index int64 1 367 | name stringlengths 17 29 | textbook stringclasses 10
values | header stringclasses 60
values | helper stringclasses 14
values | informal_stmt stringlengths 42 474 | formal_stmt stringlengths 66 545 | informal_proof stringlengths 40 3.85k | formal_proof stringclasses 1
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101 | Ireland_Rosen_exercise_4_8 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Let $p$ be an odd prime. Show that $a$ is a primitive root modulo $p$ iff $a^{(p-1) / q} \not \equiv 1(p)$ for all prime divisors $q$ of $p-1$. | theorem Ireland_Rosen_exercise_4_8 {p : ℕ} (hp : Odd p) (hp' : p.Prime) (a : (ZMod p)ˣ) :
IsPrimitiveRoot (a : ZMod p) (p - 1) ↔
∀ q : ℕ, q ∣ (p - 1) → q.Prime →
¬ ((a : ZMod p) ^ ((p - 1) / q) = 1) := by
sorry | $\bullet$ If $a$ is a primitive root, then $a^k \not \equiv 1$ for all $k, 1\leq k < p-1$, so $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$.
$\bullet$ In the other direction, suppose $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$.
Let $\delta$ the order of $a$, a... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
102 | Ireland_Rosen_exercise_5_13 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Show that any prime divisor of $x^{4}-x^{2}+1$ is congruent to 1 modulo 12 . | theorem Ireland_Rosen_exercise_5_13 {p : ℕ} {x : ℤ}
(hp : Nat.Prime p)
(hpx : (p : ℤ) ∣ (x^4 - x^2 + 1)) : (p : ℤ) ≡ 1 [ZMOD 12] := by
sorry | \newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
$\bullet$ As $a^6 +1 = (a^2+1)(a^4-a^2+1)$, $p\mid a^4 - a^2+1$ implies $p \mid a^6 + 1$, so $\legendre{-1}{p} = 1$ and $p\equiv 1 \pmod 4$.
$\bullet$ $p \mid 4a^4 - 4 a^2 +4 = (2a-1)^2 + 3$, so $\legendre{-3}{p} = 1$.
As $-3 \equiv 1 \pmod 4$, $\legendre{-3}{p} =... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
103 | Ireland_Rosen_exercise_5_37 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators
open scoped NumberTheorySymbols | Show that if $a$ is negative then $p \equiv q(4 a)$ together with $p\not | a$ imply $(a / p)=(a / q)$. | theorem Ireland_Rosen_exercise_5_37 {p q : ℕ} [Fact (p.Prime)] [Fact (q.Prime)] {a : ℤ}
(ha : a < 0) (h0 : p ≡ q [ZMOD 4*a]) (h1 : ¬ ((p : ℤ) ∣ a)) :
legendreSym p a = legendreSym q a := by
sorry | \newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
Write $a = -A, A>0$. As $p \equiv q \pmod {4a}$, we know from Prop. 5.3.3. (b) that $(A/p) = (A/q)$.
Moreover,
\begin{align*}
\legendre{a}{p}&= \legendre{-A}{p} = (-1)^{(p-1)/2} \legendre{A}{p}\\
\legendre{a}{q}&= \legendre{-A}{q} = (-1^{(q-1)/2} \legendre{A}{q}
\e... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
104 | Ireland_Rosen_exercise_18_4 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Show that 1729 is the smallest positive integer expressible as the sum of two different integral cubes in two ways. | theorem Ireland_Rosen_exercise_18_4 :
1729 = sInf (fun n => 0 < n ∧
∃ x y z w : ℕ,
0 < x ∧ 0 < y ∧ 0 < z ∧ 0 < w ∧
x ^ 3 + y ^ 3 = n ∧
z ^ 3 + w ^ 3 = n ∧
x ≠ y ∧ w ≠ z ∧
x ≠ z ∧ x ≠ w ∧
y ≠ z ∧ y ≠ w) := by
sorry | Let $n=a^3+b^3$, and suppose that $\operatorname{gcd}(a, b)=1$. If a prime $p \mid a^3+b^3$, then
$$
\left(a b^{-1}\right)^3 \equiv_p-1
$$
Thus $3 \mid \frac{p-1}{2}$, that is, $p \equiv_6 1$.
If we have $n=a^3+b^3=c^3+d^3$, then we can factor $n$ as
$$
\begin{aligned}
& n=(a+b)\left(a^2-a b+b^2\right) \\
& n=(c+d)\lef... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
105 | Munkres_exercise_13_1 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x \in A$ there is an open set $U$ containing $x$ such that $U \subset A$. Show that $A$ is open in $X$. | theorem Munkres_exercise_13_1 (X : Type*) [TopologicalSpace X] (A : Set X)
(h1 : ∀ x ∈ A, ∃ U : Set X, x ∈ U ∧ IsOpen U ∧ U ⊆ A) :
IsOpen A := by
sorry | Since, from the given hypothesis given any $x \in A$ there exists an open set containing $x$ say, $U_x$ such that $U_x \subset A$. Thus, we claim that
$$
A=\bigcup_{x \in A} U_x
$$
Observe that if we prove the above claim, then $A$ will be open, being a union of arbitrary open sets. Since, for each $x \in A, U_x \subse... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
106 | Munkres_exercise_13_4a1 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def is_topology (X : Type*) (T : Set (Set X)) :=
univ ∈ T ∧
(∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧
(∀s, (∀t ∈ s, t ∈ T) → sUnion s ∈ T) | If $\mathcal{T}_\alpha$ is a family of topologies on $X$, show that $\bigcap \mathcal{T}_\alpha$ is a topology on $X$. | theorem Munkres_exercise_13_4a1 (X I : Type*) (T : I → Set (Set X)) (h : ∀ i, is_topology X (T i)) :
is_topology X (⋂ i : I, T i) := by
sorry | Since $\emptyset$ and $X$ belong to $\mathcal{T}_\alpha$ for each $\alpha$, they belong to $\bigcap_\alpha \mathcal{T}_\alpha$. Let $\left\{V_\beta\right\}_\beta$ be a collection of open sets in $\bigcap_\alpha \mathcal{T}_\alpha$. For any fixed $\alpha$ we have $\cup_\beta V_\beta \in \mathcal{T}_\alpha$ since $\mathc... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
107 | Munkres_exercise_13_4b1 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def is_topology (X : Type*) (T : Set (Set X)) :=
univ ∈ T ∧
(∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧
(∀s, (∀t ∈ s, t ∈ T) → sUnion s ∈ T) | Let $\mathcal{T}_\alpha$ be a family of topologies on $X$. Show that there is a unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$. | theorem Munkres_exercise_13_4b1 (X I : Type*) (T : I → Set (Set X)) (h : ∀ i, is_topology X (T i)) :
∃! T', is_topology X T' ∧ (∀ i, T i ⊆ T') ∧
∀ T'', is_topology X T'' → (∀ i, T i ⊆ T'') → T' ⊆ T'' := by
sorry | (b) First we prove that there is a unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$. Uniqueness of such topology is clear. For each $\alpha$ let $\mathcal{B}_\alpha$ be a basis for $\mathcal{T}_\alpha$. Let $\mathcal{T}$ be the topology generated by the subbasis $\mathcal{S}=\bigcup_\... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
108 | Munkres_exercise_13_5a | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def is_topology (X : Type*) (T : Set (Set X)) :=
univ ∈ T ∧
(∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧
(∀s, (∀t ∈ s, t ∈ T) → sUnion s ∈ T) | Show that if $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$. | theorem Munkres_exercise_13_5a {X : Type*}
(A : Set (Set X))
(hcover : ∀ x : X, ∃ B ∈ A, x ∈ B)
(hrefine : ∀ B₁ ∈ A, ∀ B₂ ∈ A, ∀ x ∈ B₁ ∩ B₂, ∃ B₃ ∈ A, x ∈ B₃ ∧ B₃ ⊆ B₁ ∩ B₂) :
{U : Set X | @IsOpen X (generateFrom A) U}
=
sInter {T : Set (Set X) | is_topology X T ∧ A ⊆ T} := by
sorry | Let $\mathcal{T}$ be the topology generated by $\mathcal{A}$ and let $\mathcal{O}$ be the intersection of all topologies on $X$ that contains $\mathcal{A}$. Clearly $\mathcal{O} \subset \mathcal{T}$ since $\mathcal{T}$ is a topology on $X$ that contain $\mathcal{A}$. Conversely, let $U \in \mathcal{T}$, so that $U$ is ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
109 | Munkres_exercise_13_6 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def lower_limit_topology (X : Type) [Preorder X] :=
generateFrom {S : Set X | ∃ a b, a < b ∧ S = Ico a b}
def Rl := lower_limit_topology ℝ
def K : Set ℝ := {r | ∃ n : ℕ, r = 1 / (n + 1 : ℝ)}
def K_topology := generateFrom
({S : Set ℝ | ∃ a b, a < b ∧ S = Ioo a b} ∪ {S : Set ℝ | ∃ a b, a < b ∧ S = Ioo a b \ K}) | Show that the lower limit topology $\mathbb{R}_l$ and $K$-topology $\mathbb{R}_K$ are not comparable. | theorem Munkres_exercise_13_6 :
¬ (∀ U, Rl.IsOpen U → K_topology.IsOpen U) ∧ ¬ (∀ U, K_topology.IsOpen U → Rl.IsOpen U) := by
sorry | Let $\mathcal{T}_{\ell}$ and $\mathcal{T}_K$ denote the topologies of $\mathbb{R}_{\ell}$ and $\mathbb{R}_K$ respectively. Given the basis element $[0,1)$ for $\mathcal{T}_{\ell}$, there is no basis element for $\mathcal{T}_K$ containing 0 and contained in $[0,1)$, so $\mathcal{T}_{\ell} \not \subset \mathcal{T}_K$. Si... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
110 | Munkres_exercise_13_8b | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def lower_limit_topology (X : Type) [Preorder X] :=
generateFrom {S : Set X | ∃ a b, a < b ∧ S = Ico a b} | Show that the collection $\{(a,b) \mid a < b, a \text{ and } b \text{ rational}\}$ is a basis that generates a topology different from the lower limit topology on $\mathbb{R}$. | theorem Munkres_exercise_13_8b (T : Set (Set ℝ))
(hT : T = {S : Set ℝ | ∃ a b : ℚ, a < b ∧ S = Ioo ↑a ↑b}) :
(∀ x : ℝ, ∃ S ∈ T, x ∈ S) ∧
(∀ S₁ ∈ T, ∀ S₂ ∈ T, ∀ x ∈ S₁ ∩ S₂, ∃ S₃ ∈ T, x ∈ S₃ ∧ S₃ ⊆ S₁ ∩ S₂) ∧
generateFrom T ≠ lower_limit_topology ℝ := by
sorry | (b) $\mathcal{C}$ is a basis for a topology on $\mathbb{R}$ since the union of its elements is $\mathbb{R}$ and the intersection of two elements of $\mathcal{C}$ is either empty or another element of $\mathcal{C}$. Now consider $[r, s)$ where $r$ is any irrational number and $s$ is any real number greater than $r$. The... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
111 | Munkres_exercise_16_4 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | A map $f: X \rightarrow Y$ is said to be an open map if for every open set $U$ of $X$, the set $f(U)$ is open in $Y$. Show that $\pi_{1}: X \times Y \rightarrow X$ and $\pi_{2}: X \times Y \rightarrow Y$ are open maps. | theorem Munkres_exercise_16_4 {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y]
(π₁ : X × Y → X)
(π₂ : X × Y → Y)
(h₁ : π₁ = Prod.fst)
(h₂ : π₂ = Prod.snd) :
IsOpenMap π₁ ∧ IsOpenMap π₂ := by
sorry | Exercise 16.4. Let $U \times V$ be a (standard) basis element for $X \times Y$, so that $U$ is open in $X$ and $V$ is open in $Y$. Then $\pi_1(U \times V)=U$ is open in $X$ and $\pi_2(U \times V)=V$ is open in $Y$. Since arbitrary maps and unions satisfy $f\left(\bigcup_\alpha W_\alpha\right)=\bigcup_\alpha f\left(W_\a... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
112 | Munkres_exercise_17_4 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that if $U$ is open in $X$ and $A$ is closed in $X$, then $U-A$ is open in $X$, and $A-U$ is closed in $X$. | theorem Munkres_exercise_17_4 {X : Type*} [TopologicalSpace X]
(U A : Set X) (hU : IsOpen U) (hA : IsClosed A) :
IsOpen (U \ A) ∧ IsClosed (A \ U) := by
sorry | Since
$$
X \backslash(U \backslash A)=(X \backslash U) \cup A \text { and } \quad X \backslash(A \backslash U)=(X \backslash A) \cup U,
$$
it follows that $X \backslash(U \backslash A)$ is closed in $X$ and $X \backslash(A \backslash U)$ is open in $X$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
113 | Munkres_exercise_18_8b | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $Y$ be an ordered set in the order topology. Let $f, g: X \rightarrow Y$ be continuous. Let $h: X \rightarrow Y$ be the function $h(x)=\min \{f(x), g(x)\}.$ Show that $h$ is continuous. | theorem Munkres_exercise_18_8b {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y]
[LinearOrder Y] [OrderTopology Y] {f g : X → Y}
(hf : Continuous f) (hg : Continuous g) :
Continuous (λ x => min (f x) (g x)) := by
sorry | Let $A=\{x \mid f(x) \leq g(x)\}$ and $B=\{x \mid g(x) \leq f(x)\}$. Then $A$ and $B$ are closed in $X$ by (a), $A \cap B=\{x \mid f(x)=g(x)\}$, and $X=A \cup B$. Since $f$ and $g$ are continuous, their restrictions $f^{\prime}: A \rightarrow Y$ and $g^{\prime}: B \rightarrow Y$ are continuous. It follows from the past... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
114 | Munkres_exercise_19_6a | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $\mathbf{x}_1, \mathbf{x}_2, \ldots$ be a sequence of the points of the product space $\prod X_\alpha$. Show that this sequence converges to the point $\mathbf{x}$ if and only if the sequence $\pi_\alpha(\mathbf{x}_i)$ converges to $\pi_\alpha(\mathbf{x})$ for each $\alpha$. | theorem Munkres_exercise_19_6a
{ι : Type*}
{f : ι → Type*} {x : ℕ → Πa, f a}
(y : Πi, f i)
[Πa, TopologicalSpace (f a)] :
Tendsto x atTop (𝓝 y) ↔ ∀ i, Tendsto (λ j => (x j) i) atTop (𝓝 (y i)) := by
sorry | For each $n \in \mathbb{Z}_{+}$, we write $\mathbf{x}_n=\left(x_n^\alpha\right)_\alpha$, so that $\pi_\alpha\left(\mathbf{x}_n\right)=x_n^\alpha$ for each $\alpha$.
First assume that the sequence $\mathbf{x}_1, \mathbf{x}_2, \ldots$ converges to $\mathbf{x}=\left(x_\alpha\right)_\alpha$ in the product space $\prod_\alp... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
115 | Munkres_exercise_21_6a | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | abbrev I : Set ℝ := Icc 0 1 | Define $f_{n}:[0,1] \rightarrow \mathbb{R}$ by the equation $f_{n}(x)=x^{n}$. Show that the sequence $\left(f_{n}(x)\right)$ converges for each $x \in[0,1]$. | theorem Munkres_exercise_21_6a
(f : ℕ → I → ℝ)
(h : ∀ x n, f n x = ↑x ^ n) :
∀ x, ∃ y, Tendsto (λ n => f n x) atTop (𝓝 y) := by
sorry | If $0 \leq x<1$ is fixed, then $f_n(x) \rightarrow 0$ as $n \rightarrow \infty$. As $f_n(1)=1$ for all $n, f_n(1) \rightarrow 1$. Thus $\left(f_n\right)_n$ converges to $f:[0,1] \rightarrow \mathbb{R}$ given by $f(x)=0$ if $x=0$ and $f(1)=1$. The sequence | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
116 | Munkres_exercise_21_8 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $X$ be a topological space and let $Y$ be a metric space. Let $f_{n}: X \rightarrow Y$ be a sequence of continuous functions. Let $x_{n}$ be a sequence of points of $X$ converging to $x$. Show that if the sequence $\left(f_{n}\right)$ converges uniformly to $f$, then $\left(f_{n}\left(x_{n}\right)\right)$ converges... | theorem Munkres_exercise_21_8
{X : Type*} [TopologicalSpace X] {Y : Type*} [MetricSpace Y]
{f : ℕ → X → Y} {x : ℕ → X}
(hf : ∀ n, Continuous (f n))
(x₀ : X)
(hx : Tendsto x atTop (𝓝 x₀))
(f₀ : X → Y)
(hh : TendstoUniformly f f₀ atTop) :
Tendsto (λ n => f n (x n)) atTop (𝓝 (f₀ x₀)) := by
sorry | Let $d$ be the metric on $Y$. Let $V$ be a neighbourhood of $f(x)$, and let $\varepsilon>0$ be such that $f(x) \in B_d(f(x), \varepsilon) \subset V$. Since $\left(f_n\right)_n$ converges uniformly to $f$, there exists $N_1 \in \mathbb{Z}_{+}$such that $d\left(f_n(x), f(x)\right)<\varepsilon / 2$ for all $x \in X$ and a... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
117 | Munkres_exercise_22_2b | Munkres | import Mathlib
open Filter Set TopologicalSpace Topology | If $A \subset X$, a retraction of $X$ onto $A$ is a continuous map $r: X \rightarrow A$ such that $r(a)=a$ for each $a \in A$. Show that a retraction is a quotient map. | theorem Munkres_exercise_22_2b {X : Type*} [TopologicalSpace X]
{A : Set X} (r : X → A) (hr : Continuous r) (h : ∀ x : A, r x = x) :
IsQuotientMap r := by
sorry | The inclusion map $i: A \rightarrow X$ is continuous and $r \circ i=1_A$ is the identity. Thus $r$ is a quotient map by (a). | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
118 | Munkres_exercise_23_2 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $\left\{A_{n}\right\}$ be a sequence of connected subspaces of $X$, such that $A_{n} \cap A_{n+1} \neq \varnothing$ for all $n$. Show that $\bigcup A_{n}$ is connected. | theorem Munkres_exercise_23_2 {X : Type*}
[TopologicalSpace X] {A : ℕ → Set X} (hA : ∀ n, IsConnected (A n))
(hAn : ∀ n, A n ∩ A (n + 1) ≠ ∅) :
IsConnected (⋃ n, A n) := by
sorry | Suppose that $\bigcup_n A_n=B \cup C$, where $B$ and $C$ are disjoint open subsets of $\bigcup_n A_n$. Since $A_1$ is connected and a subset of $B \cup C$, by Lemma $23.2$ it lies entirely within either $B$ or $C$. Without any loss of generality, we may assume $A_1 \subset B$. Note that given $n$, if $A_n \subset B$ th... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
119 | Munkres_exercise_23_4 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
set_option checkBinderAnnotations false | Show that if $X$ is an infinite set, it is connected in the finite complement topology. | theorem Munkres_exercise_23_4 {X : Type*} [Infinite X] :
IsConnected (Set.univ : Set (CofiniteTopology X)) := by
sorry | Suppose that $A$ is a non-empty subset of $X$ that is both open and closed, i.e., $A$ and $X \backslash A$ are finite or all of $X$. Since $A$ is non-empty, $X \backslash A$ is finite. Thus $A$ cannot be finite as $X \backslash A$ is infinite, so $A$ is all of $X$. Therefore $X$ is connected. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
120 | Munkres_exercise_23_9 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $A$ be a proper subset of $X$, and let $B$ be a proper subset of $Y$. If $X$ and $Y$ are connected, show that $(X \times Y)-(A \times B)$ is connected. | theorem Munkres_exercise_23_9 {X Y : Type*}
[TopologicalSpace X] [TopologicalSpace Y]
(A : Set X) (B : Set Y)
(hA : A ⊂ (Set.univ : Set X))
(hB : B ⊂ (Set.univ : Set Y))
(hX : IsConnected (Set.univ : Set X))
(hY : IsConnected (Set.univ : Set Y)) :
IsConnected ((Set.univ : Set (X × Y)) \ (A ×ˢ B)) := by
... | This is similar to the proof of Theorem 23.6. Take $c \times d \in(X \backslash A) \times(Y \backslash B)$. For each $x \in X \backslash A$, the set
$$
U_x=(X \times\{d\}) \cup(\{x\} \times Y)
$$
is connected since $X \times\{d\}$ and $\{x\} \times Y$ are connected and have the common point $x \times d$. Then $U=\bigcu... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
121 | Munkres_exercise_24_2 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $f: S^{1} \rightarrow \mathbb{R}$ be a continuous map. Show there exists a point $x$ of $S^{1}$ such that $f(x)=f(-x)$. | theorem Munkres_exercise_24_2
{f : (Metric.sphere (0 : ℂ) 1 : Set ℂ) → ℝ}
(hf : Continuous f) : ∃ x, f x = f (-x) := by
sorry | Let $f: S^1 \rightarrow \mathbb{R}$ be continuous. Let $x \in S^1$. If $f(x)=f(-x)$ we are done, so assume $f(x) \neq f(-x)$. Define $g: S^1 \rightarrow \mathbb{R}$ by setting $g(x)=f(x)-f(-x)$. Then $g$ is continuous. Suppose $f(x)>f(-x)$, so that $g(x)>0$. Then $-x \in S^1$ and $g(-x)<0$. By the intermediate value th... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
122 | Munkres_exercise_25_4 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $X$ be locally path connected. Show that every connected open set in $X$ is path connected. | theorem Munkres_exercise_25_4 {X : Type*} [TopologicalSpace X]
[LocPathConnectedSpace X] (U : Set X) (hU : IsOpen U)
(hcU : IsConnected U) : IsPathConnected U := by
sorry | Let $U$ be a open connected set in $X$. By Theorem 25.4, each path component of $U$ is open in $X$, hence open in $U$. Thus, each path component in $U$ is both open and closed in $U$, so must be empty or all of $U$. It follows that $U$ is path-connected. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
123 | Munkres_exercise_26_11 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $X$ be a compact Hausdorff space. Let $\mathcal{A}$ be a collection of closed connected subsets of $X$ that is simply ordered by proper inclusion. Then $Y=\bigcap_{A \in \mathcal{A}} A$ is connected. | theorem Munkres_exercise_26_11
{X : Type*} [TopologicalSpace X] [CompactSpace X] [T2Space X]
(A : Set (Set X)) (hAne : A.Nonempty)
(hChain : ∀ (a b : Set X), a ∈ A → b ∈ A → a ⊆ b ∨ b ⊆ a)
(hClosed : ∀ a ∈ A, IsClosed a) (hConn : ∀ a ∈ A, IsConnected a) :
IsConnected (⋂₀ A) := by
sorry | Since each $A \in \mathcal{A}$ is closed, $Y$ is closed. Suppose that $C$ and $D$ form a separation of $Y$. Then $C$ and $D$ are closed in $Y$, hence closed in $X$. Since $X$ is compact, $C$ and $D$ are compact by Theorem 26.2. Since $X$ is Hausdorff, by Exercise 26.5, there exist $U$ and $V$ open in $X$ and disjoint c... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
124 | Munkres_exercise_27_4 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that a connected metric space having more than one point is uncountable. | theorem Munkres_exercise_27_4
{X : Type*} [MetricSpace X] [ConnectedSpace X] (hX : ∃ x y : X, x ≠ y) :
¬ Countable (univ : Set X) := by
sorry | The distance function $d: X \times X \rightarrow \mathbb{R}$ is continuous by Exercise 20.3(a), so given $x \in X$, the function $d_x: X \rightarrow \mathbb{R}$ given by $d_x(y)=d(x, y)$ is continuous by Exercise 19.11. Since $X$ is connected, the image $d_x(X)$ is a connected subspace of $\mathbb{R}$, and contains 0 s... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
125 | Munkres_exercise_28_5 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def countably_compact (X : Type*) [TopologicalSpace X] :=
∀ U : ℕ → Set X,
(∀ i, IsOpen (U i)) ∧ ((univ : Set X) ⊆ ⋃ i, U i) →
(∃ t : Finset ℕ, (univ : Set X) ⊆ ⋃ i ∈ t, U i) | Show that X is countably compact if and only if every nested sequence $C_1 \supset C_2 \supset \cdots$ of closed nonempty sets of X has a nonempty intersection. | theorem Munkres_exercise_28_5
(X : Type*) [TopologicalSpace X] :
countably_compact X ↔ ∀ (C : ℕ → Set X), (∀ n, IsClosed (C n)) ∧
(∀ n, C n ≠ ∅) ∧ (∀ n, C (n + 1) ⊆ C n) → ∃ x, ∀ n, x ∈ C n := by
sorry | We could imitate the proof of Theorem 26.9, but we prove directly each direction. First let $X$ be countable compact and let $C_1 \supset C_2 \supset \cdots$ be a nested sequence of closed nonempty sets of $X$. For each $n \in \mathbb{Z}_{+}, U_n=X \backslash C_n$ is open in $X$. Then $\left\{U_n\right\}_{n \in \mathbb... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
126 | Munkres_exercise_29_1 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that the rationals $\mathbb{Q}$ are not locally compact. | theorem Munkres_exercise_29_1 : ¬ LocallyCompactSpace ℚ := by
sorry | First, we prove that each set $\mathbb{Q} \cap[a, b]$, where $a, b$ are irrational numbers, is not compact. Indeed, since $\mathbb{Q} \cap[a, b]$ is countable, we can write $\mathbb{Q} \cap[a, b]=\left\{q_1, q_2, \ldots\right\}$. Then $\left\{U_i\right\}_{i \in \mathbb{Z}_{+}}$, where $U_i=\mathbb{Q} \cap\left[a, q_i\r... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
127 | Munkres_exercise_29_10 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that if $X$ is a Hausdorff space that is locally compact at the point $x$, then for each neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\bar{V}$ is compact and $\bar{V} \subset U$. | theorem Munkres_exercise_29_10 {X : Type*}
[TopologicalSpace X] [T2Space X] (x : X)
(hx : ∃ U : Set X, x ∈ U ∧ IsOpen U ∧ (∃ K : Set X, U ⊆ K ∧ IsCompact K))
(U : Set X) (hU : IsOpen U) (hxU : x ∈ U) :
∃ (V : Set X), IsOpen V ∧ x ∈ V ∧ IsCompact (closure V) ∧ closure V ⊆ U := by
sorry | Let $U$ be a neighbourhood of $x$. Since $X$ is locally compact at $x$, there exists a compact subspace $C$ of $X$ containing a neighbourhood $W$ of $x$. Then $U \cap W$ is open in $X$, hence in $C$. Thus, $C \backslash(U \cap W)$ is closed in $C$, hence compact. Since $X$ is Hausdorff, there exist disjoint open sets $... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
128 | Munkres_exercise_30_13 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that if $X$ has a countable dense subset, every collection of disjoint open sets in $X$ is countable. | theorem Munkres_exercise_30_13 {X : Type*} [TopologicalSpace X]
(h : ∃ (s : Set X), Countable s ∧ Dense s) (U : Set (Set X))
(hU : ∀ (x y : Set X), x ∈ U → y ∈ U → x ≠ y → x ∩ y = ∅)
(hUopen : ∀ (u : Set X), u ∈ U → IsOpen u) :
Countable U := by
sorry | Let $\mathcal{U}$ be a collection of disjoint open sets in $X$ and let $A$ be a countable dense subset of $X$.
Since $A$ is dense in $X$, every $U \in \mathcal{U}$ intersects $S$. Therefore, there exists a point $x_U \in U \cap S$.
Let $U_1, U_2 \in \mathcal{U}, U_1 \neq U_2$. Then $x_{U_1} \neq x_{U_2}$ since $U_1 \ca... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
129 | Munkres_exercise_31_2 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that if $X$ is normal, every pair of disjoint closed sets have neighborhoods whose closures are disjoint. | theorem Munkres_exercise_31_2 {X : Type*}
[TopologicalSpace X] [NormalSpace X] {A B : Set X}
(hA : IsClosed A) (hB : IsClosed B) (hAB : Disjoint A B) :
∃ (U V : Set X), IsOpen U ∧ IsOpen V ∧ A ⊆ U ∧ B ⊆ V ∧ closure U ∩ closure V = ∅ := by
sorry | Let $A$ and $B$ be disjoint closed sets. Then there exist disjoint open sets $U$ and $V$ containing $A$ and $B$ respectively.
Since $X \backslash V$ is closed and contains $U$, the closure of $U$ is contained in $X \backslash V$ hence $B$ and closure of $U$ are disjoint.
Repeat steps 1 and 2 for $B$ and $\bar{U}$ ins... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
130 | Munkres_exercise_32_1 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that a closed subspace of a normal space is normal. | theorem Munkres_exercise_32_1 {X : Type*} [TopologicalSpace X]
(hX : NormalSpace X) (A : Set X) (hA : IsClosed A) :
NormalSpace {x // x ∈ A} := by
sorry | Let $X$ be a normal space and $Y$ a closed subspace of $X$.
First we shows that $Y$ is a $T_1$-space.
Let $y \in Y$ be any point. Since $X$ is normal, $X$ is also a $T_1$ space and therefore $\{y\}$ is closed in $X$.
Then it follows that $\{y\}=\{y\} \cap Y$ is closed in $Y$ (in relative topology).
Now let's prove that... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
131 | Munkres_exercise_32_2b | Munkres | import Mathlib
open Filter Set TopologicalSpace Topology
open scoped Topology | Show that if $\prod X_\alpha$ is regular (separation property only), then so is $X_\alpha$. Assume that each $X_\alpha$ is nonempty. | theorem Munkres_exercise_32_2b
{ι : Type*} {X : ι → Type*} [∀ i, TopologicalSpace (X i)]
(h : ∀ i, Nonempty (X i)) (h2 : RegularSpace (Π i, X i)) :
∀ i, RegularSpace (X i) := by
sorry | Suppose that $X=\prod_\beta X_\beta$ is regular and let $\alpha$ be any index.
We have to prove that $X_\alpha$ satisfies the $T_1$ and the $T_3$ axiom.
Since $X$ is regular, it follows that $X$ is Hausdorff, which then implies that $X_\alpha$ is Hausdorff. However, this implies that $X_\alpha$ satisfies the $T_1$ axio... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
132 | Munkres_exercise_32_3 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that every locally compact Hausdorff space is regular. | theorem Munkres_exercise_32_3 {X : Type*} [TopologicalSpace X]
(hX : LocallyCompactSpace X) (hX' : T2Space X) :
RegularSpace X := by
sorry | Let $X$ be a LCH space.
Then it follows that for every $x \in X$ and for every open neighborhood $U \subseteq X$ of $x$ there exists an open neighborhood $V \subseteq X$ of $x$ such that $\bar{V} \subseteq U$ (and $\bar{V}$ is compact, but this is not important here).
Since $X$ is a Hausdorff space, it satisfies the $T... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
133 | Munkres_exercise_33_8 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | abbrev I : Set ℝ := Icc 0 1 | Let $X$ be completely regular, let $A$ and $B$ be disjoint closed subsets of $X$. Show that if $A$ is compact, there is a continuous function $f \colon X \rightarrow [0, 1]$ such that $f(A) = \{0\}$ and $f(B) = \{1\}$. | theorem Munkres_exercise_33_8
(X : Type*) [TopologicalSpace X] [CompletelyRegularSpace X]
(A B : Set X) (hA : IsClosed A) (hB : IsClosed B)
(hAB : Disjoint A B) (hAc : IsCompact A)
(hAn : A.Nonempty) (hBn : B.Nonempty) :
∃ (f : X → I), Continuous f ∧ f '' A = {0} ∧ f '' B = {1} := by
sorry | Since $X$ is completely regular $\forall a \in A, \exists f_a: X \rightarrow[0,1]: f_a(a)=0$ and $f_a(B)=\{1\}$. For some $\epsilon_a \in(0,1)$ we have that $U_a:=f_a^{-1}([0, \epsilon))$ is an open neighborhood of $a$ that does not intersect $B$. We therefore have an open covering $\left\{U_a \mid a \in A\right\}$ of ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
134 | Munkres_exercise_38_6 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $X$ be completely regular (separation property only). Show that $X$ is connected if and only if the Stone-Čech compactification of $X$ is connected. | theorem Munkres_exercise_38_6 {X : Type*} [TopologicalSpace X] [CompletelyRegularSpace X] :
IsConnected (univ : Set X) ↔ IsConnected (univ : Set (StoneCech X)) := by
sorry | The closure of a connected set is connected, so if $X$ is connected so is $\beta(X)$
Suppose $X$ is the union of disjoint open subsets $U, V \subset X$. Define the continuous map
$$
\begin{aligned}
& f: X \rightarrow\{0,1\} \\
& x \mapsto \begin{cases}0, & x \in U \\
1, & x \in V\end{cases}
\end{aligned}
$$
By the fact... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
135 | Pugh_exercise_2_12a | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Let $(p_n)$ be a sequence and $f:\mathbb{N}\to\mathbb{N}$. The sequence $(q_k)_{k\in\mathbb{N}}$ with $q_k=p_{f(k)}$ is called a rearrangement of $(p_n)$. Show that if $f$ is an injection, the limit of a sequence is unaffected by rearrangement. | theorem Pugh_exercise_2_12a {α : Type*} [TopologicalSpace α]
(f : ℕ → ℕ) (p : ℕ → α) (a : α)
(hf : Injective f) (hp : Tendsto p atTop (𝓝 a)) :
Tendsto (λ n => p (f n)) atTop (𝓝 a) := by
sorry | Let $\varepsilon>0$. Since $p_n \rightarrow L$, we have that, for all $n$ except $n \leq N$, $d\left(p_n, L\right)<\epsilon$. Let $S=\{n \mid f(n) \leq N\}$, let $n_0$ be the largest $n \in S$, we know there is such a largest $n$ because $f(n)$ is injective. Now we have that $\forall n>n_0 f(n)>N$ which implies that $p... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
136 | Pugh_exercise_2_29 | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Let $\mathcal{T}$ be the collection of open subsets of a metric space $\mathrm{M}$, and $\mathcal{K}$ the collection of closed subsets. Show that there is a bijection from $\mathcal{T}$ onto $\mathcal{K}$. | theorem Pugh_exercise_2_29 (M : Type*) [MetricSpace M]
(O C : Set (Set M))
(hO : O = {s | IsOpen s})
(hC : C = {s | IsClosed s}) :
∃ f : O → C, Bijective f := by
sorry | The bijection given by $x\mapsto X^C$ suffices. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
137 | Pugh_exercise_2_41 | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Let $\|\cdot\|$ be any norm on $\mathbb{R}^{m}$ and let $B=\left\{x \in \mathbb{R}^{m}:\|x\| \leq 1\right\}$. Prove that $B$ is compact. | theorem Pugh_exercise_2_41
(m : ℕ) (E : Type*) [NormedAddCommGroup E] [NormedSpace ℝ E]
[FiniteDimensional ℝ E] (hdim : Module.finrank ℝ E = m) :
IsCompact (Metric.closedBall (0 : E) 1) :=
sorry | Let us call $\|\cdot\|_E$ the Euclidean norm in $\mathbb{R}^m$. We start by claiming that there exist constants $C_1, C_2>0$ such that
$$
C_1\|x\|_E \leq\|x\| \leq C_2\|x\|_E, \forall x \in \mathbb{R}^m .
$$
Assuming (1) to be true, let us finish the problem. First let us show that $B$ is bounded w.r.t. $d_E$, which is... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
138 | Pugh_exercise_2_57 | Pugh | import Mathlib
open Filter Real Function Set Metric
open scoped Topology | Show that if $S$ is connected, it is not true in general that its interior is connected. | theorem Pugh_exercise_2_57 :
∃ S : Set (ℝ × ℝ), IsConnected S ∧ ¬ IsConnected (interior S) := by
sorry | Consider $X=\mathbb{R}^2$ and
$$
A=([-2,0] \times[-2,0]) \cup([0,2] \times[0,2])
$$
which is connected, while $\operatorname{int}(A)$ is not connected.
To see this consider the continuous function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is defined by $f(x, y)=x+y$. Let $U=f^{-1}(0,+\infty)$ which is open in $\mathbb{R... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
139 | Pugh_exercise_2_126 | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Suppose that $E$ is an uncountable subset of $\mathbb{R}$. Prove that there exists a point $p \in \mathbb{R}$ at which $E$ condenses. | theorem Pugh_exercise_2_126 {E : Set ℝ}
(hE : ¬ Set.Countable E) :
∃ p : ℝ, ∀ U ∈ 𝓝 p, ¬ Set.Countable (U ∩ E) := by
sorry | I think this is the proof by contrapositive that you were getting at.
Suppose that $E$ has no limit points at all. Pick an arbitrary point $x \in E$. Then $x$ cannot be a limit point, so there must be some $\delta>0$ such that the ball of radius $\delta$ around $x$ contains no other points of $E$ :
$$
B_\delta(x) \cap ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
140 | Pugh_exercise_3_4 | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Prove that the real function $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$. | theorem Pugh_exercise_3_4 :
Tendsto (fun n : ℝ => sqrt (n + 1) - sqrt n) atTop (𝓝 0) := by
sorry | $$
\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2 \sqrt{n}}
$$ | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
141 | Pugh_exercise_3_63b | Pugh | import Mathlib
open Filter Real Function
open scoped Topology | Prove that $\sum 1/k(\log(k))^p$ diverges when $p \leq 1$. | theorem Pugh_exercise_3_63b (p : ℝ) (f : ℕ → ℝ) (hp : p ≤ 1)
(h : f = fun k : ℕ => 1 / (k * (Real.log k) ^ p)) :
¬ Summable f := by
sorry | Using the integral test, for a set $a$, we see
$$
\lim _{b \rightarrow \infty} \int_a^b \frac{1}{x \log (x)^c} d x=\lim _{b \rightarrow \infty}\left(\frac{\log (b)^{1-c}}{1-c}-\frac{\log (a)^{1-c}}{1-c}\right)
$$
which goes to infinity if $c \leq 1$ and converges if $c>1$. Thus,
$$
\sum_{n=2}^{\infty} \frac{1}{n \log (... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
142 | Putnam_exercise_1998_a3 | Putnam | import Mathlib
open scoped BigOperators | Let $f$ be a real function on the real line with continuous third derivative. Prove that there exists a point $a$ such that $f(a) \cdot f^{\prime}(a) \cdot f^{\prime \prime}(a) \cdot f^{\prime \prime \prime}(a) \geq 0$. | theorem Putnam_exercise_1998_a3 (f : ℝ → ℝ) (hf : ContDiff ℝ 3 f) :
∃ a : ℝ, (f a) * (deriv f a) * (iteratedDeriv 2 f a) * (iteratedDeriv 3 f a) ≥ 0 := by
sorry | If at least one of $f(a)$, $f'(a)$, $f''(a)$, or $f'''(a)$ vanishes
at some point $a$, then we are done. Hence we may assume each of
$f(x)$, $f'(x)$, $f''(x)$, and $f'''(x)$ is either strictly positive
or strictly negative on the real line. By replacing $f(x)$ by $-f(x)$
if necessary, we may assume $f''(x)>0$; by rep... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
143 | Putnam_exercise_2000_a2 | Putnam | import Mathlib
open scoped BigOperators | Prove that there exist infinitely many integers $n$ such that $n, n+1, n+2$ are each the sum of the squares of two integers. | theorem Putnam_exercise_2000_a2 :
∀ N : ℕ, ∃ n : ℕ, n > N ∧ ∃ i : Fin 6 → ℕ, n = (i 0)^2 + (i 1)^2 ∧
n + 1 = (i 2)^2 + (i 3)^2 ∧ n + 2 = (i 4)^2 + (i 5)^2 := by
sorry | It is well-known that the equation $x^2-2y^2=1$ has infinitely
many solutions (the so-called ``Pell'' equation). Thus setting
$n=2y^2$ (so that $n=y^2+y^2$, $n+1=x^2+0^2$, $n+2=x^2+1^2$)
yields infinitely many $n$ with the desired property. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
144 | Putnam_exercise_2010_a4 | Putnam | import Mathlib
open scoped BigOperators | Prove that for each positive integer $n$, the number $10^{10^{10^n}}+10^{10^n}+10^n-1$ is not prime. | theorem Putnam_exercise_2010_a4 (n : ℕ) (hn : n > 0) :
¬ Nat.Prime (10^10^10^n + 10^10^n + 10^n - 1) := by
sorry | Put
\[
N = 10^{10^{10^n}} + 10^{10^n} + 10^n - 1.
\]
Write $n = 2^m k$ with $m$ a nonnegative integer and $k$ a positive odd integer.
For any nonnegative integer $j$,
\[
10^{2^m j} \equiv (-1)^j \pmod{10^{2^m} + 1}.
\]
Since $10^n \geq n \geq 2^m \geq m+1$, $10^n$ is divisible by $2^n$ and hence by $2^{m+1}$,
and simil... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
145 | Putnam_exercise_2017_b3 | Putnam | import Mathlib
set_option maxHeartbeats 400000
open scoped BigOperators | Suppose that $f(x)=\sum_{i=0}^{\infty} c_{i} x^{i}$ is a power series for which each coefficient $c_{i}$ is 0 or 1 . Show that if $f(2 / 3)=3 / 2$, then $f(1 / 2)$ must be irrational. | theorem Putnam_exercise_2017_b3 (f : ℝ → ℝ) (c : ℕ → ℝ)
(hf : f = λ x => (∑' (i : ℕ), (c i) * x^i))
(hc : ∀ n, c n = 0 ∨ c n = 1)
(hf1 : f (2/3) = 3/2) :
Irrational (f (1/2)) := by
sorry | Suppose by way of contradiction that $f(1/2)$ is rational. Then $\sum_{i=0}^{\infty} c_i 2^{-i}$ is the binary expansion of a rational number, and hence must be eventually periodic; that is, there exist some integers $m,n$ such that
$c_i = c_{m+i}$ for all $i \geq n$. We may then write
\[
f(x) = \sum_{i=0}^{n-1} c_i x^... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
146 | Putnam_exercise_2018_b2 | Putnam | import Mathlib
set_option maxHeartbeats 400000
noncomputable section
open scoped BigOperators
open BigOperators Finset Complex | Let $n$ be a positive integer, and let $f_{n}(z)=n+(n-1) z+$ $(n-2) z^{2}+\cdots+z^{n-1}$. Prove that $f_{n}$ has no roots in the closed unit disk $\{z \in \mathbb{C}:|z| \leq 1\}$. | theorem Putnam_exercise_2018_b2 (n : ℕ) (hn : n > 0) (f : ℕ → ℂ → ℂ)
(hf : ∀ n : ℕ, f n = λ (z : ℂ) => (∑ i : Fin n, (n-i)* z^(i : ℕ))) :
¬ (∃ z : ℂ, ‖z‖ ≤ 1 ∧ f n z = 0) := by
sorry | Note first that $f_n(1) > 0$, so $1$ is not a root of $f_n$.
Next, note that
\[
(z-1)f_n(z) = z^n + \cdots + z - n;
\]
however, for $\left| z \right| \leq 1$, we have
$\left| z^n + \cdots + z \right| \leq n$ by the triangle inequality;
equality can only occur if $z,\dots,z^n$ have norm 1 and the same argument, which o... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
147 | Putnam_exercise_2020_b5 | Putnam | import Mathlib
open scoped BigOperators | For $j \in\{1,2,3,4\}$, let $z_{j}$ be a complex number with $\left|z_{j}\right|=1$ and $z_{j} \neq 1$. Prove that $3-z_{1}-z_{2}-z_{3}-z_{4}+z_{1} z_{2} z_{3} z_{4} \neq 0 .$ | theorem Putnam_exercise_2020_b5 (z : Fin 4 → ℂ) (hz0 : ∀ n, ‖z n‖ = 1)
(hz1 : ∀ n, z n ≠ 1) :
3 - z 0 - z 1 - z 2 - z 3 + (z 0) * (z 1) * (z 2) * (z 3) ≠ 0 := by
sorry | It will suffice to show that for any $z_1, z_2, z_3, z_4 \in \mathbb{C}$ of modulus 1 such that $|3-z_1-z_2-z_3-z_4| = |z_1z_2z_3z_4|$, at least one of $z_1, z_2, z_3$ is equal to 1.
To this end, let $z_1=e^{\alpha i}, z_2=e^{\beta i}, z_3=e^{\gamma i}$ and
\[
f(\alpha, \beta, \gamma)=|3-z_1-z_2-z_3|^2-|1-z_1z_2z_3|^... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
148 | Rudin_exercise_1_1a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational. | theorem Rudin_exercise_1_1a
(x : ℝ) (y : ℚ) (hy : y ≠ 0) :
( Irrational x ) -> Irrational ( x + y ) := by
sorry | If $r$ and $r+x$ were both rational, then $x=r+x-r$ would also be rational. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
149 | Rudin_exercise_1_2 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Prove that there is no rational number whose square is $12$. | theorem Rudin_exercise_1_2 : ¬ ∃ (x : ℚ), ( x ^ 2 = 12 ) := by
sorry | Suppose $m^2=12 n^2$, where $m$ and $n$ have no common factor. It follows that $m$ must be even, and therefore $n$ must be odd. Let $m=2 r$. Then we have $r^2=3 n^2$, so that $r$ is also odd. Let $r=2 s+1$ and $n=2 t+1$. Then
$$
4 s^2+4 s+1=3\left(4 t^2+4 t+1\right)=12 t^2+12 t+3,
$$
so that
$$
4\left(s^2+s-3 t^2-3 t\r... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
150 | Rudin_exercise_1_5 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$. | theorem Rudin_exercise_1_5 (A minus_A : Set ℝ) (hA : A.Nonempty)
(hA_bdd_below : BddBelow A) (hminus_A : minus_A = {x | -x ∈ A}) :
sInf A = - sSup minus_A := by
sorry | We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$. For brevity, let $\alpha=-\sup (-A)$. We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$. Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$. It follows that $x \geq-... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
151 | Rudin_exercise_1_11a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $z$ is a complex number, prove that there exists an $r\geq 0$ and a complex number $w$ with $| w | = 1$ such that $z = rw$. | theorem Rudin_exercise_1_11a (z : ℂ) :
∃ (r : ℝ) (w : ℂ), r ≥ 0 ∧ ‖w‖ = 1 ∧ z = r * w := by
sorry | If $z=0$, we take $r=0, w=1$. (In this case $w$ is not unique.) Otherwise we take $r=|z|$ and $w=z /|z|$, and these choices are unique, since if $z=r w$, we must have $r=r|w|=|r w|=|z|, z / r$ | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
152 | Rudin_exercise_1_13 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $x, y$ are complex, prove that $||x|-|y|| \leq |x-y|$. | theorem Rudin_exercise_1_13 (x y : ℂ) :
|‖x‖ - ‖y‖| ≤ ‖x-y‖ := by
sorry | Since $x=x-y+y$, the triangle inequality gives
$$
|x| \leq|x-y|+|y|
$$
so that $|x|-|y| \leq|x-y|$. Similarly $|y|-|x| \leq|x-y|$. Since $|x|-|y|$ is a real number we have either ||$x|-| y||=|x|-|y|$ or ||$x|-| y||=|y|-|x|$. In either case, we have shown that ||$x|-| y|| \leq|x-y|$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
153 | Rudin_exercise_1_16a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators InnerProductSpace
set_option maxHeartbeats 400000 | Suppose $k \geq 3, x, y \in \mathbb{R}^k, |x - y| = d > 0$, and $r > 0$. Prove that if $2r > d$, there are infinitely many $z \in \mathbb{R}^k$ such that $|z-x|=|z-y|=r$. | theorem Rudin_exercise_1_16a
(n : ℕ)
(d r : ℝ)
(x y z : EuclideanSpace ℝ (Fin n)) -- R^n
(h₁ : n ≥ 3)
(h₂ : ‖x - y‖ = d)
(h₃ : d > 0)
(h₄ : r > 0)
(h₅ : 2 * r > d)
: Set.Infinite {z : EuclideanSpace ℝ (Fin n) | ‖z - x‖ = r ∧ ‖z - y‖ = r} := by
sorry | (a) Let w be any vector satisfying the following two equations:
$$
\begin{aligned}
\mathbf{w} \cdot(\mathbf{x}-\mathbf{y}) &=0, \\
|\mathbf{w}|^2 &=r^2-\frac{d^2}{4} .
\end{aligned}
$$
From linear algebra it is known that all but one of the components of a solution $\mathbf{w}$ of the first equation can be arbitrary. T... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
154 | Rudin_exercise_1_18a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators InnerProductSpace | If $k \geq 2$ and $\mathbf{x} \in R^{k}$, prove that there exists $\mathbf{y} \in R^{k}$ such that $\mathbf{y} \neq 0$ but $\mathbf{x} \cdot \mathbf{y}=0$ | theorem Rudin_exercise_1_18a
(n : ℕ)
(h : n > 1)
(x : EuclideanSpace ℝ (Fin n)) -- R^n
: ∃ (y : EuclideanSpace ℝ (Fin n)), y ≠ 0 ∧ ⟪x, y⟫_ℝ = 0 := by
sorry | If $\mathbf{x}$ has any components equal to 0 , then $\mathbf{y}$ can be taken to have the corresponding components equal to 1 and all others equal to 0 . If all the components of $\mathbf{x}$ are nonzero, $\mathbf{y}$ can be taken as $\left(-x_2, x_1, 0, \ldots, 0\right)$. This is, of course, not true when $k=1$, sinc... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
155 | Rudin_exercise_2_24 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is separable. | theorem Rudin_exercise_2_24
{X : Type*} [MetricSpace X]
(hBW : ∀ A : Set X, A.Infinite → ∃ p : X, AccPt p (𝓟 A)) :
SeparableSpace X :=
sorry | We observe that if the process of constructing $x_j$ did not terminate, the result would be an infinite set of points $x_j, j=1,2, \ldots$, such that $d\left(x_i, x_j\right) \geq \delta$ for $i \neq j$. It would then follow that for any $x \in X$, the open ball $B_{\frac{\delta}{2}}(x)$ contains at most one point of th... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
156 | Rudin_exercise_2_27a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $E\subset\mathbb{R}^k$ is uncountable, and let $P$ be the set of condensation points of $E$. Prove that $P$ is perfect. | theorem Rudin_exercise_2_27a (k : ℕ) (E P : Set (EuclideanSpace ℝ (Fin k)))
(hE : ¬ Set.Countable E)
(hP : P = {x | ∀ U ∈ 𝓝 x, ¬ Set.Countable (U ∩ E)}) :
Perfect P := by
sorry | We see that $E \cap W$ is at most countable, being a countable union of at-most-countable sets. It remains to show that $P=W^c$, and that $P$ is perfect. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
157 | Rudin_exercise_2_28 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable. | theorem Rudin_exercise_2_28 (X : Type*) [MetricSpace X] [SeparableSpace X]
(A : Set X) (hA : IsClosed A) :
∃ P₁ P₂ : Set X, A = P₁ ∪ P₂ ∧ Perfect P₁ ∧ Set.Countable P₂ := by
sorry | If $E$ is closed, it contains all its limit points, and hence certainly all its condensation points. Thus $E=P \cup(E \backslash P)$, where $P$ is perfect (the set of all condensation points of $E$ ), and $E \backslash P$ is at most countable.
Since a perfect set in a separable metric space has the same cardinality as... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
158 | Rudin_exercise_3_1a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Prove that in $\mathbb{C}$, convergence of $\left\{s_{n}\right\}$ implies convergence of $\left\{\left|s_{n}\right|\right\}$. | theorem Rudin_exercise_3_1a
(f : ℕ → ℂ)
(h : ∃ (a : ℂ), Tendsto (λ (n : ℕ) => f n) atTop (𝓝 a))
: ∃ (a : ℝ), Tendsto (λ (n : ℕ) => ‖f n‖) atTop (𝓝 a) := by
sorry | Let $\varepsilon>0$. Since the sequence $\left\{s_n\right\}$ is a Cauchy sequence, there exists $N$ such that $\left|s_m-s_n\right|<\varepsilon$ for all $m>N$ and $n>N$. We then have $\left| |s_m| - |s_n| \right| \leq\left|s_m-s_n\right|<\varepsilon$ for all $m>N$ and $n>N$. Hence the sequence $\left\{\left|s_n\right|\... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
159 | Rudin_exercise_3_3 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | noncomputable def f : ℕ → ℝ
| 0 => Real.sqrt 2
| (n + 1) => Real.sqrt (2 + Real.sqrt (f n)) | If $s_{1}=\sqrt{2}$, and $s_{n+1}=\sqrt{2+\sqrt{s_{n}}} \quad(n=1,2,3, \ldots),$ prove that $\left\{s_{n}\right\}$ converges, and that $s_{n}<2$ for $n=1,2,3, \ldots$. | theorem Rudin_exercise_3_3
: ∃ (x : ℝ), Tendsto f atTop (𝓝 x) ∧ ∀ n, f n < 2 := by
sorry | Since $\sqrt{2}<2$, it is manifest that if $s_n<2$, then $s_{n+1}<\sqrt{2+2}=2$. Hence it follows by induction that $\sqrt{2}<s_n<2$ for all $n$. In view of this fact,it also follows that $\left(s_n-2\right)\left(s_n+1\right)<0$ for all $n>1$, i.e., $s_n>s_n^2-2=s_{n-1}$. Hence the sequence is an increasing sequence th... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
160 | Rudin_exercise_3_6a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | noncomputable def g (n : ℕ) : ℝ := Real.sqrt (n + 1) - Real.sqrt n | Prove that $\lim_{n \rightarrow \infty} \sum_{i < n} a_i = \infty$, where $a_i = \sqrt{i + 1} -\sqrt{i}$. | theorem Rudin_exercise_3_6a
: Tendsto (λ (n : ℕ) => (∑ i ∈ range n, g i)) atTop atTop := by
sorry | (a) Multiplying and dividing $a_n$ by $\sqrt{n+1}+\sqrt{n}$, we find that $a_n=\frac{1}{\sqrt{n+1}+\sqrt{n}}$, which is larger than $\frac{1}{2 \sqrt{n+1}}$. The series $\sum a_n$ therefore diverges by comparison with the $p$ series $\left(p=\frac{1}{2}\right)$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
161 | Rudin_exercise_3_8 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset Metric
open scoped BigOperators | If $\Sigma a_{n}$ converges, and if $\left\{b_{n}\right\}$ is monotonic and bounded, prove that $\Sigma a_{n} b_{n}$ converges. | theorem Rudin_exercise_3_8
(a b : ℕ → ℝ)
(h1 : ∃ y, Tendsto (fun n => ∑ i ∈ range n, a i) atTop (𝓝 y))
(h2 : Monotone b ∨ Antitone b)
(h3 : Bornology.IsBounded (Set.range b)) :
∃ y, Tendsto (fun n => ∑ i ∈ range n, a i * b i) atTop (𝓝 y) := by
sorry | We shall show that the partial sums of this series form a Cauchy sequence, i.e., given $\varepsilon>0$ there exists $N$ such that $\left|\sum_{k=m+1}^n a_k b_k\right|\langle\varepsilon$ if $n\rangle$ $m \geq N$. To do this, let $S_n=\sum_{k=1}^n a_k\left(S_0=0\right)$, so that $a_k=S_k-S_{k-1}$ for $k=1,2, \ldots$ Let ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
162 | Rudin_exercise_3_20 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $\left\{p_{n}\right\}$ is a Cauchy sequence in a metric space $X$, and some subsequence $\left\{p_{n l}\right\}$ converges to a point $p \in X$. Prove that the full sequence $\left\{p_{n}\right\}$ converges to $p$. | theorem Rudin_exercise_3_20 {X : Type*} [MetricSpace X]
(p : ℕ → X) (φ : ℕ → ℕ) (r : X)
(hp : CauchySeq p)
(hφ : StrictMono φ)
(hpl : Tendsto (p ∘ φ) atTop (𝓝 r)) :
Tendsto p atTop (𝓝 r) := by
sorry | Let $\varepsilon>0$. Choose $N_1$ so large that $d\left(p_m, p_n\right)<\frac{\varepsilon}{2}$ if $m>N_1$ and $n>N_1$. Then choose $N \geq N_1$ so large that $d\left(p_{n_k}, p\right)<\frac{\varepsilon}{2}$ if $k>N$. Then if $n>N$, we have
$$
d\left(p_n, p\right) \leq d\left(p_n, p_{n_{N+1}}\right)+d\left(p_{n_{N+1}}, ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
163 | Rudin_exercise_3_22 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $X$ is a nonempty complete metric space, and $\left\{G_{n}\right\}$ is a sequence of dense open sets of $X$. Prove Baire's theorem, namely, that $\bigcap_{1}^{\infty} G_{n}$ is not empty. | theorem Rudin_exercise_3_22 (X : Type*) [MetricSpace X] [CompleteSpace X] [Nonempty X]
(G : ℕ → Set X) (hG : ∀ n, IsOpen (G n) ∧ Dense (G n)) :
∃ x, ∀ n, x ∈ G n := by
sorry | Let $F_n$ be the complement of $G_n$, so that $F_n$ is closed and contains no open sets. We shall prove that any nonempty open set $U$ contains a point not in any $F_n$, hence in all $G_n$. To this end, we note that $U$ is not contained in $F_1$, so that there is a point $x_1 \in U \backslash F_1$. Since $U \backslash ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
164 | Rudin_exercise_4_2a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that $f(\overline{E}) \subset \overline{f(E)}$ for every set $E \subset X$. ($\overline{E}$ denotes the closure of $E$). | theorem Rudin_exercise_4_2a
{α : Type} [MetricSpace α]
{β : Type} [MetricSpace β]
(f : α → β)
(h₁ : Continuous f)
: ∀ (x : Set α), f '' (closure x) ⊆ closure (f '' x) := by
sorry | Let $x \in \bar{E}$. We need to show that $f(x) \in \overline{f(E)}$. To this end, let $O$ be any neighborhood of $f(x)$. Since $f$ is continuous, $f^{-1}(O)$ contains (is) a neighborhood of $x$. Since $x \in \bar{E}$, there is a point $u$ of $E$ in $f^{-1}(O)$. Hence $\frac{f(u)}{f(E)} \in O \cap f(E)$. Since $O$ was ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
165 | Rudin_exercise_4_4a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | 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 $f(E)$ is dense in $f(X)$. | theorem Rudin_exercise_4_4a
{α : Type} [MetricSpace α]
{β : Type} [MetricSpace β]
(f : α → β)
(s : Set α)
(h₁ : Continuous f)
(h₂ : Dense s)
: f '' Set.univ ⊆ closure (f '' s) := by
sorry | To prove that $f(E)$ is dense in $f(X)$, simply use that $f(X)=f(\bar{E}) \subseteq \overline{f(E)}$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
166 | Rudin_exercise_4_5a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $f$ is a real continuous function defined on a closed set $E \subset \mathbb{R}$, prove that there exist continuous real functions $g$ on $\mathbb{R}$ such that $g(x)=f(x)$ for all $x \in E$. | theorem Rudin_exercise_4_5a
(f : ℝ → ℝ)
(E : Set ℝ)
(h₁ : IsClosed E)
(h₂ : ContinuousOn f E)
: ∃ (g : ℝ → ℝ), Continuous g ∧ ∀ x ∈ E, f x = g x := by
sorry | Following the hint, let the complement of $E$ consist of a countable collection of finite open intervals $\left(a_k, b_k\right)$ together with possibly one or both of the the semi-infinite intervals $(b,+\infty)$ and $(-\infty, a)$. The function $f(x)$ is already defined at $a_k$ and $b_k$, as well as at $a$ and $b$ (i... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
167 | Rudin_exercise_4_6 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | If $f$ is defined on $E$, the graph of $f$ is the set of points $(x, f(x))$, for $x \in E$. In particular, if $E$ is a set of real numbers, and $f$ is real-valued, the graph of $f$ is a subset of the plane. Suppose $E$ is compact, and prove that $f$ is continuous on $E$ if and only if its graph is compact. | theorem Rudin_exercise_4_6
(f : ℝ → ℝ)
(E : Set ℝ)
(G : Set (ℝ × ℝ))
(h₁ : IsCompact E)
(h₂ : G = {(x, f x) | x ∈ E})
: ContinuousOn f E ↔ IsCompact G := by
sorry | Let $Y$ be the co-domain of the function $f$. We invent a new metric space $E \times Y$ as the set of pairs of points $(x, y), x \in E, y \in Y$, with the metric $\rho\left(\left(x_1, y_1\right),\left(x_2, y_2\right)\right)=d_E\left(x_1, x_2\right)+d_Y\left(y_1, y_2\right)$. The function $\varphi(x)=(x, f(x))$ is then ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
168 | Rudin_exercise_4_8a | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Let f be a real uniformly continuous function on the bounded set E in $R^{1}$. Prove that f is bounded on E. | theorem Rudin_exercise_4_8a
(E : Set ℝ) (h : Bornology.IsBounded E) :
∀ f : ℝ → ℝ, UniformContinuousOn f E → Bornology.IsBounded (Set.image f E) := by
sorry | Since $E$ is bounded in $\mathbb{R}$, it is totally bounded. A uniformly continuous function maps totally bounded sets to totally bounded sets. In $\mathbb{R}$, totally bounded sets are bounded, so $f(E)$ is bounded. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
169 | Rudin_exercise_4_12 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | A uniformly continuous function of a uniformly continuous function is uniformly continuous. | theorem Rudin_exercise_4_12
{α β γ : Type*} [UniformSpace α] [UniformSpace β] [UniformSpace γ]
{f : α → β} {g : β → γ}
(hf : UniformContinuous f) (hg : UniformContinuous g) :
UniformContinuous (g ∘ f) := by
sorry | Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be uniformly continuous. Then $g \circ f: X \rightarrow Z$ is uniformly continuous, where $g \circ f(x)=g(f(x))$ for all $x \in X$.
To prove this fact, let $\varepsilon>0$ be given. Then, since $g$ is uniformly continuous, there exists $\eta>0$ such that $d_Z(g(u), g(v)... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
170 | Rudin_exercise_4_19 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $f$ is a real function with domain $R^{1}$ which has the intermediate value property: if $f(a) < c < f(b)$, then $f(x)=c$ for some $x$ between $a$ and $b$. Suppose also, for every rational $r$, that the set of all $x$ with $f(x)=r$ is closed. Prove that $f$ is continuous. | theorem Rudin_exercise_4_19
{f : ℝ → ℝ}
(hf : ∀ a b c, a < b →
((f a < c ∧ c < f b) ∨ (f b < c ∧ c < f a)) →
∃ x, a < x ∧ x < b ∧ f x = c)
(hg : ∀ r : ℚ, IsClosed {x | f x = r}) : Continuous f := by
sorry | The contradiction is evidently that $x_0$ is a limit point of the set of $t$ such that $f(t)=r$, yet, $x_0$ does not belong to this set. This contradicts the hypothesis that the set is closed. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
171 | Rudin_exercise_4_24 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Assume that $f$ is a continuous real function defined in $(a, b)$ such that $f\left(\frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}$ for all $x, y \in(a, b)$. Prove that $f$ is convex. | theorem Rudin_exercise_4_24 {f : ℝ → ℝ} {a b : ℝ}
(hf : ContinuousOn f (Set.Ioo a b)) (hab : a < b)
(h : ∀ x y : ℝ, a < x → x < b → a < y → y < b → f ((x + y) / 2) ≤ (f x + f y) / 2) :
ConvexOn ℝ (Set.Ioo a b) f := by
sorry | We shall prove that
$$
f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)
$$
for all "dyadic rational" numbers, i.e., all numbers of the form $\lambda=\frac{k}{2^n}$, where $k$ is a nonnegative integer not larger than $2^n$. We do this by induction on $n$. The case $n=0$ is trivial (since $\lambda=0$ or $\la... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
172 | Rudin_exercise_5_2 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $f^{\prime}(x)>0$ in $(a, b)$. Prove that $f$ is strictly increasing in $(a, b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that $g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \quad(a < x < b)$. | theorem Rudin_exercise_5_2 :
∀ (a b : ℝ) (f g : ℝ → ℝ),
(∀ x ∈ Set.Ioo a b, deriv f x > 0) →
(∀ x ∈ Set.Ioo a b, g (f x) = x) →
StrictMonoOn f (Set.Ioo a b) ∧
DifferentiableOn ℝ g (f '' Set.Ioo a b) ∧
∀ x ∈ Set.Ioo a b, deriv g (f x) = 1 / deriv f x := by
sorry | For any $c, d$ with $a<c<d<b$ there exists a point $p \in(c, d)$ such that $f(d)-f(c)=f^{\prime}(p)(d-c)>0$. Hence $f(c)<f(d)$
We know from Theorem $4.17$ that the inverse function $g$ is continuous. (Its restriction to each closed subinterval $[c, d]$ is continuous, and that is sufficient.) Now observe that if $f(x)=... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
173 | Rudin_exercise_5_4 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators Polynomial | If $C_{0}+\frac{C_{1}}{2}+\cdots+\frac{C_{n-1}}{n}+\frac{C_{n}}{n+1}=0,$ where $C_{0}, \ldots, C_{n}$ are real constants, prove that the equation $C_{0}+C_{1} x+\cdots+C_{n-1} x^{n-1}+C_{n} x^{n}=0$ has at least one real root strictly between 0 and 1. | theorem Rudin_exercise_5_4 {n : ℕ}
(C : ℕ → ℝ)
(hC : ∑ i ∈ (range (n + 1)), (C i) / (i + 1) = 0) :
∃ x, x ∈ (Set.Ioo (0 : ℝ) 1) ∧ ∑ i ∈ range (n + 1), (C i) * (x^i) = 0 := by
sorry | Consider the polynomial
$$
p(x)=C_0 x+\frac{C_1}{2} x^2+\cdots+\frac{C_{n-1}}{n} x^n+\frac{C_n}{n+1} x^{n+1},
$$
whose derivative is
$$
p^{\prime}(x)=C_0+C_1 x+\cdots+C_{n-1} x^{n-1}+C_n x^n .
$$
It is obvious that $p(0)=0$, and the hypothesis of the problem is that $p(1)=0$. Hence Rolle's theorem implies that $p^{\pri... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
174 | Rudin_exercise_5_6 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose (a) $f$ is continuous for $x \geq 0$, (b) $f^{\prime}(x)$ exists for $x>0$, (c) $f(0)=0$, (d) $f^{\prime}$ is monotonically increasing. Put $g(x)=\frac{f(x)}{x} \quad(x>0)$ and prove that $g$ is monotonically increasing. | theorem Rudin_exercise_5_6
{f : ℝ → ℝ}
(hf1 : ContinuousOn f (Set.Ici 0))
(hf2 : DifferentiableOn ℝ f (Set.Ioi 0))
(hf3 : f 0 = 0)
(hf4 : MonotoneOn (deriv f) (Set.Ioi 0)) :
MonotoneOn (λ x => f x / x) (Set.Ioi 0) := by
sorry | Put
$$
g(x)=\frac{f(x)}{x} \quad(x>0)
$$
and prove that $g$ is monotonically increasing.
By the mean-value theorem
$$
f(x)=f(x)-f(0)=f^{\prime}(c) x
$$
for some $c \in(0, x)$. Since $f^{\prime}$ is monotonically increasing, this result implies that $f(x)<x f^{\prime}(x)$. It therefore follows that
$$
g^{\prime}(x)=\fra... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
175 | Rudin_exercise_5_15 | Rudin | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators | Suppose $a \in R^{1}, f$ is a twice-differentiable real function on $(a, \infty)$, and $M_{0}, M_{1}, M_{2}$ are the least upper bounds of $|f(x)|,\left|f^{\prime}(x)\right|,\left|f^{\prime \prime}(x)\right|$, respectively, on $(a, \infty)$. Prove that $M_{1}^{2} \leq 4 M_{0} M_{2} .$ | theorem Rudin_exercise_5_15
{f : ℝ → ℝ} {a M0 M1 M2 : ℝ}
(hf' : DifferentiableOn ℝ f (Set.Ioi a))
(hf'' : DifferentiableOn ℝ (deriv f) (Set.Ioi a))
(hbdd0 : BddAbove {(|f x|) | x ∈ (Set.Ioi a)})
(hbdd1 : BddAbove {(|deriv f x|) | x ∈ (Set.Ioi a)})
(hbdd2 : BddAbove {(|deriv (deriv f) x|) | x ∈ (... | The inequality is obvious if $M_0=+\infty$ or $M_2=+\infty$, so we shall assume that $M_0$ and $M_2$ are both finite. We need to show that
$$
\left|f^{\prime}(x)\right| \leq 2 \sqrt{M_0 M_2}
$$
for all $x>a$. We note that this is obvious if $M_2=0$, since in that case $f^{\prime}(x)$ is constant, $f(x)$ is a linear fun... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
176 | Shakarchi_exercise_1_13a | Shakarchi | import Mathlib
import Mathlib.Analysis.Complex.OpenMapping
open Complex Filter Function Metric Finset
open scoped BigOperators Topology | Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $\text{Re}(f)$ is constant, then $f$ is constant. | theorem Shakarchi_exercise_1_13a {f : ℂ → ℂ} (Ω : Set ℂ) (a b : Ω) (hΩ : IsOpen Ω)
(hconn : IsPreconnected Ω) (hf : DifferentiableOn ℂ f Ω)
(hc : ∃ c : ℝ, ∀ z ∈ Ω, (f z).re = c) : f a = f b := by
sorry | Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.
Since $\operatorname{Re}(f)=$ constant,
$$
\frac{\partial u}{\partial x}=0, \frac{\partial u}{\partial y}=0 .
$$
By the Cauchy-Riemann equations,
$$
\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=0 .
$$
Thus, in $\Omega$,
$$
f^{\prime}(z)=\frac{\parti... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
177 | Shakarchi_exercise_1_13c | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology | Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $|f|$ is constant, then $f$ is constant. | theorem Shakarchi_exercise_1_13c {f : ℂ → ℂ} {Ω : Set ℂ} (a b : Ω)
(hΩ : IsOpen Ω) (hconn : IsPreconnected Ω)
(hf : DifferentiableOn ℂ f Ω) (hc : ∃ c : ℝ, ∀ z ∈ Ω, ‖f z‖ = c) :
f a = f b := by
sorry | Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.
We first give a mostly correct argument; the reader should pay attention to find the difficulty. Since $|f|=\sqrt{u^2+v^2}$ is constant,
$$
\left\{\begin{array}{l}
0=\frac{\partial\left(u^2+v^2\right)}{\partial x}=2 u \frac{\partial u}{\partial x}+2 v \frac{\partia... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
178 | Shakarchi_exercise_1_19b | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology | Prove that the power series $\sum z^n/n^2$ converges at every point of the unit circle. | theorem Shakarchi_exercise_1_19b (z : ℂ) (hz : ‖z‖ = 1) (s : ℕ → ℂ)
(h : s = (λ n => ∑ i ∈ (range n), z ^ (i + 1) / (i + 1) ^ 2)) :
∃ y, Tendsto s atTop (𝓝 y) := by
sorry | Since $\left|z^n / n^2\right|=1 / n^2$ for all $|z|=1$, then $\sum z^n / n^2$ converges at every point in the unit circle as $\sum 1 / n^2$ does ( $p$-series $p=2$.) | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
179 | Shakarchi_exercise_1_26 | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset Set
open scoped BigOperators Topology | Suppose $f$ is continuous in a region $\Omega$. Prove that any two primitives of $f$ (if they exist) differ by a constant. | theorem Shakarchi_exercise_1_26 :
∀ (f F₁ F₂ : ℂ → ℂ) (Ω : Set ℂ),
Nonempty Ω → IsOpen Ω → IsConnected Ω →
ContinuousOn f Ω → DifferentiableOn ℂ F₁ Ω → DifferentiableOn ℂ F₂ Ω →
(∀ x ∈ Ω, deriv F₁ x = f x) → (∀ x ∈ Ω, deriv F₂ x = f x) →
∃ c : ℂ, ∀ x ∈ Ω, F₁ x = F₂ x + c := by
sorry | Suppose $F_1$ and $F_2$ are primitives of $F$. Then $(F_1-F_2)^\prime = f - f = 0$, therefore $F_1$ and $F_2$ differ by a constant. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
180 | Shakarchi_exercise_2_9 | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
noncomputable section | Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_{0} \in \Omega$ such that $\varphi\left(z_{0}\right)=z_{0} \quad \text { and } \quad \varphi^{\prime}\left(z_{0}\right)=1$ then $\varphi$ is linear. | theorem Shakarchi_exercise_2_9
{f : ℂ → ℂ} (Ω : Set ℂ) (b : Bornology.IsBounded Ω) (hΩ : IsOpen Ω)
(hconn : IsPreconnected Ω)
(hf : DifferentiableOn ℂ f Ω)
(h_maps : Set.MapsTo f Ω Ω)
(z₀ : Ω) (hz : f z₀ = z₀) (h'z : deriv f z₀ = 1) :
∃ (f_lin : ℂ →L[ℂ] ℂ), ∀ x ∈ Ω, f x = f_lin x := by
sorry | When $\Omega$ is connected, if $\varphi$ is not linear, then there exists $n \geq 2$ and $a_n \neq 0$, such that
$$
\varphi(z)=z+a_n\left(z-z_0\right)^n+O\left(\left(z-z_0\right)^{n+1}\right) .
$$
As you have noticed, by induction, it follows that for every $k \geq 1$,
$$
\varphi^k(z)=z+k a_n\left(z-z_0\right)^n+O\left... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
181 | Shakarchi_exercise_3_3 | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset MeasureTheory Real Set
open scoped BigOperators Topology FourierTransform RealInnerProductSpace Complex | Show that $ \int_{-\infty}^{\infty} \frac{\cos x}{x^2 + a^2} dx = \pi \frac{e^{-a}}{a}$ for $a > 0$. | theorem Shakarchi_exercise_3_3 (a : ℝ) (ha : 0 < a) :
Tendsto (λ y => ∫ x in -y..y, Real.cos x / (x ^ 2 + a ^ 2))
atTop (𝓝 (Real.pi * (Real.exp (-a) / a))) := by
sorry | $\cos x=\frac{e^{i x}+e^{-i x}}{2}$. changing $x \rightarrow-x$ we see that we can just integrate $e^{i x} /\left(x^2+a^2\right)$ and we'll get the same answer. Again, we use the same semicircle and part of the real line. The only pole is $x=i a$, it has order 1 and the residue at it is $\lim _{x \rightarrow i a} \frac... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
182 | Shakarchi_exercise_3_9 | Shakarchi | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology | Show that $\int_0^1 \log(\sin \pi x) dx = - \log 2$. | theorem Shakarchi_exercise_3_9 : ∫ x in (0 : ℝ)..(1 : ℝ), Real.log (Real.sin (Real.pi * x)) = - Real.log 2 := by
sorry | Consider
$$
\begin{gathered}
f(z)=\log \left(1-e^{2 \pi z i}\right)=\log \left(e^{\pi z i}\left(e^{-\pi z i}-e^{\pi z i}\right)\right)=\log (-2 i)+\pi z i+\log \\
(\sin (\pi z))
\end{gathered}
$$
Then we have
$$
\begin{aligned}
\int_0^1 f(z) d z & =\log (-2 i)+\frac{i \pi}{2}+\int_0^1 \log (\sin (\pi z)) d z \\
& =\int... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
183 | Shakarchi_exercise_3_22 | Shakarchi | import Mathlib
open Complex Function Metric Finset
open scoped BigOperators Topology Bornology | Show that there is no holomorphic function $f$ in the unit disc $D$ that extends continuously to $\partial D$ such that $f(z) = 1/z$ for $z \in \partial D$. | theorem Shakarchi_exercise_3_22 (D : Set ℂ) (hD : D = ball 0 1) (f : ℂ → ℂ)
(hf : DifferentiableOn ℂ f D) (hfc : ContinuousOn f (closure D)) :
¬ ∀ z ∈ (sphere (0 : ℂ) 1), f z = 1 / z := by
sorry | Consider $g(r)=\int_{|z|=r} f(z) d z$. Cauchy theorem implies that $g(r)=0$ for all $r<1$. Now since $\left.f\right|_{\partial D}=1 / z$ we have $\lim _{r \rightarrow 1} \int_{|z|=r} f(z) d z=\int_{|z|=1} \frac{1}{z} d z=\frac{2}{\pi i} \neq 0$. Contradiction. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
184 | Artin_exercise_2_3_2 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that the products $a b$ and $b a$ are conjugate elements in a group. | theorem Artin_exercise_2_3_2 {G : Type*} [Group G] (a b : G) :
∃ g : G, b* a = g * a * b * g⁻¹ := by
sorry | We have that $(a^{-1})ab(a^{-1})^{-1} = ba$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
185 | Artin_exercise_2_8_6 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that the center of the product of two groups is the product of their centers. | theorem Artin_exercise_2_8_6 {G H : Type*} [Group G] [Group H] :
Subgroup.center (G × H) = (Subgroup.center G).prod (Subgroup.center H) := by
sorry | 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$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
186 | Artin_exercise_3_2_7 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd RingHom
open scoped BigOperators | Prove that every homomorphism of fields is injective. | theorem Artin_exercise_3_2_7 {F : Type*} [Field F] {G : Type*} [Field G]
(φ : F →+* G) : Injective φ := by
sorry | 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$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
187 | Artin_exercise_3_7_2 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Let $V$ be a vector space over an infinite field $F$. Prove that $V$ is not the union of finitely many proper subspaces. | theorem Artin_exercise_3_7_2 {K V : Type*} [Field K] [Infinite K] [AddCommGroup V]
[Module K V] {ι : Type*} [Fintype ι] (γ : ι → Submodule K V)
(h : ∀ i : ι, γ i ≠ ⊤) :
(⋃ (i : ι), (γ i : Set V)) ≠ ⊤ := by
sorry | If $V$ is the set-theoretic union of $n$ proper subspaces $W_i$ ( $1 \leq i \leq n$ ), then $|F| \leq n-1$.
Proof. 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$.
Then $(v+F u) \cap W_i=\varnothing$ and $(v+F u) \cap W_... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
188 | Artin_exercise_6_4_2 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that no group of order $p q$, where $p$ and $q$ are prime, is simple. | theorem Artin_exercise_6_4_2 {G : Type*} [Group G] [Fintype G] {p q : ℕ}
(hp : Nat.Prime p) (hq : Nat.Prime q) (hG : card G = p*q) :
IsSimpleGroup G → false := by
sorry | If $|G|=n=p q$ then the only two Sylow subgroups are of order $p$ and $q$.
From Sylow's third theorem we know that $n_p \mid q$ which means that $n_p=1$ or $n_p=q$.
If $n_p=1$ then we are done (by a corollary of Sylow's theorem)
If $n_p=q$ then we have accounted for $q(p-1)=p q-q$ elements of $G$ and so there is only o... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
189 | Artin_exercise_6_4_12 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that no group of order 224 is simple. | theorem Artin_exercise_6_4_12 {G : Type*} [Group G] [Fintype G]
(hG : card G = 224) :
IsSimpleGroup G → false := by
sorry | The following proves there must exist a normal Sylow 2 -subgroup of order 32 ,
Suppose 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 $... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
190 | Artin_exercise_10_1_13 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | 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$. | theorem Artin_exercise_10_1_13 {R : Type*} [Ring R] {x : R}
(hx : IsNilpotent x) : IsUnit (1 + x) := by
sorry | If $x^n=0$, then
$$
(1+x)\left(\sum_{k=0}^{n-1}(-1)^k x^k\right)=1+(-1)^{n-1} x^n=1 .
$$ | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
191 | Artin_exercise_10_4_7a | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Let $I, J$ be ideals of a ring $R$ such that $I+J=R$. Prove that $I J=I \cap J$. | theorem Artin_exercise_10_4_7a {R : Type*} [CommRing R]
(I J : Ideal R) (hIJ : I + J = ⊤) : I * J = I ⊓ J := by
sorry | 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
\[ k = ik+kj. \]
The first term i... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
192 | Artin_exercise_10_6_7 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that every nonzero ideal in the ring of Gauss integers contains a nonzero integer. | theorem Artin_exercise_10_6_7 {I : Ideal GaussianInt}
(hI : I ≠ ⊥) : ∃ (z : I), z ≠ 0 ∧ (z : GaussianInt).im = 0 := by
sorry | 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:
1. If $b=0$, then $z=a$, so $z \in \mathbb{Z} \cap I$, and $z \neq 0$, so the statement of the exercise holds.
2. If $a=0$, then $z=... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
193 | Artin_exercise_11_2_13 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | If $a, b$ are integers and if $a$ divides $b$ in the ring of Gauss integers, then $a$ divides $b$ in $\mathbb{Z}$. | theorem Artin_exercise_11_2_13 (a b : ℤ) :
(ofInt a : GaussianInt) ∣ ofInt b → a ∣ b := by
sorry | 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 $y=0$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
194 | Artin_exercise_11_4_6a | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that $x^2+x+1$ is irreducible in the field $\mathbb{F}_2$. | theorem Artin_exercise_11_4_6a {F : Type*} [Field F] [Fintype F] (hF : card F = 2) :
Irreducible (X ^ 2 + X + 1 : Polynomial F) := by
sorry | 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. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
195 | Artin_exercise_11_4_6c | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that $x^3 - 9$ is irreducible in $\mathbb{F}_{31}$. | theorem Artin_exercise_11_4_6c : Irreducible (X^3 - 9 : Polynomial (ZMod 31)) := by
sorry | 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 $x \in \mathbb{F}_31$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
196 | Artin_exercise_11_13_3 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Prove that there are infinitely many primes congruent to $-1$ (modulo $4$). | theorem Artin_exercise_11_13_3 (N : ℕ):
∃ p ≥ N, Nat.Prime p ∧ p + 1 ≡ 0 [MOD 4] := by
sorry | First we show a lemma: if $a \equiv 3(\bmod 4)$ then there exists a prime $p$ such that $p \mid a$ and $p \equiv 3(\bmod 4)$.
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.
To prove the m... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
197 | Artin_exercise_13_6_10 | Artin | import Mathlib
open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd
open scoped BigOperators | Let $K$ be a finite field. Prove that the product of the nonzero elements of $K$ is $-1$. | theorem Artin_exercise_13_6_10 {K : Type*} [Field K] [Fintype Kˣ] :
(∏ x : Kˣ, x) = -1 := by
sorry | Since we are working with a finite field with $q$ elements, anyone of them is a root of the following polynomial
$$
x^q-x=0 .
$$
In particular if we rule out the 0 element, any $a_i \neq 0$ is a root of
$$
x^{q-1}-1=0 .
$$
This polynomial splits completely in $\mathbb{F}_q$ so we find
$$
\left(x-a_1\right) \cdots\left(... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
198 | Axler_exercise_1_2 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | Show that $\frac{-1 + \sqrt{3}i}{2}$ is a cube root of 1 (meaning that its cube equals 1). | theorem Axler_exercise_1_2 :
(⟨-1/2, Real.sqrt 3 / 2⟩ : ℂ) ^ 3 = 1 := by
sorry | $$
\left(\frac{-1+\sqrt{3} i}{2}\right)^2=\frac{-1-\sqrt{3} i}{2},
$$
hence
$$
\left(\frac{-1+\sqrt{3} i}{2}\right)^3=\frac{-1-\sqrt{3} i}{2} \cdot \frac{-1+\sqrt{3} i}{2}=1
$$
This means $\frac{-1+\sqrt{3} i}{2}$ is a cube root of 1. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
199 | Axler_exercise_1_4 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | Prove that if $a \in \mathbf{F}$, $v \in V$, and $av = 0$, then $a = 0$ or $v = 0$. | theorem Axler_exercise_1_4 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] (v : V) (a : F): a • v = 0 → a = 0 ∨ v = 0 := by
sorry | If $a=0$, then we immediately have our result. So suppose $a \neq 0$. Then, because $a$ is some nonzero real or complex number, it has a multiplicative inverse $\frac{1}{a}$. Now suppose that $v$ is some vector such that
$$
a v=0
$$
Multiply by $\frac{1}{a}$ on both sides of this equation to get
$$
\begin{aligned}
\fra... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
200 | Axler_exercise_1_9 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | 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. | theorem Axler_exercise_1_9 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] (U W : Submodule F V):
(∃ U' : Submodule F V, U'.carrier = ↑U ∪ ↑W) ↔ (U ≤ W ∨ W ≤ U) := by
sorry | 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 $v \in U_1$ an... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
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