < x\u2081 \u2227 x\u2081 < b \u2227 c < x\u2082 \u2227 x\u2082 < d})) :\n  is_topological_basis S :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that the countable collection \\[\\{(a, b) \\times (c, d) \\mid a < b \\text{ and } c < d, \\text{ and } a, b, c, d \\text{ are rational}\\}\\] is a basis for $\\mathbb{R}^2$.", "nl_proof": "\\begin{proof}\n\n    We know that $\\mathcal{B}=\\{(a,b)|a<b, a \\text{ and } b \\text rational\\}$ is a basis for $\\mathcal{R}$, therefore the set we are concerned with in the above question is a basis for $\\mathcal{R}^2$. \n\n\\end{proof}"}
{"id": "Munkres|exercise_18_8a", "formal_statement": "theorem exercise_18_8a {X Y : Type*} [topological_space X] [topological_space Y]\n  [linear_order Y] [order_topology Y] {f g : X \u2192 Y}\n  (hf : continuous f) (hg : continuous g) :\n  is_closed {x | f x \u2264 g x} :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $Y$ be an ordered set in the order topology. Let $f, g: X \\rightarrow Y$ be continuous. Show that the set $\\{x \\mid f(x) \\leq g(x)\\}$ is closed in $X$.", "nl_proof": "\\begin{proof}\n\n    We prove that $U=\\{x \\mid g(x)<f(x)\\}$ is open in $X$. Let $a \\in U$, so that $g(a)<f(a)$. If there is an element $c$ between $g(a)$ and $f(a)$, then $a \\in g^{-1}((-\\infty, c)) \\cap f^{-1}((c,+\\infty))$. If there are no elements between $g(a)$ and $f(a)$, then $a \\in g^{-1}\\left((-\\infty, f(a)) \\cap f^{-1}((g(a),+\\infty))\\right.$. Note that all these preimages are open since $f$ and $g$ are continuous. Thus $U=V \\cup W$, where\n\n$$\n\nV=\\bigcup_{c \\in X} g^{-1}((-\\infty, c)) \\cap f^{-1}((c,+\\infty)) \\quad \\text { and } \\quad W=\\bigcup_{\\substack{g(a)<f(a) \\\\(g(a), f(a))=\\emptyset}} g^{-1}\\left((-\\infty, f(a)) \\cap f^{-1}((g(a),+\\infty))\\right.\n\n$$\n\nare open in $X$. So $U$ is open in $X$ and therefore $X \\backslash U=\\{x \\mid f(x) \\leq g(x)\\}$ is closed in $X$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_18_13", "formal_statement": "theorem exercise_18_13\n  {X : Type*} [topological_space X] {Y : Type*} [topological_space Y]\n  [t2_space Y] {A : set X} {f : A \u2192 Y} (hf : continuous f)\n  (g : closure A \u2192 Y)\n  (g_con : continuous g) :\n  \u2200 (g' : closure A \u2192 Y), continuous g' \u2192  (\u2200 (x : closure A), g x = g' x) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $A \\subset X$; let $f: A \\rightarrow Y$ be continuous; let $Y$ be Hausdorff. Show that if $f$ may be extended to a continuous function $g: \\bar{A} \\rightarrow Y$, then $g$ is uniquely determined by $f$.", "nl_proof": "\\begin{proof}\n\n    Let $h, g: \\bar{A} \\rightarrow Y$ be continuous extensions of $f$. Suppose that there is a point $x \\in \\bar{A}$ such that $h(x) \\neq g(x)$. Since $h=g$ on $A$, we must have $x \\in A^{\\prime}$. Since $Y$ is Hausdorff, there is a neighbourhood $U$ of $h(x)$ and a neighbourhood $V$ of $g(x)$ such that $U \\cap V=\\emptyset$. Since $h$ and $g$ are continuous, $h^{-1}(U) \\cap g^{-1}(V)$ is a neighbourhood of $x$. Since $x \\in A^{\\prime}$, there is a point $y \\in h^{-1}(U) \\cap g^{-1}(V) \\cap A$ different from $x$. But $h=g$ on $A$, so $g^{-1}(V) \\cap A=h^{-1}(V) \\cap A$ and hence $y \\in h^{-1}(U) \\cap h^{-1}(V)=h^{-1}(U \\cap V)=\\emptyset$, a contradiction. It follows that $h=g$ on $\\bar{A}$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_20_2", "formal_statement": "theorem exercise_20_2\n  [topological_space (\u211d \u00d7\u2097 \u211d)] [order_topology (\u211d \u00d7\u2097 \u211d)]\n  : metrizable_space (\u211d \u00d7\u2097 \u211d) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that $\\mathbb{R} \\times \\mathbb{R}$ in the dictionary order topology is metrizable.", "nl_proof": "\\begin{proof}\n\n    The dictionary order topology on $\\mathbb{R} \\times \\mathbb{R}$ is the same as the product topology $\\mathbb{R}_d \\times \\mathbb{R}$, where $\\mathbb{R}_d$ denotes $\\mathbb{R}$ with the discrete topology. We know that $\\mathbb{R}_d$ and $\\mathbb{R}$ are metrisable. Thus, it suffices to show that the product of two metrisable spaces is metrisable.\n\nSo let $X$ and $Y$ be metrisable spaces, with metrics $d$ and $d^{\\prime}$ respectively. On $X \\times Y$, define\n\n$$\n\n\\rho(x \\times y, w \\times z)=\\max \\left\\{d(x, w), d^{\\prime}(y, z)\\right\\} .\n\n$$\n\nThen $\\rho$ is a metric on $X \\times Y$; it remains to prove that it induces the product topology on $X \\times Y$. If $B_d\\left(x, r_1\\right) \\times B_d\\left(y, r_2\\right)$ is a basis element for the product space $X \\times Y$, and $r=\\min \\left\\{r_1, r_2\\right\\}$, then $x \\times y \\in B_\\rho(x \\times y, r) \\subset B_d\\left(x, r_1\\right) \\times B_d\\left(y, r_2\\right)$, so the product topology is coarser than the $\\rho$-topology. Conversely, if $B_\\rho(x \\times y, \\delta)$ is a basis element for the $\\rho$-topology, then $x \\times y \\in B_d(x, \\delta) \\times B_{d^{\\prime}}(y, \\delta) \\subset$ $B_\\rho(x \\times y, \\delta)$, so the product topology is finer than the $\\rho$-topology. It follows that both topologies are equal, so the product space $X \\times Y$ is metrisable.\n\n\\end{proof}"}
{"id": "Munkres|exercise_21_6b", "formal_statement": "theorem exercise_21_6b\n  (f : \u2115 \u2192 I \u2192 \u211d )\n  (h : \u2200 x n, f n x = x ^ n) :\n  \u00ac \u2203 f\u2080, tendsto_uniformly f f\u2080 at_top :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Define $f_{n}:[0,1] \\rightarrow \\mathbb{R}$ by the equation $f_{n}(x)=x^{n}$. Show that the sequence $\\left(f_{n}\\right)$ does not converge uniformly.", "nl_proof": "\\begin{proof}\n\n    The sequence $\\left(f_n\\right)_n$ does not converge uniformly, since given $0<\\varepsilon<1$ and $N \\in \\mathbb{Z}_{+}$, for $x=\\varepsilon^{1 / N}$ we have $d\\left(f_N(x), f(x)\\right)=\\varepsilon$. We can also apply Theorem 21.6: the convergence is not uniform since $f$ is not continuous.\n\n\\end{proof}"}
{"id": "Munkres|exercise_22_2a", "formal_statement": "theorem exercise_22_2a {X Y : Type*} [topological_space X]\n  [topological_space Y] (p : X \u2192 Y) (h : continuous p) :\n  quotient_map p \u2194 \u2203 (f : Y \u2192 X), continuous f \u2227 p \u2218 f = id :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $p: X \\rightarrow Y$ be a continuous map. Show that if there is a continuous map $f: Y \\rightarrow X$ such that $p \\circ f$ equals the identity map of $Y$, then $p$ is a quotient map.", "nl_proof": "\\begin{proof}\n\nLet $1_Y: Y \\rightarrow Y$ be the identity map in $Y$. If $U$ is a subset of $Y$ and $p^{-1}(U)$ is open in $X$, then $f^{-1}\\left(p^{-1}(U)\\right)=1_Y^{-1}(U)=U$ is open in $Y$ by continuity of $f$. Thus $p$ is a quotient map.\n\n\\end{proof}"}
{"id": "Munkres|exercise_22_5", "formal_statement": "theorem exercise_22_5 {X Y : Type*} [topological_space X]\n  [topological_space Y] (p : X \u2192 Y) (hp : is_open_map p)\n  (A : set X) (hA : is_open A) : is_open_map (p \u2218 subtype.val : A \u2192 Y) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $p \\colon X \\rightarrow Y$ be an open map. Show that if $A$ is open in $X$, then the map $q \\colon A \\rightarrow p(A)$ obtained by restricting $p$ is an open map.", "nl_proof": "\\begin{proof}\n\nLet $U$ be open in $A$. Since $A$ is open in $X, U$ is open in $X$ as well, so $p(U)$ is open in $Y$. Since $q(U)=p(U)=p(U) \\cap p(A)$, the set $q(U)$ is open in $p(A)$. Thus $q$ is an open map.\n\n\\end{proof}"}
{"id": "Munkres|exercise_23_3", "formal_statement": "theorem exercise_23_3 {X : Type*} [topological_space X]\n  [topological_space X] {A : \u2115 \u2192 set X}\n  (hAn : \u2200 n, is_connected (A n))\n  (A\u2080 : set X)\n  (hA : is_connected A\u2080)\n  (h : \u2200 n, A\u2080 \u2229 A n \u2260 \u2205) :\n  is_connected (A\u2080 \u222a (\u22c3 n, A n)) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $\\left\\{A_{\\alpha}\\right\\}$ be a collection of connected subspaces of $X$; let $A$ be a connected subset of $X$. Show that if $A \\cap A_{\\alpha} \\neq \\varnothing$ for all $\\alpha$, then $A \\cup\\left(\\bigcup A_{\\alpha}\\right)$ is connected.", "nl_proof": "\\begin{proof}\n\n    For each $\\alpha$ we have $A \\cap A_\\alpha \\neq \\emptyset$, so each $A \\cup A_\\alpha$ is connected by Theorem 23.3. In turn $\\left\\{A \\cup A_\\alpha\\right\\}_\\alpha$ is a collection of connected spaces that have a point in common (namely any point in $A)$, so $\\bigcup_\\alpha\\left(A \\cup A_\\alpha\\right)=A \\cup\\left(\\bigcup_\\alpha A_\\alpha\\right)$ is connected. \n\n\\end{proof}"}
{"id": "Munkres|exercise_23_6", "formal_statement": "theorem exercise_23_6 {X : Type*}\n  [topological_space X] {A C : set X} (hc : is_connected C)\n  (hCA : C \u2229 A \u2260 \u2205) (hCXA : C \u2229 A\u1d9c \u2260 \u2205) :\n  C \u2229 (frontier A) \u2260 \u2205 :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $A \\subset X$. Show that if $C$ is a connected subspace of $X$ that intersects both $A$ and $X-A$, then $C$ intersects $\\operatorname{Bd} A$.", "nl_proof": "\\begin{proof}\n\n    Suppose that $C \\cap B d A=C \\cap \\bar{A} \\cap \\overline{X-A}=\\emptyset$. Then $C \\cap A$ and $C \\cap(X \\backslash A)$ are a pair of disjoint non-empty sets whose union is all of $C$, neither of which contains a limit point of the other. Indeed, if $C \\cap(X-A)$ contains a limit point $x$ of $C \\cap A$, then $x \\in C \\cap(X-A) \\cap A^{\\prime} \\subset C \\cap \\bar{A} \\cap \\overline{X-A}=\\emptyset$, a contradiction, and similarly $C \\cap A$ does not contain a limit point of $C \\cap(X-A)$. Then $C \\cap A$ and $C \\cap(X-A)$ constitute a separation of $C$, contradicting the fact that $C$ is connected (Lemma 23.1).\n\n\\end{proof}"}
{"id": "Munkres|exercise_23_11", "formal_statement": "theorem exercise_23_11 {X Y : Type*} [topological_space X] [topological_space Y]\n  (p : X \u2192 Y) (hq : quotient_map p)\n  (hY : connected_space Y) (hX : \u2200 y : Y, is_connected (p \u207b\u00b9' {y})) :\n  connected_space X :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $p: X \\rightarrow Y$ be a quotient map. Show that if each set $p^{-1}(\\{y\\})$ is connected, and if $Y$ is connected, then $X$ is connected.", "nl_proof": "\\begin{proof}\n\n    Suppose that $U$ and $V$ constitute a separation of $X$. If $y \\in p(U)$, then $y=p(x)$ for some $x \\in U$, so that $x \\in p^{-1}(\\{y\\})$. Since $p^{-1}(\\{y\\})$ is connected and $x \\in U \\cap p^{-1}(\\{y\\})$, we have $p^{-1}(\\{y\\}) \\subset U$. Thus $p^{-1}(\\{y\\}) \\subset U$ for all $y \\in p(U)$, so that $p^{-1}(p(U)) \\subset U$. The inclusion $U \\subset p^{-1}(p(U))$ if true for any subset and function, so we have the equality $U=p^{-1}(p(U))$ and therefore $U$ is saturated. Similarly, $V$ is saturated. Since $p$ is a quotient map, $p(U)$ and $p(V)$ are disjoint non-empty open sets in $Y$. But $p(U) \\cup p(V)=Y$ as $p$ is surjective, so $p(U)$ and $p(V)$ constitute a separation of $Y$, contradicting the fact that $Y$ is connected. We conclude that $X$ is connected.\n\n\\end{proof}"}
{"id": "Munkres|exercise_24_3a", "formal_statement": "theorem exercise_24_3a [topological_space I] [compact_space I]\n  (f : I \u2192 I) (hf : continuous f) :\n  \u2203 (x : I), f x = x :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $f \\colon X \\rightarrow X$ be continuous. Show that if $X = [0, 1]$, there is a point $x$ such that $f(x) = x$. (The point $x$ is called a fixed point of $f$.)", "nl_proof": "\\begin{proof}\n\n    If $f(0)=0$ or $f(1)=1$ we are done, so suppose $f(0)>0$ and $f(1)<1$. Let $g:[0,1] \\rightarrow[0,1]$ be given by $g(x)=f(x)-x$. Then $g$ is continuous, $g(0)>0$ and $g(1)<0$. Since $[0,1]$ is connected and $g(1)<0<g(0)$, by the intermediate value theorem there exists $x \\in(0,1)$ such that $g(x)=0$, that is, such that $f(x)=x$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_25_9", "formal_statement": "theorem exercise_25_9 {G : Type*} [topological_space G] [group G]\n  [topological_group G] (C : set G) (h : C = connected_component 1) :\n  is_normal_subgroup C :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $G$ be a topological group; let $C$ be the component of $G$ containing the identity element $e$. Show that $C$ is a normal subgroup of $G$.", "nl_proof": "\\begin{proof}\n\n    Given $x \\in G$, the maps $y \\mapsto x y$ and $y \\mapsto y x$ are homeomorphisms of $G$ onto itself. Since $C$ is a component, $x C$ and $C x$ are both components that contain $x$, so they are equal. Hence $x C=C x$ for all $x \\in G$, so $C$ is a normal subgroup of $G$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_26_12", "formal_statement": "theorem exercise_26_12 {X Y : Type*} [topological_space X] [topological_space Y]\n  (p : X \u2192 Y) (h : function.surjective p) (hc : continuous p) (hp : \u2200 y, is_compact (p \u207b\u00b9' {y}))\n  (hY : compact_space Y) : compact_space X :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $p: X \\rightarrow Y$ be a closed continuous surjective map such that $p^{-1}(\\{y\\})$ is compact, for each $y \\in Y$. (Such a map is called a perfect map.) Show that if $Y$ is compact, then $X$ is compact.", "nl_proof": "\\begin{proof}\n\n    We first show that if $U$ is an open set containing $p^{-1}(\\{y\\})$, then there is a neighbourhood $W$ of $y$ such that $p^{-1}(W)$ is contained in $U$. Since $X-U$ is closed in $X$, $p(X-U)$ is closed in $Y$ and does not contain $y$, so $W=Y \\backslash p(X \\backslash U)$ is a neighbourhood of $y$. Moreover, since $X \\backslash U \\subset p^{-1}(p(X \\backslash U)$ ) (by elementary set theory), we have\n\n$$\n\np^{-1}(W)=p^{-1}(Y \\backslash p(X \\backslash U))=p^{-1}(Y) \\backslash p^{-1}(p(X \\backslash U)) \\subset X \\backslash(X \\backslash U)=U .\n\n$$\n\nNow let $\\mathcal{A}$ be an open covering of $X$. For each $y \\in Y$, let $\\mathcal{A}_y$ be a subcollection of $\\mathcal{A}$ such that\n\n$$\n\np^{-1}(\\{y\\}) \\subset \\bigcup_{A \\in \\mathcal{A}_y} A .\n\n$$\n\nSince $p^{-1}(\\{y\\})$ is compact, there exists a finite subcollection of $\\mathcal{A}_y$ that also covers $p^{-1}(\\{y\\})$, say $\\left\\{A_y^1, \\ldots, A_y^{n_y}\\right\\}$. Thus $\\bigcup_{i=1}^{n_y} A_y^i$ is open and contains $p^{-1}(\\{y\\})$, so there exists a neighbourhood $W_y$ of $y$ such that $p^{-1}\\left(W_y\\right)$ is contained in $\\bigcup_{i=1}^{n_y} A_y^i$. Then $\\left\\{W_y\\right\\}_{y \\in Y}$ is an open covering of $Y$, so there exist $y_1, \\ldots, y_k \\in Y$ such that $\\left\\{W_{y_j}\\right\\}_{j=1}^k$ also covers $Y$. Then\n\n$$\n\nX=p^{-1}(Y) \\subset p^{-1}\\left(\\bigcup_{j=1}^k W_{y_j}\\right)=\\bigcup_{j=1}^k p^{-1}\\left(W_{y_j}\\right) \\subset \\bigcup_{j=1}^k\\left(\\bigcup_{i=1}^{n_{y_j}} A_{y_j}^i\\right)\n\n$$\n\nso\n\n$$\n\n\\left\\{A_{y_j}^i\\right\\}_{\\substack{j=1, \\ldots, k . \\\\ i=1, \\ldots, n_{y_j}}}\n\n$$\n\nis a finite subcollection of $\\mathcal{A}$ that also covers $X$. Therefore, $X$ is compact.\n\n\\end{proof}"}
{"id": "Munkres|exercise_28_4", "formal_statement": "theorem exercise_28_4 {X : Type*}\n  [topological_space X] (hT1 : t1_space X) :\n  countably_compact X \u2194 limit_point_compact X :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "A space $X$ is said to be countably compact if every countable open covering of $X$ contains a finite subcollection that covers $X$. Show that for a $T_1$ space $X$, countable compactness is equivalent to limit point compactness.", "nl_proof": "\\begin{proof}\n\n    First let $X$ be a countable compact space. Note that if $Y$ is a closed subset of $X$, then $Y$ is countable compact as well, for if $\\left\\{U_n\\right\\}_{n \\in \\mathbb{Z}_{+}}$is a countable open covering of $Y$, then $\\left\\{U_n\\right\\}_{n \\in \\mathbb{Z}_{+}} \\cup(X \\backslash Y)$ is a countable open covering of $X$; there is a finite subcovering of $X$, hence a finite subcovering of $Y$. Now let $A$ be an infinite subset. We show that $A$ has a limit point. Let $B$ be a countable infinite subset of $A$. Suppose that $B$ has no limit point, so that $B$ is closed in $X$. Then $B$ is countable compact. Since $B$ has no limit point, for each $b \\in B$ there is a neighbourhood $U_b$ of $b$ that intersects $B$ in the point $b$ alone. Then $\\left\\{U_b\\right\\}_{b \\in B}$ is an open covering of $B$ with no finite subcovering, contradicting the fact that $B$ is countable compact. Hence $B$ has a limit point, so that $A$ has a limit point as well. Since $A$ was arbitrary, we deduce that $X$ is limit point compact. (Note that the $T_1$ property is not necessary in this direction.)\n\n\n\nNow assume that $X$ is a limit point compact $T_1$ space. We show that $X$ is countable compact. Suppose, on the contrary, that $\\left\\{U_n\\right\\}_{n \\in \\mathbb{Z}_{+}}$is a countable open covering of $X$ with no finite subcovering. For each $n$, take a point $x_n$ in $X$ not in $U_1 \\cup \\cdots \\cup U_n$. By assumption, the infinite set $A=\\left\\{x_n \\mid n \\in \\mathbb{Z}_{+}\\right\\}$has a limit point $y \\in X$. Since $\\left\\{U_n\\right\\}_{n \\in \\mathbb{Z}_{+}}$covers $X$, there exists $N \\in \\mathbb{Z}_{+}$such that $y \\in U_1 \\cup \\cdots \\cup U_N$. Now $X$ is $T_1$, so for each $i=1, \\ldots, N$ there exists a neighbourhood $V_i$ of $y$ that does not contain $x_i$. Then\n\n$$\n\nV=\\left(V_1 \\cap \\cdots \\cap V_N\\right) \\cap\\left(U_1 \\cup \\cdots \\cup U_N\\right)\n\n$$\n\nis a neighbourhood of $y$ that does not contain any of the points $x_i$, contradicting the fact that $y$ is a limit point of $A$. It follows that every countable open covering of $X$ must have a finite subcovering, so $X$ is countable compact.\n\n\\end{proof}"}
{"id": "Munkres|exercise_28_6", "formal_statement": "theorem exercise_28_6 {X : Type*} [metric_space X]\n  [compact_space X] {f : X \u2192 X} (hf : isometry f) :\n  function.bijective f :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $(X, d)$ be a metric space. If $f: X \\rightarrow X$ satisfies the condition $d(f(x), f(y))=d(x, y)$ for all $x, y \\in X$, then $f$ is called an isometry of $X$. Show that if $f$ is an isometry and $X$ is compact, then $f$ is bijective and hence a homeomorphism.", "nl_proof": "\\begin{proof}\n\n    Note that $f$ is an imbedding. It remains to prove that $f$ is surjective. Suppose it is not, and let $a \\in f(X)$. Since $X$ is compact, $f(X)$ is compact and hence closed (every metric space is Hausdorff). Thus, there exists $\\varepsilon>0$ such that the $\\varepsilon^{-}$ neighbourhood of $a$ is contained in $X \\backslash f(X)$. Set $x_1=a$, and inductively $x_{n+1}=f\\left(x_n\\right)$ for $n \\in \\mathbb{Z}_{+}$. We show that $d\\left(x_n, x_m\\right) \\geq \\varepsilon$ for $n \\neq m$. Indeed, we may assume $n<m$. If $n \\geq 1$, then $d\\left(x_n, x_m\\right)=$ $d\\left(f^{-1}\\left(x_n\\right), f^{-1}\\left(x_m\\right)\\right)=d\\left(x_{n-1}, x_{m-1}\\right)$. By induction it follows that $d\\left(x_n, x_m\\right)=d\\left(x_{n-i}, x_{m-i}\\right)$ for all $i \\geq 1$, and hence $d\\left(x_n, x_m\\right)=d\\left(a, x_{m-n}\\right)=d\\left(a, f\\left(x_{m-n-1}\\right)\\right)$. Since $f\\left(x_{m-n-1}\\right) \\in f(X)$ and $B(a, \\varepsilon) \\cap f(X)=\\emptyset$, we have $d\\left(x_n, x_m\\right) \\geq \\varepsilon$, as claimed. Thus $\\left\\{x_n\\right\\}_{n \\in \\mathbb{Z}_{+}}$is a sequence with no convergent subsequence, so $X$ is not sequentially compact. This contradicts the fact that $X$ is compact. Therefore $f$ is surjective and hence a homeomorphism.\n\n\\end{proof}"}
{"id": "Munkres|exercise_29_4", "formal_statement": "theorem exercise_29_4 [topological_space (\u2115 \u2192 I)] :\n  \u00ac locally_compact_space (\u2115 \u2192 I) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that $[0, 1]^\\omega$ is not locally compact in the uniform topology.", "nl_proof": "\\begin{proof}\n\n    Consider $\\mathbf{0} \\in[0,1]^\\omega$ and suppose that $[0,1]^\\omega$ is locally compact at $\\mathbf{0}$. Then there exists a compact $C$ containing an open ball $B=B_\\rho(\\mathbf{0}, \\varepsilon) \\subset[0,1]^\\omega$. Note that $\\bar{B}=[0, \\varepsilon]^\\omega$. Then $[0, \\varepsilon]^\\omega$ is closed and contained in the compact $C$, so it is compact. But $[0, \\varepsilon]^\\omega$ is homeomorphic to $[0,1]^\\omega$, which is not compact by Exercise 28.1. This contradiction proves that $[0,1]^\\omega$ is not locally compact in the uniform topology.\n\n\\end{proof}"}
{"id": "Munkres|exercise_30_10", "formal_statement": "theorem exercise_30_10\n  {X : \u2115 \u2192 Type*} [\u2200 i, topological_space (X i)]\n  (h : \u2200 i, \u2203 (s : set (X i)), countable s \u2227 dense s) :\n  \u2203 (s : set (\u03a0 i, X i)), countable s \u2227 dense s :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $X$ is a countable product of spaces having countable dense subsets, then $X$ has a countable dense subset.", "nl_proof": "\\begin{proof}\n\n    Let $\\left(X_n\\right)$ be spaces having countable dense subsets $\\left(A_n\\right)$. For each $n$, fix an arbitrary $x_n \\in X_n$. Consider the subset $A$ of $X$ defined by\n\n$$\n\nA=\\bigcup\\left\\{\\prod U_n: U_n=A_n \\text { for finitely many } n \\text { and is }\\left\\{x_n\\right\\} \\text { otherwise }\\right\\} .\n\n$$\n\nThis set is countable because the set of finite subsets of $\\mathbb{N}$ is countable and each of the inner sets is countable. Now, let $x \\in X$ and $V=\\prod V_n$ be a basis element containing $x$ such that each $V_n$ is open in $X_n$ and $V_n=X_n$ for all but finitely many $n$. For each $n$, if $V_n \\neq X_n$, choose a $y_n \\in\\left(A_n \\cap V_n\\right)$ (such a $y_n$ exists since $A_n$ is dense in $\\left.X_n\\right)$. Otherwise, let $y_n=x_n$. Then $\\left(y_n\\right) \\in(A \\cap V)$, proving that $A$ is dense in $X$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_31_1", "formal_statement": "theorem exercise_31_1 {X : Type*} [topological_space X]\n  (hX : regular_space X) (x y : X) :\n  \u2203 (U V : set X), is_open U \u2227 is_open V \u2227 x \u2208 U \u2227 y \u2208 V \u2227 closure U \u2229 closure V = \u2205 :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $X$ is regular, every pair of points of $X$ have neighborhoods whose closures are disjoint.", "nl_proof": "\\begin{proof}\n\n    Let $x, y \\in X$ be two points such that $x \\neq y$. Since $X$ is regular (and thus Hausdorff), there exist disjoint open sets $U, V \\subseteq X$ such that $x \\in U$ and $y \\in V$.\n\nNote that $y \\notin \\bar{U}$. Otherwise $V$ must intersect $U$ in a point different from $y$ since $V$ is an open neighborhood of $y$, which is a contradiction since $U$ and $V$ are disjoint.\n\nSince $X$ is regular and $\\bar{U}$ is closed, there exist disjoint open sets $U^{\\prime}, V^{\\prime} \\subseteq X$ such that $\\bar{U} \\subseteq U^{\\prime}$ and $y \\in V^{\\prime}$.\n\n\n\nAnd now $U$ and $V^{\\prime}$ are neighborhoods of $x$ and $y$ whose closures are disjoint. If $\\bar{U} \\cap \\overline{V^{\\prime}} \\neq \\emptyset$, then it follows that $U^{\\prime} \\supseteq U$ intersects $\\overline{V^{\\prime}}$. Since $U^{\\prime}$ is open, it follows that $U^{\\prime}$ intersects $V^{\\prime}$, which is a contradiction.\n\n\\end{proof}"}
{"id": "Munkres|exercise_31_3", "formal_statement": "theorem exercise_31_3 {\u03b1 : Type*} [partial_order \u03b1]\n  [topological_space \u03b1] (h : order_topology \u03b1) : regular_space \u03b1 :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that every order topology is regular.", "nl_proof": "\\begin{proof}\n\n    Let $X$ be an ordered set.\n\nFirst we show that $X$ is a $T_1$-space. For $x \\in X$ we have that\n\n$$\n\nX \\backslash\\{x\\}=\\langle-\\infty, x\\rangle \\cup\\langle x,+\\infty\\rangle\n\n$$\n\nwhich is an open set as an union of two open intervals. Therefore, the set $\\{x\\}$ is closed.\n\nStep 2\n\n2 of 3\n\nNow to prove that $X$ is regular we use Lemma $\\mathbf{3 1 . 1 .}$\n\nLet $x \\in X$ be any point and $U \\subseteq X$ any open neighborhood of $x$. Then there exist $a, b \\in X$ such that $x \\in\\langle a, b\\rangle \\subseteq U$. Now we have four possibilities.\n\n1. If there exist $x_1, x_2 \\in U$ such that $a<x_1<x<x_2<b$, then\n\n$$\n\nx \\in\\left\\langle x_1, x_2\\right\\rangle \\subseteq \\overline{\\left\\langle x_1, x_2\\right\\rangle} \\subseteq\\left[x_1, x_2\\right] \\subseteq\\langle a, b\\rangle \\subseteq U\n\n$$\n\n2. If there exists $x_1 \\in U$ such that $a<x_1<x$, but there's no $x_2 \\in U$ such that $x<x_2<b$, then\n\n$$\n\nx \\in\\left\\langle x_1, b\\right\\rangle=\\left(x_1, x\\right] \\subseteq \\overline{\\left(x_1, x\\right]} \\subseteq\\left[x_1, x\\right] \\subseteq\\langle a, b\\rangle \\subseteq U\n\n$$\n\n3. If there exists $x_2 \\in U$ such that $x<x_2<b$, but there's no $x_1 \\in U$ such that $a<x_1<x$, then\n\n$$\n\nx \\in\\left\\langle a, x_2\\right\\rangle=\\left[x, x_2\\right) \\subseteq \\overline{\\left[x, x_2\\right)} \\subseteq\\left[x, x_2\\right] \\subseteq\\langle a, b\\rangle \\subseteq U\n\n$$\n\n4. If there's no $x_1 \\in U$ such that $a<x_1<x$ and no $x_2 \\in U$ such that $x<x_2<b$, then\n\n$$\n\nx \\in\\langle a, b\\rangle=\\{x\\}=\\overline{\\{x\\}}=\\{x\\} \\subseteq U\n\n$$\n\nWe have that $\\overline{\\{x\\}}=\\{x\\}$ because $X$ is a $T_1$-space.\n\nIn all four cases we proved that there exists an open interval $V$ such that $x \\in V \\subseteq \\bar{V} \\subseteq U$, so $X$ is regular.\n\n\\end{proof}"}
{"id": "Munkres|exercise_32_2a", "formal_statement": "theorem exercise_32_2a\n  {\u03b9 : Type*} {X : \u03b9 \u2192 Type*} [\u2200 i, topological_space (X i)]\n  (h : \u2200 i, nonempty (X i)) (h2 : t2_space (\u03a0 i, X i)) :\n  \u2200 i, t2_space (X i) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $\\prod X_\\alpha$ is Hausdorff, then so is $X_\\alpha$. Assume that each $X_\\alpha$ is nonempty.", "nl_proof": "\\begin{proof}\n\n    Suppose that $X=\\prod_\\beta X_\\beta$ is Hausdorff and let $\\alpha$ be any index.\n\nLet $x, y \\in X_\\alpha$ be any points such that $x \\neq y$. Since all $X_\\beta$ are nonempty, there exist points $\\mathbf{x}, \\mathbf{y} \\in X$ such that $x_\\beta=y_\\beta$ for every $\\beta \\neq \\alpha$ and $x_\\alpha=x, y_\\alpha=y$.\n\nSince $x \\neq y$, it follows that $\\mathbf{x} \\neq \\mathbf{y}$. Since $X$ is Hausdorff, there exist open disjoint sets $U, V \\subseteq X$ such that $\\mathbf{x} \\in U$ and $\\mathbf{y} \\in V$.\n\nFor $\\beta \\neq \\alpha$ we have that $x_\\beta=y_\\beta \\in \\pi_\\beta(U) \\cap \\pi_\\beta(V)$, hence $\\pi_\\beta(U)$ and $\\pi_\\beta(V)$ are not disjoint.\n\nSince $U$ and $V$ are disjoint, it follows that $\\pi_\\alpha(U) \\cap \\pi_\\alpha(V)=\\emptyset$.\n\nWe also have that $x \\in \\pi_\\alpha(U)$ and $y \\in \\pi_\\alpha(V)$ and since the projections are open maps, it follows that the sets $\\pi_\\alpha(U)$ and $\\pi_\\alpha(V)$ are open.\n\nThis proves that $x$ and $y$ can be separated by open sets, so $X_\\alpha$ is Hausdorff.\n\n\\end{proof}"}
{"id": "Munkres|exercise_32_2c", "formal_statement": "theorem exercise_32_2c\n  {\u03b9 : Type*} {X : \u03b9 \u2192 Type*} [\u2200 i, topological_space (X i)]\n  (h : \u2200 i, nonempty (X i)) (h2 : normal_space (\u03a0 i, X i)) :\n  \u2200 i, normal_space (X i) :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that if $\\prod X_\\alpha$ is normal, then so is $X_\\alpha$. Assume that each $X_\\alpha$ is nonempty.", "nl_proof": "\\begin{proof}\n\n    Suppose that $X=\\prod_\\beta X_\\beta$ is normal and let $\\alpha$ be any index.\n\nSince $X$ is normal, it follows that $X$ is Hausdorff (or regular), which then implies that $X_\\alpha$ is Hausdorff (or regular). This imples that $X_\\alpha$ satisfies the $T_1$ axiom.\n\nNow the proof that $X_\\alpha$ satisfies the $T_4$ axiom is the same as for regular spaces.\n\nIf $F, G \\subseteq X_\\alpha$ are disjoint closed sets, then $\\prod_\\beta F_\\beta$ and $\\prod_\\beta G_\\beta$, where $F_\\alpha=F, G_\\alpha=G$ and $F_\\beta=G_\\beta=X_\\beta$ for $\\beta \\neq \\alpha$, are disjoint closed sets in $X$.\n\nSince $X$ is normal (and therefore satisfies the $T_4$ axiom), there exist disjoint open sets $U, V \\subseteq X$ such that $\\prod_\\beta F_\\beta \\subseteq U$ and $\\prod_\\beta G_\\beta \\subseteq V$\n\nThen $\\pi_\\alpha(U)$ and $\\pi_\\alpha(V)$ are disjoint open sets in $X_\\alpha$ such that $F \\subseteq \\pi_\\alpha(U)$ and $G \\subseteq \\pi_\\alpha(V)$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_33_7", "formal_statement": "theorem exercise_33_7 {X : Type*} [topological_space X]\n  (hX : locally_compact_space X) (hX' : t2_space X) :\n  \u2200 x A, is_closed A \u2227 \u00ac x \u2208 A \u2192\n  \u2203 (f : X \u2192 I), continuous f \u2227 f x = 1 \u2227 f '' A = {0}\n  :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Show that every locally compact Hausdorff space is completely regular.", "nl_proof": "\\begin{proof}\n\n    $X$ is a subspace of a compact Hausdorff space $Y$, its one-point compactification. $Y$ is normal, and so by the Urysohn lemma $Y$ is completely regular. Therefore by corollary $X$ is completely regular.\n\n\\end{proof}"}
{"id": "Munkres|exercise_34_9", "formal_statement": "theorem exercise_34_9\n  (X : Type*) [topological_space X] [compact_space X]\n  (X1 X2 : set X) (hX1 : is_closed X1) (hX2 : is_closed X2)\n  (hX : X1 \u222a X2 = univ) (hX1m : metrizable_space X1)\n  (hX2m : metrizable_space X2) : metrizable_space X :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $X$ be a compact Hausdorff space that is the union of the closed subspaces $X_1$ and $X_2$. If $X_1$ and $X_2$ are metrizable, show that $X$ is metrizable.", "nl_proof": "\\begin{proof}\n\n    Both $X_1$ and $X_2$ are compact, Hausdorff and metrizable, so by exercise 3 they are second countable, i.e. there are countable bases $\\left\\{U_{i, n} \\subset X_i \\mid n \\in \\mathbb{N}\\right\\}$ for $i \\in\\{1,2\\}$. By the same exercise it is enough to show that $X$ is second countable. If $X_1 \\cap X_2=\\emptyset$ both $X_1$ and $X_2$ are open and the union $\\left\\{U_{i, n} \\mid i \\in\\{1,2\\} ; n \\in \\mathbb{N}\\right\\}$ of their countable bases form a countable base for $X$.\n\n\n\nSuppose now $X_1 \\cap X_2 \\neq \\emptyset$. Let $x \\in X$ and $U \\subset X$ be an open neighborhood of $x$. If $x \\in X_i-X_j=X-$ $X_j$ then $U \\cap X_i$ is open in $X_i$ and there is a basis neighborhood $U_{i, n}$ of $x$ such that $x \\in U_{i, n} \\cap X-X_j$ is an open neighborhood of $x$ in the open subset $X-X_j$, so $U_{i, n} \\cap X-X_j$ is also open in $X$.\n\nSuppose now that $x \\in X_1 \\cap X_2$. We have that $U \\cap X_i$ is open in $X_i$ so there is a basis neighborhood $U_{i, n_i}$ contained in $U \\cap X_i$. By definition of sub-space topology there is some open subset $V_{i, n_i} \\subset X$ such that $U_{i, n_i}=$ $X_i \\cap V_{i, n_i}$. Then\n\n$$\n\nx \\in V_{1, n_1} \\cap V_{2, n_2}=\\left(V_{1, n_1} \\cap V_{2, n_2} \\cap X_1\\right) \\cup\\left(V_{1, n_1} \\cap V_{2, n_2} \\cap X_2\\right)=\\left(U_{1, n_1} \\cap V_{2, n_2}\\right) \\cup\\left(V_{1, n_1} \\cap U_{2, n_2}\\right) \\subset U\n\n$$\n\nTherefore the open subsets $U_{i, n} \\cap X-X_j$ and $V_{1, n_1} \\cap V_{2, n_2}$ form a countable base for $X$.\n\n\\end{proof}"}
{"id": "Munkres|exercise_43_2", "formal_statement": "theorem exercise_43_2 {X : Type*} [metric_space X]\n  {Y : Type*} [metric_space Y] [complete_space Y] (A : set X)\n  (f : X \u2192 Y) (hf : uniform_continuous_on f A) :\n  \u2203! (g : X \u2192 Y), continuous_on g (closure A) \u2227\n  uniform_continuous_on g (closure A) \u2227 \u2200 (x : A), g x = f x :=", "src_header": "import .common \n\nopen set topological_space filter \nopen_locale classical topology filter \nnoncomputable theory \n\n", "nl_statement": "Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces; let $Y$ be complete. Let $A \\subset X$. Show that if $f \\colon A \\rightarrow Y$ is uniformly continuous, then $f$ can be uniquely extended to a continuous function $g \\colon \\bar{A} \\rightarrow Y$, and $g$ is uniformly continuous.", "nl_proof": "\\begin{proof}\n\n    Let $\\left(X, d_X\\right)$ and $\\left(Y, d_Y\\right)$ be metric spaces; let $Y$ be complete. Let $A \\subset X$. It is also given that $f: A \\longrightarrow X$ is a uniformly continuous function. Then we have to show that $f$ can be uniquely extended to a continuous function $g: \\bar{A} \\longrightarrow Y$, and $g$ is uniformly coninuous.\n\nWe define the function $g$ as follows:\n\n$$\n\ng(x)= \\begin{cases}f(x) & \\text { if } x \\in A \\\\ \\lim _{n \\rightarrow \\infty} f\\left(x_n\\right) & \\text { if } x \\in \\bar{A},\\end{cases}\n\n$$\n\nwhere $\\left\\{x_n\\right\\} \\subset A$ is some sequence in $A$ such that $\\lim _{n \\rightarrow \\infty} x_n=x$. Now we have to check that the above definition is well defined. We first note that since $\\left\\{x_n\\right\\} \\subset A$ is convergent in $X,\\left\\{x_n\\right\\}$ is Cauchy in $A$ and since $f$ is uniformly continuous on $A,\\left\\{f\\left(x_n\\right)\\right\\}$ is Cauchy in $Y$. Further, since $Y$ is complete $\\left\\{f\\left(x_n\\right)\\right\\}$ is convergent. So let us now consider two sequences $\\left\\{x_n\\right\\}$ and $\\left\\{y_n\\right\\}$ such that $\\lim _{n \\rightarrow \\infty} x_n=\\lim _{n \\rightarrow \\infty} y_n=x$. Then we need to prove that $\\lim _{n \\rightarrow \\infty} f\\left(x_n\\right)=\\lim _{n \\rightarrow \\infty} f\\left(y_n\\right)=f(x)$. Let\n\n$$\n\n\\lim _{n \\rightarrow \\infty} f\\left(x_n\\right)=a \\text { and } b=\\lim _{n \\rightarrow \\infty} f\\left(y_n\\right)\n\n$$\n\nNow since $f$ is uniformly continuous for any given $\\epsilon>0$ there exists $\\delta>0$ such that\n\n$$\n\nd_Y(f(x), f(y))<\\epsilon \\text { whenever } d_X(x, y)<\\delta \\text { and } x, y \\in A\n\n$$\n\nSo for this $\\delta>0$ there exists $N \\in \\mathbb{N}$ such that\n\n$$\n\nd_X\\left(x_n, x\\right)<\\frac{\\delta}{2} \\text { and } d_X\\left(y_n, x\\right)<\\frac{\\delta}{2}, \\text { foe all } n \\geq N .\n\n$$\n\nTherefore, we have that for all $n \\geq N$,\n\n$$\n\nd_X\\left(x_n, y_n\\right)<\\delta\n\n$$\n\nThus the equation (1) yields us that\n\n$$\n\nd_Y\\left(f\\left(x_n\\right), f\\left(y_n\\right)\\right)<\\epsilon \\text { for all } n \\geq N .\n\n$$\n\nNow since $\\lim _{n \\rightarrow \\infty} f\\left(x_n\\right)=a$ and $b=\\lim _{n \\rightarrow \\infty} f\\left(y_n\\right)$, so for the above $\\epsilon>0$ we have a natural number $K \\geq N$ such that\n\n$$\n\n\\begin{gathered}\n\nd_Y\\left(f\\left(x_n\\right), a\\right)<\\epsilon \\text { for all } n \\geq K \\text { and } \\\\\n\nd_Y\\left(f\\left(y_n\\right), b\\right)<\\epsilon \\text { for all } n \\geq K .\n\n\\end{gathered}\n\n$$\n\nMoreover, since $K \\geq N$, from $(2)$ we get\n\n$$\n\nd_Y\\left(f\\left(x_n\\right), f\\left(y_n\\right)\\right)<\\epsilon \\text { for all } n \\geq K .\n\n$$\n\nNow we calculate the following, for $n \\geq K$,\n\n$$\n\n\\begin{array}{rlr}\n\nd_Y(a, b) & \\leq & d_Y\\left(a, f\\left(x_n\\right)\\right)+d_Y\\left(f\\left(x_n\\right), f\\left(y_n\\right)\\right)+d\\left(f\\left(y_n\\right), b\\right) \\\\\n\n& < & \\epsilon+\\epsilon+\\epsilon \\text { by }(3),(4) \\text { and }(5) \\\\\n\n& = & 3 \\epsilon\n\n\\end{array}\n\n$$\n\nwhere the first inequality holds because of triangular inequality. Since $\\epsilon>0$ is arbitrary the above calculation shows that $d_Y(a, b)=0$. Thus, the above definition is independent of the choice of the sequence $\\left\\{x_n\\right\\}$ and hence the map $g$ is well defined. Moreover, from the construction it follows that $g$ is continuous on $\\bar{A}$.\n\nMoreover, we observe that $g$ is unique extension of $f$ by the construction.\n\nSo it remains to show that $g$ is uniformly continuous. In order to that we take a Cauchy sequence $\\left\\{a_n\\right\\} \\subset \\bar{A}$. Then since $\\bar{A}$ is a closed set so the sequence $\\left\\{a_n\\right\\}$ is convergent and hence $\\left\\{g\\left(a_n\\right)\\right\\}$ is also a convergent sequence as $g$ is continuous on $\\bar{A}$. So $\\left\\{g\\left(a_n\\right)\\right\\}$ is a Cauchy sequence in $Y$. Since a function is uniformly continuous if and only if it sends Cauchy sequences to Cauchy sequences, we conclude that $g$ is uniformly continuous.\n\n\\end{proof}"}
{"id": "Axler|exercise_1_2", "formal_statement": "theorem exercise_1_2 :\n  (\u27e8-1/2, real.sqrt 3 / 2\u27e9 : \u2102) ^ 3 = -1 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Show that $\\frac{-1 + \\sqrt{3}i}{2}$ is a cube root of 1 (meaning that its cube equals 1).", "nl_proof": "\\begin{proof}\n\n$$\n\n\\left(\\frac{-1+\\sqrt{3} i}{2}\\right)^2=\\frac{-1-\\sqrt{3} i}{2},\n\n$$\n\nhence\n\n$$\n\n\\left(\\frac{-1+\\sqrt{3} i}{2}\\right)^3=\\frac{-1-\\sqrt{3} i}{2} \\cdot \\frac{-1+\\sqrt{3} i}{2}=1\n\n$$\n\nThis means $\\frac{-1+\\sqrt{3} i}{2}$ is a cube root of 1.\n\n\\end{proof}"}
{"id": "Axler|exercise_1_4", "formal_statement": "theorem exercise_1_4 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] (v : V) (a : F): a \u2022 v = 0 \u2194 a = 0 \u2228 v = 0 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that if $a \\in \\mathbf{F}$, $v \\in V$, and $av = 0$, then $a = 0$ or $v = 0$.", "nl_proof": "\\begin{proof}\n\n    If $a=0$, then we immediately have our result. So suppose $a \\neq 0$. Then, because $a$ is some nonzero real or complex number, it has a multiplicative inverse $\\frac{1}{a}$. Now suppose that $v$ is some vector such that\n\n$$\n\na v=0\n\n$$\n\nMultiply by $\\frac{1}{a}$ on both sides of this equation to get\n\n$$\n\n\\begin{aligned}\n\n\\frac{1}{a}(a v) & =\\frac{1}{a} 0 & & \\\\\n\n\\frac{1}{a}(a v) & =0 & & \\\\\n\n\\left(\\frac{1}{a} \\cdot a\\right) v & =0 & & \\text { (associativity) } \\\\\n\n1 v & =0 & & \\text { (definition of } 1/a) \\\\\n\nv & =0 & & \\text { (multiplicative identity) }\n\n\\end{aligned}\n\n$$\n\nHence either $a=0$ or, if $a \\neq 0$, then $v=0$.\n\n\\end{proof}"}
{"id": "Axler|exercise_1_7", "formal_statement": "theorem exercise_1_7 : \u2203 U : set (\u211d \u00d7 \u211d),\n  (U \u2260 \u2205) \u2227\n  (\u2200 (c : \u211d) (u : \u211d \u00d7 \u211d), u \u2208 U \u2192 c \u2022 u \u2208 U) \u2227\n  (\u2200 U' : submodule \u211d (\u211d \u00d7 \u211d), U \u2260 \u2191U') :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Give an example of a nonempty subset $U$ of $\\mathbf{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\\mathbf{R}^2$.", "nl_proof": "\\begin{proof}\n\n$$\n\nU=\\left\\{(x, y) \\in \\mathbf{R}^2:|x|=|y|\\right\\}\n\n$$\n\nFor $(x, y) \\in U$ and $\\lambda \\in \\mathbb{R}$, it follows $\\lambda(x, y)=$ $(\\lambda x, \\lambda y)$, so $|\\lambda x|=|\\lambda||x|=|\\lambda||y|=|\\lambda y|$. Therefore, $\\lambda(x, y) \\in U$.\n\n\n\nOn the other hand, consider $a=(1,-1), b=$ $(1,1) \\in U$. Then, $a+b=(1,-1)+(1,1)=$ $(2,0) \\notin U$. So, $U$ is not a subspace of $\\mathbb{R}^2$.\n\n\\end{proof}"}
{"id": "Axler|exercise_1_9", "formal_statement": "theorem exercise_1_9 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] (U W : submodule F V):\n  \u2203 U' : submodule F V, (U'.carrier = \u2191U \u2229 \u2191W \u2194 (U \u2264 W \u2228 W \u2264 U)) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that the union of two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other.", "nl_proof": "\\begin{proof}\n\n    To prove this one way, suppose for purposes of contradiction that for $U_1$ and $U_2$, which are subspaces of $V$, that $U_1 \\cup U_2$ is a subspace and neither is completely contained within the other. In other words, $U_1 \\nsubseteq U_2$ and $U_2 \\nsubseteq U_1$. We will show that you can pick a vector $v \\in U_1$ and a vector $u \\in U_2$ such that $v+u \\notin U_1 \\cup U_2$, proving that if $U_1 \\cup U_2$ is a subspace, one must be completely contained inside the other.\n\n\n\nIf $U_1 \\nsubseteq U_2$, we can pick a $v \\in U_1$ such that $v \\notin U_2$. Since $v$ is in the subspace $U_1$, then $(-v)$ must also be, by definition. Similarly, if $U_2 \\nsubseteq U_1$, then we can pick a $u \\in U_2$ such that $u \\notin U_1$. Since $u$ is in the subspace $U_2$, then $(-u)$ must also be, by definition.\n\n\n\nIf $v+u \\in U_1 \\cup U_2$, then $v+u$ must be in $U_1$ or $U_2$. But, $v+u \\in U_1 \\Rightarrow v+u+(-v) \\in U_1 \\Rightarrow u \\in U_1$\n\nSimilarly,\n\n$$\n\nv+u \\in U_2 \\Rightarrow v+u+(-u) \\in U_2 \\Rightarrow v \\in U_2\n\n$$\n\nThis is clearly a contradiction, as each element was defined to not be in these subspaces. Thus our initial assumption must have been wrong, and $U_1 \\subseteq U_2$ or $U_2 \\subseteq U_1$\n\nTo prove the other way, Let $U_1 \\subseteq U_2$ (WLOG). $U_1 \\subseteq U_2 \\Rightarrow U_1 \\cup U_2=U_2$. Since $U_2$ is a subspace, $U_1 \\cup U_2$ is as well. QED.\n\n\\end{proof}"}
{"id": "Axler|exercise_3_8", "formal_statement": "theorem exercise_3_8 {F V W : Type*}  [add_comm_group V]\n  [add_comm_group W] [field F] [module F V] [module F W]\n  (L : V \u2192\u2097[F] W) :\n  \u2203 U : submodule F V, U \u2293 L.ker = \u22a5 \u2227\n  linear_map.range L = range (dom_restrict L U):=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose that $V$ is finite dimensional and that $T \\in \\mathcal{L}(V, W)$. Prove that there exists a subspace $U$ of $V$ such that $U \\cap \\operatorname{null} T=\\{0\\}$ and range $T=\\{T u: u \\in U\\}$.", "nl_proof": "\\begin{proof}\n\n    The point here is to note that every subspace of a vector space has a complementary subspace.\n\nIn this example, $U$ will precisely turn out to be the complementary subspace of null $T$. That is, $V=U \\oplus \\operatorname{null} T$\n\nHow should we characterize $U$ ? This can be achieved by extending a basis $B_1=\\left\\{v_1, v_2, \\ldots, v_m\\right\\}$ of null $T$ to a basis of $V$. Let $B_2=\\left\\{u_1, u_2, \\ldots, u_n\\right\\}$ be such that $B=B_1 \\cup B_2$ is a basis of $V$.\n\n\n\nDefine $U=\\operatorname{span}\\left(B_2\\right)$. Now, since $B_1$ and $B_2$ are complementary subsets of the basis $B$ of $V$, their spans will turn out to be complementary subspaces of $V$. Let's prove that $V=U \\oplus$ null $T$.\n\n\n\nLet $v \\in V$. Then, $v$ can be expressed as a linear combination of the vectors in $B$.\n\nLet $v=a_1 u_1+\\cdots+a_n u_n+c_1 v_1+\\cdots+c_m v_m$. However, since $\\left\\{u_1, u_2, \\ldots, u_n\\right\\}$ is a basis of $U, a_1 u_1+$ $\\cdots+a_n u_n=u \\in U$ and since $\\left\\{v_1, v_2, \\ldots, v_m\\right\\}$ is a basis of null $T, c_1 v_1+\\cdots+c_m v_m=w \\in$ null $T$.\n\nHence, $v=u+w \\in U+\\operatorname{null} T$. This shows that\n\n$$\n\nV=U+\\operatorname{null} T\n\n$$\n\nNow, let $v \\in U \\cap \\operatorname{null} T$.\n\nSince $v \\in U, u$ can be expressed as a linear combination of basis vectors of $U$. Let\n\n$$\n\nv=a_1 u_1+\\cdots+a_n u_n\n\n$$\n\nSimilarly, since $v \\in \\operatorname{null} T$, it can also be expressed as a tinear combination of the basis vectors of null $T$. Let\n\n$$\n\nv=c_1 v_1+\\cdots+c_m v_m\n\n$$\n\nThe left hand sides of the above two equations are equal. Therefore, we can equate the right hand sides.\n\n$$\n\n\\begin{aligned}\n\n& a_1 u_1+\\cdots+a_n u_n=v=c_1 v_1+\\cdots+c_m v_m \\\\\n\n& a_1 u_1+\\cdots+a_n u_n-c_1 v_1-\\cdots-c_m v_m=0\n\n\\end{aligned}\n\n$$\n\nWe have found a linear combination of $u_i^{\\prime}$ 's and $v_i$ 's which is equal to zero. However, they are basis vectors of $V$. Hence, all the multipliers $c_i$ 's and $a_i$ 's must be zero implying that $v=0$.\n\nTherefore, if $v \\in U \\cap$ null $T$, then $v=0$. this means that\n\n$$\n\nU \\cap \\operatorname{null} T=\\{0\\}\n\n$$\n\nThe above shows that $U$ satisfies the first of the required conditions.\n\nNow let $w \\in$ range $T$. Then, there exists $v \\in V$ such that $T v=w$. This allows us to write $v=u+w$ where $u \\in U$ and $w \\in$ null $T$. This implies\n\n$$\n\n\\begin{aligned}\n\nw & =T v \\\\\n\n& =T(u+w) \\\\\n\n& =T u+T w \\\\\n\n& =T u+0 \\quad \\quad(\\text { since } w \\in \\operatorname{null} T) \\\\\n\n& =T u\n\n\\end{aligned}\n\n$$\n\nThis shows that if $w \\in$ range $T$ then $w=T u$ for some $u \\in U$. Therefore, range $T \\subseteq\\{T u \\mid u \\in U\\}$.\n\nSince $U$ is a subspace of $V$, it follows that $T u \\in$ range $T$ for all $u \\in U$. Thus, $\\{T u \\mid u \\in U\\} \\subseteq$ range $T$.\n\nTherefore, range $T=\\{T u \\mid u \\in U\\}$.\n\nThis shows that $U$ satisfies the second required condition as well.\n\n\\end{proof}"}
{"id": "Axler|exercise_5_1", "formal_statement": "theorem exercise_5_1 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] {L : V \u2192\u2097[F] V} {n : \u2115} (U : fin n \u2192 submodule F V)\n  (hU : \u2200 i : fin n, map L (U i) = U i) :\n  map L (\u2211 i : fin n, U i : submodule F V) =\n  (\u2211 i : fin n, U i : submodule F V) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $T \\in \\mathcal{L}(V)$. Prove that if $U_{1}, \\ldots, U_{m}$ are subspaces of $V$ invariant under $T$, then $U_{1}+\\cdots+U_{m}$ is invariant under $T$.", "nl_proof": "\\begin{proof}\n\n    First off, assume that $U_1, \\ldots, U_m$ are subspaces of $V$ invariant under $T$. Now, consider a vector $u \\in$ $U_1+\\ldots+U_m$. There does exist $u_1 \\in U_1, \\ldots, u_m \\in U_m$ such that $u=u_1+\\ldots+u_m$.\n\n\n\nOnce you apply $T$ towards both sides of the previous equation, we would then get $T u=T u_1+\\ldots+$ $T u_m$.\n\n\n\nSince each $U_j$ is invariant under $T$, then we would have $T u_1 \\in U_1+\\ldots+T u_m$. This would then make the equation shows that $T u \\in U_1+\\ldots+T u_m$, which does imply that $U_1+. .+U_m$ is invariant under $T$\n\n\\end{proof}"}
{"id": "Axler|exercise_5_11", "formal_statement": "theorem exercise_5_11 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] (S T : End F V) :\n  (S * T).eigenvalues = (T * S).eigenvalues  :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $S, T \\in \\mathcal{L}(V)$. Prove that $S T$ and $T S$ have the same eigenvalues.", "nl_proof": "\\begin{proof}\n\n    To start, let $\\lambda \\in F$ be an eigenvalue of $S T$. Now, we would want $\\lambda$ to be an eigenvalue of $T S$. Since $\\lambda$, by itself, is an eigenvalue of $S T$, then there has to be a nonzero vector $v \\in V$ such that $(S T) v=\\lambda v$.\n\nNow, With a given reference that $(S T) v=\\lambda v$, you will then have the following: $(T S)(T v)=$ $T(S T v)=T(\\lambda v)=\\lambda T v$\n\nIf $T v \\neq 0$, then the listed equation above shows that $\\lambda$ is an eigenvalue of $T S$.\n\nIf $T v=0$, then $\\lambda=0$, since $S(T v)=\\lambda T v$. This also means that $T$ isn't invertible, which would imply that $T S$ isn't invertible, which can also be implied that $\\lambda$, which equals 0 , is an eigenvalue of $T S$.\n\nStep 3\n\n3 of 3\n\nNow, regardless of whether $T v=0$ or not, we would have shown that $\\lambda$ is an eigenvalue of $T S$. Since $\\lambda$ (was) an arbitrary eigenvalue of $S T$, we have shown that every single eigenvalue of $S T$ is an eigenvalue of $T S$. When you do reverse the roles of both $S$ and $T$, then we can conclude that that every single eigenvalue of $T S$ is also an eigenvalue of $S T$. Therefore, both $S T$ and $T S$ have the exact same eigenvalues.\n\n\\end{proof}"}
{"id": "Axler|exercise_5_13", "formal_statement": "theorem exercise_5_13 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] [finite_dimensional F V] {T : End F V}\n  (hS : \u2200 U : submodule F V, finrank F U = finrank F V - 1 \u2192\n  map T U = U) : \u2203 c : F, T = c \u2022 id :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $T \\in \\mathcal{L}(V)$ is such that every subspace of $V$ with dimension $\\operatorname{dim} V-1$ is invariant under $T$. Prove that $T$ is a scalar multiple of the identity operator.", "nl_proof": "\\begin{proof}\n\n    First off, let $T$ isn't a scalar multiple of the identity operator. So, there does exists that $v \\in V$ such that $u$ isn't an eigenvector of $T$. Therefore, $(u, T u)$ is linearly independent.\n\n\n\nNext, you should extend $(u, T u)$ to a basis of $\\left(u, T u, v_1, \\ldots, v_n\\right)$ of $V$. So, let $U=\\operatorname{span}\\left(u, v_1, \\ldots, v_n\\right)$. Then, $U$ is a subspace of $V$ and $\\operatorname{dim} U=\\operatorname{dim} V-1$. However, $U$ isn't invariant under $T$ since both $u \\in U$ and $T u \\in U$. This given contradiction to our hypothesis about $T$ actually shows us that our guess that $T$ is not a scalar multiple of the identity must have been false.\n\n\\end{proof}"}
{"id": "Axler|exercise_5_24", "formal_statement": "theorem exercise_5_24 {V : Type*} [add_comm_group V]\n  [module \u211d V] [finite_dimensional \u211d V] {T : End \u211d V}\n  (hT : \u2200 c : \u211d, eigenspace T c = \u22a5) {U : submodule \u211d V}\n  (hU : map T U = U) : even (finrank U) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $V$ is a real vector space and $T \\in \\mathcal{L}(V)$ has no eigenvalues. Prove that every subspace of $V$ invariant under $T$ has even dimension.", "nl_proof": "\\begin{proof}\n\n    First off, let us assume that $U$ is a subspace of $V$ that is invariant under $T$. Therefore, $\\left.T\\right|_U \\in \\mathcal{L}(U)$. If $\\operatorname{dim}$ $U$ were odd, then $\\left.T\\right|_U$ would have an eigenvalue $\\lambda \\in \\mathbb{R}$, so there would exist a nonzero vector $u \\in U$ such that\n\n$$\n\n\\left.T\\right|_U u=\\lambda u .\n\n$$\n\nSo, this would imply that $T_u=\\lambda u$, which would imply that $\\lambda$ is an eigenvalue of $T$. But $T$ has no eigenvalues, so $\\operatorname{dim} U$ must be even.\n\n\\end{proof}"}
{"id": "Axler|exercise_6_3", "formal_statement": "theorem exercise_6_3 {n : \u2115} (a b : fin n \u2192 \u211d) :\n  (\u2211 i, a i * b i) ^ 2 \u2264 (\u2211 i : fin n, i * a i ^ 2) * (\u2211 i, b i ^ 2 / i) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that $\\left(\\sum_{j=1}^{n} a_{j} b_{j}\\right)^{2} \\leq\\left(\\sum_{j=1}^{n} j a_{j}{ }^{2}\\right)\\left(\\sum_{j=1}^{n} \\frac{b_{j}{ }^{2}}{j}\\right)$ for all real numbers $a_{1}, \\ldots, a_{n}$ and $b_{1}, \\ldots, b_{n}$.", "nl_proof": "\\begin{proof}\n\n    Let $a_1, a_2, \\ldots, a_n, b_1, b_2, \\ldots, b_n \\in R$.\n\nWe have that\n\n$$\n\n\\left(\\sum_{j=1}^n a_j b_j\\right)^2\n\n$$\n\nis equal to the\n\n$$\n\n\\left(\\sum_{j=1}^n a_j b_j \\frac{\\sqrt{j}}{\\sqrt{j}}\\right)^2=\\left(\\sum_{j=1}^n\\left(\\sqrt{j} a_j\\right)\\left(b_j \\frac{1}{\\sqrt{j}}\\right)\\right)^2\n\n$$\n\nThis can be observed as an inner product, and using the Cauchy-Schwarz Inequality, we get\n\n$$\n\n\\begin{aligned}\n\n&\\left(\\sum_{j=1}^n a_j b_j\\right)^2=\\left(\\sum_{j=1}^n\\left(\\sqrt{j} a_j\\right)\\left(b_j \\frac{1}{\\sqrt{j}}\\right)\\right)^2 \\\\\n\n&=\\left\\langle\\left(a, \\sqrt{2} a_2, \\ldots, \\sqrt{n} a_n\\right),\\left(b_1, \\frac{b_2}{\\sqrt{2}}, \\ldots, \\frac{b_n}{\\sqrt{n}}\\right)\\right\\rangle \\\\\n\n& \\leq\\left\\|\\left(a, \\sqrt{2} a_2, \\ldots, \\sqrt{n} a_n\\right)\\right\\|^2\\left\\|\\left(b_1, \\frac{b_2}{\\sqrt{2}}, \\ldots, \\frac{b_n}{\\sqrt{n}}\\right)\\right\\|^2 \\\\\n\n&=\\left(\\sum_{j=1}^n j a_j^2\\right)\\left(\\sum_{j=1}^n \\frac{b_j^2}{j}\\right) \\\\\n\n& \\text { Hence, }\\left(\\sum_{j=1}^n a_j b_j\\right)^2=\\left(\\sum_{j=1}^n j a_j^2\\right)\\left(\\sum_{j=1}^n \\frac{b_j^2}{j}\\right) .\n\n\\end{aligned}\n\n$$\n\n\\end{proof}"}
{"id": "Axler|exercise_6_13", "formal_statement": "theorem exercise_6_13 {V : Type*} [inner_product_space \u2102 V] {n : \u2115}\n  {e : fin n \u2192 V} (he : orthonormal \u2102 e) (v : V) :\n  \u2016v\u2016^2 = \u2211 i : fin n, \u2016\u27eav, e i\u27eb_\u2102\u2016^2 \u2194 v \u2208 span \u2102 (e '' univ) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $\\left(e_{1}, \\ldots, e_{m}\\right)$ is an or thonormal list of vectors in $V$. Let $v \\in V$. Prove that $\\|v\\|^{2}=\\left|\\left\\langle v, e_{1}\\right\\rangle\\right|^{2}+\\cdots+\\left|\\left\\langle v, e_{m}\\right\\rangle\\right|^{2}$ if and only if $v \\in \\operatorname{span}\\left(e_{1}, \\ldots, e_{m}\\right)$.", "nl_proof": "\\begin{proof}\n\nIf $v \\in \\operatorname{span}\\left(e_1, \\ldots, e_m\\right)$, it means that\n\n$$\n\nv=\\alpha_1 e_1+\\ldots+\\alpha_m e_m .\n\n$$\n\nfor some scalars $\\alpha_i$. We know that $\\alpha_k=\\left\\langle v, e_k\\right\\rangle, \\forall k \\in\\{1, \\ldots, m\\}$. Therefore,\n\n$$\n\n\\begin{aligned}\n\n\\|v\\|^2 & =\\langle v, v\\rangle \\\\\n\n& =\\left\\langle\\alpha_1 e_1+\\ldots+\\alpha_m e_m, \\alpha_1 e_1+\\ldots+\\alpha_m e_m\\right\\rangle \\\\\n\n& =\\left|\\alpha_1\\right|^2\\left\\langle e_1, e_1\\right\\rangle+\\ldots+\\left|\\alpha_m\\right|^2\\left\\langle e_m, e_m\\right\\rangle \\\\\n\n& =\\left|\\alpha_1\\right|^2+\\ldots+\\left|\\alpha_m\\right|^2 \\\\\n\n& =\\left|\\left\\langle v, e_1\\right\\rangle\\right|^2+\\ldots+\\left|\\left\\langle v, e_m\\right\\rangle\\right|^2 .\n\n\\end{aligned}\n\n$$\n\n$\\Rightarrow$ Assume that $v \\notin \\operatorname{span}\\left(e_1, \\ldots, e_m\\right)$. Then, we must have\n\n$$\n\nv=v_{m+1}+\\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0,\n\n$$\n\nwhere $v_0=\\alpha_1 e_1+\\ldots+\\alpha_m e_m, \\alpha_k=\\left\\langle v, e_k\\right\\rangle, \\forall k \\in\\{1, \\ldots, m\\}$, and $v_{m+1}=v-$ $\\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0 \\neq 0$.\n\n\n\nWe have $\\left\\langle v_0, v_{m+1}\\right\\rangle=0$ (from which we get $\\left\\langle v, v_0\\right\\rangle=\\left\\langle v_0, v_0\\right\\rangle$ and $\\left\\langle v, v_{m+1}\\right\\rangle=$ $\\left.\\left\\langle v_{m+1}, v_{m+1}\\right\\rangle\\right)$. Now,\n\n$$\n\n\\begin{aligned}\n\n\\|v\\|^2 & =\\langle v, v\\rangle \\\\\n\n& =\\left\\langle v, v_{m+1}+\\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0\\right\\rangle \\\\\n\n& =\\left\\langle v, v_{m+1}\\right\\rangle+\\left\\langle v, \\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0\\right\\rangle \\\\\n\n& =\\left\\langle v_{m+1}, v_{m+1}\\right\\rangle+\\frac{\\left\\langle v_0, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2}\\left\\langle v_0, v_0\\right\\rangle \\\\\n\n& =\\left\\|v_{m+1}\\right\\|^2+\\left\\|v_0\\right\\|^2 \\\\\n\n& >\\left\\|v_0\\right\\|^2 \\\\\n\n& =\\left|\\alpha_1\\right|^2+\\ldots+\\left|\\alpha_m\\right|^2 \\\\\n\n& =\\left|\\left\\langle v, e_1\\right\\rangle\\right|^2+\\ldots+\\left|\\left\\langle v, e_m\\right\\rangle\\right|^2 .\n\n\\end{aligned}\n\n$$\n\nBy contrapositive, if $\\left\\|v_1\\right\\|^2=\\left|\\left\\langle v, e_1\\right\\rangle\\right|^2+\\ldots+\\left|\\left\\langle v, e_m\\right\\rangle\\right|^2$, then $v \\in \\operatorname{span}\\left(e_1, \\ldots, e_m\\right)$.\n\n\\end{proof}"}
{"id": "Axler|exercise_7_5", "formal_statement": "theorem exercise_7_5 {V : Type*} [inner_product_space \u2102 V] \n  [finite_dimensional \u2102 V] (hV : finrank V \u2265 2) :\n  \u2200 U : submodule \u2102 (End \u2102 V), U.carrier \u2260\n  {T | T * T.adjoint = T.adjoint * T} :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Show that if $\\operatorname{dim} V \\geq 2$, then the set of normal operators on $V$ is not a subspace of $\\mathcal{L}(V)$.", "nl_proof": "\\begin{proof}\n\n    First off, suppose that $\\operatorname{dim} V \\geq 2$. Next let $\\left(e_1, \\ldots, e_n\\right)$ be an orthonormal basis of $V$. Now, define $S, T \\in L(V)$ by both $S\\left(a_1 e_1+\\ldots+a_n e_n\\right)=a_2 e_1-a_1 e_2$ and $T\\left(a_1 e_1+\\ldots+\\right.$ $\\left.a_n e_n\\right)=a_2 e_1+a_1 e_2$. So, just by now doing a simple calculation verifies that $S^*\\left(a_1 e_1+\\right.$ $\\left.\\ldots+a_n e_n\\right)=-a_2 e_1+a_1 e_2$\n\n\n\nNow, based on this formula, another calculation would show that $S S^*=S^* S$. Another simple calculation would that that $T$ is self-adjoint. Therefore, both $S$ and $T$ are normal. However, $S+T$ is given by the formula of $(S+T)\\left(a_1 e_1+\\ldots+a_n e_n\\right)=2 a_2 e_1$. In this case, a simple calculator verifies that $(S+T)^*\\left(a_1 e_1+\\ldots+a_n e_n\\right)=2 a_1 e_2$.\n\n\n\nTherefore, there is a final simple calculation that shows that $(S+T)(S+T)^* \\neq(S+$ $T)^*(S+T)$. So, in other words, $S+T$ isn't normal. Thereofre, the set of normal operators on $V$ isn't closed under addition and hence isn't a subspace of $L(V)$.\n\n\\end{proof}"}
{"id": "Axler|exercise_7_9", "formal_statement": "theorem exercise_7_9 {V : Type*} [inner_product_space \u2102 V]\n  [finite_dimensional \u2102 V] (T : End \u2102 V)\n  (hT : T * T.adjoint = T.adjoint * T) :\n  is_self_adjoint T \u2194 \u2200 e : T.eigenvalues, (e : \u2102).im = 0 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that a normal operator on a complex inner-product space is self-adjoint if and only if all its eigenvalues are real.", "nl_proof": "\\begin{proof}\n\n    First off, suppose $V$ is a complex inner product space and $T \\in L(V)$ is normal. If $T$ is self-adjoint, then all its eigenvalues are real. So, conversely, let all of the eigenvalues of $T$ be real. By the complex spectral theorem, there's an orthonormal basis $\\left(e_1, \\ldots, e_n\\right)$ of $V$ consisting of eigenvectors of $T$. Thus, there exists real numbers $\\lambda_1, \\ldots, \\lambda_n$ such that $T e_j=\\lambda_j e_j$ for $j=$ $1, \\ldots, n$.\n\nThe matrix of $T$ with respect to the basis of $\\left(e_1, \\ldots, e_n\\right)$ is the diagonal matrix with $\\lambda_1, \\ldots, \\lambda_n$ on the diagonal. So, the matrix equals its conjugate transpose. Therefore, $T=T^*$. In other words, $T$ s self-adjoint.\n\n\\end{proof}"}
{"id": "Axler|exercise_7_11", "formal_statement": "theorem exercise_7_11 {V : Type*} [inner_product_space \u2102 V]\n  [finite_dimensional \u2102 V] {T : End \u2102 V} (hT : T*T.adjoint = T.adjoint*T) :\n  \u2203 (S : End \u2102 V), S ^ 2 = T :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $V$ is a complex inner-product space. Prove that every normal operator on $V$ has a square root. (An operator $S \\in \\mathcal{L}(V)$ is called a square root of $T \\in \\mathcal{L}(V)$ if $S^{2}=T$.)", "nl_proof": "\\begin{proof}\n\n    Let $V$ be a complex inner product space.\n\nIt is known that an operator $S \\in \\mathcal{L}(V)$ is called a square root of $T \\in \\mathcal{L}(V)$ if\n\n$$\n\nS^2=T\n\n$$\n\nNow, suppose that $T$ is a normal operator on $V$.\n\nBy the Complex Spectral Theorem, there is $e_1, \\ldots, e_n$ an orthonormal basis of $V$ consisting of eigenvalues of $T$ and let $\\lambda_1, \\ldots, \\lambda_n$ denote their corresponding eigenvalues.\n\nDefine $S$ by\n\n$$\n\nS e_j=\\sqrt{\\lambda_j} e_j,\n\n$$\n\nfor each $j=1, \\ldots, n$.\n\nObviously, $S^2 e_j=\\lambda_j e_j=T e_j$.\n\nHence, $S^2=T$ so there exist a square root of $T$.\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_1_30", "formal_statement": "theorem exercise_1_30 {n : \u2115} : \n  \u00ac \u2203 a : \u2124, \u2211 (i : fin n), (1 : \u211a) / (n+2) = a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $\\frac{1}{2}+\\frac{1}{3}+\\cdots+\\frac{1}{n}$ is not an integer.", "nl_proof": "\\begin{proof}\n\nLet $2^s$ be the largest power of 2 occuring as a denominator in $H_n$, say $2^s=k \\leqslant n$. Write $H_n=$ $\\frac{1}{2^s}+\\left(1+1 / 2+\\ldots+1 /(k-1)+1 /(k+1)+\\ldots+1 / n\\right.$. The sum in parentheses can be written as $1 / 2^{s-1}$ times sum of fractions with odd denominators, so the denominator of the sum in parentheses will not be divisible by $2^s$, but it must equal $2^s$ by Ex $1.29$.\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_2_4", "formal_statement": "theorem exercise_2_4 {a : \u2124} (ha : a \u2260 0) \n  (f_a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a$ is a nonzero integer, then for $n>m$ show that $\\left(a^{2^{n}}+1, a^{2^{m}}+1\\right)=1$ or 2 depending on whether $a$ is odd or even.", "nl_proof": "\\begin{proof}    \n\n\\begin{align*}\n\n\\operatorname{ord}_p\\, n!  &= \\sum_{k\\geq 1} \\left \\lfloor \\frac{n}{p^{k}}\\right \\rfloor \\leq  \\sum_{k\\geq 1}  \\frac{n}{p^{k}} = \\frac{n}{p} \\frac{1}{1 - \\frac{1}{p}} = \\frac{n}{p-1}\n\n\\end{align*}\n\n\n\nThe decomposition of $n!$ in prime factors is\n\n\n\n$n! = p_1^{\\alpha_1}p_2^{\\alpha_2}\\cdots p_k^{\\alpha_k}$ \n\nwhere $\\alpha_i = \\operatorname{ord}_{p_i}\\, n! \\leq \\frac{n}{p_i-1}$, and $p_i \\leq n, \\ i=1,2,\\cdots,k$.\n\n\n\nThen\n\n\\begin{align*}\n\nn! &\\leq p_1^{\\frac{n}{p_1-1}}p_2^{\\frac{n}{p_2-1}}\\cdots p_k^{\\frac{n}{p_n-1}}\\\\\n\n\\sqrt[n]{n!} &\\leq p_1^{\\frac{1}{p_1-1}}p_2^{\\frac{1}{p_2-1}}\\cdots p_k^{\\frac{1}{p_n-1}}\\\\\n\n&\\leq \\prod_{p\\leq n} p^{\\frac{1}{p-1}}\n\n\\end{align*}\n\n(the values of $p$ in this product describe all prime numbers $p\\leq n$.)\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_2_27a", "formal_statement": "theorem exercise_2_27a : \n  \u00ac summable (\u03bb i : {p : \u2124 // squarefree p}, (1 : \u211a) / i) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that $\\sum^{\\prime} 1 / n$, the sum being over square free integers, diverges.", "nl_proof": "\\begin{proof}\n\n    \n\nLet $S \\subset \\mathbb{N}^*$ the set of square free integers.\n\n\n\nLet $N \\in \\mathbb{N}^*$. Every integer $n, \\, 1\\leq n \\leq N$ can be written as $n = a b^2$, where $a,b$ are integers and $a$ is square free. Then $1\\leq a \\leq N$, and $1\\leq b \\leq \\sqrt{N}$, so\n\n$$\\sum_{n\\leq N} \\frac{1}{n} \\leq \\sum_{a \\in S, a\\leq N}\\  \\sum_{1\\leq b \\leq \\sqrt{N}} \\frac{1}{ab^2} \\leq  \\sum_{a \\in S, a\\leq N}\\ \\frac{1}{a} \\, \\sum_{b=1}^\\infty  \\frac{1}{b^2} = \\frac{\\pi^2}{6} \\sum_{a \\in S, a\\leq N}\\ \\frac{1}{a}.$$\n\nSo $$\\sum_{a \\in S, a\\leq N} \\frac{1}{a}  \\geq \\frac{6}{\\pi^2} \\sum_{n\\leq N} \\frac{1}{n}.$$\n\nAs $\\sum_{n=1}^\\infty \\frac{1}{n}$ diverges, $\\lim\\limits_{N \\to \\infty} \\sum\\limits_{a \\in S, a\\leq N} \\frac{1}{a} = +\\infty$, so the family $\\left(\\frac{1}{a}\\right)_{a\\in S}$ of the inverse of square free integers is not summable.\n\n\n\nLet $S_N = \\prod_{p<N}(1+1/p)$ , and $p_1,p_2,\\ldots, p_l\\ (l = l(N))$ all prime integers less than $N$. Then\n\n\\begin{align*}\n\nS_N &= \\left(1+\\frac{1}{p_1}\\right) \\cdots \\left(1+\\frac{1}{p_l}\\right)\\\\\n\n&=\\sum_{(\\varepsilon_1,\\cdots,\\varepsilon_l) \\in \\{0,1\\}^l } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}}\n\n\\end{align*}\n\nWe prove this last formula  by induction. This is true for $l=1$ : $\\sum_{\\varepsilon \\in \\{0,1\\}} 1/p_1^\\varepsilon = 1 + 1/p_1$.\n\n\n\nIf it is true for the integer $l$, then \n\n\\begin{align*}\n\n\\left(1+\\frac{1}{p_1}\\right) \\cdots \\left(1+\\frac{1}{p_l}\\right)\\left(1+\\frac{1}{p_{l+1}}\\right) &= \\sum_{(\\varepsilon_1,\\ldots,\\varepsilon_l) \\in \\{0,1\\}^l } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}} \\left(1+\\frac{1}{p_{l+1}}\\right)\\\\\n\n&=\\sum_{(\\varepsilon_1,\\ldots,\\varepsilon_l) \\in \\{0,1\\}^l } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}} + \\sum_{(\\varepsilon_1,\\ldots,\\varepsilon_l) \\in \\{0,1\\}^l } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}p_{l+1}}\\\\\n\n&=\\sum_{(\\varepsilon_1,\\ldots,\\varepsilon_l,\\varepsilon_{l+1}) \\in \\{0,1\\}^{l+1} } \\frac{1}{p_1^{\\varepsilon_1} \\cdots p_l^{\\varepsilon_l}p_{l+1}^{\\varepsilon_{l+1}}} \n\n\\end{align*}\n\nSo it is true for all $l$. \n\n\n\nThus $S_N = \\sum_{n\\in \\Delta} \\frac{1}{n}$, where $\\Delta$ is the set of square free integers whose prime factors are less than $N$.\n\n\n\nAs $\\sum 1/n$, the sum being over square free integers, diverges, $\\lim\\limits_{N\\to \\infty} S_N = + \\infty$ :\n\n$$\\lim_{N \\to \\infty} \\prod_{p<N} \\left(1+\\frac{1}{p}\\right) = +\\infty.$$\n\n $e^x \\geq 1+x, x \\geq \\log (1+x)$ for $x>0$, so\n\n$$\\log S_N = \\sum_{k=1}^{l(N)} \\log\\left(1+\\frac{1}{p_k}\\right) \\leq \\sum_{k=1}^{l(N)} \\frac{1}{p_k}.$$\n\n$\\lim\\limits_{N\\to \\infty} \\log S_N = +\\infty$ and $\\lim\\limits_{N\\to \\infty} l(N) = +\\infty$, so\n\n$$\\lim_{N\\to \\infty} \\sum_{p<N} \\frac{1}{p} = +\\infty.$$\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_3_4", "formal_statement": "theorem exercise_3_4 : \u00ac \u2203 x y : \u2124, 3*x^2 + 2 = y^2 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that the equation $3 x^{2}+2=y^{2}$ has no solution in integers.", "nl_proof": "\\begin{proof}    \n\nIf $3x^2+2 = y^2$, then  $\\overline{y}^2 = \\overline{2}$ in $\\mathbb{Z}/3\\mathbb{Z}$.\n\n\n\n\n\nAs $\\{-1,0,1\\}$ is a complete set of residues modulo $3$, the squares in $\\mathbb{Z}/3\\mathbb{Z}$ are $\\overline{0} = \\overline{0}^2$ and  $\\overline{1} = \\overline{1}^2 = (\\overline{-1})^2$, so $\\overline{2}$ is not a square in $\\mathbb{Z}/3\\mathbb{Z}$ : $\\overline{y}^2 = \\overline{2}$ is impossible in $\\mathbb{Z}/3\\mathbb{Z}$.\n\n\n\nThus $3x^2+2 = y^2$ has no solution in integers.\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_3_10", "formal_statement": "theorem exercise_3_10 {n : \u2115} (hn0 : \u00ac n.prime) (hn1 : n \u2260 4) : \n  factorial (n-1) \u2261 0 [MOD n] :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $n$ is not a prime, show that $(n-1) ! \\equiv 0(n)$, except when $n=4$.", "nl_proof": "\\begin{proof}    \n\nSuppose that $n >1$ is not a prime. Then $n = uv$, where $2 \\leq u \\leq v \\leq n-1$.\n\n\n\n$\\bullet$ If $u \\neq v$, then $n = uv \\mid (n-1)! = 1\\times 2 \\times\\cdots \\times u \\times\\cdots \\times v \\times \\cdots \\times (n-1)$ (even if $u\\wedge v \\neq 1$ !).\n\n\n\n$\\bullet$ If $u=v$, $n = u^2$ is a square.\n\n\n\nIf $u$ is not prime, $u =st,\\ 2\\leq s \\leq t \\leq u-1 \\leq n-1$, and $n = u' v'$, where $u' =s,v' =st^2$ verify  $2 \\leq u' < v' \\leq n-1$. As in the first case, $n = u'v' \\mid (n-1)!$.  \n\n\n\nIf $u = p$ is a prime, then $n =p^2$.\n\n\n\nIn the case $p = 2$, $n = 4$ and $n=4  \\nmid (n-1)! = 6$. In the other case $p >2$, and $(n-1)! = (p^2 - 1)!$ contains the factors $p < 2p < p^2$, so $p^2 \\mid (p^2-1)!, n \\mid (n-1)!$.\n\n\n\nConclusion : if $n$ is not a prime, $(n - 1)! \\equiv 0 \\pmod n$, except when $n=4$.\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_4_4", "formal_statement": "theorem exercise_4_4 {p t: \u2115} (hp0 : p.prime) (hp1 : p = 4*t + 1) \n  (a : zmod p) : \n  is_primitive_root a p \u2194 is_primitive_root (-a) p :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Consider a prime $p$ of the form $4 t+1$. Show that $a$ is a primitive root modulo $p$ iff $-a$ is a primitive root modulo $p$.", "nl_proof": "\\begin{proof}\n\n    Suppose that $a$ is a primitive root modulo $p$. As $p-1$ is even, $(-a)^{p-1}=a^{p-1} \\equiv 1$ $(\\bmod p)$\n\nIf $(-a)^n \\equiv 1(\\bmod p)$, with $n \\in \\mathbb{N}$, then $a^n \\equiv(-1)^n(\\bmod p)$.\n\nTherefore $a^{2 n} \\equiv 1(\\bmod p)$. As $a$ is a primitive root modulo $p, p-1|2 n, 2 t| n$, so $n$ is even.\n\n\n\nHence $a^n \\equiv 1(\\bmod p)$, and $p-1 \\mid n$. So the least $n \\in \\mathbb{N}^*$ such that $(-a)^n \\equiv 1$ $(\\bmod p)$ is $p-1:$ the order of $-a$ modulo $p$ is $p-1,-a$ is a primitive root modulo $p$. Conversely, if $-a$ is a primitive root modulo $p$, we apply the previous result at $-a$ to to obtain that $-(-a)=a$ is a primitive root.\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_4_6", "formal_statement": "theorem exercise_4_6 {p n : \u2115} (hp : p.prime) (hpn : p = 2^n + 1) : \n  is_primitive_root 3 p :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $p=2^{n}+1$ is a Fermat prime, show that 3 is a primitive root modulo $p$.", "nl_proof": "\\begin{proof}\n\n \\newcommand{\\legendre}[2]{\\genfrac{(}{)}{}{}{#1}{#2}}\n\nWrite $p = 2^k + 1$, with $k = 2^n$.\n\n\n\nWe suppose that $n>0$, so $k\\geq 2, p \\geq 5$. As $p$ is prime, $3^{p-1} \\equiv 1 \\pmod p$. \n\n\n\nIn other words, $3^{2^k} \\equiv 1 \\pmod p$ : the order of $3$ is a divisor of $2^k$, a power of $2$.\n\n\n\n$3$ has order $2^k$ modulo $p$ iff $3^{2^{k-1}} \\not \\equiv 1 \\pmod p$. As $\\left (3^{2^{k-1}} \\right)^2 \\equiv 1 \\pmod p$, where $p$ is prime, this is equivalent to $3^{2^{k-1}}  \\equiv -1 \\pmod p$, which remains to prove.\n\n\n\n$3^{2^{k-1}} = 3^{(p-1)/2} \\equiv \\legendre{3}{p} \\pmod p$.\n\n\n\nAs the result is true for $p=5$, we can suppose $n\\geq 2$.\n\nFrom the law of quadratic reciprocity :\n\n$$\\legendre{3}{p} \\legendre{p}{3} = (-1)^{(p-1)/2} = (-1)^{2^{k-1}} = 1.$$\n\nSo $\\legendre{3}{p} = \\legendre{p}{3}$\n\n \n\n\\begin{align*}\n\np = 2^{2^n}+1 &\\equiv (-1)^{2^n} + 1 \\pmod 3\\\\\n\n&\\equiv 2 \\equiv -1 \\pmod 3,\n\n\\end{align*}\n\nso $\\legendre{3}{p} = \\legendre {p}{3} = -1$, that is to say\n\n$$3^{2^{k-1}}  \\equiv -1 \\pmod p.$$\n\nThe order of $3$ modulo $p = 2^{2^n} + 1$ is $p-1 = 2^{2^n}$ : $3$ is a primitive root modulo $p$.\n\n\n\n(On the other hand, if $3$ is of order $p-1$ modulo $p$, then $p$ is prime, so\n\n$$ F_n = 2^{2^n} + 1 \\ \\mathrm{is}\\ \\mathrm{prime}\\ \\iff 3^{(F_n-1)/2} = 3^{2^{2^n - 1}} \\equiv -1 \\pmod {F_n}.)$$\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_4_11", "formal_statement": "theorem exercise_4_11 {p : \u2115} (hp : p.prime) (k s: \u2115) \n  (s :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $1^{k}+2^{k}+\\cdots+(p-1)^{k} \\equiv 0(p)$ if $p-1 \\nmid k$ and $-1(p)$ if $p-1 \\mid k$.", "nl_proof": "\\begin{proof}    \n\nLet $S_k = 1^k+2^k+\\cdots+(p-1)^k$.\n\n\n\nLet $g$ a primitive root modulo $p$ : $\\overline{g}$ a generator of $\\mathbb{F}_p^*$.\n\n\n\nAs $(\\overline{1},\\overline{g},\\overline{g}^{2}, \\ldots, \\overline{g}^{p-2}) $ is a permutation of $ (\\overline{1},\\overline{2}, \\ldots,\\overline{p-1})$,\n\n\\begin{align*}\n\n\\overline{S_k} &= \\overline{1}^k + \\overline{2}^k+\\cdots+ \\overline{p-1}^k\\\\\n\n&= \\sum_{i=0}^{p-2} \\overline{g}^{ki} =\n\n\\left\\{\n\n\\begin{array}{ccc}\n\n\\overline{ p-1} = -\\overline{1} &  \\mathrm{if} &  p-1 \\mid k  \\\\\n\n \\frac{ \\overline{g}^{(p-1)k} -1}{ \\overline{g}^k -1} = \\overline{0}&  \\mathrm{if}  &   p-1 \\nmid k\n\n\\end{array}\n\n\\right.\n\n\\end{align*}\n\nsince $p-1 \\mid k \\iff \\overline{g}^k = \\overline{1}$.\n\n\n\nConclusion :\n\n\\begin{align*}\n\n1^k+2^k+\\cdots+(p-1)^k&\\equiv 0 \\pmod p\\ \\mathrm{if} \\ p-1 \\nmid k\\\\\n\n1^k+2^k+\\cdots+(p-1)^k&\\equiv -1 \\pmod p\\ \\mathrm{if} \\ p-1 \\mid k\\\\\n\n\\end{align*}\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_5_28", "formal_statement": "theorem exercise_5_28 {p : \u2115} (hp : p.prime) (hp1 : p \u2261 1 [MOD 4]): \n  \u2203 x, x^4 \u2261 2 [MOD p] \u2194 \u2203 A B, p = A^2 + 64*B^2 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that $x^{4} \\equiv 2(p)$ has a solution for $p \\equiv 1(4)$ iff $p$ is of the form $A^{2}+64 B^{2}$.", "nl_proof": "\\begin{proof}    \n\nIf  $p\\equiv 1\\ [4]$ and if there exists $x \\in \\mathbb{Z}$ such that $x^4 \\equiv 2\\ [p]$, then\n\n$$2^{\\frac{p-1}{4} }\\equiv  x^{p-1} \\equiv 1 \\ [p].$$ \n\n\n\nFrom Ex. 5.27, where $p = a^2 +b^2, a$ odd,  we know that $$f^{\\frac{ab}{2}} \\equiv 2^{\\frac{p-1}{4} } \\equiv 1 \\ [p].$$\n\n\n\nSince $f^2 \\equiv -1\\ [p]$, the order of $f$ modulo $p$ is 4, thus $4 \\mid \\frac{ab}{2}$, so $8\\mid ab$.\n\n\n\nAs $a$ is odd, $8 | b$, then $p = A^2 + 64 B^2$ (with $A = a, B = b/8$).\n\n\n\n\\bigskip\n\n\n\nConversely, if $p=A^2+64 B^2$, then $p\\equiv A^2 \\equiv 1 \\ [4]$.\n\n\n\nLet $a=A,b=8B$. Then $$2^{\\frac{p-1}{4} } \\equiv f^{\\frac{ab}{2}} \\equiv f^{4AB} \\equiv (-1)^{2AB} \\equiv 1 \\ [p].$$\n\n\n\nAs $2^{\\frac{p-1}{4} } \\equiv 1 \\ [p]$, $x^4 \\equiv 2 \\ [p]$ has a solution in $\\mathbb{Z}$ (Prop. 4.2.1) : $2$ is a biquadratic residue modulo $p$.\n\n\n\nConclusion : \n\n\n\n$$\\exists A \\in \\mathbb{Z}, \\exists B \\in \\mathbb{Z}\\,, p = A^2+64 B^2 \\iff( p\\equiv 1 \\ [4] \\ \\mathrm{and} \\ \\exists x \\in \\mathbb{Z}, \\, x^4 \\equiv 2 \\ [p]).$$\n\n\\end{proof}"}
{"id": "Ireland-Rosen|exercise_12_12", "formal_statement": "theorem exercise_12_12 : is_algebraic \u211a (sin (real.pi/12)) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen zsqrtd gaussian_int char_p nat.arithmetic_function \n\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that $\\sin (\\pi / 12)$ is an algebraic number.", "nl_proof": "\\begin{proof}\n\n$$\n\n\\begin{aligned}\n\n    \\sin \\pi/12=\\sin \\left(\\pi/4-\\pi/6\\right) & =\\sin \\pi/4 \\cos \\pi/6-\\cos \\pi/4 \\sin \\pi/6 \\\\\n\n& =\\frac{\\sqrt{3}}{2 \\sqrt{2}}-\\frac{1}{2 \\sqrt{2}} \\\\\n\n& =\\frac{\\sqrt{3}-1}{2 \\sqrt{2}}\n\n\\end{aligned}\n\n$$\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_1_13b", "formal_statement": "theorem exercise_1_13b {f : \u2102 \u2192 \u2102} (\u03a9 : set \u2102) (a b : \u03a9) (h : is_open \u03a9)\n  (hf : differentiable_on \u2102 f \u03a9) (hc : \u2203 (c : \u211d), \u2200 z \u2208 \u03a9, (f z).im = c) :\n  f a = f b :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Suppose that $f$ is holomorphic in an open set $\\Omega$. Prove that if $\\text{Im}(f)$ is constant, then $f$ is constant.", "nl_proof": "\\begin{proof}\n\nLet $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.\n\nSince $\\operatorname{Im}(f)=$ constant,\n\n$$\n\n\\frac{\\partial v}{\\partial x}=0, \\frac{\\partial v}{\\partial y}=0 .\n\n$$\n\nBy the Cauchy-Riemann equations,\n\n$$\n\n\\frac{\\partial u}{\\partial x}=\\frac{\\partial v}{\\partial y}=0 .\n\n$$\n\nThus in $\\Omega$,\n\n$$\n\nf^{\\prime}(z)=\\frac{\\partial f}{\\partial x}=\\frac{\\partial u}{\\partial x}+i \\frac{\\partial v}{\\partial x}=0+0=0 .\n\n$$\n\nThus $f$ is constant.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_1_19a", "formal_statement": "theorem exercise_1_19a (z : \u2102) (hz : abs z = 1) (s : \u2115 \u2192 \u2102)\n    (h : s = (\u03bb n, \u2211 i in (finset.range n), i * z ^ i)) :\n    \u00ac \u2203 y, tendsto s at_top (\ud835\udcdd y) :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Prove that the power series $\\sum nz^n$ does not converge on any point of the unit circle.", "nl_proof": "\\begin{proof}\n\n    For $z \\in S:=\\{z \\in \\mathbb{C}:|z|=1\\}$ it also holds $z^n \\in S$ for all $n \\in \\mathbb{N}$ (since in this case $\\left.\\left|z^n\\right|=|z|^n=1^n=1\\right)$\n\nThus, the sequence $\\left(a_n\\right)_{n \\in \\mathbb{N}}$ with $a_n=n z^n$ does not converge to zero which is necessary for the corresponding sum $\\sum_{n \\in \\mathbb{N}} a_n$ to be convergent. Hence this sum does not converge.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_1_19c", "formal_statement": "theorem exercise_1_19c (z : \u2102) (hz : abs z = 1) (hz2 : z \u2260 1) (s : \u2115 \u2192 \u2102)\n    (h : s = (\u03bb n, \u2211 i in (finset.range n), i * z / i)) :\n    \u2203 z, tendsto s at_top (\ud835\udcdd z) :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Prove that the power series $\\sum zn/n$ converges at every point of the unit circle except $z = 1$.", "nl_proof": "\\begin{proof}\n\n    If $z=1$ then $\\sum z^n / n=\\sum 1 / n$ is divergent (harmonic series). If $|z|=1$ and $z \\neq 1$, write $z=e^{2 \\pi i t}$ with $t \\in(0,1)$ and apply Dirichlet's test: if $\\left\\{a_n\\right\\}$ is a sequence of real numbers and $\\left\\{b_n\\right\\}$ a sequence of complex numbers satisfying\n\n- $a_{n+1} \\leq a_n$\n\n- $\\lim _{n \\rightarrow \\infty} a_n=0$\n\n- $\\left|\\sum_{n=1}^N b_n\\right| \\leq M$ for every positive integer $N$ and some $M>0$,\n\nthen $\\sum a_n b_n$ converges. Let $a_n=1 / n$, so $a_n$ satisfies $a_{n+1} \\leq a_n$ and $\\lim _{n \\rightarrow \\infty} a_n=0$. Let $b_n=e^{2 \\pi i n t}$, then\n\n$$\n\n\\left|\\sum_{n=1}^N b_n\\right|=\\left|\\sum_{n=1}^N e^{2 \\pi i n t}\\right|=\\left|\\frac{e^{2 \\pi i t}-e^{2 \\pi i(N+1) t}}{1-e^{2 \\pi i t}}\\right| \\leq \\frac{2}{\\left|1-e^{2 \\pi i t}\\right|}=M \\text { for all } N\n\n$$\n\nThus $\\sum a_n b_n=\\sum z^n / n$ converges for every point in the unit circle except $z=1$.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_2_2", "formal_statement": "theorem exercise_2_2 :\n  tendsto (\u03bb y, \u222b x in 0..y, real.sin x / x) at_top (\ud835\udcdd (real.pi / 2)) :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Show that $\\int_{0}^{\\infty} \\frac{\\sin x}{x} d x=\\frac{\\pi}{2}$.", "nl_proof": "\\begin{proof}\n\n    We have $\\int_0^{\\infty} \\frac{\\sin x}{x} d x=\\frac{1}{2 * i} \\int_0^{\\infty} \\frac{e^{i * x}-e^{-i * x}}{x} d x=\\frac{1}{2 * i}\\left(\\int_0^{\\infty} \\frac{e^{i * x}-1}{x} d x-\\int_0^{\\infty} \\frac{e^{-i * x}-1}{x} d x=\\right.$ $\\frac{1}{2 * i} \\int_{-\\infty}^{\\infty} \\frac{e^{i * x}-1}{x} d x$. Now integrate along the big and small semicircles $C_0$ and $C_1$ shown below. For $C_0$ : we have that $\\int_{C_0} \\frac{1}{x} d x=\\pi * i$ and $\\left|\\int_{C_0} \\frac{e^{i * x}}{x} d x\\right| \\leq$ $2 *\\left|\\int_{C_{00}} \\frac{e^{i * x}}{x} d x\\right|+\\left|\\int_{C_{01}} \\frac{e^{i * x}}{x} d x\\right|$ where $C_{00}$ and $C_{01}$ are shown below $\\left(C_{01}\\right.$ contains the part of $C_0$ that has points with imaginary parts more than $a$ and $C_{00}$ is one of the other 2 components). We have $\\left|\\int_{C_{00}} \\frac{e^{i * x}}{x} d x\\right| \\leq$ $\\sup _{x \\in C_{00}}\\left(e^{i * x}\\right) / R * \\int_{C_{00}}|d x| \\leq e^{-a} * \\pi$ and $\\left|\\int_{C_{01}} \\frac{e^{i * x}}{x} d x\\right| \\leq\\left|\\int_{C_{01}} \\frac{1}{x} d x\\right| \\leq$ $\\frac{1}{R} * C * a$ for some constant $C$ (the constant $C$ exists because the length of the curve approaches $a$ as $a / R \\rightarrow 0)$. Thus, the integral of $e^{i * x} / x$ over $C_0$ is bounded by $A * e^{-a}+B * a / R$ for some constants $A$ and $B$. Pick $R$ large and $a=\\sqrt{R}$ and note that the above tends to 0 . About the integral over $C_1$ : We have $e^{i * x}-1=1+O(x)$ for $x \\rightarrow 0$ (this is again from $\\sin (x) / x \\rightarrow 1$ ),\n\n4\n\nso $\\left|\\int_{C_1} \\frac{e^{i * x}-1}{x} d x\\right| \\leq O(1) *\\left|\\int_{C_1} d x\\right| \\rightarrow 0$ as $x \\rightarrow 0$. Thus, we only care about the integral over $C_{00}$ which is $-\\pi * i$. Using Cauchy's theorem we get that our integral equals $\\frac{1}{2 * i}(-(\\pi * i))=\\pi / 2$.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_2_13", "formal_statement": "theorem exercise_2_13 {f : \u2102 \u2192 \u2102}\n    (hf : \u2200 z\u2080 : \u2102, \u2203 (s : set \u2102) (c : \u2115 \u2192 \u2102), is_open s \u2227 z\u2080 \u2208 s \u2227\n      \u2200 z \u2208 s, tendsto (\u03bb n, \u2211 i in finset.range n, (c i) * (z - z\u2080)^i) at_top (\ud835\udcdd (f z\u2080))\n      \u2227 \u2203 i, c i = 0) :\n    \u2203 (c : \u2115 \u2192 \u2102) (n : \u2115), f = \u03bb z, \u2211 i in finset.range n, (c i) * z ^ n :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Suppose $f$ is an analytic function defined everywhere in $\\mathbb{C}$ and such that for each $z_0 \\in \\mathbb{C}$ at least one coefficient in the expansion $f(z) = \\sum_{n=0}^\\infty c_n(z - z_0)^n$ is equal to 0. Prove that $f$ is a polynomial.", "nl_proof": "\\begin{proof}\n\nSay that at least one of the coefficients of the Taylor series vanishes is the same as saying that for every $a \\in \\mathbb{C}$ there is $m \\in \\mathbb{N}$ such that $f^{(m)}(a)=0$.\n\nConsider $A_n:=\\left\\{z \\in \\mathbb{C}: f^{(n)}(z)=0\\right\\}$ for each $n \\in \\mathbb{N}$. Note that:\n\n$f$ is polynomial iff $A_n$ is not countable for some $n \\in \\mathbb{N}$.\n\nIndeed, if $f$ is polynomial of degree $n$, then $f^{(n+1)}(z)=0$ for all $z \\in \\mathbb{C}$, then $A_{n+1}=\\mathbb{C}$, so, $A_{n+1}$ is not countable. Conversely, if there is $n \\in \\mathbb{C}$ such that $A_n$ is not countable, then $A_n$ has a limit point, then by Identity principle we have $f^{(n)}(z)=0$ for all $z \\in \\mathbb{C}$, so, $f$ is a polynomial of degree at most $n-1$.\n\n\n\nTherefore, tt suffices to show that there is $n \\in \\mathbb{N}$ such that $A_n$ is not countable. Indeed, consider $\\bigcup_{n \\in \\mathbb{N}} A_n$, by hypothesis for each $a \\in \\mathbb{C}$ there is $m \\in \\mathbb{N}$ such that $f^{(m)}(a)=0$, then $\\mathbb{C} \\subseteq \\bigcup_{n \\in \\mathbb{N}} A_n$. Therefore, $\\bigcup_{n \\in \\mathbb{N}} A_n$ is not countable, then there is $n \\in \\mathbb{N}$ such that $A_n$ is not countable.\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_3_4", "formal_statement": "theorem exercise_3_4 (a : \u211d) (ha : 0 < a) :\n    tendsto (\u03bb y, \u222b x in -y..y, x * real.sin x / (x ^ 2 + a ^ 2))\n    at_top (\ud835\udcdd (real.pi * (real.exp (-a)))) :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Show that $ \\int_{-\\infty}^{\\infty} \\frac{x \\sin x}{x^2 + a^2} dx = \\pi e^{-a}$ for $a > 0$.", "nl_proof": "\\begin{proof}\n\n$$\n\nx /\\left(x^2+a^2\\right)=x / 2 i a(1 /(x-i a)-1 /(x+i a))=1 / 2 i a(i a /(x-i a)+i a /(x+\n\n$$\n\n$i a))=(1 /(x-i a)+1 /(x+i a)) / 2$. So we care about $\\sin (x)(1 /(x-i a)+$ $1 /(x+i a)) / 2$. Its residue at $x=i a$ is $\\sin (i a) / 2=\\left(e^{-a}-e^a\\right) / 4 i$.?\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_3_14", "formal_statement": "theorem exercise_3_14 {f : \u2102 \u2192 \u2102} (hf : differentiable \u2102 f)\n    (hf_inj : function.injective f) :\n    \u2203 (a b : \u2102), f = (\u03bb z, a * z + b) \u2227 a \u2260 0 :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Prove that all entire functions that are also injective take the form $f(z) = az + b$, $a, b \\in \\mathbb{C}$ and $a \\neq 0$.", "nl_proof": "\\begin{proof}\n\nLook at $f(1 / z)$. If it has an essential singularity at 0 , then pick any $z_0 \\neq 0$. Now we know that the range of $f$ is dense as $z \\rightarrow 0$. We also know that the image of $f$ in some small ball around $z_0$ contains a ball around $f\\left(z_0\\right)$. But this means that the image of $f$ around this ball intersects the image of $f$ in any arbitrarily small ball around 0 (because of the denseness). Thus, $f$ cannot be injective. So the singularity at 0 is not essential, so $f(1 / z)$ is some polynomial of $1 / z$, so $f$ is some polynomial of $z$. If its degree is more than 1 it is not injective (fundamental theorem of algebra), so the degree of $f$ is 1 .\n\n\\end{proof}"}
{"id": "Shakarchi|exercise_5_1", "formal_statement": "theorem exercise_5_1 (f : \u2102 \u2192 \u2102) (hf : differentiable_on \u2102 f (ball 0 1))\n  (hb : bounded (set.range f)) (h0 : f \u2260 0) (zeros : \u2115 \u2192 \u2102) (hz : \u2200 n, f (zeros n) = 0)\n  (hzz : set.range zeros = {z | f z = 0 \u2227 z \u2208 (ball (0 : \u2102) 1)}) :\n  \u2203 (z : \u2102), tendsto (\u03bb n, (\u2211 i in finset.range n, (1 - zeros i))) at_top (\ud835\udcdd z) :=", "src_header": "import .common \n\nopen complex filter function interval_integral metric\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology\n\n", "nl_statement": "Prove that if $f$ is holomorphic in the unit disc, bounded and not identically zero, and $z_{1}, z_{2}, \\ldots, z_{n}, \\ldots$ are its zeros $\\left(\\left|z_{k}\\right|<1\\right)$, then $\\sum_{n}\\left(1-\\left|z_{n}\\right|\\right)<\\infty$.", "nl_proof": "\\begin{proof}\n\n    Fix $\\mathrm{N}$ and let $D(0, R)$ contains the first $\\mathrm{N}$ zeroes of f. Let $S_N=\\sum_{k=1}^N\\left(1-\\left|z_k\\right|\\right)=$ $\\sum_{k=1}^N \\int_{\\left|z_k\\right|}^1 1 d r$. Let $\\eta_k$ be the characteristic function of the interval $\\left.\\| z_k \\mid, 1\\right]$. We have $S_N=\\sum_{k=1}^N \\int_0^1 \\eta(r) d r=\\int_0^1\\left(\\sum_{k=1}^N \\eta_k(r)\\right) d r \\leq \\int_0^1 n(r) d r$, where $n(r)$ is the number of zeroes of $f$ at the disk $D(0, r)$. For $r \\leq 1$ we have $n(r) \\leq \\frac{n(r)}{r}$. This means that $S_N \\leq \\int_0^1 n(r) \\frac{d r}{r}$. If $f(0)=0$ then we have $f(z)=z^m g(z)$ for some integer $\\mathrm{m}$ and some holomorphic $g$ with $g(0) \\neq 0$. The other zeroes of $\\mathrm{f}$ are precisely the zeroes of $g$. Thus we have reduced the problem to $f(0) \\neq 0$. By the Corollary of the Jensen's equality we get $S_N \\leq \\int_0^1 n(r) \\frac{d r}{r}=\\frac{1}{2 \\pi} \\int_0^{2 \\pi} \\log \\left|f\\left(R e^{i \\pi}\\right)\\right| d \\phi-\\log |f(0)|<M$ since $f$ is bounded. The partial sums of the series are boundend and therefore the series converges.\n\n\\end{proof}"}
{"id": "Putnam|exercise_2018_a5", "formal_statement": "theorem exercise_2018_a5 (f : \u211d \u2192 \u211d) (hf : cont_diff \u211d \u22a4 f)\n  (hf0 : f 0 = 0) (hf1 : f 1 = 1) (hf2 : \u2200 x, f x \u2265 0) :\n  \u2203 (n : \u2115) (x : \u211d), iterated_deriv n f x = 0 :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Let $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ be an infinitely differentiable function satisfying $f(0)=0, f(1)=1$, and $f(x) \\geq 0$ for all $x \\in$ $\\mathbb{R}$. Show that there exist a positive integer $n$ and a real number $x$ such that $f^{(n)}(x)<0$.", "nl_proof": "\\begin{proof}\n\n    Call a function $f\\colon \\mathbb{R} \\to \\mathbb{R}$ \\textit{ultraconvex} if $f$ is infinitely differentiable and $f^{(n)}(x) \\geq 0$ for all $n \\geq 0$ and all $x \\in \\mathbb{R}$, where $f^{(0)}(x) = f(x)$;\n\nnote that if $f$ is ultraconvex, then so is $f'$.\n\nDefine the set\n\n\\[\n\nS = \\{ f :\\thinspace \\mathbb{R} \\to \\mathbb{R} \\,|\\,f \\text{ ultraconvex and } f(0)=0\\}.\n\n\\]\n\nFor $f \\in S$, we must have $f(x) = 0$ for all $x < 0$: if $f(x_0) > 0$ for some $x_0 < 0$, then\n\nby the mean value theorem there exists $x \\in (0,x_0)$ for which $f'(x) = \\frac{f(x_0)}{x_0} < 0$.\n\nIn particular, $f'(0) = 0$, so $f' \\in S$ also.\n\n\n\nWe show by induction that for all $n \\geq 0$,\n\n\\[\n\nf(x) \\leq \\frac{f^{(n)}(1)}{n!} x^n \\qquad (f \\in S, x \\in [0,1]).\n\n\\]\n\nWe induct with base case $n=0$, which holds because any $f \\in S$ is nondecreasing. Given the claim for $n=m$,\n\nwe apply the induction hypothesis to $f' \\in S$ to see that\n\n\\[\n\nf'(t) \\leq \\frac{f^{(n+1)}(1)}{n!} t^n \\qquad (t \\in [0,1]),\n\n\\]\n\nthen integrate both sides from $0$ to $x$ to conclude.\n\n\n\nNow for $f \\in S$, we have $0 \\leq f(1) \\leq \\frac{f^{(n)}(1)}{n!}$ for all $n \\geq 0$. \n\nOn the other hand, by Taylor's theorem with remainder,\n\n\\[\n\nf(x) \\geq \\sum_{k=0}^n \\frac{f^{(k)}(1)}{k!}(x-1)^k \\qquad (x \\geq 1).\n\n\\]\n\nApplying this with $x=2$, we obtain $f(2) \\geq \\sum_{k=0}^n \\frac{f^{(k)}(1)}{k!}$ for all $n$;\n\nthis implies that $\\lim_{n\\to\\infty}  \\frac{f^{(n)}(1)}{n!} = 0$.\n\nSince $f(1) \\leq \\frac{f^{(n)}(1)}{n!}$, we must have $f(1) = 0$.\n\n\n\nFor $f \\in S$, we proved earlier that $f(x) = 0$ for all $x\\leq 0$, as well as for $x=1$. Since\n\nthe function $g(x) = f(cx)$ is also ultraconvex for $c>0$, we also have $f(x) = 0$ for all $x>0$;\n\nhence $f$ is identically zero.\n\n\n\nTo sum up, if $f\\colon \\mathbb{R} \\to \\mathbb{R}$ is infinitely differentiable, $f(0)=0$, and $f(1) = 1$,\n\nthen $f$ cannot be ultraconvex. This implies the desired result.\n\n\\end{proof}"}
{"id": "Putnam|exercise_2018_b4", "formal_statement": "theorem exercise_2018_b4 (a : \u211d) (x : \u2115 \u2192 \u211d) (hx0 : x 0 = a)\n  (hx1 : x 1 = a) \n  (hxn : \u2200 n : \u2115, n \u2265 2 \u2192 x (n+1) = 2*(x n)*(x (n-1)) - x (n-2))\n  (h : \u2203 n, x n = 0) : \n  \u2203 c, function.periodic x c :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Given a real number $a$, we define a sequence by $x_{0}=1$, $x_{1}=x_{2}=a$, and $x_{n+1}=2 x_{n} x_{n-1}-x_{n-2}$ for $n \\geq 2$. Prove that if $x_{n}=0$ for some $n$, then the sequence is periodic.", "nl_proof": "\\begin{proof}\n\n    We first rule out the case $|a|>1$. In this case, we prove that $|x_{n+1}| \\geq |x_n|$ for all $n$, meaning that we cannot have $x_n = 0$. We proceed by induction; the claim is true for $n=0,1$ by hypothesis. To prove the claim for  $n \\geq 2$, write\n\n\\begin{align*}\n\n|x_{n+1}| &= |2x_nx_{n-1}-x_{n-2}| \\\\\n\n&\\geq 2|x_n||x_{n-1}|-|x_{n-2}| \\\\\n\n&\\geq |x_n|(2|x_{n-1}|-1) \\geq |x_n|,\n\n\\end{align*} \n\nwhere the last step follows from $|x_{n-1}| \\geq |x_{n-2}| \\geq \\cdots \\geq |x_0| = 1$.\n\n\n\nWe may thus assume hereafter that $|a|\\leq 1$. We can then write $a = \\cos b$ for some $b \\in [0,\\pi]$. \n\nLet $\\{F_n\\}$ be the Fibonacci sequence, defined as usual by $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1}$. We show by induction that\n\n\\[\n\nx_n = \\cos(F_n b) \\qquad (n \\geq 0).\n\n\\]\n\nIndeed, this is true for $n=0,1,2$; given that it is true for $n \\leq m$, then\n\n\\begin{align*}\n\n2x_mx_{m-1}&=2\\cos(F_mb)\\cos(F_{m-1}b) \\\\\n\n&= \\cos((F_m-F_{m-1})b)+\\cos((F_m+F_{m-1})b) \\\\\n\n&= \\cos(F_{m-2}b)+\\cos(F_{m+1}b)\n\n\\end{align*}\n\nand so \n\n$x_{m+1} = 2x_mx_{m-1}-x_{m-2} = \\cos(F_{m+1}b)$. This completes the induction.\n\n\n\n\n\nSince $x_n = \\cos(F_n b)$, if $x_n=0$ for some $n$ then $F_n b = \\frac{k}{2} \\pi$ for some odd integer $k$. In particular, we can write $b = \\frac{c}{d}(2\\pi)$ where $c = k$ and $d = 4F_n$ are integers.\n\n\n\n\n\nLet $x_n$ denote the pair $(F_n,F_{n+1})$, where each entry in this pair is viewed as an element of $\\mathbb{Z}/d\\mathbb{Z}$. Since there are only finitely many possibilities for $x_n$, there must be some $n_2>n_1$ such that $x_{n_1}=x_{n_2}$. Now $x_n$ uniquely determines both $x_{n+1}$ and $x_{n-1}$, and it follows that the sequence $\\{x_n\\}$ is periodic: for $\\ell = n_2-n_1$, $x_{n+\\ell} = x_n$ for all $n \\geq 0$. In particular, $F_{n+\\ell} \\equiv F_n \\pmod{d}$ for all $n$. But then $\\frac{F_{n+\\ell}c}{d}-\\frac{F_n c}{d}$ is an integer, and so\n\n\\begin{align*}\n\nx_{n+\\ell} &= \\cos\\left(\\frac{F_{n+\\ell}c}{d}(2\\pi)\\right)\\\\\n\n& = \\cos\\left(\\frac{F_n c}{d}(2\\pi)\\right) = x_n\n\n\\end{align*}\n\nfor all $n$. Thus the sequence $\\{x_n\\}$ is periodic, as desired.\n\n\\end{proof}"}
{"id": "Putnam|exercise_2014_a5", "formal_statement": "theorem exercise_2014_a5 (P : \u2115 \u2192 polynomial \u2124) \n  (hP : \u2200 n, P n = \u2211 (i : fin n), (n+1) * X ^ n) : \n  \u2200 (j k : \u2115), j \u2260 k \u2192 is_coprime (P j) (P k) :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Let", "nl_proof": ""}
{"id": "Putnam|exercise_2001_a5", "formal_statement": "theorem exercise_2001_a5 : \n  \u2203! a n : \u2115, a > 0 \u2227 n > 0 \u2227 a^(n+1) - (a+1)^n = 2001 :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Prove that there are unique positive integers $a, n$ such that $a^{n+1}-(a+1)^n=2001$.", "nl_proof": "\\begin{proof}\n\n    Suppose $a^{n+1} - (a+1)^n = 2001$.\n\nNotice that $a^{n+1} + [(a+1)^n - 1]$ is a multiple of $a$; thus\n\n$a$ divides $2002 = 2 \\times 7 \\times 11 \\times 13$.\n\n\n\nSince $2001$ is divisible by 3, we must have $a \\equiv 1 \\pmod{3}$,\n\notherwise one of $a^{n+1}$ and $(a+1)^n$ is a multiple of 3 and the\n\nother is not, so their difference cannot be divisible by 3. Now\n\n$a^{n+1} \\equiv 1 \\pmod{3}$, so we must have $(a+1)^n \\equiv 1\n\n\\pmod{3}$, which forces $n$ to be even, and in particular at least 2.\n\n\n\nIf $a$ is even, then $a^{n+1} - (a+1)^n \\equiv -(a+1)^n \\pmod{4}$.\n\nSince $n$ is even, $-(a+1)^n \\equiv -1 \\pmod{4}$. Since $2001 \\equiv 1\n\n\\pmod{4}$, this is impossible. Thus $a$ is odd, and so must divide\n\n$1001 = 7 \\times 11 \\times 13$. Moreover, $a^{n+1} - (a+1)^n \\equiv a\n\n\\pmod{4}$, so $a \\equiv 1 \\pmod{4}$.\n\n\n\nOf the divisors of $7 \\times 11 \\times 13$, those congruent to 1 mod 3\n\nare precisely those not divisible by 11 (since 7 and 13 are both\n\ncongruent to 1 mod 3). Thus $a$ divides $7 \\times 13$. Now\n\n$a \\equiv 1 \\pmod{4}$ is only possible if $a$ divides $13$.\n\n\n\nWe cannot have $a=1$, since $1 - 2^n \\neq 2001$ for any $n$. Thus\n\nthe only possibility is $a = 13$. One easily checks that $a=13, n=2$ is a\n\nsolution; all that remains is to check that no other $n$ works. In fact,\n\nif $n > 2$, then $13^{n+1} \\equiv 2001 \\equiv 1 \\pmod{8}$.\n\nBut $13^{n+1} \\equiv 13 \\pmod{8}$ since $n$ is even, contradiction.\n\nThus $a=13, n=2$ is the unique solution.\n\n\n\nNote: once one has that $n$ is even, one can use that $2002\n\n=a^{n+1} + 1 - (a+1)^n$ is divisible by $a+1$ to rule out cases.\n\n\\end{proof}"}
{"id": "Putnam|exercise_1999_b4", "formal_statement": "theorem exercise_1999_b4 (f : \u211d \u2192 \u211d) (hf: cont_diff \u211d 3 f) \n  (hf1 : \u2200 (n \u2264 3) (x : \u211d), iterated_deriv n f x > 0) \n  (hf2 : \u2200 x : \u211d, iterated_deriv 3 f x \u2264 f x) : \n  \u2200 x : \u211d, deriv f x < 2 * f x :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Let $f$ be a real function with a continuous third derivative such that $f(x), f^{\\prime}(x), f^{\\prime \\prime}(x), f^{\\prime \\prime \\prime}(x)$ are positive for all $x$. Suppose that $f^{\\prime \\prime \\prime}(x) \\leq f(x)$ for all $x$. Show that $f^{\\prime}(x)<2 f(x)$ for all $x$.", "nl_proof": "\\begin{proof}    \n\n\\setcounter{equation}{0}\n\nWe make repeated use of the following fact: if $f$ is a differentiable function on all of\n\n$\\mathbb{R}$, $\\lim_{x \\to -\\infty} f(x) \\geq 0$, and $f'(x) > 0$ for all $x \\in \\mathbb{R}$, then\n\n$f(x) > 0$ for all $x \\in \\mathbb{R}$. (Proof: if $f(y) < 0$ for some $x$, then $f(x)< f(y)$ for all\n\n$x<y$ since $f'>0$, but then $\\lim_{x \\to -\\infty} f(x) \\leq f(y) < 0$.)\n\n\n\nFrom the inequality $f'''(x) \\leq f(x)$ we obtain\n\n\\[\n\nf'' f'''(x) \\leq f''(x) f(x) < f''(x) f(x) + f'(x)^2\n\n\\]\n\nsince $f'(x)$ is positive. Applying the fact to the difference between the right and left sides,\n\nwe get\n\n\\begin{equation}\n\n\\frac{1}{2} (f''(x))^2 < f(x) f'(x).\n\n\\end{equation}\n\n\n\nOn the other hand, since $f(x)$ and $f'''(x)$ are both positive for all $x$,\n\nwe have\n\n\\[\n\n2f'(x) f''(x) < 2f'(x)f''(x) + 2f(x) f'''(x).\n\n\\]\n\nApplying the fact to the difference between the sides yields\n\n\\begin{equation}\n\nf'(x)^2 \\leq 2f(x) f''(x).\n\n\\end{equation}\n\nCombining (1) and (2), we obtain\n\n\\begin{align*}\n\n\\frac{1}{2} \\left( \\frac{f'(x)^2}{2f(x)} \\right)^2\n\n&< \\frac{1}{2} (f''(x))^2 \\\\\n\n&< f(x) f'(x),\n\n\\end{align*}\n\nor $(f'(x))^3 < 8 f(x)^3$. We conclude $f'(x) < 2f(x)$, as desired.\n\n\\end{proof}"}
{"id": "Putnam|exercise_1998_b6", "formal_statement": "theorem exercise_1998_b6 (a b c : \u2124) : \n  \u2203 n : \u2124, n > 0 \u2227 \u00ac \u2203 m : \u2124, sqrt (n^3 + a*n^2 + b*n + c) = m :=", "src_header": "import .common \n\nopen real topological_space filter polynomial\nopen_locale topology big_operators complex_conjugate filter ennreal \n\n", "nl_statement": "Prove that, for any integers $a, b, c$, there exists a positive integer $n$ such that $\\sqrt{n^3+a n^2+b n+c}$ is not an integer.", "nl_proof": "\\begin{proof}\n\n    We prove more generally that for any polynomial $P(z)$ with integer\n\ncoefficients which is not a perfect square, there exists a positive\n\ninteger $n$ such that $P(n)$ is not a perfect square. Of course it\n\nsuffices to assume $P(z)$ has no repeated factors, which is to say $P(z)$\n\nand its derivative $P'(z)$ are relatively prime.\n\n\n\nIn particular, if we carry out the Euclidean algorithm on $P(z)$ and $P'(z)$\n\nwithout dividing, we get an integer $D$ (the discriminant of $P$) such that\n\nthe greatest common divisor of $P(n)$ and $P'(n)$ divides $D$ for any $n$.\n\nNow there exist infinitely many primes $p$ such that $p$ divides $P(n)$ for\n\nsome $n$: if there were only finitely many, say, $p_1, \\dots, p_k$, then\n\nfor any $n$ divisible by $m = P(0) p_1 p_2 \\cdots p_k$, we have $P(n)\n\n\\equiv P(0) \\pmod{m}$, that is, $P(n)/P(0)$ is not divisible by $p_1,\n\n\\dots, p_k$, so must be $\\pm 1$, but then $P$ takes some value infinitely\n\nmany times, contradiction. In particular, we can choose some such $p$ not\n\ndividing $D$, and choose $n$ such that $p$ divides $P(n)$. Then $P(n+kp)\n\n\\equiv P(n) + kp P'(n) (\\mathrm{mod}\\,p)$\n\n(write out the Taylor series of the left side);\n\nin particular, since $p$ does not divide $P'(n)$, we can find some $k$\n\nsuch that $P(n+kp)$ is divisible by $p$ but not by $p^2$, and so\n\nis not a perfect square.\n\n\\end{proof}"}
{"id": "Pugh|exercise_2_26", "formal_statement": "theorem exercise_2_26 {M : Type*} [topological_space M]\n  (U : set M) : is_open U \u2194 \u2200 x \u2208 U, \u00ac cluster_pt x (\ud835\udcdf U\u1d9c) :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Prove that a set $U \\subset M$ is open if and only if none of its points are limits of its complement.", "nl_proof": "\\begin{proof}\n\n    Assume that none of the points of $U$ are limits of its complement, and let us prove that $U$ is open. Assume by contradiction that $U$ is not open, so there exists $p \\in M$ so that $\\forall r>0$ there exists $q \\in M$ with $d(p, q)<r$ but $q \\notin U$. Applying this to $r=1 / n$ we obtain $q_n \\in U^c$ such that $d\\left(q_n, p\\right)<1 / n$. But then $q_n \\rightarrow p$ and $p$ is a limit of a sequence of points in $U^c$, a contradiction.\n\n\n\nAssume now that $U$ is open. Assume by contradiction there exists $p \\in U$ and $p_n \\in U^c$ such that $p_n \\rightarrow p$. Since $U$ is open, there exists $r>0$ such that $d(p, x)<r$ for $x \\in M$ implies $x \\in U$. But since $p_n \\rightarrow p$, there exists $n_0 \\in \\mathbb{N}$ such that $n \\geq n_0$ implies $d\\left(p_n, p\\right)<r$, therefore $p_n \\in U$ for $n \\geq n_0$, a contradiction since $p_n \\in U^c$.\n\n\\end{proof}"}
{"id": "Pugh|exercise_2_32a", "formal_statement": "theorem exercise_2_32a (A : set \u2115) : is_clopen A :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Show that every subset of $\\mathbb{N}$ is clopen.", "nl_proof": "\\begin{proof}\n\n    32. The one-point set $\\{n\\} \\subset \\mathbb{N}$ is open, since it contains all $m \\in \\mathbb{N}$ that satisfy $d(m, n)<\\frac{1}{2}$. Every subset of $\\mathbb{N}$ is a union of one-point sets, hence is open. Then every set it closed, since its complement is necessarily open.\n\n\\end{proof}"}
{"id": "Pugh|exercise_2_46", "formal_statement": "theorem exercise_2_46 {M : Type*} [metric_space M]\n  {A B : set M} (hA : is_compact A) (hB : is_compact B)\n  (hAB : disjoint A B) (hA\u2080 : A \u2260 \u2205) (hB\u2080 : B \u2260 \u2205) :\n  \u2203 a\u2080 b\u2080, a\u2080 \u2208 A \u2227 b\u2080 \u2208 B \u2227 \u2200 (a : M) (b : M),\n  a \u2208 A \u2192 b \u2208 B \u2192 dist a\u2080 b\u2080 \u2264 dist a b :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Assume that $A, B$ are compact, disjoint, nonempty subsets of $M$. Prove that there are $a_0 \\in A$ and $b_0 \\in B$ such that for all $a \\in A$ and $b \\in B$ we have $d(a_0, b_0) \\leq d(a, b)$.", "nl_proof": "\\begin{proof}\n\nLet $A$ and $B$ be compact, disjoint and non-empty subsets of $M$. We want to show that there exist $a_0 \\in A, b_0 \\in B$ such that for all $a \\in A, b \\in B$,\n\n$$\n\nd\\left(a_0, b_0\\right) \\leq d(a, b) .\n\n$$\n\nWe saw in class that the distance function $d: M \\times M \\rightarrow \\mathbb{R}$ is continuous. We also saw in class that any continuous, real-valued function assumes maximum and minimum values on a compact set. Since $A$ and $B$ are compact, $A \\times B$ is (non-empty) compact in $M \\times M$. Therefore there exists $\\left(a_0, b_0\\right) \\in A \\times B$ such that $d\\left(a_0, b_0\\right) \\leq d(a, b), \\forall(a, b) \\in A \\times B$.\n\n\\end{proof}"}
{"id": "Pugh|exercise_2_92", "formal_statement": "theorem exercise_2_92 {\u03b1 : Type*} [topological_space \u03b1]\n  {s : \u2115 \u2192 set \u03b1}\n  (hs : \u2200 i, is_compact (s i))\n  (hs : \u2200 i, (s i).nonempty)\n  (hs : \u2200 i, (s i) \u2283 (s (i + 1))) :\n  (\u22c2 i, s i).nonempty :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Give a direct proof that the nested decreasing intersection of nonempty covering compact sets is nonempty.", "nl_proof": "\\begin{proof}\n\n    Let\n\n$$\n\nA_1 \\supset A_2 \\supset \\cdots \\supset A_n \\supset \\cdots\n\n$$\n\nbe a nested decreasing sequence of compacts. Suppose that $\\bigcap A_n=\\emptyset$. Take $U_n=A_n^c$, then\n\n$$\n\n\\bigcup U_n=\\bigcup A_n^c=\\left(\\bigcap A_n\\right)^c=A_1 .\n\n$$\n\nHere, I'm thinking of $A_1$ as the main metric space. Since $\\left\\{U_n\\right\\}$ is an open covering of $A_1$, we can extract a finite subcovering, that is,\n\n$$\n\nA_{\\alpha_1}^c \\cup A_{\\alpha_2}^c \\cup \\cdots \\cup A_{\\alpha_m}^c \\supset A_1\n\n$$\n\nor\n\n$$\n\n\\left(A_1 \\backslash A_{\\alpha_1}\\right) \\cup\\left(A_1 \\backslash A_{\\alpha_2}\\right) \\cup \\cdots \\cup\\left(A_1 \\backslash A_{\\alpha_m}\\right) \\supset A_1 .\n\n$$\n\nBut, this is true only if $A_{\\alpha_i}=\\emptyset$ for some $i$, a contradiction.\n\n\\end{proof}"}
{"id": "Pugh|exercise_3_1", "formal_statement": "theorem exercise_3_1 {f : \u211d \u2192 \u211d}\n  (hf : \u2200 x y, |f x - f y| \u2264 |x - y| ^ 2) :\n  \u2203 c, f = \u03bb x, c :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Assume that $f \\colon \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfies $|f(t)-f(x)| \\leq|t-x|^{2}$ for all $t, x$. Prove that $f$ is constant.", "nl_proof": "\\begin{proof}\n\n    We have $|f(t)-f(x)| \\leq|t-x|^2, \\forall t, x \\in \\mathbb{R}$. Fix $x \\in \\mathbb{R}$ and let $t \\neq x$. Then\n\n$$\n\n\\left|\\frac{f(t)-f(x)}{t-x}\\right| \\leq|t-x| \\text {, hence } \\lim _{t \\rightarrow x}\\left|\\frac{f(t)-f(x)}{t-x}\\right|=0 \\text {, }\n\n$$\n\nso $f$ is differentiable in $\\mathbb{R}$ and $f^{\\prime}=0$. This implies that $f$ is constant, as seen in class.\n\n\\end{proof}"}
{"id": "Pugh|exercise_3_63a", "formal_statement": "theorem exercise_3_63a (p : \u211d) (f : \u2115 \u2192 \u211d) (hp : p > 1)\n  (h : f = \u03bb k, (1 : \u211d) / (k * (log k) ^ p)) :\n  \u2203 l, tendsto f at_top (\ud835\udcdd l) :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "Prove that $\\sum 1/k(\\log(k))^p$ converges when $p > 1$.", "nl_proof": "\\begin{proof}\n\n    Using the integral test, for a set $a$, we see\n\n$$\n\n\\lim _{b \\rightarrow \\infty} \\int_a^b \\frac{1}{x \\log (x)^c} d x=\\lim _{b \\rightarrow \\infty}\\left(\\frac{\\log (b)^{1-c}}{1-c}-\\frac{\\log (a)^{1-c}}{1-c}\\right)\n\n$$\n\nwhich goes to infinity if $c \\leq 1$ and converges if $c>1$. Thus,\n\n$$\n\n\\sum_{n=2}^{\\infty} \\frac{1}{n \\log (n)^c}\n\n$$\n\nconverges if and only if $c>1$. \n\n\\end{proof}"}
{"id": "Pugh|exercise_4_15a", "formal_statement": "theorem exercise_4_15a {\u03b1 : Type*}\n  (a b : \u211d) (F : set (\u211d \u2192 \u211d)) :\n  (\u2200 (x : \u211d) (\u03b5 > 0), \u2203 (U \u2208 (\ud835\udcdd x)),\n  (\u2200 (y z \u2208 U) (f : \u211d \u2192 \u211d), f \u2208 F \u2192 (dist (f y) (f z) < \u03b5)))\n  \u2194\n  \u2203 (\u03bc : \u211d \u2192 \u211d), \u2200 (x : \u211d), (0 : \u211d) \u2264 \u03bc x \u2227 tendsto \u03bc (\ud835\udcdd 0) (\ud835\udcdd 0) \u2227\n  (\u2200 (s t : \u211d) (f : \u211d \u2192 \u211d), f \u2208 F \u2192 |(f s) - (f t)| \u2264 \u03bc (|s - t|)) :=", "src_header": "import .common\n\nopen set real filter function ring_hom topological_space\n\nopen_locale big_operators\nopen_locale filter\nopen_locale topology \nnoncomputable theory \n\n", "nl_statement": "A continuous, strictly increasing function $\\mu \\colon (0, \\infty) \\rightarrow (0, \\infty)$ is a modulus of continuity if $\\mu(s) \\rightarrow 0$ as $s \\rightarrow 0$. A function $f \\colon [a, b] \\rightarrow \\mathbb{R}$ has modulus of continuity $\\mu$ if $|f(s) - f(t)| \\leq \\mu(|s - t|)$ for all $s, t \\in [a, b]$. Prove that a function is uniformly continuous if and only if it has a modulus of continuity.", "nl_proof": "\\begin{proof}\n\n    Suppose there exists a modulus of continuity $w$ for $f$, then fix $\\varepsilon>0$, since $\\lim _{s \\rightarrow 0} w(s)=0$, there exists $\\delta>0$ such that for any $|s|<\\delta$, we have $w(s)<\\varepsilon$, then we have for any $x, z \\in X$ such that $d_X(x, z)<\\delta$, we have $d_Y(f(x), f(z)) \\leq w\\left(d_X(x, z)\\right)<\\varepsilon$, which means $f$ is uniformly continuous.\n\n\n\nSuppose $f:\\left(X, d_X\\right) \\rightarrow\\left(Y, d_Y\\right)$ is uniformly continuous.\n\nLet $\\delta_1>0$ be such that $d_X(a, b)<\\delta_1$ implies $d_Y(f(a), f(b))<1$.\n\nDefine $w:[0, \\infty) \\rightarrow[0, \\infty]$ by\n\n$$\n\nw(s)= \\begin{cases}\\left.\\sup \\left\\{d_Y(f(a), f(b))\\right\\} \\mid d_X(a, b) \\leq s\\right\\} & \\text { if } s \\leq \\delta_1 \\\\ \\infty & \\text { if } s>\\delta_1\\end{cases}\n\n$$\n\nWe'll show that $w$ is a modulus of continuity for $f \\ldots$\n\nBy definition of $w$, it's immediate that $w(0)=0$ and it's clear that\n\n$$\n\nd_Y(f(a), f(b)) \\leq w\\left(d_X(a, b)\\right)\n\n$$\n\nfor all $a, b \\in X$.\n\nIt remains to show $\\lim _{s \\rightarrow 0^{+}} w(s)=0$.\n\nIt's easily seen that $w$ is nonnegative and non-decreasing, hence $\\lim _{s \\rightarrow 0^{+}}=L$ for some $L \\geq 0$, where $L=\\inf w((0, \\infty))$\n\nLet $\\epsilon>0$.\n\nBy uniform continuity of $f$, there exists $\\delta>0$ such that $d_X(a, b)<\\delta$ implies $d_Y(f(a), f(b))<\\epsilon$, hence by definition of $w$, we get $w(\\delta) \\leq \\epsilon$.\n\nThus $L \\leq \\epsilon$ for all $\\epsilon>0$, hence $L=0$.\n\nThis completes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_1_18", "formal_statement": "theorem exercise_2_1_18 {G : Type*} [group G] \n  [fintype G] (hG2 : even (fintype.card G)) :\n  \u2203 (a : G), a \u2260 1 \u2227 a = a\u207b\u00b9 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $G$ is a finite group of even order, show that there must be an element $a \\neq e$ such that $a=a^{-1}$.", "nl_proof": "\\begin{proof}\n\n    First note that $a=a^{-1}$ is the same as saying $a^2=e$, where $e$ is the identity. I.e. the statement is that there exists at least one element of order 2 in $G$.\n\nEvery element $a$ of $G$ of order at least 3 has an inverse $a^{-1}$ that is not itself -- that is, $a \\neq a^{-1}$. So the subset of all such elements has an even cardinality (/size). There's exactly one element with order 1 : the identity $e^1=e$. So $G$ contains an even number of elements -call it $2 k$-- of which an even number are elements of order 3 or above -- call that $2 n$ where $n<k$-- and exactly one element of order 1 . Hence the number of elements of order 2 is\n\n$$\n\n2 k-2 n-1=2(k-n)-1\n\n$$\n\nThis cannot equal 0 as $2(k-n)$ is even and 1 is odd. Hence there's at least one element of order 2 in $G$, which concludes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_1_26", "formal_statement": "theorem exercise_2_1_26 {G : Type*} [group G] \n  [fintype G] (a : G) : \u2203 (n : \u2115), a ^ n = 1 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $G$ is a finite group, prove that, given $a \\in G$, there is a positive integer $n$, depending on $a$, such that $a^n = e$.", "nl_proof": "\\begin{proof}\n\n    Because there are only a finite number of elements of $G$, it's clear that the set $\\left\\{a, a^2, a^3, \\ldots\\right\\}$ must be a finite set and in particular, there should exist some $i$ and $j$ such that $i \\neq j$ and $a^i=a^j$. WLOG suppose further that $i>j$ (just reverse the roles of $i$ and $j$ otherwise). Then multiply both sides by $\\left(a^j\\right)^{-1}=a^{-j}$ to get\n\n$$\n\na^i * a^{-j}=a^{i-j}=e\n\n$$\n\nThus the $n=i-j$ is a positive integer such that $a^n=e$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_2_3", "formal_statement": "theorem exercise_2_2_3 {G : Type*} [group G]\n  {P : \u2115 \u2192 Prop} {hP : P = \u03bb i, \u2200 a b : G, (a*b)^i = a^i * b^i}\n  (hP1 : \u2203 n : \u2115, P n \u2227 P (n+1) \u2227 P (n+2)) : comm_group G :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $G$ is a group in which $(a b)^{i}=a^{i} b^{i}$ for three consecutive integers $i$, prove that $G$ is abelian.", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a group, $a, b \\in G$ and $i$ be any integer. Then from given condition,\n\n$$\n\n\\begin{aligned}\n\n(a b)^i & =a^i b^i \\\\\n\n(a b)^{i+1} & =a^{i+1} b^{i+1} \\\\\n\n(a b)^{i+2} & =a^{i+2} b^{i+2}\n\n\\end{aligned}\n\n$$\n\nFrom first and second, we get\n\n$$\n\na^{i+1} b^{i+1}=(a b)^i(a b)=a^i b^i a b \\Longrightarrow b^i a=a b^i\n\n$$\n\nFrom first and third, we get\n\n$$\n\na^{i+2} b^{i+2}=(a b)^i(a b)^2=a^i b^i a b a b \\Longrightarrow a^2 b^{i+1}=b^i a b a\n\n$$\n\nThis gives\n\n$$\n\na^2 b^{i+1}=a\\left(a b^i\\right) b=a b^i a b=b^i a^2 b\n\n$$\n\nFinally, we get\n\n$$\n\nb^i a b a=b^i a^2 b \\Longrightarrow b a=a b\n\n$$\n\nThis shows that $G$ is Abelian.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_2_6c", "formal_statement": "theorem exercise_2_2_6c {G : Type*} [group G] {n : \u2115} (hn : n > 1) \n  (h : \u2200 (a b : G), (a * b) ^ n = a ^ n * b ^ n) :\n  \u2200 (a b : G), (a * b * a\u207b\u00b9 * b\u207b\u00b9) ^ (n * (n - 1)) = 1 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a group in which $(a b)^{n}=a^{n} b^{n}$ for some fixed integer $n>1$ for all $a, b \\in G$. For all $a, b \\in G$, prove that $\\left(a b a^{-1} b^{-1}\\right)^{n(n-1)}=e$.", "nl_proof": "\\begin{proof}\n\n    We start with the following two intermediate results.\n\n(1) $(a b)^{n-1}=b^{n-1} a^{n-1}$.\n\n(2) $a^n b^{n-1}=b^{n-1} a^n$.\n\nTo prove (1), notice by the given condition for all $a, b \\in G$\n\n$(b a)^n=b^n a^n$, for some fixed integers $n>1$.\n\nThen,\n\n$(b a)^n=b^n a^n \\Longrightarrow b .(a b)(a b) \\ldots .(a b) . a=b\\left(b^{n-1} a^{n-1}\\right) a$, where $(a b)$ occurs $n-1$ times $\\Longrightarrow(a b)^{n-1}=b^{n-1} a^{n-1}$, by cancellation law.\n\nHence, for all $a, b \\in G$\n\n$$\n\n(a b)^{n-1}=b^{n-1} a^{n-1} .\n\n$$\n\nTo prove (2), notice by the given condition for all $a, b \\in G$\n\n$(b a)^n=b^n a^n$, for some fixed integers $n>1$.\n\nThen we have\n\n$$\n\n\\begin{aligned}\n\n& (b a)^n=b^n a^n \\\\\n\n\\Longrightarrow & b \\cdot(a b)(a b) \\ldots(a b) \\cdot a=b\\left(b^{n-1} a^{n-1}\\right) a, \\text { where }(a b) \\text { occurs } n-1 \\text { times } \\\\\n\n\\Longrightarrow & (a b)^{n-1}=b^{n-1} a^{n-1}, \\text { by cancellation law } \\\\\n\n\\Longrightarrow & (a b)^{n-1}(a b)=\\left(b^{n-1} a^{n-1}\\right)(a b) \\\\\n\n\\Longrightarrow & (a b)^n=b^{n-1} a^n b \\\\\n\n\\Longrightarrow & a^n b^n=b^{n-1} a^n b, \\text { given condition } \\\\\n\n\\Longrightarrow & a^n b^{n-1}=b^{n-1} a^n, \\text { by cancellation law. }\n\n\\end{aligned}\n\n$$\n\nTherefore for all $a, b \\in G$ we have\n\n$$\n\na^n b^{n-1}=b^{n-1} a^n\n\n$$\n\nIn order to show that\n\n$$\n\n\\left(a b a^{-1} b^{-1}\\right)^{n(n-1)}=e, \\text { for all } a, b \\in G\n\n$$\n\nit is enough to show that\n\n$$\n\n(a b)^{n(n-1)}=(b a)^{n(n-1)}, \\quad \\forall x, y \\in G .\n\n$$\n\nStep 3\n\nThis is because of\n\n$$\n\n\\begin{aligned}\n\n(a b)^{n(n-1)}=(b a)^{n(n-1)} & \\left.\\Longrightarrow(b a)^{-1}\\right)^{n(n-1)}(a b)^{n(n-1)}=e \\\\\n\n& \\Longrightarrow\\left(a^{-1} b^{-1}\\right)^{n(n-1)}(a b)^{n(n-1)}=e \\\\\n\n& \\Longrightarrow\\left(\\left(a^{-1} b^{-1}\\right)^n\\right)^{n-1}\\left((a b)^n\\right)(n-1)=e \\\\\n\n& \\Longrightarrow\\left((a b)^n\\left(a^{-1} b^{-1}\\right)^n\\right)^{n-1}=e, \\text { by }(1) \\\\\n\n& \\Longrightarrow\\left(a b a^{-1} b^{-1}\\right)^{n(n-1)}=e, \\text { ( given condition) }\n\n\\end{aligned}\n\n$$\n\nNow, it suffices to show that\n\n$$\n\n(a b)^{n(n-1)}=(b a)^{n(n-1)}, \\quad \\forall x, y \\in G .\n\n$$\n\nNow, we have\n\n$$\n\n\\begin{aligned}\n\n(a b)^{n(n-1)} & =\\left(a^n b^n\\right)^{n-1}, \\text { by the given condition } \\\\\n\n& =\\left(a^n b^{n-1} b\\right)^{n-1} \\\\\n\n& =\\left(b^{n-1} a^n b\\right)^{n-1}, \\text { by }(2) \\\\\n\n& =\\left(a^n b\\right)^{n-1}\\left(b^{n-1}\\right)^{n-1}, \\text { by }(1) \\\\\n\n& =b^{n-1}\\left(a^n\\right)^{n-1}\\left(b^{n-1}\\right)^{n-1}, \\text { by }(1) \\\\\n\n& =\\left(b^{n-1}\\left(a^{n-1}\\right)^n\\right)\\left(b^{n-1}\\right)^{n-1} \\\\\n\n& =\\left(a^{n-1}\\right)^n b^{n-1}\\left(b^{n-1}\\right)^{n-1}, \\text { by }(2) \\\\\n\n& =\\left(a^{n-1}\\right)^n\\left(b^{n-1}\\right)^n \\\\\n\n& =\\left(a^{n-1} b^{n-1}\\right)^n, \\text { by }(1) \\\\\n\n& =(b a)^{n(n-1)}, \\text { by }(1) .\n\n\\end{aligned}\n\n$$\n\nThis completes our proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_3_16", "formal_statement": "theorem exercise_2_3_16 {G : Type*} [group G]\n  (hG : \u2200 H : subgroup G, H = \u22a4 \u2228 H = \u22a5) :\n  is_cyclic G \u2227 \u2203 (p : \u2115) (fin : fintype G), nat.prime p \u2227 @card G fin = p :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If a group $G$ has no proper subgroups, prove that $G$ is cyclic of order $p$, where $p$ is a prime number.", "nl_proof": "\\begin{proof}\n\n    Case-1: $G=(e), e$ being the identity element in $G$. Then trivially $G$ is cyclic.\n\n    Case-2: $G \\neq(e)$. Then there exists an non-identity element in $G.$ Let us consider an non-identity element in $G$, say $a\\neq (e)$. Now look at the cyclic subgroup generated by $a$, that is, $\\langle a\\rangle$. Since\n\n    $a\\neq (e) \\in G,\\langle a\\rangle$ is a subgroup of $G$.\n\nIf $G \\neq\\langle a\\rangle$ then $\\langle a\\rangle$ is a proper non-trivial subgroup of $G$, which is an impossibility. Therfore we must have\n\n$$\n\nG=\\langle a\\rangle .\n\n$$\n\nThis implies, $G$ is a cyclic group generated by $a$. Then it follows that every non-identity element of $G$ is a generator of $G$. Now we claim that $G$ is finite.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_5_23", "formal_statement": "theorem exercise_2_5_23 {G : Type*} [group G] \n  (hG : \u2200 (H : subgroup G), H.normal) (a b : G) :\n  \u2203 (j : \u2124) , b*a = a^j * b:=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a group such that all subgroups of $G$ are normal in $G$. If $a, b \\in G$, prove that $ba = a^jb$ for some $j$.", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a group where each subgroup is normal in $G$. let $a, b \\in G$.\n\n$$\n\n\\begin{aligned}\n\n    \\langle a\\rangle\\triangleright  G  &\\Rightarrow b \\cdot\\langle a\\rangle=\\langle a\\rangle \\cdot b . \\\\\n\n& \\Rightarrow \\quad b \\cdot a=a^j \\cdot b \\text { for some } j \\in \\mathbb{Z}.\n\n\\end{aligned}\n\n$$\n\n(hence for $a_1 b \\in G \\quad a^j b=b \\cdot a$ ).\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_5_31", "formal_statement": "theorem exercise_2_5_31 {G : Type*} [comm_group G] [fintype G]\n  {p m n : \u2115} (hp : nat.prime p) (hp1 : \u00ac p \u2223 m) (hG : card G = p^n*m)\n  {H : subgroup G} [fintype H] (hH : card H = p^n) : \n  characteristic H :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Suppose that $G$ is an abelian group of order $p^nm$ where $p \\nmid m$ is a prime.  If $H$ is a subgroup of $G$ of order $p^n$, prove that $H$ is a characteristic subgroup of $G$.", "nl_proof": "\\begin{proof}\n\n    Let $G$ be an abelian group of order $p^n m$, such that $p \\nmid m$. Now, Given that $H$ is a subgroup of order $p^n$. Since $G$ is abelian $H$ is normal. Now we want to prove that $H$ is a characterestic subgroup, that is $\\phi(H)=H$ for any automorphism $\\phi$ of $G$. Now consider $\\phi(H)$. Clearly $|\\phi(H)|=p^n$. Suppose $\\phi(H) \\neq H$, then $|H \\cap \\phi(H)|=p^s$, where $s<n$. Consider $H \\phi(H)$, this is a subgroup of $G$ as $H$ is normal. Also $|H \\phi(H)|=\\frac{|H||\\phi(H)|}{|H \\cap \\phi(H)|}=\\frac{p^{2 n}}{p^s}=p^{2 n-s}$, where $2 n-s>n$. By lagrange's theorem then $p^{2 n-s}\\left|p^n m \\Longrightarrow p^{n-s}\\right| m \\Longrightarrow p \\mid m$-contradiction. So $\\phi(H)=H$, and $H$ is characterestic subgroup of $G$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_5_43", "formal_statement": "theorem exercise_2_5_43 (G : Type*) [group G] [fintype G]\n  (hG : card G = 9) :\n  comm_group G :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that a group of order 9 must be abelian.", "nl_proof": "\\begin{proof}\n\n    We use the result from problem 40 which is as follows: Suppose $G$ is a group, $H$ is a subgroup and $|G|=n$ and $n \\nmid\\left(i_G(H)\\right) !$. Then there exists a normal subgroup $\\$ K \\backslash$ neq $\\{$ e $\\} \\$$ and $K \\subseteq H$.\n\nSo, we have now a group $G$ of order 9. Suppose that $G$ is cyclic, then $G$ is abelian and there is nothing more to prove. Suppose that $G$ s not cyclic,then there exists an element $a$ of order 3 , and $A=\\langle a\\rangle$. Now $i_G(A)=3$, now $9 \\nmid 3$ !, hence by the above result there is a normal subgroup $K$, non-trivial and $K \\subseteq A$. But $|A|=3$, a prime order subgroup, hence has no non-trivial subgroup, so $K=A$. So $A$ is normal subgroup. Now since $G$ is not cyclic any non-identity element is of order 3.So Let $a(\\neq$ $e) \\in G$.Consider $A=\\langle a\\rangle$. As shown before $A$ is normal. $a$ commutes with any if its powers. Now Let $b \\in G$ such that $b \\notin A$. Then $b a b^{-1} \\in A$ and hence $b a b^{-1}=a^i$.This implies $a=b^3 a b^{-3}=a^{i^3} \\Longrightarrow a^{i^3-1}=e$. So, 3 divides $i^3-1$. Also by fermat's little theorem 3 divides $i^2-1$.So 3 divides $i-1$. But $0 \\leq i \\leq 2$. So $i=1$, is the only possibility and hence $a b=b a$. So $a \\in Z(G)$ as $b$ was arbitrary. Since $a$ was arbitrary $G=Z(G)$. Hence $G$ is abelian.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_5_52", "formal_statement": "theorem exercise_2_5_52 {G : Type*} [group G] [fintype G]\n  (\u03c6 : G \u2243* G) {I : finset G} (hI : \u2200 x \u2208 I, \u03c6 x = x\u207b\u00b9)\n  (hI1 : (0.75 : \u211a) * card G \u2264 card I) : \n  \u2200 x : G, \u03c6 x = x\u207b\u00b9 \u2227 \u2200 x y : G, x*y = y*x :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a finite group and $\\varphi$ an automorphism of $G$ such that $\\varphi(x) = x^{-1}$ for more than three-fourths of the elements of $G$. Prove that $\\varphi(y) = y^{-1}$ for all $y \\in G$, and so $G$ is abelian.", "nl_proof": "\\begin{proof}\n\nLet us start with considering $b$ to be an arbitrary element in $A$. \n\n\n\n1. Show that $\\left|A \\cap\\left(b^{-1} A\\right)\\right|>\\frac{|G|}{2}$, where\n\n$$\n\nb^{-1} A=\\left\\{b^{-1} a \\mid a \\in A\\right\\}\n\n$$\n\nFirst notice that if we consider a map $f: A \\rightarrow b^{-1} A$ defined by $f(a)=b^{-1} a$, for all $a \\in A$, then $f$ is a 1-1 map and so $\\left|b^{-1} A\\right| \\geq|A|>\\frac{3}{4}|G|$. Now using inclusion-exclusion principle we have\n\n$$\n\n\\left|A \\cap\\left(b^{-1} A\\right)\\right|=|A|+\\left|b^{-1} A\\right|-\\left|A \\cup\\left(b^{-1} A\\right)\\right|>\\frac{3}{4}|G|+\\frac{3}{4}|G|-|G|=\\frac{1}{2}|G|\n\n$$\n\n2. Argue that $A \\cap\\left(b^{-1} A\\right) \\subseteq C(b)$, where $C(b)$ is the centralizer of $b$ in $G$.\n\n\n\nSuppose $x \\in A \\cap\\left(b^{-1} A\\right)$, that means, $x \\in A$ and $x \\in b^{-1} A$. Thus there exist an element $a \\in A$ such that $x=$ $b^{-1} a$, which gives us $x b=a \\in A$. Now notice that $x, b \\in A$ and $x b \\in A$, therefore we get\n\n$$\n\n\\phi(x b)=(x b)^{-1} \\Longrightarrow \\phi(x) \\phi(b)=(x b)^{-1} \\Longrightarrow x^{-1} b^{-1}=b^{-1} x^{-1} \\Longrightarrow x b=b x\n\n$$\n\nTherefore, we get $x b=b x$, for any $x \\in A \\cap\\left(b^{-1} A\\right)$, that means, $x \\in C(b)$.\n\n\n\n3. Argue that $C(b)=G$.\n\nWe know that centralizer of an element in a group $G$ is a subgroup (See Page 53). Therefore $C(b)$ is a subgroup of $G$. From statements $\\mathbf{1}$ and $\\mathbf{2}$, we have\n\n$$\n\n|C(b)| \\geq\\left|A \\cap\\left(b^{-1} A\\right)\\right|>\\frac{|G|}{2}\n\n$$\n\nWe need to use the following remark to argue $C(b)=G$ from the above step.\n\nRemark. Let $G$ be a finite group and $H$ be a subgroup with more then $|G| / 2$ elements then $H=G$.\n\n\n\nProof of Remark. Suppose $|H|=p$ Then by Lagrange Theorem, there exist an $n \\in \\mathbb{N}$, such that $|G|=n p$, as $|H|$ divide $|G|$. Now by hypothesis $p>\\frac{G]}{2}$ gives us,\n\n$$\n\np>\\frac{|G|}{2} \\Longrightarrow n p>\\frac{n|G|}{2} \\Longrightarrow n<2 \\Longrightarrow n=1\n\n$$\n\nTherefore we get $H=G$.\n\n\n\nNow notice that $C(b)$ is a subgroup of $G$ with $C(b)$ having more than $|G| / 2$ elements. Therefore, $C(b)=G$.\n\n\n\n4. Show that $A \\in Z(G)$.\n\n\n\nWe know that $x \\in Z(G)$ if and only if $C(a)=G$. Now notice that, for any $b \\in A$ we have $C(b)=G$. Therefore, every element of $A$ is in the center of $G$, that means, $A \\subseteq Z(G)$.\n\n\n\n5. 5how that $Z(G)=G$.\n\n\n\nAs it is given that $|A|>\\frac{3|G|}{4}$ and $A \\leq|Z(G)|$, therefore we get\n\n$$\n\n|Z(G)|>\\frac{3}{4}|G|>\\frac{1}{2}|G| .\n\n$$\n\nAs $Z(G)$ is a subgroup of $G$, so by the above Remark we have $Z(G)=G$. Hence $G$ is abelian.\n\n\n\n6. Finally show that $A=G$.\n\n\n\nFirst notice that $A$ is a subgroup of $G$. To show this let $p, q \\in A$. Then we have\n\n$$\n\n\\phi(p q)=\\phi(p) \\phi(q)=p^{-1} q^{-1}=(q p)^{-1}=(p q)^{-1}, \\quad \\text { As } G \\text { is abelian. }\n\n$$\n\nTherefore, $p q \\in A$ and so we have $A$ is a subgroup of $G$. Again by applying the above remark. we get $A=G$. Therefore we have\n\n$$\n\n\\phi(y)=y^{-1}, \\quad \\text { for all } y \\in G\n\n$$\n\n\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_7_7", "formal_statement": "theorem exercise_2_7_7 {G : Type*} [group G] {G' : Type*} [group G']\n  (\u03c6 : G \u2192* G') (N : subgroup G) [N.normal] : \n  (map \u03c6 N).normal  :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $\\varphi$ is a homomorphism of $G$ onto $G'$ and $N \\triangleleft G$, show that $\\varphi(N) \\triangleleft G'$.", "nl_proof": "\\begin{proof}\n\nWe first claim that $\\varphi(N)$ is a subgroup of $G'$. To see this, note that since $N$ is a subgroup of $G$, the identity element $e_G$ of $G$ belongs to $N$. Therefore, the element $\\varphi(e_G) \\in \\varphi(N)$, so $\\varphi(N)$ is a non-empty subset of $G'$.\n\n\n\nNow, let $a', b' \\in \\varphi(N)$. Then there exist elements $a, b \\in N$ such that $\\varphi(a) = a'$ and $\\varphi(b) = b'$. Since $N$ is a subgroup of $G$, we have $a, b \\in N$, so $ab^{-1} \\in N$. Thus, we have\n\n$$\\varphi(ab^{-1}) = \\varphi(a) \\varphi(b^{-1}) = a'b'^{-1} \\in \\varphi(N),$$\n\nwhich shows that $a', b' \\in \\varphi(N)$ implies $a'b'^{-1} \\in \\varphi(N)$. Therefore, $\\varphi(N)$ is a subgroup of $G'$.\n\n\n\nNext, we will show that $\\varphi(N)$ is a normal subgroup of $G'$. Let $\\varphi(N) = N'$, a subgroup of $G'$. Let $x' \\in G'$ and $h' \\in N'$. Since $\\varphi$ is onto, there exist elements $x \\in G$ and $h \\in N$ such that $\\varphi(x) = x'$ and $\\varphi(h) = h'$.\n\n\n\nSince $N$ is a normal subgroup of $G$, we have $xhx^{-1} \\in N$. Thus,\n\n$$\\varphi(xhx^{-1}) = \\varphi(x)\\varphi(h)\\varphi(x^{-1}) = x'h'x'^{-1} \\in \\varphi(N),$$\n\nwhich shows that $x' \\in G'$ and $h' \\in N'$ implies $x'h'x'^{-1} \\in \\varphi(N)$. Therefore, $\\varphi(N)$ is a normal subgroup of $G'$. This completes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_8_15", "formal_statement": "theorem exercise_2_8_15 {G H: Type*} [fintype G] [group G] [fintype H]\n  [group H] {p q : \u2115} (hp : nat.prime p) (hq : nat.prime q) \n  (h : p > q) (h1 : q \u2223 p - 1) (hG : card G = p*q) (hH : card G = p*q) :\n  G \u2243* H :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $p > q$ are two primes such that $q \\mid p - 1$, then any two nonabelian groups of order $pq$ are isomorphic.", "nl_proof": "\\begin{proof}\n\n    For a nonabelian group of order $p q$, the structure of the group $G$ is set by determining the relation $a b a^{-1}=b^{k^{\\frac{p-1}{q}}}$ for some generator $k$ of the cyclic group. Here we are using the fact that $k^{\\frac{p-1}{q}}$ is a generator for the unique subgroup of order $q$ in $U_p$ (a cyclic group of order $m$ has a unique subgroup of order $d$ for each divisor $d$ of $m$ ). The other possible generators of this subgroup are $k^{\\frac{l(p-1)}{q}}$ for each $1 \\leq l \\leq q-1$, so these give potentially new group structures. Let $G^{\\prime}$ be a group with an element $c$ of order $q$, an element $d$ of order $p$ with structure defined by the relation $c d c^{-1}=d^{k^{\\frac{l(p-1)}{q}}}$. We may then define\n\n$$\n\n\\begin{aligned}\n\n\\phi: G^{\\prime} & \\rightarrow G \\\\\n\nc & \\mapsto a^l \\\\\n\nd & \\mapsto b\n\n\\end{aligned}\n\n$$\n\nsince $c$ and $a^l$ have the same order and $b$ and $d$ have the same order this is a well defined function.\n\nSince\n\n$$\n\n\\begin{aligned}\n\n\\phi(c) \\phi(d) \\phi(c)^{-1} & =a^l b a^{-l} \\\\\n\n& =b^{\\left(k^{\\frac{p-1}{q}}\\right)^l} \\\\\n\n& =b^{k^{\\frac{l(p-1)}{q}}} \\\\\n\n& =\\phi(d)^{k^{\\frac{l(p-1)}{q}}}\n\n\\end{aligned}\n\n$$\n\n$\\phi\\left(c^i d^j\\right)=a^{l i} b^j=e$ only if $i=j=0$, so $\\phi$ is 1-to-l. Therefore $G$ and $G^{\\prime}$ are isomorphic and so up to isomorphism there is only one nonabelian group of order $p q$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_10_1", "formal_statement": "theorem exercise_2_10_1 {G : Type*} [group G] (A : subgroup G) \n  [A.normal] {b : G} (hp : nat.prime (order_of b)) :\n  A \u2293 (closure {b}) = \u22a5 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $A$ be a normal subgroup of a group $G$, and suppose that $b \\in G$ is an element of prime order $p$, and that $b \\not\\in A$. Show that $A \\cap (b) = (e)$.", "nl_proof": "\\begin{proof}\n\nIf $b \\in G$ has order $p$, then $(b)$ is a cyclic group of order $p$. Since $A$ is a subgroup of $G$, we have $A \\cap (b)$ is a subgroup of $G$. Also, $A \\cap (b) \\subseteq (b)$. So $A \\cap (b)$ is a subgroup of $(b)$. Since $(b)$ is a cyclic group of order $p$, the only subgroups of $(b)$ are $(e)$ and $(b)$ itself.\n\n\n\nTherefore, either $A \\cap (b) = (e)$ or $A \\cap (b) = (b)$. If $A \\cap (b) = (e)$, then we are done. Otherwise, if $A \\cap (b) = (b)$, then $A \\subseteq (b)$. Since $A$ is a subgroup of $G$ and $A \\subseteq (b)$, it follows that $A$ is a subgroup of $(b)$.\n\n\n\nSince the only subgroups of $(b)$ are $(e)$ and $(b)$ itself, we have either $A = (e)$ or $A = (b)$. If $A = (e)$, then $A \\cap (b) = (e)$ and we are done. But if $A = (b)$, then $b \\in A$ as $b \\in (b)$, which contradicts our hypothesis that $b \\notin A$. So $A \\neq (b)$.\n\n\n\nHence $A \\cap (b) \\neq (b)$. Therefore, $A \\cap (b) = (e)$. This completes our proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_2_11_7", "formal_statement": "theorem exercise_2_11_7 {G : Type*} [group G] {p : \u2115} (hp : nat.prime p)\n  {P : sylow p G} (hP : P.normal) : \n  characteristic (P : subgroup G) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $P \\triangleleft G$, $P$ a $p$-Sylow subgroup of $G$, prove that $\\varphi(P) = P$ for every automorphism $\\varphi$ of $G$.", "nl_proof": "\\begin{proof}\n\n    Let $\\phi$ be an automorphism of $G$. Let $P$ be a normal sylow p-subgroup. $\\phi(P)$ is also a sylow-p subgroup. But since $P$ is normal, it is unique. Hence $\\phi(P)=P$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_3_2_21", "formal_statement": "theorem exercise_3_2_21 {\u03b1 : Type*} [fintype \u03b1] {\u03c3 \u03c4: equiv.perm \u03b1} \n  (h1 : \u2200 a : \u03b1, \u03c3 a = a \u2194 \u03c4 a \u2260 a) (h2 : \u03c4 \u2218 \u03c3 = id) : \n  \u03c3 = 1 \u2227 \u03c4 = 1 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $\\sigma, \\tau$ are two permutations that disturb no common element and $\\sigma \\tau = e$, prove that $\\sigma = \\tau = e$.", "nl_proof": "\\begin{proof}\n\n    Note that $\\sigma \\tau=e$ can equivalentnly be phrased as $\\tau$ being the inverse of $\\sigma$. Our statement is then equivalent to the statement that an inverse of a nonidentity permutation disturbs at least one same element as that permutation. To prove this, let $\\sigma$ be a nonidentity permutation, then let $\\left(i_1 \\cdots i_n\\right)$ be a cycle in $\\sigma$. Then we have that\n\n$$\n\n\\sigma\\left(i_1\\right)=i_2, \\sigma\\left(i_2\\right)=i_2, \\ldots, \\sigma\\left(i_{n-1}\\right)=i_n, \\sigma\\left(i_n\\right)=i_1,\n\n$$\n\nbut then also\n\n$$\n\ni_1=\\tau\\left(i_2\\right), i_2=\\tau\\left(i_3\\right), \\ldots, i_{n-1}=\\tau\\left(i_n\\right), i_n=\\tau\\left(i_1\\right),\n\n$$\n\ni.e. its inverse disturbs $i_1, \\ldots, i_n$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_1_34", "formal_statement": "theorem exercise_4_1_34 : equiv.perm (fin 3) \u2243* general_linear_group (fin 2) (zmod 2) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $T$ be the group of $2\\times 2$ matrices $A$ with entries in the field $\\mathbb{Z}_2$ such that $\\det A$ is not equal to 0. Prove that $T$ is isomorphic to $S_3$, the symmetric group of degree 3.", "nl_proof": "\\begin{proof}\n\n    The order of $T$ is $2^4-2^3-2^2+2=6$; we now find those six matrices:\n\n$$\n\n\\begin{array}{ll}\n\nA_1=\\left(\\begin{array}{ll}\n\n1 & 0 \\\\\n\n0 & 1\n\n\\end{array}\\right), & A_2=\\left(\\begin{array}{ll}\n\n0 & 1 \\\\\n\n1 & 0\n\n\\end{array}\\right) \\\\\n\nA_3=\\left(\\begin{array}{ll}\n\n1 & 0 \\\\\n\n1 & 1\n\n\\end{array}\\right), & A_4=\\left(\\begin{array}{ll}\n\n1 & 1 \\\\\n\n0 & 1\n\n\\end{array}\\right) \\\\\n\nA_5=\\left(\\begin{array}{ll}\n\n0 & 1 \\\\\n\n1 & 1\n\n\\end{array}\\right), & A_6=\\left(\\begin{array}{ll}\n\n1 & 1 \\\\\n\n1 & 0\n\n\\end{array}\\right)\n\n\\end{array}\n\n$$\n\nwith orders $1,2,2,2,3,3$ respectively.\n\nNote that $S_3$ is composed of elements\n\n$$\n\n\\text{ id, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)} \n\n$$\n\nwith orders 1, 2, 2, 2, 3, 3 respectively. Also note that, by Problem 17 of generate $S_3$. We also have that $\\left(\\begin{array}{llll}1 & 3 & 2\\end{array}\\right)=\\left(\\begin{array}{llll}1 & 2 & 3\\end{array}\\right)\\left(\\begin{array}{lll}1 & 2 & 3\\end{array}\\right)$, that $\\left(\\begin{array}{lll}1 & 3\\end{array}\\right)=\\left(\\begin{array}{lll}1 & 2 & 3\\end{array}\\right)\\left(\\begin{array}{ll}1 & 2\\end{array}\\right)$, $\\left(\\begin{array}{ll}1 & 2\\end{array}\\right)\\left(\\begin{array}{lll}1 & 2 & 3\\end{array}\\right)=\\left(\\begin{array}{ll}2 & 3\\end{array}\\right)$ and $\\left(\\begin{array}{lll}1 & 2\\end{array}\\right)\\left(\\begin{array}{ll}1 & 2\\end{array}\\right)=\\mathrm{id}$\n\n\n\nNow we can check that $\\tau\\left(A_2\\right)=\\left(\\begin{array}{ll}1 & 2\\end{array}\\right), \\tau\\left(A_5\\right)=\\left(\\begin{array}{lll}1 & 2 & 3\\end{array}\\right)$ induces an isomorphism. We compute\n\n$$\n\n\\begin{aligned}\n\n& \\tau\\left(A_1\\right)=\\tau\\left(A_2 A_2\\right)=\\tau\\left(A_2\\right) \\tau\\left(A_2\\right)=\\mathrm{id} \\\\\n\n& \\tau\\left(A_3\\right)=\\tau\\left(A_5 A_2\\right)=\\tau\\left(A_5\\right) \\tau\\left(A_2\\right)=\\left(\\begin{array}{llll}\n\n1 & 2 & 3\n\n\\end{array}\\right)\\left(\\begin{array}{lll}\n\n1 & 2\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\n1 & 3\n\n\\end{array}\\right) \\\\\n\n& \\tau\\left(A_4\\right)=\\tau\\left(A_2 A_5\\right)=\\tau\\left(A_2\\right) \\tau\\left(A_5\\right)=\\left(\\begin{array}{lll}\n\n1 & 2\n\n\\end{array}\\right)\\left(\\begin{array}{lll}\n\n1 & 2 & 3\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\n2 & 3\n\n\\end{array}\\right) \\\\\n\n& \\tau\\left(A_6\\right)=\\tau\\left(A_5 A_5\\right)=\\tau\\left(A_5\\right) \\tau\\left(A_5\\right)=\\left(\\begin{array}{lll}\n\n1 & 3 & 2\n\n\\end{array}\\right)\n\n\\end{aligned}\n\n$$\n\nThus we see that $\\tau$ extendeds to an isomorphism, since $A_2$ and $A_5$ generate $T$, so that $\\tau\\left(A_i A_j\\right)=\\tau\\left(A_i\\right) \\tau\\left(A_j\\right)$ follows from writing $A_i$ and $A_j$ in terms of $A_2$ and $A_5$ and using the equlities and relations shown above.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_2_6", "formal_statement": "theorem exercise_4_2_6 {R : Type*} [ring R] (a x : R) \n  (h : a ^ 2 = 0) : a * (a * x + x * a) = (x + x * a) * a :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a^2 = 0$ in $R$, show that $ax + xa$ commutes with $a$.", "nl_proof": "\\begin{proof}\n\nWe need to show that\n\n$$\n\na(a x+x a)=(a x+x a) a \\text { for } a, x \\in R .\n\n$$\n\nNow,\n\n$$\n\n\\begin{gathered}\n\na(a x+x a)=a(a x)+a(x a) \\\\\n\n=a^2 x+a x a \\\\\n\n=0+a x a=a x a .\n\n\\end{gathered}\n\n$$\n\nAgain,\n\n$$\n\n\\begin{gathered}\n\n(a x+x a) a=(a x) a+(x a) a \\\\\n\n=a x a+x a^2 \\\\\n\n=a x a+0=a x a .\n\n\\end{gathered}\n\n$$\n\nIt follows that,\n\n$$\n\na(a x+x a)=(a x+x a) a, \\text { for } x, a \\in R .\n\n$$\n\nThis shows that $a x+x a$ commutes with $a$. This completes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_3_1", "formal_statement": "theorem exercise_4_3_1 {R : Type*} [comm_ring R] (a : R) :\n  \u2203 I : ideal R, {x : R | x*a=0} = I :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $R$ is a commutative ring and $a \\in R$, let $L(a) = \\{x \\in R \\mid xa = 0\\}$. Prove that $L(a)$ is an ideal of $R$.", "nl_proof": "\\begin{proof}\n\n    First, note that if $x \\in L(a)$ and $y \\in L(a)$ then $x a=0$ and $y a=0$, so that\n\n$$\n\n\\begin{aligned}\n\nx a-y a & =0 \\\\\n\n(x-y) a & =0,\n\n\\end{aligned}\n\n$$\n\ni.e. $L(a)$ is an additive subgroup of $R$. (We have used the criterion that $H$ is a subgroup of $G$ if for any $h_1, h_2 \\in H$ we have that $h_1 h_2^{-1} \\in H$. \n\n\n\nNow we prove the conclusion. Let $r \\in R$ and $b \\in L(a)$, then $b a=0$, and so $x b a=0$ which by associativity of multiplication in $R$ is equivalent to\n\n$$\n\n(x b) a=0,\n\n$$\n\nso that $x b \\in L(a)$. Since $R$ is commutative, (1) implies that $(bx)a=0$, so that $b x \\in L(a)$, which concludes the proof that $L(a)$ is an ideal.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_4_9", "formal_statement": "theorem exercise_4_4_9 (p : \u2115) (hp : nat.prime p) :\n  (\u2203 S : finset (zmod p), S.card = (p-1)/2 \u2227 \u2203 x : zmod p, x^2 = p) \u2227 \n  (\u2203 S : finset (zmod p), S.card = (p-1)/2 \u2227 \u00ac \u2203 x : zmod p, x^2 = p) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that $(p - 1)/2$ of the numbers $1, 2, \\ldots, p - 1$ are quadratic residues and $(p - 1)/2$ are quadratic nonresidues $\\mod p$.", "nl_proof": "\\begin{proof}\n\n    To find all the quadratic residues $\\bmod p$ among the integers $1,2, \\ldots, p-1$, we compute the least positive residues modulo $p$ of the squares of the integers $1,2, \\ldots, p-1\\}$.\n\n\n\nSince there are $p-1$ squares to consider, and since each congruence $x^2 \\equiv a (\\bmod p)$ has either zero or two solutions, there must be exactly $\\frac{(p-1)}{2}$ quadratic residues mod $p$ among the integers $1,2, \\ldots, p-1$.\n\nThe remaining\n\n$$\n\n(p-1)-\\frac{(p-1)}{2}=\\frac{(p-1)}{2}\n\n$$\n\npositive integers less than $p-1$ are quadratic non-residues of $\\bmod p$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_5_23", "formal_statement": "theorem exercise_4_5_23 {p q: polynomial (zmod 7)} \n  (hp : p = X^3 - 2) (hq : q = X^3 + 2) : \n  irreducible p \u2227 irreducible q \u2227 \n  (nonempty $ polynomial (zmod 7) \u29f8 ideal.span ({p} : set $ polynomial $ zmod 7) \u2243+*\n  polynomial (zmod 7) \u29f8 ideal.span ({q} : set $ polynomial $ zmod 7)) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $F = \\mathbb{Z}_7$ and let $p(x) = x^3 - 2$ and $q(x) = x^3 + 2$ be in $F[x]$. Show that $p(x)$ and $q(x)$ are irreducible in $F[x]$ and that the fields $F[x]/(p(x))$ and $F[x]/(q(x))$ are isomorphic.", "nl_proof": "\\begin{proof}\n\n    We have that $p(x)$ and $q(x)$ are irreducible if they have no roots in $\\mathbb{Z}_7$, which can easily be checked. E.g. for $p(x)$ we have that $p(0)=5, p(1)=6, p(2)=6, p(3)=4, p(4)=6$, $p(5)=4, p(6)=4$, and similarly for $q(x)$.\n\n\n\nWe have that every element of $F[x] /(p(x))$ is equal to $a x^2+b x+c+(p(x))$, and likewise for $F[x] /(q(x))$. We consider a map $\\tau$ : $F[x] /(p(x)) \\rightarrow F[x] /(q(x))$ given by\n\n$$\n\n\\tau\\left(a x^2+b x+c+(p(x))\\right)=a x^2-b x+c+(q(x)) .\n\n$$\n\nThis map is obviously onto, and since $|F[x] /(p(x))|=|F[x] /(q(x))|=7^3$ by Problem 16, it is also one-to-one. We claim that it is a homomorphism. Additivity of $\\tau$ is immediate by the linearity of addition of polynomial coefficient, so we just have to check the multiplicativity; if $n=a x^2+b x+$ $c+(p(x))$ and $m=d x^2+e x+f+(p(x))$ then\n\n$$\n\n\\begin{aligned}\n\n\\tau(n m) & =\\tau\\left(a d x^4+(a e+b d) x^3+(a f+b e+c d) x^2+(b f+c e) x+c f+(p(x))\\right) \\\\\n\n& =\\tau\\left(2 a d x+2(a e+b d)+(a f+b e+c d) x^2+(b f+c e) x+c f+(p(x))\\right) \\\\\n\n& =\\tau\\left((a f+b e+c d) x^2+(b f+c e+2 a d) x+(c f+2 a e+2 b d)+(p(x))\\right) \\\\\n\n& =(a f+b e+c d) x^2-(b f+c e+2 a d) x+c f+2 a e+2 b d+(q(x)) \\\\\n\n& =a d x^4-(a e+b d) x^3+(a f+b e+c d) x^2-(b f+c e) x+c f+(q(x)) \\\\\n\n& =\\left(a x^2-b x+c+(q(x))\\right)\\left(d x^2-e x+f+(q(x))\\right) \\\\\n\n& =\\tau(n) \\tau(m) .\n\n\\end{aligned}\n\n$$\n\nwhere in the second equality we used that $x^3+p(x)=2+p(x)$ and in the fifth we used that $x^3+$ $q(x)=-2+q(x)$\n\n\\end{proof}"}
{"id": "Herstein|exercise_4_6_2", "formal_statement": "theorem exercise_4_6_2 : irreducible (X^3 + 3*X + 2 : polynomial \u211a) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $f(x) = x^3 + 3x + 2$ is irreducible in $Q[x]$.", "nl_proof": "\\begin{proof}    \n\nLet us assume that $f(x)$ is reducible over $\\mathbb{Q}[x]$.\n\nThen there exists a rational root of $f(x)$.\n\nLet $p / q$ be a rational root of $f(x)$, where $\\operatorname{gcd}(p, q)=1$.\n\nThen $f(p / q)=0$.\n\nNow,\n\n$$\n\n\\begin{aligned}\n\n& f(p / q)=(p / q)^3+3(p / q)+2 \\\\\n\n\\Longrightarrow & (p / q)^3+3(p / q)+2=0 \\\\\n\n\\Longrightarrow & p^3+3 p q^2=-2 q^3 \\\\\n\n\\Longrightarrow & p\\left(p^2+3 q^2\\right)=-q^3\n\n\\end{aligned}\n\n$$\n\nIt follows that, $p$ divides $q$ which is a contradiction to the fact that $\\operatorname{gcd}(p, q)=1$.\n\nThis implies that $f(x)$ has no rational root.\n\nNow we know that, a polynomial of degree two or three over a field $F$ is reducible if and only if it has a root in $F$.\n\nNow $f(x)$ is a 3 degree polynomial having no root in $\\mathbb{Q}$.\n\nSo, $f(x)$ is irreducible in $\\mathbb{Q}[x]$.\n\nThis completes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_5_1_8", "formal_statement": "theorem exercise_5_1_8 {p m n: \u2115} {F : Type*} [field F] \n  (hp : nat.prime p) (hF : char_p F p) (a b : F) (hm : m = p ^ n) : \n  (a + b) ^ m = a^m + b^m :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $F$ is a field of characteristic $p \\neq 0$, show that $(a + b)^m = a^m + b^m$, where $m = p^n$, for all $a, b \\in F$ and any positive integer $n$.", "nl_proof": "\\begin{proof}\n\n    Since $F$ is of characteristic $p$ and we have considered arbitrary two elements $a, b$ in $F$ we have\n\n$$\n\n\\begin{aligned}\n\n& p a=p b=0 \\\\\n\n& \\Longrightarrow p^n a=p^n b=0 \\\\\n\n& \\Longrightarrow m a=m b=0 \\text {. } \\\\\n\n&\n\n\\end{aligned}\n\n$$\n\nNow we know from Binomial Theorem that\n\n$$\n\n(a+b)^m=\\sum_{i=0}^m\\left(\\begin{array}{c}\n\nm \\\\\n\ni\n\n\\end{array}\\right) a^i b^{m-i}\n\n$$\n\nHere\n\n$$\n\n\\left(\\begin{array}{c}\n\nm \\\\\n\ni\n\n\\end{array}\\right)=\\frac{m !}{i !(m-i) !} .\n\n$$\n\nNow we know that for any integer $n$ and any integer $k$ satisfying $1 \\leq k<n, n$ always divides $\\left(\\begin{array}{l}n \\\\ k\\end{array}\\right)$. So in our case for $i$ in the range $1 \\leq i<m, m$ divides $\\left(\\begin{array}{c}m \\\\ i\\end{array}\\right)$. It follows that $p$ divides $\\left(\\begin{array}{c}m \\\\ i\\end{array}\\right)$, for $i$ satisfying $1 \\leq i<m$, since $m=p^n$ for any integer $n$. Therefore other than the terms $a^m$ and $b^m$ in the expansion $\\sum_{i=0}^m\\left(\\begin{array}{c}m \\\\ i\\end{array}\\right) a^i b^{m-i}$ will vanish due to char $p$ nature of $F$.\n\nHence we have\n\n$$\n\n\\sum_{i=0}^m\\left(\\begin{array}{c}\n\nm \\\\\n\ni\n\n\\end{array}\\right) a^i b^{m-i}=a^m+b^m\n\n$$\n\nThis follows that, for all $a, b \\in F$\n\n$$\n\n(a+b)^m=a^m+b^m .\n\n$$\n\nThis completes the proof.\n\n\\end{proof}"}
{"id": "Herstein|exercise_5_3_7", "formal_statement": "theorem exercise_5_3_7 {K : Type*} [field K] {F : subfield K} \n  {a : K} (ha : is_algebraic F (a ^ 2)) : is_algebraic F a :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a \\in K$ is such that $a^2$ is algebraic over the subfield $F$ of $K$, show that a is algebraic over $F$.", "nl_proof": "\\begin{proof}\n\n    Since $a^2$ is algebraic over $F$, there exist a non-zero polynomial $f(x)$ in $F[x]$ such that $f\\left(a^2\\right)=0$. Consider a new polynomial $g(x)$ defined as $g(x)=f\\left(x^2\\right)$. Clearly $g(x) \\in F[x]$ and $g(a)=f\\left(a^2\\right)= 0$.\n\n\\end{proof}"}
{"id": "Herstein|exercise_5_4_3", "formal_statement": "theorem exercise_5_4_3 {a : \u2102} {p : \u2102 \u2192 \u2102} \n  (hp : p = \u03bb x, x^5 + real.sqrt 2 * x^3 + real.sqrt 5 * x^2 + \n  real.sqrt 7 * x + 11)\n  (ha : p a = 0) : \n  \u2203 p : polynomial \u2102 , p.degree < 80 \u2227 a \u2208 p.roots \u2227 \n  \u2200 n : p.support, \u2203 a b : \u2124, p.coeff n = a / b :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a \\in C$ is such that $p(a) = 0$, where $p(x) = x^5 + \\sqrt{2}x^3 + \\sqrt{5}x^2 + \\sqrt{7}x + \\sqrt{11}$, show that $a$ is algebraic over $\\mathbb{Q}$ of degree at most 80.", "nl_proof": "\\begin{proof}\n\n    Given $a \\in \\mathbb{C}$ such that $p(a)=0$, where\n\n$$\n\np(x)=x^5+\\sqrt{2} x^3+\\sqrt{5} x^2+\\sqrt{7} x+\\sqrt{11}\n\n$$\n\nHere, we note that $p(x) \\in \\mathbb{Q}(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})$ and\n\n$$\n\n\\begin{aligned}\n\n    {[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}): \\mathbb{Q}] } & =[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}): Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7})] \\cdot[\\mathbb{Q}(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}): \\mathbb{Q}(\\sqrt{2}, \\sqrt{5})] \\\\\n\n& \\cdot[\\mathbb{Q}(\\sqrt{2}, \\sqrt{5}): \\mathbb{Q}(\\sqrt{2})] \\cdot[\\mathbb{Q}(\\sqrt{2}): \\mathbb{Q}] \\\\\n\n& =2 \\cdot 2 \\cdot 2 \\cdot 2 \\\\\n\n& =16\n\n\\end{aligned}\n\n$$\n\nHere, we note that $p(x)$ is of degree 5 over $\\mathbb{Q}(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})$. If $a$ is root of $p(x)$, then\n\n$$\n\n[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}, a): \\mathbb{Q}]=[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}): Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})] \\cdot 15\n\n$$\n\nand $[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}): Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})] \\leq 5$. We get equality if $p(x)$ is irreducible over $Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11})$. This gives\n\n$$\n\n[Q(\\sqrt{2}, \\sqrt{5}, \\sqrt{7}, \\sqrt{11}, a): \\mathbb{Q}] \\leq 16 \\cdot 5=80\n\n$$\n\n\\end{proof}"}
{"id": "Herstein|exercise_5_6_14", "formal_statement": "theorem exercise_5_6_14 {p m n: \u2115} (hp : nat.prime p) {F : Type*} \n  [field F] [char_p F p] (hm : m = p ^ n) : \n  card (root_set (X ^ m - X : polynomial F) F) = m :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $F$ is of characteristic $p \\neq 0$, show that all the roots of $x^m - x$, where $m = p^n$, are distinct.", "nl_proof": "\\begin{proof}\n\n    Let us consider $f(x)=x^m-x$. Then $f \\in F[x]$.\n\nClaim: $f(x)$ has a multiple root in some extension of $F$ if and only if $f(x)$ is not relatively prime to its formal derivative, $f^{\\prime}(x)$. \n\n\n\nProof of the Claim: Let us assume that $f(x)$ has a multiple root in some extension of $F$. Let $y$ be a multiple root of $f(x)$. Then over a splitting field, we have\n\n$$\n\nf(x)=(x-y)^n g(x), \\text { for some integer } n \\geq 2 .\n\n$$\n\nHere $g(x)$ is a polynomial such that $g(y) \\neq 0$. Now taking derivative of $f$ we get\n\n$$\n\nf^{\\prime}(x)=n \\cdot(x-y)^{n-1} g(x)+(x-y)^n g^{\\prime}(x)\n\n$$\n\nhere $g^{\\prime}(x)$ implies derivative of $g$ with respect to $x$. Since we have $n \\geq 2$, this implies $(n-1) \\geq 1$. Hence, (1) shows that $f^{\\prime}(x)$ has $y$ as a root. Therefore, $f(x)$ is not relatively prime to $f^{\\prime}(x)$. We now prove the other direction.\n\nConversely, let us assume that $f(x)$ is not relatively prime to $f^{\\prime}(x)$. Let $y$ is a root of both $f(x)$ and $f^{\\prime}(x)$. Since $y$ is a root of $f(x)$, we can write\n\n$$\n\nf(x)=(x-y) \\cdot g(x)\n\n$$\n\nfor some polynomial $g(x)$. then taking derivative of $f(x)$ we have\n\n$$\n\nf^{\\prime}(x)=g(x)+(x-y) \\cdot g^{\\prime}(x)\n\n$$\n\nwhere $g^{\\prime}(x)$ is the derivative of $g(x)$ with respect to $x$. Since $y$ is a root of $f^{\\prime}(x)$ also we have\n\n$$\n\nf^{\\prime}(y)=0\n\n$$\n\nThen we have\n\n$$\n\n\\begin{aligned}\n\n& f^{\\prime}(y)=g(y)+(y-y) \\cdot g^{\\prime}(y) \\\\\n\n\\Longrightarrow & f^{\\prime}(y)=g(y) \\\\\n\n\\Longrightarrow & g(y)=0 .\n\n\\end{aligned}\n\n$$\n\nThis implies $y$ is a root of $g(x)$ also. Therefore we have\n\n$$\n\ng(x)=(x-y) \\cdot h(x)\n\n$$\n\nfor some polynomial $h(x)$. Now form (2) we have\n\n$$\n\nf(x)=(x-y)^2 \\cdot h(x) .\n\n$$\n\nThis follows that $y$ is a multiple root of $f(x)$. Therefore, $f(x)$ has a multiple root in some extension of the field $F$. This completes the proof of the Claim.\n\n\n\nIn our case, $f(x)=x^m-x$, where $m=p^n$. Now we calculate the derivative of $f$. That is\n\n$$\n\nf^{\\prime}(x)=m x^{m-1}-1=-1(\\bmod p) .\n\n$$\n\nBy the above condition it follows that, $f^{\\prime}$ has no root same as $f$, that is, $f(x)$ and $f^{\\prime}(x)$ are relatively prime. Hence, $f(x)$ has no multiple root in $F$. Since $f(x)=x^m-x$ is a polynomial of degree $m$, it follows that $f(x)$ has $m$ distinct roots in $F$, where $m=p^n$. This completes the proof.\n\n\\end{proof}"}
{"id": "Artin|exercise_2_3_2", "formal_statement": "theorem exercise_2_3_2 {G : Type*} [group G] (a b : G) :\n  \u2203 g : G, b* a = g * a * b * g\u207b\u00b9 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the products $a b$ and $b a$ are conjugate elements in a group.", "nl_proof": "\\begin{proof}\n\n    We have that $(a^{-1})ab(a^{-1})^{-1} = ba$. \n\n\\end{proof}"}
{"id": "Artin|exercise_2_8_6", "formal_statement": "theorem exercise_2_8_6 {G H : Type*} [group G] [group H] :\n  center (G \u00d7 H) \u2243* (center G) \u00d7 (center H) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the center of the product of two groups is the product of their centers.", "nl_proof": "\\begin{proof}\n\n    We have that $(g_1, g_2)\\cdot (h_1, h_2) = (h_1, h_2)\\cdot (g_1, g_2)$ if and only if $g_1h_1 = h_1g_1$ and $g_2h_2 = h_2g_2$. \n\n\\end{proof}"}
{"id": "Artin|exercise_3_2_7", "formal_statement": "theorem exercise_3_2_7 {F : Type*} [field F] {G : Type*} [field G]\n  (\u03c6 : F \u2192+* G) : injective \u03c6 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that every homomorphism of fields is injective.", "nl_proof": "\\begin{proof}\n\n    Suppose $f(a)=f(b)$, then $f(a-b)=0=f(0)$. If $u=(a-b) \\neq 0$, then $f(u) f\\left(u^{-1}\\right)=f(1)=1$, but that means that $0 f\\left(u^{-1}\\right)=1$, which is impossible. Hence $a-b=0$ and $a=b$.\n\n\\end{proof}"}
{"id": "Artin|exercise_3_7_2", "formal_statement": "theorem exercise_3_7_2 {K V : Type*} [field K] [add_comm_group V]\n  [module K V] {\u03b9 : Type*} [fintype \u03b9] (\u03b3 : \u03b9 \u2192 submodule K V)\n  (h : \u2200 i : \u03b9, \u03b3 i \u2260 \u22a4) :\n  (\u22c2 (i : \u03b9), (\u03b3 i : set V)) \u2260 \u22a4 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $V$ be a vector space over an infinite field $F$. Prove that $V$ is not the union of finitely many proper subspaces.", "nl_proof": "\\begin{proof}\n\n    If $V$ is the set-theoretic union of $n$ proper subspaces $W_i$ ( $1 \\leq i \\leq n$ ), then $|F| \\leq n-1$.\n\nProof. We may suppose no $W_i$ is contained in the union of the other subspaces. Let $u \\in W_i, \\quad u \\notin \\bigcup_{j \\neq i} W_j$ and $v \\notin W_i$.\n\nThen $(v+F u) \\cap W_i=\\varnothing$ and $(v+F u) \\cap W_j(j \\neq i)$ contains at most one vector since otherwise $W_j$ would contain $u$. Hence\n\n$$\n\n|v+F u|=|F| \\leq n-1 .\n\n$$\n\nCorollary: Avoidance lemma for vector spaces.\n\nLet $E$ be a vector space over an infinite field. If a subspace is contained in a finite union of subspaces, it is contained in one of them.\n\n\\end{proof}"}
{"id": "Artin|exercise_6_4_2", "formal_statement": "theorem exercise_6_4_2 {G : Type*} [group G] [fintype G] {p q : \u2115}\n  (hp : prime p) (hq : prime q) (hG : card G = p*q) :\n  is_simple_group G \u2192 false :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that no group of order $p q$, where $p$ and $q$ are prime, is simple.", "nl_proof": "\\begin{proof}\n\n    If $|G|=n=p q$ then the only two Sylow subgroups are of order $p$ and $q$.\n\nFrom Sylow's third theorem we know that $n_p \\mid q$ which means that $n_p=1$ or $n_p=q$.\n\nIf $n_p=1$ then we are done (by a corollary of Sylow's theorem)\n\nIf $n_p=q$ then we have accounted for $q(p-1)=p q-q$ elements of $G$ and so there is only one group of order $q$ and again we are done.\n\n\\end{proof}"}
{"id": "Artin|exercise_6_4_12", "formal_statement": "theorem exercise_6_4_12 {G : Type*} [group G] [fintype G]\n  (hG : card G = 224) :\n  is_simple_group G \u2192 false :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that no group of order 224 is simple.", "nl_proof": "\\begin{proof}\n\n    The following proves there must exist a normal Sylow 2 -subgroup of order 32 ,\n\nSuppose there are $n_2=7$ Sylow 2 -subgroups in $G$. Making $G$ act on the set of these Sylow subgroups by conjugation (Mitt wrote about this but on the set of the other Sylow subgroups, which gives no contradiction), we get a homomorphism $G \\rightarrow S_7$ which must be injective if $G$ is simple (why?).\n\n\n\nBut this cannot be since then we would embed $G$ into $S_7$, which is impossible since $|G| \\nmid 7 !=\\left|S_7\\right|$ (why?)\n\n\\end{proof}"}
{"id": "Artin|exercise_10_1_13", "formal_statement": "theorem exercise_10_1_13 {R : Type*} [ring R] {x : R}\n  (hx : is_nilpotent x) : is_unit (1 + x) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "An element $x$ of a ring $R$ is called nilpotent if some power of $x$ is zero. Prove that if $x$ is nilpotent, then $1+x$ is a unit in $R$.", "nl_proof": "\\begin{proof}\n\n    If $x^n=0$, then\n\n$$\n\n(1+x)\\left(\\sum_{k=0}^{n-1}(-1)^k x^k\\right)=1+(-1)^{n-1} x^n=1 .\n\n$$\n\n\\end{proof}"}
{"id": "Artin|exercise_10_6_7", "formal_statement": "theorem exercise_10_6_7 {I : ideal gaussian_int}\n  (hI : I \u2260 \u22a5) : \u2203 (z : I), z \u2260 0 \u2227 (z : gaussian_int).im = 0 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that every nonzero ideal in the ring of Gauss integers contains a nonzero integer.", "nl_proof": "\\begin{proof}\n\n    Let $I$ be some nonzero ideal. Then there exists some $z \\in \\mathbb{Z}[i], z \\neq 0$ such that $z \\in I$. We know that $z=a+b i$, for some $a, b \\in \\mathbb{Z}$. We consider three cases:\n\n1. If $b=0$, then $z=a$, so $z \\in \\mathbb{Z} \\cap I$, and $z \\neq 0$, so the statement of the exercise holds.\n\n2. If $a=0$, then $z=i b$. Since $z \\neq 0$, we conclude that $b \\neq 0$. Since $I$ is an ideal in $\\mathbb{Z}[i]$, and $i \\in \\mathbb{Z}[i]$, we conclude that $i z \\in I$. Furthermore, $i z=-b \\in \\mathbb{Z}$. Thus, $i z$ is a nonzero integer which is in $I$.\n\n3. Let $a \\neq 0$ and $b \\neq 0$. Since $I$ is an ideal and $z \\in I$, we conclude that $z^2 \\in I$; that is,\n\n$$\n\n(a+b i)^2=a^2-b^2+2 a b i \\in I\n\n$$\n\nFurthermore, since $-2 a \\in \\mathbb{Z}[i]$, and $z \\in I$ and $I$ is an ideal, $-2 a z \\in I$; that is,\n\n$$\n\n-2 a z=-2 a(a+b i)=-2 a^2-2 a b i \\in I\n\n$$\n\nSince $I$ is closed under addition,\n\n$$\n\n\\left(a^2-b^2+2 a b i\\right)+\\left(-2 a^2-2 a b i\\right) \\in I \\Longrightarrow-a^2-b^2 \\in I\n\n$$\n\nNotice that $-a^2-b^2 \\neq 0$ since $a^2>0$ and $b^2>0$, so $-a^2-b^2<0$. Furthermore, it is an integer. Thus, we have found a nonzero integer in $I$.\n\n\\end{proof}"}
{"id": "Artin|exercise_10_4_7a", "formal_statement": "theorem exercise_10_4_7a {R : Type*} [comm_ring R] [no_zero_divisors R]\n  (I J : ideal R) (hIJ : I + J = \u22a4) : I * J = I \u2293 J :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $I, J$ be ideals of a ring $R$ such that $I+J=R$. Prove that $I J=I \\cap J$.", "nl_proof": "\\begin{proof}\n\n    We have seen that $IJ \\subset I \\cap J$, so it remains to show that $I \\cap J \\subset IJ$.  Since $I+J = (1)$, there are elements $i \\in I$ and $j \\in J$ such that $i+j = 1$.  Let $k \\in I \\cap J$, and multiply $i+j=1$ through by $k$ to get $ki+kj = k$.  Write this more suggestively as\n\n\\[ k = ik+kj. \\]\n\nThe first term is in $IJ$ because $k \\in J$, and the second term is in $IJ$ because $k \\in I$, so $k \\in IJ$ as desired.\n\n\\end{proof}"}
{"id": "Artin|exercise_11_2_13", "formal_statement": "theorem exercise_11_2_13 (a b : \u2124) :\n  (of_int a : gaussian_int) \u2223 of_int b \u2192 a \u2223 b :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $a, b$ are integers and if $a$ divides $b$ in the ring of Gauss integers, then $a$ divides $b$ in $\\mathbb{Z}$.", "nl_proof": "\\begin{proof}\n\n    Suppose $a|b$ in $\\mathbb{Z}[i]$ and $a,b\\in\\mathbb{Z}$. Then $a(x+yi)=b$ for $x,y\\in\\mathbb{Z}$. Expanding this we get $ax+ayi=b$, and equating imaginary parts gives us $ay=0$, implying $y=0$. \n\n\\end{proof}"}
{"id": "Artin|exercise_11_4_6a", "formal_statement": "theorem exercise_11_4_6a {F : Type*} [field F] [fintype F] (hF : card F = 7) :\n  irreducible (X ^ 2 + 1 : polynomial F) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^2+x+1$ is irreducible in the field $\\mathbb{F}_2$.", "nl_proof": "\\begin{proof}\n\n    If $x^2+x+1$ were reducible in $\\mathbb{F}_2$, its factors must be linear. But we neither have that $0^2+0+1=$ nor $1^2+1+1=0$, therefore $x^2+x+1$ is irreducible.  \n\n\\end{proof}"}
{"id": "Artin|exercise_11_4_6c", "formal_statement": "theorem exercise_11_4_6c : irreducible (X^3 - 9 : polynomial (zmod 31)) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^3 - 9$ is irreducible in $\\mathbb{F}_{31}$.", "nl_proof": "\\begin{proof}\n\n    If $p(x) = x^3-9$ were reducible, it would have a linear factor, since it either has a linear factor and a quadratic factor or three linear factors. We can then verify by brute force that $p(x)\\neq 0$ for $x \\in \\mathbb{F}_31$. \n\n\\end{proof}"}
{"id": "Artin|exercise_11_13_3", "formal_statement": "theorem exercise_11_13_3 (N : \u2115):\n  \u2203 p \u2265 N, nat.prime p \u2227 p + 1 \u2261 0 [MOD 4] :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that there are infinitely many primes congruent to $-1$ (modulo $4$).", "nl_proof": "\\begin{proof}\n\n    First we show a lemma: if $a \\equiv 3(\\bmod 4)$ then there exists a prime $p$ such that $p \\mid a$ and $p \\equiv 3(\\bmod 4)$.\n\n\n\n    Clearly, all primes dividing $a$ are odd. Suppose all of them would be $\\equiv 1(\\bmod 4)$. Then their product would also be $a \\equiv 1(\\bmod 4)$, which is a contradiction.\n\n\n\nTo prove the main claim, suppose that $p_1, \\ldots, p_n$ would be all such primes. (In particular, we have $p_1=3$.) Consider $a=4 p_2 \\cdots p_n+3$. (Or you can take $a=4 p_2 \\cdots p_n-1$.) Show that $p_i \\nmid a$ for $i=1, \\ldots, n$. (The case $3 \\nmid a$ is solved differently than the other primes - this is the reason for omitting $p_1$ in the definition of $a$.) Then use the above lemma to get a contradiction.\n\n\\end{proof}"}
{"id": "Artin|exercise_13_6_10", "formal_statement": "theorem exercise_13_6_10 {K : Type*} [field K] [fintype K\u02e3] :\n  \u220f (x : K\u02e3), x = -1 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $K$ be a finite field. Prove that the product of the nonzero elements of $K$ is $-1$.", "nl_proof": "\\begin{proof}\n\n    Since we are working with a finite field with $q$ elements, anyone of them is a root of the following polynomial\n\n$$\n\nx^q-x=0 .\n\n$$\n\nIn particular if we rule out the 0 element, any $a_i \\neq 0$ is a root of\n\n$$\n\nx^{q-1}-1=0 .\n\n$$\n\nThis polynomial splits completely in $\\mathbb{F}_q$ so we find\n\n$$\n\n\\left(x-a_1\\right) \\cdots\\left(x-a_{q-1}\\right)=0\n\n$$\n\nin particular\n\n$$\n\nx^{q-1}-1=\\left(x-a_1\\right) \\cdots\\left(x-a_{q-1}\\right)\n\n$$\n\nThus $a_1 \\cdots a_{q-1}=-1$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_2a", "formal_statement": "theorem exercise_1_1_2a : \u2203 a b : \u2124, a - b \u2260 b - a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove the the operation $\\star$ on $\\mathbb{Z}$ defined by $a\\star b=a-b$ is not commutative.", "nl_proof": "\\begin{proof}\n\n    Not commutative since\n\n$$\n\n1 \\star(-1)=1-(-1)=2\n\n$$\n\n$$\n\n(-1) \\star 1=-1-1=-2 .\n\n$$\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_4", "formal_statement": "theorem exercise_1_1_4 (n : \u2115) : \n  \u2200 (a b c : \u2115), (a * b) * c \u2261 a * (b * c) [ZMOD n] :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the multiplication of residue class $\\mathbb{Z}/n\\mathbb{Z}$ is associative.", "nl_proof": "\\begin{proof}\n\n    We have\n\n$$\n\n\\begin{aligned}\n\n(\\bar{a} \\cdot \\bar{b}) \\cdot \\bar{c} &=\\overline{a \\cdot b} \\cdot \\bar{c} \\\\\n\n&=\\overline{(a \\cdot b) \\cdot c} \\\\\n\n&=\\overline{a \\cdot(b \\cdot c)} \\\\\n\n&=\\bar{a} \\cdot \\overline{b \\cdot c} \\\\\n\n&=\\bar{a} \\cdot(\\bar{b} \\cdot \\bar{c})\n\n\\end{aligned}\n\n$$\n\nsince integer multiplication is associative.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_15", "formal_statement": "theorem exercise_1_1_15 {G : Type*} [group G] (as : list G) :\n  as.prod\u207b\u00b9 = (as.reverse.map (\u03bb x, x\u207b\u00b9)).prod :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $(a_1a_2\\dots a_n)^{-1} = a_n^{-1}a_{n-1}^{-1}\\dots a_1^{-1}$ for all $a_1, a_2, \\dots, a_n\\in G$.", "nl_proof": "\\begin{proof}\n\n    For $n=1$, note that for all $a_1 \\in G$ we have $a_1^{-1}=a_1^{-1}$.\n\nNow for $n \\geq 2$ we proceed by induction on $n$. For the base case, note that for all $a_1, a_2 \\in G$ we have\n\n$$\n\n\\left(a_1 \\cdot a_2\\right)^{-1}=a_2^{-1} \\cdot a_1^{-1}\n\n$$\n\nsince\n\n$$\n\na_1 \\cdot a_2 \\cdot a_2^{-1} a_1^{-1}=1 .\n\n$$\n\nFor the inductive step, suppose that for some $n \\geq 2$, for all $a_i \\in G$ we have\n\n$$\n\n\\left(a_1 \\cdot \\ldots \\cdot a_n\\right)^{-1}=a_n^{-1} \\cdot \\ldots \\cdot a_1^{-1} .\n\n$$\n\nThen given some $a_{n+1} \\in G$, we have\n\n$$\n\n\\begin{aligned}\n\n\\left(a_1 \\cdot \\ldots \\cdot a_n \\cdot a_{n+1}\\right)^{-1} &=\\left(\\left(a_1 \\cdot \\ldots \\cdot a_n\\right) \\cdot a_{n+1}\\right)^{-1} \\\\\n\n&=a_{n+1}^{-1} \\cdot\\left(a_1 \\cdot \\ldots \\cdot a_n\\right)^{-1} \\\\\n\n&=a_{n+1}^{-1} \\cdot a_n^{-1} \\cdot \\ldots \\cdot a_1^{-1},\n\n\\end{aligned}\n\n$$\n\nusing associativity and the base case where necessary.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_17", "formal_statement": "theorem exercise_1_1_17 {G : Type*} [group G] {x : G} {n : \u2115}\n  (hxn: order_of x = n) :\n  x\u207b\u00b9 = x ^ (n - 1 : \u2124) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $x$ be an element of $G$. Prove that if $|x|=n$ for some positive integer $n$ then $x^{-1}=x^{n-1}$.", "nl_proof": "\\begin{proof}\n\n    We have $x \\cdot x^{n-1}=x^n=1$, so by the uniqueness of inverses $x^{-1}=x^{n-1}$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_20", "formal_statement": "theorem exercise_1_1_20 {G : Type*} [group G] {x : G} :\n  order_of x = order_of x\u207b\u00b9 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "For $x$ an element in $G$ show that $x$ and $x^{-1}$ have the same order.", "nl_proof": "\\begin{proof}\n\n    Recall that the order of a group element is either a positive integer or infinity.\n\nSuppose $|x|$ is infinite and that $\\left|x^{-1}\\right|=n$ for some $n$. Then\n\n$$\n\nx^n=x^{(-1) \\cdot n \\cdot(-1)}=\\left(\\left(x^{-1}\\right)^n\\right)^{-1}=1^{-1}=1,\n\n$$\n\na contradiction. So if $|x|$ is infinite, $\\left|x^{-1}\\right|$ must also be infinite. Likewise, if $\\left|x^{-1}\\right|$ is infinite, then $\\left|\\left(x^{-1}\\right)^{-1}\\right|=|x|$ is also infinite.\n\nSuppose now that $|x|=n$ and $\\left|x^{-1}\\right|=m$ are both finite. Then we have\n\n$$\n\n\\left(x^{-1}\\right)^n=\\left(x^n\\right)^{-1}=1^{-1}=1,\n\n$$\n\nso that $m \\leq n$. Likewise, $n \\leq m$. Hence $m=n$ and $x$ and $x^{-1}$ have the same order.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_22b", "formal_statement": "theorem exercise_1_1_22b {G: Type*} [group G] (a b : G) : \n  order_of (a * b) = order_of (b * a) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Deduce that $|a b|=|b a|$ for all $a, b \\in G$.", "nl_proof": "\\begin{proof}\n\n    Let $a$ and $b$ be arbitrary group elements. Letting $x=a b$ and $g=a$, we see that\n\n$$\n\n|a b|=\\left|a^{-1} a b a\\right|=|b a| .\n\n$$\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_1_29", "formal_statement": "theorem exercise_1_1_29 {A B : Type*} [group A] [group B] :\n  \u2200 x y : A \u00d7 B, x*y = y*x \u2194 (\u2200 x y : A, x*y = y*x) \u2227 \n  (\u2200 x y : B, x*y = y*x) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $A \\times B$ is an abelian group if and only if both $A$ and $B$ are abelian.", "nl_proof": "\\begin{proof}\n\n    $(\\Rightarrow)$ Suppose $a_1, a_2 \\in A$ and $b_1, b_2 \\in B$. Then\n\n$$\n\n\\left(a_1 a_2, b_1 b_2\\right)=\\left(a_1, b_1\\right) \\cdot\\left(a_2, b_2\\right)=\\left(a_2, b_2\\right) \\cdot\\left(a_1, b_1\\right)=\\left(a_2 a_1, b_2 b_1\\right) .\n\n$$\n\nSince two pairs are equal precisely when their corresponding entries are equal, we have $a_1 a_2=a_2 a_1$ and $b_1 b_2=b_2 b_1$. Hence $A$ and $B$ are abelian.\n\n$(\\Leftarrow)$ Suppose $\\left(a_1, b_1\\right),\\left(a_2, b_2\\right) \\in A \\times B$. Then we have\n\n$$\n\n\\left(a_1, b_1\\right) \\cdot\\left(a_2, b_2\\right)=\\left(a_1 a_2, b_1 b_2\\right)=\\left(a_2 a_1, b_2 b_1\\right)=\\left(a_2, b_2\\right) \\cdot\\left(a_1, b_1\\right) .\n\n$$\n\nHence $A \\times B$ is abelian.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_3_8", "formal_statement": "theorem exercise_1_3_8 : infinite (equiv.perm \u2115) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $\\Omega=\\{1,2,3, \\ldots\\}$ then $S_{\\Omega}$ is an infinite group", "nl_proof": "\\begin{proof}\n\n    Recall that the codomain of an injective function must be at least as large (in cardinality) as the domain of the function. With that in mind, define the function\n\n$$\n\n\\begin{gathered}\n\nf: \\mathbb{N} \\rightarrow S_{\\mathbb{N}} \\\\\n\nf(n)=(1 n)\n\n\\end{gathered}\n\n$$\n\nwhere $(1 n)$ is the cycle decomposition of an element of $S_{\\mathbb{N}}$ (specifically it's the function given by $g(1)=n, g(2)=2, g(3)=3, \\ldots)$. The function $f$ maps every natural number to a distinct one of these functions. Hence $f$ is injective. Hence $\\infty=|\\mathbb{N}| \\leq\\left|S_{\\mathbb{N}}\\right|$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_6_11", "formal_statement": "theorem exercise_1_6_11 {A B : Type*} [group A] [group B] : \n  A \u00d7 B \u2243* B \u00d7 A :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $A$ and $B$ be groups. Prove that $A \\times B \\cong B \\times A$.", "nl_proof": "\\begin{proof}\n\n    We know from set theory that the mapping $\\varphi: A \\times B \\rightarrow B \\times A$ given by $\\varphi((a, b))=(b, a)$ is a bijection with inverse $\\psi: B \\times A \\rightarrow A \\times B$ given by $\\psi((b, a))=(a, b)$. Also $\\varphi$ is a homomorphism, as we show below.\n\nLet $a_1, a_2 \\in A$ and $b_1, b_2 \\in B$. Then\n\n$$\n\n\\begin{aligned}\n\n\\varphi\\left(\\left(a_1, b_1\\right) \\cdot\\left(a_2, b_2\\right)\\right) &=\\varphi\\left(\\left(a_1 a_2, b_1 b_2\\right)\\right) \\\\\n\n&=\\left(b_1 b_2, a_1 a_2\\right) \\\\\n\n&=\\left(b_1, a_1\\right) \\cdot\\left(b_2, a_2\\right) \\\\\n\n&=\\varphi\\left(\\left(a_1, b_1\\right)\\right) \\cdot \\varphi\\left(\\left(a_2, b_2\\right)\\right)\n\n\\end{aligned}\n\n$$\n\nHence $A \\times B \\cong B \\times A$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_1_6_23", "formal_statement": "theorem exercise_1_6_23 {G : Type*} \n  [group G] (\u03c3 : mul_aut G) (hs : \u2200 g : G, \u03c3 g = 1 \u2192 g = 1) \n  (hs2 : \u2200 g : G, \u03c3 (\u03c3 g) = g) :\n  \u2200 x y : G, x*y = y*x :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a finite group which possesses an automorphism $\\sigma$ such that $\\sigma(g)=g$ if and only if $g=1$. If $\\sigma^{2}$ is the identity map from $G$ to $G$, prove that $G$ is abelian.", "nl_proof": "\\begin{proof}\n\n    Solution: We define a mapping $f: G \\rightarrow G$ by $f(x)=x^{-1} \\sigma(x)$.\n\nClaim: $f$ is injective.\n\nProof of claim: Suppose $f(x)=f(y)$. Then $y^{-1} \\sigma(y)=x^{-1} \\sigma(x)$, so that $x y^{-1}=\\sigma(x) \\sigma\\left(y^{-1}\\right)$, and $x y^{-1}=\\sigma\\left(x y^{-1}\\right)$. Then we have $x y^{-1}=1$, hence $x=y$. So $f$ is injective.\n\n\n\nSince $G$ is finite and $f$ is injective, $f$ is also surjective. Then every $z \\in G$ is of the form $x^{-1} \\sigma(x)$ for some $x$. Now let $z \\in G$ with $z=x^{-1} \\sigma(x)$. We have\n\n$$\n\n\\sigma(z)=\\sigma\\left(x^{-1} \\sigma(x)\\right)=\\sigma(x)^{-1} x=\\left(x^{-1} \\sigma(x)\\right)^{-1}=z^{-1} .\n\n$$\n\nThus $\\sigma$ is in fact the inversion mapping, and we assumed that $\\sigma$ is a homomorphism. By a previous example, then, $G$ is abelian.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_2_1_13", "formal_statement": "theorem exercise_2_1_13 (H : add_subgroup \u211a) {x : \u211a} \n  (hH : x \u2208 H \u2192 (1 / x) \u2208 H):\n  H = \u22a5 \u2228 H = \u22a4 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $H$ be a subgroup of the additive group of rational numbers with the property that $1 / x \\in H$ for every nonzero element $x$ of $H$. Prove that $H=0$ or $\\mathbb{Q}$.", "nl_proof": "\\begin{proof}\n\n    Solution: First, suppose there does not exist a nonzero element in $H$. Then $H=0$.\n\nNow suppose there does exist a nonzero element $a \\in H$; without loss of generality, say $a=p / q$ in lowest terms for some integers $p$ and $q$ - that is, $\\operatorname{gcd}(p, q)=1$. Now $q \\cdot \\frac{p}{q}=p \\in H$, and since $q / p \\in H$, we have $p \\cdot \\frac{q}{p} \\in H$. There exist integers $x, y$ such that $q x+p y=1$; note that $q x \\in H$ and $p y \\in H$, so that $1 \\in H$. Thus $n \\in H$ for all $n \\in \\mathbb{Z}$. Moreover, if $n \\neq 0,1 / n \\in H$. Then $m / n \\in H$ for all integers $m, n$ with $n \\neq 0$; hence $H=\\mathbb{Q}$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_2_4_16a", "formal_statement": "theorem exercise_2_4_16a {G : Type*} [group G] {H : subgroup G}  \n  (hH : H \u2260 \u22a4) : \n  \u2203 M : subgroup G, M \u2260 \u22a4 \u2227\n  \u2200 K : subgroup G, M \u2264 K \u2192 K = M \u2228 K = \u22a4 \u2227 \n  H \u2264 M :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "A subgroup $M$ of a group $G$ is called a maximal subgroup if $M \\neq G$ and the only subgroups of $G$ which contain $M$ are $M$ and $G$. Prove that if $H$ is a proper subgroup of the finite group $G$ then there is a maximal subgroup of $G$ containing $H$.", "nl_proof": "\\begin{proof}\n\nIf $H$ is maximal, then we are done. If $H$ is not maximal, then there is a subgroup $K_1$ of $G$ such that $H<K_1<G$. If $K_1$ is maximal, we are done. But if $K_1$ is not maximal, there is a subgroup $K_2$ with $H<K_1<K_2<G$. If $K_2$ is maximal, we are done, and if not, keep repeating the procedure. Since $G$ is finite, this process must eventually come to an end, so that $K_n$ is maximal for some positive integer $n$. Then $K_n$ is a maximal subgroup containing $H$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_2_4_16c", "formal_statement": "theorem exercise_2_4_16c {n : \u2115} (H : add_subgroup (zmod n)) : \n  \u2203 p : \u2115, nat.prime p \u2227 H = add_subgroup.closure {p} \u2194 \n  H \u2260 \u22a4 \u2227 \u2200 K : add_subgroup (zmod n), H \u2264 K \u2192 K = H \u2228 K = \u22a4 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that if $G=\\langle x\\rangle$ is a cyclic group of order $n \\geq 1$ then a subgroup $H$ is maximal if and only $H=\\left\\langle x^{p}\\right\\rangle$ for some prime $p$ dividing $n$.", "nl_proof": "\\begin{proof}\n\n    Suppose $H$ is a maximal subgroup of $G$. Then $H$ is cyclic, and we may write $H=\\left\\langle x^k\\right\\rangle$ for some integer $k$, with $k>1$. Let $d=(n, k)$. Since $H$ is a proper subgroup, we know by Proposition 6 that $d>1$. Choose a prime factor $p$ of $d$. If $k=p=d$ then $k \\mid n$ as required.\n\n\n\nIf, however, $k$ is not prime, then consider the subgroup $K=\\left\\langle x^p\\right\\rangle$. Since $p$ is a proper divisor of $k$, it follows that $H<K$. But $H$ is maximal, so we must have $K=G$. Again by Proposition 6 , we must then have $(p, n)=1$. However, $p$ divides $d$ which divides $n$, so $p \\mid n$ and $(p, n)=p>1$, a contradiction. Therefore $k=p$ and the left-to-right implication holds.\n\nNow, for the converse, suppose $H=\\left\\langle x^p\\right\\rangle$ for $p$ a prime dividing $n$. If $H$ is not maximal then the first part of this exercise shows that there is a maximal subgroup $K$ containing $H$. Then $K=\\left\\langle x^q\\right\\rangle$. So $x^p \\in\\left\\langle x^q\\right\\rangle$, which implies $q \\mid p$. But the only divisors of $p$ are 1 and $p$. If $q=1$ then $K=G$ and $K$ cannot be a proper subgroup, and if $q=p$ then $H=K$ and $H$ cannot be a proper subgroup of $K$. This contradiction shows that $H$ is maximal.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_1_22a", "formal_statement": "theorem exercise_3_1_22a (G : Type*) [group G] (H K : subgroup G) \n  [subgroup.normal H] [subgroup.normal K] :\n  subgroup.normal (H \u2293 K) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $H$ and $K$ are normal subgroups of a group $G$ then their intersection $H \\cap K$ is also a normal subgroup of $G$.", "nl_proof": "\\begin{proof}\n\n    Suppose $H$ and $K$ are normal subgroups of $G$. We already know that $H \\cap K$ is a subgroup of $G$, so we need to show that it is normal. Choose any $g \\in G$ and any $x \\in H \\cap K$. Since $x \\in H$ and $H \\unlhd G$, we know $g x g^{-1} \\in H$. Likewise, since $x \\in K$ and $K \\unlhd G$, we have $g x g^{-1} \\in K$. Therefore $g x g^{-1} \\in H \\cap K$. This shows that $g(H \\cap K) g^{-1} \\subseteq H \\cap K$, and this is true for all $g \\in G$. By Theorem 6 (5) (which we will prove in Exercise 3.1.25), this is enough to show that $H \\cap K \\unlhd G$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_2_8", "formal_statement": "theorem exercise_3_2_8 {G : Type*} [group G] (H K : subgroup G)\n  [fintype H] [fintype K] \n  (hHK : nat.coprime (fintype.card H) (fintype.card K)) : \n  H \u2293 K = \u22a5 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $H$ and $K$ are finite subgroups of $G$ whose orders are relatively prime then $H \\cap K=1$.", "nl_proof": "\\begin{proof}\n\n    Solution: Let $|H|=p$ and $|K|=q$. We saw in a previous exercise that $H \\cap K$ is a subgroup of both $H$ and $K$; by Lagrange's Theorem, then, $|H \\cap K|$ divides $p$ and $q$. Since $\\operatorname{gcd}(p, q)=1$, then, $|H \\cap K|=1$. Thus $H \\cap K=1$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_2_16", "formal_statement": "theorem exercise_3_2_16 (p : \u2115) (hp : nat.prime p) (a : \u2115) :\n  nat.coprime a p \u2192 a ^ p \u2261 a [ZMOD p] :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Use Lagrange's Theorem in the multiplicative group $(\\mathbb{Z} / p \\mathbb{Z})^{\\times}$to prove Fermat's Little Theorem: if $p$ is a prime then $a^{p} \\equiv a(\\bmod p)$ for all $a \\in \\mathbb{Z}$.", "nl_proof": "\\begin{proof}\n\n    Solution: If $p$ is prime, then $\\varphi(p)=p-1$ (where $\\varphi$ denotes the Euler totient). Thus\n\n$$\n\n\\mid\\left((\\mathbb{Z} /(p))^{\\times} \\mid=p-1 .\\right.\n\n$$\n\nSo for all $a \\in(\\mathbb{Z} /(p))^{\\times}$, we have $|a|$ divides $p-1$. Hence\n\n$$\n\na=1 \\cdot a=a^{p-1} a=a^p \\quad(\\bmod p) .\n\n$$\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_3_3", "formal_statement": "theorem exercise_3_3_3 {p : primes} {G : Type*} [group G] \n  {H : subgroup G} [hH : H.normal] (hH1 : H.index = p) : \n  \u2200 K : subgroup G, K \u2264 H \u2228 H \u2294 K = \u22a4 \u2228 (K \u2293 H).relindex K = p :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $H$ is a normal subgroup of $G$ of prime index $p$ then for all $K \\leq G$ either $K \\leq H$, or $G=H K$ and $|K: K \\cap H|=p$.", "nl_proof": "\\begin{proof}\n\n    Solution: Suppose $K \\backslash N \\neq \\emptyset$; say $k \\in K \\backslash N$. Now $G / N \\cong \\mathbb{Z} /(p)$ is cyclic, and moreover is generated by any nonidentity- in particular by $\\bar{k}$\n\n\n\nNow $K N \\leq G$ since $N$ is normal. Let $g \\in G$. We have $g N=k^a N$ for some integer a. In particular, $g=k^a n$ for some $n \\in N$, hence $g \\in K N$. We have $[K: K \\cap N]=p$ by the Second Isomorphism Theorem.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_4_4", "formal_statement": "theorem exercise_3_4_4 {G : Type*} [comm_group G] [fintype G] {n : \u2115}\n    (hn : n \u2223 (fintype.card G)) :\n    \u2203 (H : subgroup G) (H_fin : fintype H), @card H H_fin = n  :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Use Cauchy's Theorem and induction to show that a finite abelian group has a subgroup of order $n$ for each positive divisor $n$ of its order.", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a finite abelian group. We use induction on $|G|$. Certainly the result holds for the trivial group. And if $|G|=p$ for some prime $p$, then the positive divisors of $|G|$ are 1 and $p$ and the result is again trivial.\n\n\n\nNow assume that the statement is true for all groups of order strictly smaller than $|G|$, and let $n$ be a positive divisor of $|G|$ with $n>1$. First, if $n$ is prime then Cauchy's Theorem allows us to find an element $x \\in G$ having order $n$. Then $\\langle x\\rangle$ is the desired subgroup. On the other hand, if $n$ is not prime, then $n$ has a prime divisor $p$, so that $n=k p$ for some integer $k$. Cauchy's Theorem allows us to find an element $x$ having order $p$. Set $N=\\langle x\\rangle$. By Lagrange's Theorem,\n\n$$\n\n|G / N|=\\frac{|G|}{|N|}<|G| .\n\n$$\n\nNow, by the inductive hypothesis, the group $G / N$ must have a subgroup of order $k$. And by the Lattice Isomorphism Theorem, this subgroup has the form $H / N$ for some subgroup $H$ of $G$. Then $|H|=k|N|=k p=n$, so that $H$ has order $n$. This completes the inductive step.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_3_4_5b", "formal_statement": "theorem exercise_3_4_5b {G : Type*} [group G] [is_solvable G] \n  (H : subgroup G) [subgroup.normal H] : \n  is_solvable (G \u29f8 H) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that quotient groups of a solvable group are solvable.", "nl_proof": "\\begin{proof}\n\n    Next, note that\n\n$$\n\nH_i=G_i \\cap H=\\left(G_i \\cap G_{i+1}\\right) \\cap H=G_i \\cap H_{i+1} .\n\n$$\n\nBy the Second Isomorphism Theorem, we then have\n\n$$\n\nH_{i+1} / H_i=H_{i+1} /\\left(H_{i+1} \\cap G_i\\right) \\cong H_{i+1} G_i / G_i \\leq G_{i+1} / G_i .\n\n$$\n\nSince $H_{i+1} / H_i$ is isomorphic to a subgroup of the abelian group $G_{i+1} / G_i$, it follows that $H_{i+1} / H_i$ is also abelian. This completes the proof that $H$ is solvable.\n\nNext, let $N \\unlhd G$. For each $i$, define\n\n$$\n\nN_i=G_i N, \\quad 0 \\leq i \\leq n .\n\n$$\n\nNow let $g \\in N_{i+1}$, where $g=g_0 n_0$ with $g_0 \\in G_{i+1}$ and $n_0 \\in N$. Also let $x \\in N_i$, where $x=g_1 n_1$ with $g_1 \\in G_i$ and $n_1 \\in N$. Then\n\n$$\n\ng x g^{-1}=g_0 n_0 g_1 n_1 n_0^{-1} g_0^{-1} .\n\n$$\n\nNow, since $N$ is normal in $G, N g=g N$, so $n_0 g_1=g_1 n_2$ for some $n_2 \\in N$. Then\n\n$$\n\ng x g^{-1}=g_0 g_1\\left(n_2 n_1 n_0^{-1}\\right) g_0^{-1}=g_0 g_1 n_3 g_0^{-1}\n\n$$\n\nfor some $n_3 \\in N$. Then $n_3 g_0^{-1}=g_0^{-1} n_4$ for some $n_4 \\in N$. And $g_0 g_1 g_0^{-1} \\in G_i$ since $G_i \\unlhd G_{i+1}$, so\n\n$$\n\ng x g^{-1}=g_0 g_1 g_0^{-1} n_4 \\in N_i .\n\n$$\n\nThis shows that $N_i \\unlhd N_{i+1}$. So by the Lattice Isomorphism Theorem, we have $N_{i+1} / N \\unlhd N_i / N$. This shows that\n\n$$\n\n1=N_0 / N \\unlhd N_1 / N \\unlhd N_2 / N \\unlhd \\cdots \\unlhd N_n / N=G / N .\n\n$$\n\nMoreover, the Third Isomorphism Theorem says that\n\n$$\n\n\\left(N_{i+1} / N\\right) /\\left(N_i / N\\right) \\cong N_{i+1} / N_i,\n\n$$\n\nso the proof will be complete if we can show that $N_{i+1} / N_i$ is abelian.\n\nLet $x, y \\in N_{i+1} / N_i$. Then\n\n$$\n\nx=\\left(g_0 n_0\\right) N_i \\quad \\text { and } \\quad y=\\left(g_1 n_1\\right) N_i\n\n$$\n\nfor some $g_0, g_1 \\in G_{i+1}$ and $n_0, n_1 \\in N$. We have\n\n$$\n\n\\begin{aligned}\n\nx y x^{-1} y^{-1} & =\\left(g_0 n_0\\right)\\left(g_1 n_1\\right)\\left(g_0 n_0\\right)^{-1}\\left(g_1 n_1\\right)^{-1} N_i \\\\\n\n& =g_0 n_0 g_1 n_1 n_0^{-1} g_0^{-1} n_1^{-1} g_1^{-1} N_i .\n\n\\end{aligned}\n\n$$\n\nSince $N \\unlhd G, g N=N g$ for any $g \\in G$, so we can find $n_2 \\in N$ such that\n\n$$\n\nx y x^{-1} y^{-1}=g_0 g_1 g_0^{-1} g^{-1} n_2 N_i .\n\n$$\n\nNow $N_i=G_i N=N G_i$ since $N \\unlhd G$ (see Proposition 14 and its corollary). Therefore\n\n$$\n\nn_2 N_i=n_2 N G_i=N G_i=G_i N\n\n$$\n\nand we get\n\n$$\n\nx y x^{-1} y^{-1}=g_0 g_1 g_0^{-1} g^{-1} G_i N=G_i N .\n\n$$\n\nSo $x y x^{-1} y^{-1}=1 N_i$ or $x y=y x$. This completes the proof that $G / N$ is solvable.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_2_8", "formal_statement": "theorem exercise_4_2_8 {G : Type*} [group G] {H : subgroup G} \n  {n : \u2115} (hn : n > 0) (hH : H.index = n) : \n  \u2203 K \u2264 H, K.normal \u2227 K.index \u2264 n.factorial :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $H$ has finite index $n$ then there is a normal subgroup $K$ of $G$ with $K \\leq H$ and $|G: K| \\leq n!$.", "nl_proof": "\\begin{proof}\n\n    Solution: $G$ acts on the cosets $G / H$ by left multiplication. Let $\\lambda: G \\rightarrow S_{G / H}$ be the permutation representation induced by this action, and let $K$ be the kernel of the representation.\n\nNow $K$ is normal in $G$, and $K \\leq \\operatorname{stab}_G(H)=H$. By the First Isomorphism Theorem, we have an injective group homomorphism $\\bar{\\lambda}: G / K \\rightarrow S_{G / H}$. Since $\\left|S_{G / H}\\right|=n !$, we have $[G: K] \\leq n !$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_2_9a", "formal_statement": "theorem exercise_4_2_9a {G : Type*} [fintype G] [group G] {p \u03b1 : \u2115} \n  (hp : p.prime) (ha : \u03b1 > 0) (hG : card G = p ^ \u03b1) : \n  \u2200 H : subgroup G, H.index = p \u2192 H.normal :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $p$ is a prime and $G$ is a group of order $p^{\\alpha}$ for some $\\alpha \\in \\mathbb{Z}^{+}$, then every subgroup of index $p$ is normal in $G$.", "nl_proof": "\\begin{proof}\n\n    Solution: Let $G$ be a group of order $p^k$ and $H \\leq G$ a subgroup with $[G: H]=p$. Now $G$ acts on the conjugates $g H g^{-1}$ by conjugation, since\n\n$$\n\ng_1 g_2 \\cdot H=\\left(g_1 g_2\\right) H\\left(g_1 g_2\\right)^{-1}=g_1\\left(g_2 H g_2^{-1}\\right) g_1^{-1}=g_1 \\cdot\\left(g_2 \\cdot H\\right)\n\n$$\n\nand $1 \\cdot H=1 H 1=H$. Moreover, under this action we have $H \\leq \\operatorname{stab}(H)$. By Exercise 3.2.11, we have\n\n$$\n\n[G: \\operatorname{stab}(H)][\\operatorname{stab}(H): H]=[G: H]=p,\n\n$$\n\na prime.\n\nIf $[G: \\operatorname{stab}(H)]=p$, then $[\\operatorname{stab}(H): H]=1$ and we have $H=\\operatorname{stab}(H)$; moreover, $H$ has exactly $p$ conjugates in $G$. Let $\\varphi: G \\rightarrow S_p$ be the permutation representation induced by the action of $G$ on the conjugates of $H$, and let $K$ be the kernel of this representation. Now $K \\leq \\operatorname{stab}(H)=H$. By the first isomorphism theorem, the induced map $\\bar{\\varphi}: G / K \\rightarrow S_p$ is injective, so that $|G / K|$ divides $p$ !. Note, however, that $|G / K|$ is a power of $p$ and that the only powers of $p$ that divide $p$ ! are 1 and $p$. So $[G: K]$ is 1 or $p$. If $[G: K]=1$, then $G=K$ so that $g H g^{-1}=H$ for all $g \\in G$; then $\\operatorname{stab}(H)=G$ and we have $[G: \\operatorname{stab}(H)]=1$, a contradiction. Now suppose $[G: K]=p$. Again by Exercise $3.2$.11 we have $[G: K]=[G: H][H: K]$, so that $[H: K]=1$, hence $H=K$. Again, this implies that $H$ is normal so that $g H g^{-1}=H$ for all $g \\in G$, and we have $[G: \\operatorname{stab}(H)]=1$, a contradiction. Thus $[G: \\operatorname{stab}(H)] \\neq p$\n\nIf $[G: \\operatorname{stab}(H)]=1$, then $G=\\operatorname{stab}(H)$. That is, $g H g^{-1}=H$ for all $g \\in G$; thus $H \\leq G$ is normal.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_4_2", "formal_statement": "theorem exercise_4_4_2 {G : Type*} [fintype G] [group G] \n  {p q : nat.primes} (hpq : p \u2260 q) (hG : card G = p*q) : \n  is_cyclic G :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $G$ is an abelian group of order $p q$, where $p$ and $q$ are distinct primes, then $G$ is cyclic.", "nl_proof": "\\begin{proof}\n\n    Let $G$ be an abelian group of order $p q$. We need to prove that if $p$ and $q$ are distinct primes than $G$ is cyclic. By Cauchy's theorem there are $a, b \\in G$ with $a$ of order $p$ and $b$ of order $q$. Since $(|a|,|b|)=1$ and $a b=b a$ then $|a b|=|a| \\cdot|b|=p q$. Therefore $a b$ is an element of order $p q$, the order of $G$, which means $G$ is cyclic.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_4_6b", "formal_statement": "theorem exercise_4_4_6b : \n  \u2203 (G : Type*) (hG : group G) (H : @subgroup G hG), @characteristic G hG H  \u2227 \u00ac @subgroup.normal G hG H :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that there exists a normal subgroup that is not characteristic.", "nl_proof": "\\begin{proof}\n\n    We have to produce a group $G$ and a subgroup $H$ such that $H$ is normal in $G$, but not characterestic. Consider the Klein's four group $G=\\{ e, a, b, a b\\}$. This is an abelian group with each element having order 2. Consider $H=\\{ e, a\\}$. $H$ is normal in $G$. Define $\\sigma: G \\rightarrow G$ as $\\sigma(a)=b, \\sigma(b)=a, \\sigma(a b)=a b$. Clearly $\\sigma$ does not fix $H$. So, $H$ is not characterestic.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_4_8a", "formal_statement": "theorem exercise_4_4_8a {G : Type*} [group G] (H K : subgroup G)  \n  (hHK : H \u2264 K) [hHK1 : (H.subgroup_of K).normal] [hK : K.normal] : \n  H.normal :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a group with subgroups $H$ and $K$ with $H \\leq K$. Prove that if $H$ is characteristic in $K$ and $K$ is normal in $G$ then $H$ is normal in $G$.", "nl_proof": "\\begin{proof}\n\nWe prove that $H$ is invariant under every inner automorphism of $G$. Consider a inner automorphism $\\phi_g$ of $G$. Now, $\\left.\\phi_g\\right|_K$ is a automorphism of $K$ because $K$ is normal in $G$. But $H$ is a characterestic subgroup of $K$, so $\\left.\\phi_g\\right|_K(H) \\subset H$, so in general $\\phi_g(H) \\subset H$. Hence $H$ is characteretstic in $G$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_13", "formal_statement": "theorem exercise_4_5_13 {G : Type*} [group G] [fintype G]\n  (hG : card G = 56) :\n  \u2203 (p : \u2115) (P : sylow p G), P.normal :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that a group of order 56 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.", "nl_proof": "\\begin{proof}    \n\nSince $|G|=56=2^{3}.7$, $G$ has $2-$Sylow subgroup of order $8$, as well as $7-$Sylow subgroup of order $7$. Now, we count the number of such subgroups. Let $n_{7}$ be the number of  $7-$Sylow subgroup and $n_{2}$ be the number of  $2-$Sylow subgroup. Now $n_{7}=1+7k$ where $1+7k|8$. The choices for $k$ are $0$ or $1$. If $k=0$, there is only one $7-$Sylow subgroup and hence normal. So, assume now, that there are $8$ $7-$Sylow subgroup(for $k=1$). Now we look at $2-$ Sylow subgroups. $n_{2}=1+2k| 7$. So choice for $k$ are $0$ and $3$. If $k=0$, there is only one $2-$Sylow subgroup and hence normal. So, assume now, that there are $7$ $2-$Sylow subgroup (for $k=3$). Now we claim that simultaneously, there cannot be $8$ $7-$Sylow subgroup and $7$ $2-$Sylow subgroup. So, either $7-$Sylow subgroup is normal being unique, or  the $2-$Sylow subgroup is normal. Now, to prove the claim, we observe that there are 48 elements of order $7$. Let $H_{1}$ and $H_{2}$ be two distinct  $2-$Sylow subgroup. Now $|H_{1}|=8$. So we already get $48+8=56$ distinct elements in the group. Now $H_{2}$ being distinct from $H_{1}$, has at least one element which is not in $H_{1}$. This adds one more element in the group, at the least. Now already we have number of elements in the group exceeding the number of element in $G$. This gives a contradiction and proves the claim.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_15", "formal_statement": "theorem exercise_4_5_15 {G : Type*} [group G] [fintype G] \n  (hG : card G = 351) : \n  \u2203 (p : \u2115) (P : sylow p G), P.normal :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that a group of order 351 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.", "nl_proof": "\\begin{proof}\n\n    Since $|G|=351=3^{2}.13$, $G$ has $3-$Sylow subgroup of order $9$, as well as $13-$Sylow subgroup of order $13$. Now, we count the number of such subgroups. Let $n_{13}$ be the number of $13-$Sylow subgroup and $n_{3}$ be the number of  $3-$Sylow subgroup. Now $n_{13}=1+13k$ where $1+13k|9$. The choices for $k$ is $0$. Hence, there is a unique $13-$Sylow subgroup and hence is normal.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_17", "formal_statement": "theorem exercise_4_5_17 {G : Type*} [fintype G] [group G] \n  (hG : card G = 105) : \n  nonempty(sylow 5 G) \u2227 nonempty(sylow 7 G) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $|G|=105$ then $G$ has a normal Sylow 5 -subgroup and a normal Sylow 7-subgroup.", "nl_proof": "\\begin{proof}    \n\nSince $|G|=105=3.5.7$, $G$ has $3-$Sylow subgroup of order $3$, as well as $5-$Sylow subgroup of order $5$ and, $7-$Sylow subgroup of order 7. Now, we count the number of such subgroups. Let $n_{3}$ be the number of $3-$Sylow subgroup, $n_{5}$ be the number of  $5-$Sylow subgroup, and $n_{7}$ be the number of $7-$Sylow subgroup. Now $n_{7}=1+7k$ where $1+7k|15$. The choices for $k$ are $0$ or $1$. If $k=0$, there is only one $7-$Sylow subgroup and hence normal. So, assume now, that there are $15$ $7-$Sylow subgroup(for $k=1$). Now we look at $5-$ Sylow subgroups. $n_{5}=1+5k| 21$. So choice for $k$ are $0$ and $4$. If $k=0$, there is only one $5-$Sylow subgroup and hence normal. So, assume now, that there are $24$ $5-$Sylow subgroup (for $k=4$). Now we claim that simultaneously, there cannot be $15$ $7-$Sylow subgroup and $24$ $5-$Sylow subgroup. So, either $7-$Sylow subgroup is normal being unique, or  the $5-$Sylow subgroup is normal. Now, to prove the claim, we observe that there are 90 elements of order $7$. Also, see that there are $24\\times 4=96$ number of elements of order 5. So we get $90+94=184$ number of elements which exceeds the order of the group. This gives a contradiction and proves the claim. So, now we have proved that there is either a normal $5-$Sylow subgroup or a normal $7-$Sylow subgroup.\n\n    Now we prove that indeed both $5-$ Sylow subgroup and 7 -Sylow subgroup are normal. Assume that 7 -Sylow subgroup is normal. So, there is a unique 7 -Sylow subgroup, say $H$. Now assume that there are 245 -Sylow subgroups. So, we get again $24 \\times 4=96$ elements of order 5 . From $H$ we get 7 elements which gives us total of $96+7=103$ elements. Now consider the number of 3 -Sylow subgroups. $n_3=1+3 k \\mid 35$. Then the possibilities for $k$ are 0 and 2 . But we can rule out $k=2$ because having 73 -Sylow subgroup, will mean we have 14 elements of order 3 . So we get $103+14=117$ elements in total which exceeds the order of the group. So we have now that there is a unique 3 -Sylow subgroup and hence normal. Call that subgroup $K$. Now take any one 5 -Sylow subgroup, call it $L$. Now observe $L K$ is a subgroup of $G$ with order 15 . We know that a group of order 15 is cyclic by an example in Page-143 of the book. So, there is an element of order 15. Actually we have $\\phi(15)=8$ number of elements of order 15. But then again we already had 103 elements and then we actually get at least $103+8=111$ elements which exceeds the order of the group. So, there can't be 24 5-Sylow subgroups, and hence there is a unique 5-Sylow subgroup, and hence normal.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_19", "formal_statement": "theorem exercise_4_5_19 {G : Type*} [fintype G] [group G] \n  (hG : card G = 6545) : \u00ac is_simple_group G :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $|G|=6545$ then $G$ is not simple.", "nl_proof": "\\begin{proof}    \n\nSince $|G|=132=2^{2}.3.11$, $G$ has $2-$Sylow subgroup of order $4$, as well as $11-$Sylow subgroup of order $11$, and $3-$Sylow subgroup of order $3$. Now, we count the number of such subgroups. Let $n_{11}$ be the number of  $11-$Sylow subgroup and $n_{3}$ be the number of  $3-$Sylow subgroup. Now $n_{11}=1+11k$ where $1+11k|12$. The choices for $k$ are $0$ or $1$. If $k=0$, there is only one $11-$Sylow subgroup and hence normal. So, assume now, that there are $12$ $11-$Sylow subgroup(for $k=1$). Now we look at $3-$ Sylow subgroups. $n_{3}=1+3k| 44$. So choice for $k$ are $0$, $1$, and $7$. If $k=0$, there is only one $3-$Sylow subgroup and hence normal. So, assume now, that there are $4$ $2-$Sylow subgroup (for $k=3$). Now we claim that simultaneously, there cannot be $12$ $11-$Sylow subgroup and $4$ $3-$Sylow subgroups provided there is more than one $2-$Sylow subgroups. So, either $2-$Sylow subgroup is normal or if not, then, either $11-$Sylow subgroup is normal being unique, or  the $3-$Sylow subgroup is normal(We don't consider the possibility of $22$ $3-$Sylow subgroup because of obvious reason). Now, to prove the claim, we observe that there are $120$ elements of order $11$. Also there are $8$ elements of order $3$. So we already get $120+8+1=129$ distinct elements in the group. Let us count the number of $2-$Sylow subgroups in $G$. $n_{2}=1+2k|33$. The possibilities for $k$ are $0$, $1$, $5$, $16$. Now, assume there is more than one $2-$Sylow subgroups. Let $H_{1}$ and $H_{2}$ be two distinct  $2-$Sylow subgroup. Now $|H_{1}|=4$. So we already get $129+3=132$ distinct elements in the group. Now $H_{2}$ being distinct from $H_{1}$, has at least one element which is not in $H_{1}$. This adds one more element in the group, at the least. Now already we have number of elements in the group exceeding the number of element in $G$. This gives a contradiction and proves the claim.\n\nHence $G$ is not simple.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_21", "formal_statement": "theorem exercise_4_5_21 {G : Type*} [fintype G] [group G]\n  (hG : card G = 2907) : \u00ac is_simple_group G :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $|G|=2907$ then $G$ is not simple.", "nl_proof": "\\begin{proof}    \n\nSince $|G|=2907=3^{2}.17.19$, $G$ has $19-$Sylow subgroup of order $19$. Now, we count the number of such subgroups. Let $n_{19}$ be the number of $19-$Sylow subgroup. Now $n_{19}=1+19k$ where $1+19k|3^{2}.17$. The choices for $k$ is $0$. Hence, there is a unique $19-$Sylow subgroup and hence is normal. so $G$ is not simple.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_23", "formal_statement": "theorem exercise_4_5_23 {G : Type*} [fintype G] [group G]\n  (hG : card G = 462) : \u00ac is_simple_group G :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $|G|=462$ then $G$ is not simple.", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a group of order $462=11 \\cdot 42$. Note that 11 is a prime not dividing 42 . Let $P \\in$ $S y l_{11}(G)$. [We know $P$ exists since $S y l_{11}(G) \\neq \\emptyset$]. Note that $|P|=11^1=11$ by definition. \n\n\n\nThe number of Sylow 11-subgroups of $G$ is of the form $1+k \\cdot 11$, i.e., $n_{11} \\equiv 1$ (mod 11) and $n_{11}$ divides 42 . The only such number that divides 42 and equals 1 (mod 11) is 1 so $n_{11}=1$. Hence $P$ is the unique Sylow 11-subgroup.\n\n\n\nSince $P$ is the unique Sylow Il-subgroup, this implies that $P$ is normal in $G$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_4_5_33", "formal_statement": "theorem exercise_4_5_33 {G : Type*} [group G] [fintype G] {p : \u2115} \n  (P : sylow p G) [hP : P.normal] (H : subgroup G) [fintype H] : \n  \u2200 R : sylow p H, R.to_subgroup = (H \u2293 P.to_subgroup).subgroup_of H \u2227\n  nonempty (sylow p H) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $P$ be a normal Sylow $p$-subgroup of $G$ and let $H$ be any subgroup of $G$. Prove that $P \\cap H$ is the unique Sylow $p$-subgroup of $H$.", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a group and $P$ is a normal $p$-Sylow subgroup of $G .|G|=p^a . m$ where $p \\nmid m$. Then $|P|=p^a$. Let $H$ be a subgroup of $G$. Now if $|H|=k$ such that $p \\nmid k$. Then $P \\cap H=\\{e\\}$. There is nothing to prove in this case. Let $|H|=p^b . n$, where $b \\leq a$, and $p \\nmid n$. Now consider $P H$ which is a subgroup of $G$, as $P$ is normal. Now $|P H|=\\frac{|P||H|}{|P \\cap H|}=\\frac{p^{a+b} \\cdot n}{|P \\cap H|}$. Now since $P H \\leq G$, so $|P H|=p^a$.l, as $P \\leq P H$. This forces $|P \\cap H|=p^b$. So by order consideration we have $P \\cap H$ is a sylow $-p$ subgroup of $H$. Now we know $P$ is unique $p$ - Sylow subgroup. Suppose $H$ has a sylow-p subgroup distinct from $P \\cap H$, call it $H_1$. Now $H_1$ is a p-subgroup of $G$. So, $H_1$ is contained in some Sylow-p subgroup of $G$, call it $P_1$. Clearly $P_1$ is distinct from $P$, which is a contradiction. So $P \\cap H$ is the only $p$-Sylow subgroup of $H$, and hence normal in $H$\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_7_1_2", "formal_statement": "theorem exercise_7_1_2 {R : Type*} [ring R] {u : R}\n  (hu : is_unit u) : is_unit (-u) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $u$ is a unit in $R$ then so is $-u$.", "nl_proof": "\\begin{proof}\n\n    Solution: Since $u$ is a unit, we have $u v=v u=1$ for some $v \\in R$. Thus, we have\n\n$$\n\n(-v)(-u)=v u=1\n\n$$\n\nand\n\n$$\n\n(-u)(-v)=u v=1 .\n\n$$\n\nThus $-u$ is a unit.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_7_1_12", "formal_statement": "theorem exercise_7_1_12 {F : Type*} [field F] {K : subring F}\n  (hK : (1 : F) \u2208 K) : is_domain K :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that any subring of a field which contains the identity is an integral domain.", "nl_proof": "\\begin{proof}\n\n    Solution: Let $R \\subseteq F$ be a subring of a field. (We need not yet assume that $1 \\in R$ ). Suppose $x, y \\in R$ with $x y=0$. Since $x, y \\in F$ and the zero element in $R$ is the same as that in $F$, either $x=0$ or $y=0$. Thus $R$ has no zero divisors. If $R$ also contains 1 , then $R$ is an integral domain.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_7_2_2", "formal_statement": "theorem exercise_7_2_2 {R : Type*} [ring R] (p : polynomial R) :\n  p \u2223 0 \u2194 \u2203 b : R, b \u2260 0 \u2227 b \u2022 p = 0 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{1} x+a_{0}$ be an element of the polynomial ring $R[x]$. Prove that $p(x)$ is a zero divisor in $R[x]$ if and only if there is a nonzero $b \\in R$ such that $b p(x)=0$.", "nl_proof": "\\begin{proof}\n\n    Solution: If $b p(x)=0$ for some nonzero $b \\in R$, then it is clear that $p(x)$ is a zero divisor.\n\nNow suppose $p(x)$ is a zero divisor; that is, for some $q(x)=\\sum_{i=0}^m b_i x^i$, we have $p(x) q(x)=0$. We may choose $q(x)$ to have minimal degree among the nonzero polynomials with this property.\n\nWe will now show by induction that $a_i q(x)=0$ for all $0 \\leq i \\leq n$.\n\nFor the base case, note that\n\n$$\n\np(x) q(x)=\\sum_{k=0}^{n+m}\\left(\\sum_{i+j=k} a_i b_j\\right) x^k=0 .\n\n$$\n\nThe coefficient of $x^{n+m}$ in this product is $a_n b_m$ on one hand, and 0 on the other. Thus $a_n b_m=0$. Now $a_n q(x) p(x)=0$, and the coefficient of $x^m$ in $q$ is $a_n b_m=0$. Thus the degree of $a_n q(x)$ is strictly less than that of $q(x)$; since $q(x)$ has minimal degree among the nonzero polynomials which multiply $p(x)$ to 0 , in fact $a_n q(x)=0$. More specifically, $a_n b_i=0$ for all $0 \\leq i \\leq m$.\n\nFor the inductive step, suppose that for some $0 \\leq t<n$, we have $a_r q(x)=0$ for all $t<r \\leq n$. Now\n\n$$\n\np(x) q(x)=\\sum_{k=0}^{n+m}\\left(\\sum_{i+j=k} a_i b_j\\right) x^k=0 .\n\n$$\n\nOn one hand, the coefficient of $x^{m+t}$ is $\\sum_{i+j=m+t} a_i b_j$, and on the other hand, it is 0 . Thus\n\n$$\n\n\\sum_{i+j=m+t} a_i b_j=0 .\n\n$$\n\nBy the induction hypothesis, if $i \\geq t$, then $a_i b_j=0$. Thus all terms such that $i \\geq t$ are zero. If $i<t$, then we must have $j>m$, a contradiction. Thus we have $a_t b_m=0$. As in the base case,\n\n$$\n\na_t q(x) p(x)=0\n\n$$\n\nand $a_t q(x)$ has degree strictly less than that of $q(x)$, so that by minimality, $a_t q(x)=0$.\n\nBy induction, $a_i q(x)=0$ for all $0 \\leq i \\leq n$. In particular, $a_i b_m=0$. Thus $b_m p(x)=0$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_7_3_16", "formal_statement": "theorem exercise_7_3_16 {R S : Type*} [ring R] [ring S] \n  {\u03c6 : R \u2192+* S} (hf : surjective \u03c6) : \n  \u03c6 '' (center R) \u2282 center S :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $\\varphi: R \\rightarrow S$ be a surjective homomorphism of rings. Prove that the image of the center of $R$ is contained in the center of $S$.", "nl_proof": "\\begin{proof}\n\n    Suppose $r \\in \\varphi[Z(R)]$. Then $r=\\varphi(z)$ for some $z \\in Z(R)$. Now let $x \\in S$. Since $\\varphi$ is surjective, we have $x=\\varphi y$ for some $y \\in R$. Now\n\n$$\n\nx r=\\varphi(y) \\varphi(z)=\\varphi(y z)=\\varphi(z y)=\\varphi(z) \\varphi(y)=r x .\n\n$$\n\nThus $r \\in Z(S)$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_7_4_27", "formal_statement": "theorem exercise_7_4_27 {R : Type*} [comm_ring R] (hR : (0 : R) \u2260 1) \n  {a : R} (ha : is_nilpotent a) (b : R) : \n  is_unit (1-a*b) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $R$ be a commutative ring with $1 \\neq 0$. Prove that if $a$ is a nilpotent element of $R$ then $1-a b$ is a unit for all $b \\in R$.", "nl_proof": "\\begin{proof}\n\n    $\\mathfrak{N}(R)$ is an ideal of $R$. Thus for all $b \\in R,-a b$ is nilpotent. Hence $1-a b$ is a unit in $R$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_8_2_4", "formal_statement": "theorem exercise_8_2_4 {R : Type*} [ring R][no_zero_divisors R] \n  [cancel_comm_monoid_with_zero R] [gcd_monoid R]\n  (h1 : \u2200 a b : R, a \u2260 0 \u2192 b \u2260 0 \u2192 \u2203 r s : R, gcd a b = r*a + s*b)\n  (h2 : \u2200 a : \u2115 \u2192 R, (\u2200 i j : \u2115, i < j \u2192 a i \u2223 a j) \u2192 \n  \u2203 N : \u2115, \u2200 n \u2265 N, \u2203 u : R, is_unit u \u2227 a n = u * a N) : \n  is_principal_ideal_ring R :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain: (i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $r a+s b$ for some $r, s \\in R$, and (ii) if $a_{1}, a_{2}, a_{3}, \\ldots$ are nonzero elements of $R$ such that $a_{i+1} \\mid a_{i}$ for all $i$, then there is a positive integer $N$ such that $a_{n}$ is a unit times $a_{N}$ for all $n \\geq N$.", "nl_proof": "\\begin{proof}\n\n    Let $I \\leq R$ be a nonzero ideal and let $I / \\sim$ be the set of equivalence classes of elements of $I$ with regards to the relation of being associates. We can equip $I / \\sim$ with a partial order with $[x] \\leq[y]$ if $y \\mid x$. Condition (ii) implies all chains in $I / \\sim$ have an upper bound, so By Zorn's lemma $I / \\sim$ contains a maximal element, i.e. $I$ contains a class of associated elements which are minimal with respect to divisibility.\n\n\n\nNow let $a, b \\in I$ be two elements such that $[a]$ and $[b]$ are minimal with respect to divisibility. By condition (i) $a$ and $b$ have a greatest common divisor $d$ which can be expressed as $d=$ $a x+b y$ for some $x, y \\in R$. In particular, $d \\in I$. Since $a$ and $b$ are minimal with respect to divisibility, we have that $[a]=[b]=[d]$. Therefore $I$ has at least one element $a$ that is minimal with regard to divisibility and all such elements are associate, and we have $I=\\langle a\\rangle$ and so $I$ is principal. We conclude $R$ is a principal ideal domain.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_8_3_5a", "formal_statement": "theorem exercise_8_3_5a {n : \u2124} (hn0 : n > 3) (hn1 : squarefree n) : \n  irreducible (2 :zsqrtd $ -n) \u2227 \n  irreducible (\u27e80, 1\u27e9 : zsqrtd $ -n) \u2227 \n  irreducible (1 + \u27e80, 1\u27e9 : zsqrtd $ -n) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $R=\\mathbb{Z}[\\sqrt{-n}]$ where $n$ is a squarefree integer greater than 3. Prove that $2, \\sqrt{-n}$ and $1+\\sqrt{-n}$ are irreducibles in $R$.", "nl_proof": "\\begin{proof}\n\n    Suppose $a=a_1+a_2 \\sqrt{-n}, b=b_1+b_2 \\sqrt{-n} \\in R$ are such that $2=a b$, then $N(a) N(b)=4$. Without loss of generality we can assume $N(a) \\leq N(b)$, so $N(a)=1$ or $N(a)=2$. Suppose $N(a)=2$, then $a_1^2+n a_2^2=2$ and since $n>3$ we have $a_2=0$, which implies $a_1^2=2$, a contradiction. So $N(a)=1$ and $a$ is a unit. Therefore 2 is irreducible in $R$.\n\n\n\nSuppose now $\\sqrt{-n}=a b$, then $N(a) N(b)=n$ and we can assume $N(a)<$ $N(b)$ since $n$ is square free. Suppose $N(a)>1$, then $a_1^2+n a_2^2>1$ and $a_1^2+n a_2^2 \\mid n$, so $a_2=0$, and therefore $a_1^2 \\mid n$. Since $n$ is squarefree, $a_1=\\pm 1$, a contradiction. Therefore $N(a)=1$ and so $a$ is a unit and $\\sqrt{-n}$ is irreducible.\n\n\n\nSuppose $1+\\sqrt{-n}=a b$, then $N(a) N(b)=n+1$ and we can assume $N(a) \\leq N(b)$. Suppose $N(a)>1$, then $a_1^2+n a_2^2>1$ and $a_1^2+n a_2^2 \\mid n+1$. If $\\left|a_2\\right| \\geq 2$, then since $n>3$ we have a contradiction since $N(a)$ is too large. If $\\left|a_2\\right|=1$, then $a_1^2+n$ divides $1+n$ and so $a_1=\\pm 1$, and in either case $N(a)=n+1$ which contradicts $N(a) \\leq N(b)$. If $a_2=0$ then $a_1^2\\left(b_1^2+n b_2^2\\right)=\\left(a_1 b_1\\right)^2+n\\left(a_1 b_2\\right)^2=n+1$. If $\\left|a_1 b_2\\right| \\geq 2$ we have a contradiction. If $\\left|a_1 b_2\\right|=1$ then $a_1=\\pm 1$ which contradicts $N(a)>1$. If $\\left|a_1 b_2\\right|=0$, then $b_2=0$ and so $a_1 b_1=\\sqrt{-n}$, a contradiction. Therefore $N(a)=1$ and so $a$ is a unit and $1+\\sqrt{-n}$ is irreducible.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_8_3_6b", "formal_statement": "theorem exercise_8_3_6b {q : \u2115} (hq0 : q.prime) \n  (hq1 : q \u2261 3 [ZMOD 4]) {R : Type*} [ring R]\n  (hR : R = (gaussian_int \u29f8 ideal.span ({q} : set gaussian_int))) : \n  is_field R \u2227 \u2203 finR : fintype R, @card R finR = q^2 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $q \\in \\mathbb{Z}$ be a prime with $q \\equiv 3 \\bmod 4$. Prove that the quotient ring $\\mathbb{Z}[i] /(q)$ is a field with $q^{2}$ elements.", "nl_proof": "\\begin{proof}\n\n    The division algorithm gives us that every element of $\\mathbb{Z}[i] /\\langle q\\rangle$ is represented by an element $a+b i$ such that $0 \\leq a, b<q$. Each such choice is distinct since if $a_1+b_1 i+\\langle q\\rangle=a_2+b_2 i+\\langle q\\rangle$, then $\\left(a_1-a_2\\right)+\\left(b_1-b_2\\right) i$ is divisible by $q$, so $a_1 \\equiv a_2 \\bmod q$ and $b_1 \\equiv b_2 \\bmod q$. So $\\mathbb{Z}[i] /\\langle q\\rangle$ has order $q^2$.\n\n\n\nSince $q \\equiv 3 \\bmod 4, q$ is irreducible, hence prime in $\\mathbb{Z}[i]$. Therefore $\\langle q\\rangle$ is a prime ideal in $\\mathbb{Z}[i]$, and so $\\mathbb{Z}[i] /\\langle q\\rangle$ is an integral domain. So $\\mathbb{Z}[i] /\\langle q\\rangle$ is a field.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_9_1_10", "formal_statement": "theorem exercise_9_1_10 {f : \u2115 \u2192 mv_polynomial \u2115 \u2124} \n  (hf : f = \u03bb i, X i * X (i+1)): \n  infinite (minimal_primes (mv_polynomial \u2115 \u2124 \u29f8 ideal.span (range f))) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the ring $\\mathbb{Z}\\left[x_{1}, x_{2}, x_{3}, \\ldots\\right] /\\left(x_{1} x_{2}, x_{3} x_{4}, x_{5} x_{6}, \\ldots\\right)$ contains infinitely many minimal prime ideals.", "nl_proof": "\\begin{proof}\n\n    Let $R=\\mathbb{Z}\\left[x_1, x_2, \\ldots, x_n\\right]$ and consider the ideal $K=\\left(x_{2 k+1} x_{2 k+2} \\mid k \\in \\mathbb{Z}_{+}\\right)$in $R$.\n\nConsider the family of subsets $X=\\left\\{\\left\\{x_{2 k+1}, x_{2 k+2}\\right\\} \\mid k \\in \\mathbb{Z}_{+}\\right\\}$, and $Y$ the set of choice function on $X$, ie the set of functions $\\lambda: \\mathbb{Z}_{+} \\rightarrow \\cup_{\\mathbb{Z}_{+}}\\left\\{x_{2 k+1}, x_{2 k+2}\\right\\}$ with $\\lambda(a) \\in$ $\\left\\{x_{2 a+1}, x_{2 a+2}\\right\\}$\n\nFor each $\\lambda \\in Y$ we have the ideal $I_\\lambda=(\\lambda(0), \\lambda(1), \\ldots)$.\n\nAll these ideals are distinct, ie for $\\lambda \\neq \\lambda^{\\prime}$ we have $I_\\lambda \\neq I_{\\lambda^{\\prime}}$.\n\nWe also have that by construction $K \\subset I_\\lambda$ for all $\\lambda \\in Y$.\n\nBy the Third Isomorphism Treorem\n\n$$\n\n(R / K) /\\left(I_\\lambda / K\\right) \\cong R / I_\\lambda\n\n$$\n\nNote also that $R / I_\\lambda$ is isomorphic to the polynomial ring over $R$ with indeterminates the $x_i$ not in the image of $\\lambda$, and since there is a countably infinite number of them we can conclude $R / I_\\lambda \\cong R$, an integral domain. Therefore $I_\\lambda / K$ is a prime ideal of $R / K$\n\n\n\nWe prove now that $I_\\lambda / K$ is a minimal prime ideal. Let $J / K \\subseteq I_\\lambda / K$ be a prime ideal. For each pair $\\left(x_{2 k+1}, x_{2 k+2}\\right)$ we have that $x_{2 k+1} x_{2 k+2} \\in K$ so $x_{2 k+1} x_{2 k+2} \\bmod K \\in J / K$ so $J$ must contain one of the elements in $\\left\\{x_{2 k+1}, x_{2 k+2}\\right\\}$. But since $J / K \\subseteq I_\\lambda / K$ it must be $\\lambda(k)$ for all $k \\in \\mathbb{Z}_{+}$. Therefore $J / K=I_\\lambda / K$\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_9_4_2a", "formal_statement": "theorem exercise_9_4_2a : irreducible (X^4 - 4*X^3 + 6 : polynomial \u2124) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^4-4x^3+6$ is irreducible in $\\mathbb{Z}[x]$.", "nl_proof": "\\begin{proof}\n\n$$\n\nx^4-4 x^3+6\n\n$$\n\nThe polynomial is irreducible by Eisenstiens Criterion since the prime $2$ doesnt divide the leading coefficient 2 divide coefficients of the low order term $-4,0,0$ but 6 is not divided by the square of 2.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_9_4_2c", "formal_statement": "theorem exercise_9_4_2c : irreducible \n  (X^4 + 4*X^3 + 6*X^2 + 2*X + 1 : polynomial \u2124) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^4+4x^3+6x^2+2x+1$ is irreducible in $\\mathbb{Z}[x]$.", "nl_proof": "\\begin{proof}\n\n$$\n\np(x)=x^4+6 x^3+4 x^2+2 x+1\n\n$$\n\nWe calculate $p(x-1)$\n\n$$\n\n\\begin{aligned}\n\n(x-1)^4 & =x^4-4 x^3+6 x^2-4 x+1 \\\\\n\n6(x-1)^3 & =6 x^3-18 x^2+18 x-6 \\\\\n\n4(x-1)^2 & =4 x^2-8 x+4 \\\\\n\n2(x-1) & =2 x-2 \\\\\n\n1 & =1\n\n\\end{aligned}\n\n$$\n\n$$\n\n\\begin{aligned}\n\n& p(x-1)=(x-1)^4+6(x-1)^3+4(x-1)^2+2(x-1)+1=x^4+2 x^3-8 x^2+ \\\\\n\n& 8 x-2 \\\\\n\n& q(x)=x^4+2 x^3-8 x^2+8 x-2\n\n\\end{aligned}\n\n$$\n\n$q(x)$ is irreducible by Eisenstiens Criterion since the prime $\\$ 2 \\$$ divides the lower coefficient but $\\$ 2^{\\wedge} 2 \\$$ doesnt divide constant $-2$. Any factorization of $p(x)$ would provide a factor of $p(x)(x-1)$\n\nSince:\n\n$$\n\n\\begin{aligned}\n\n& p(x)=a(x) b(x) \\\\\n\n& q(x)=p(x)(x-1)=a(x-1) b(x-1)\n\n\\end{aligned}\n\n$$\n\nWe get a contradiction with the irreducibility of $p(x-1)$, so $p(x)$ is irreducible in $Z[x]$\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_9_4_9", "formal_statement": "theorem exercise_9_4_9 : \n  irreducible (X^2 - C sqrtd : polynomial (zsqrtd 2)) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the polynomial $x^{2}-\\sqrt{2}$ is irreducible over $\\mathbb{Z}[\\sqrt{2}]$. You may assume that $\\mathbb{Z}[\\sqrt{2}]$ is a U.F.D.", "nl_proof": "\\begin{proof}\n\n$Z[\\sqrt{2}]$ is an Euclidean domain, and so a unique factorization domain.\n\nWe have to prove $p(x)=x^2-\\sqrt{2}$ irreducible.\n\nSuppose to the contrary.\n\nif $p(x)$ is reducible then it must have root.\n\nLet $a+b \\sqrt{2}$ be a root of $x^2-\\sqrt{2}$.\n\nNow we have\n\n$$\n\na^2+2 b^2+2 a b \\sqrt{2}=\\sqrt{2}\n\n$$\n\nBy comparing the coefficients we get $2 a b=1$ for some pair of integers $a$ and $b$, a contradiction.\n\nSo $p(x)$ is irredicible over $Z[\\sqrt{2}]$.\n\n\\end{proof}"}
{"id": "Dummit-Foote|exercise_11_1_13", "formal_statement": "theorem exercise_11_1_13 {\u03b9 : Type*} [fintype \u03b9] : \n  (\u03b9 \u2192 \u211d) \u2243\u2097[\u211a] \u211d :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that as vector spaces over $\\mathbb{Q}, \\mathbb{R}^n \\cong \\mathbb{R}$, for all $n \\in \\mathbb{Z}^{+}$.", "nl_proof": "\\begin{proof}    \n\nSince $B$ is a basis of $V$, every element of $V$ can be written uniquely as a finite linear combination of elements of $B$. Let $X$ be the set of all such finite linear combinations. Then $X$ has the same cardinality as $V$, since the map from $X$ to $V$ that takes each linear combination to the corresponding element of $V$ is a bijection.\n\n\n\nWe will show that $X$ has the same cardinality as $B$. Since $B$ is countable and $X$ is a union of countable sets, it suffices to show that each set $X_n$, consisting of all finite linear combinations of $n$ elements of $B$, is countable.\n\n\n\nLet $P_n(X)$ be the set of all subsets of $X$ with cardinality $n$. Then we have $X_n \\subseteq P_n(B)$. Since $B$ is countable, we have $\\mathrm{card}(P_n(B)) \\leq \\mathrm{card}(B^n) = \\mathrm{card}(B)$, where $B^n$ is the Cartesian product of $n$ copies of $B$.\n\n\n\nThus, we have $\\mathrm{card}(X_n) \\leq \\mathrm{card}(P_n(B)) \\leq \\mathrm{card}(B)$, so $X_n$ is countable. It follows that $X$ is countable, and hence has the same cardinality as $B$.\n\n\n\nTherefore, we have shown that the cardinality of $V$ is equal to the cardinality of $B$. Since $F$ is countable, it follows that the cardinality of $V$ is countable as well.\n\n\n\nNow let $Q$ be a countable field, and let $R$ be a vector space over $Q$. Let $n$ be a positive integer. Then any basis of $R^n$ over $Q$ has the same cardinality as $R^n$, which is countable. Since $R$ is a direct sum of $n$ copies of $R^n$, it follows that any basis of $R$ over $Q$ has the same cardinality as $R$. Hence, the cardinality of $R$ is countable.\n\n\n\nFinally, since $R$ is a countable vector space and $Q$ is a countable field, it follows that $R$ and $Q^{\\oplus \\mathrm{card}(R)}$ are isomorphic as additive abelian groups. Therefore, we have $R \\cong_Q Q^{\\oplus \\mathrm{card}(R)}$, and in particular $R \\cong_Q R^n$ for any positive integer $n$.\n\n\\end{proof}"}