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 $\mathcal{T}_\alpha$ is a topology on $X$, so $\bigcup_\beta V_\beta \in \bigcap_\alpha \mathcal{T}_\alpha$. Similarly, if $U_1, \ldots, U_n$ are elements of $\bigcap_\alpha \mathcal{T}_\alpha$, then for each $\alpha$ we have $\bigcup_{i=1}^n U_i \in \mathcal{T}_\alpha$ and therefore $\bigcup_{i=1}^n U_i \in \bigcap_\alpha \mathcal{T}_\alpha$. It follows that $\bigcap_\alpha \mathcal{T}_\alpha$ is a topology on $X$.


import Mathlib

open Filter Set TopologicalSpace
open scoped Topology

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 $\mathcal{S}=\bigcup_\alpha \mathcal{B}_\alpha$. Then the collection $\mathcal{B}$ of all finite intersections of elements of $\mathcal{S}$ is a basis for $\mathcal{T}$. Clearly $\mathcal{T}_\alpha \subset \mathcal{T}$ for all $\alpha$. We now prove that if $\mathcal{O}$ is a topology on $X$ such that $\mathcal{T}_\alpha \subset \mathcal{O}$ for all $\alpha$, then $\mathcal{T} \subset \mathcal{O}$. Given such $\mathcal{O}$, we have $\mathcal{B}_\alpha \subset \mathcal{O}$ for all $\alpha$, so $\mathcal{S} \subset \mathcal{O}$. Since $\mathcal{O}$ is a topology, it must contain all finite intersections of elements of $\mathcal{S}$, so $\mathcal{B} \subset \mathcal{O}$ and hence $\mathcal{T} \subset \mathcal{O}$. We conclude that the topology $\mathcal{T}$ generated by the subbasis $\mathcal{S}=\cup_\alpha \mathcal{B}_\alpha$ is the unique smallest topology on $X$ containing all the collections $\mathcal{T}_\alpha$.

to avoid data contamination, the formal proof is in proofnet_verified_gt.zip


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 a union of elements of $\mathcal{A}$. Since each of this elements is also an element of $\mathcal{O}$, their union $U$ belongs to $\mathcal{O}$. Thus $\mathcal{T} \subset \mathcal{O}$ and the equality holds.


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 $\mathcal{T}_{\ell} \not \subset \mathcal{T}_K$. Similarly, given the basis element $(-1,1) \backslash K$ for $\mathcal{T}_K$, there is no basis element for $\mathcal{T}_{\ell}$ containing 0 contained in $(-1,1) \backslash K$, so $\mathcal{T}_K \not \subset \mathcal{T}_{\ell}$.


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 $r$. Then $[r, s)$ is a basis element for the topology of $\mathbb{R}_{\ell}$, but $[r, s)$ is not a union of elements of $\mathcal{C}$. Indeed, suppose that $[r, s)=\cup_\alpha\left[a_\alpha, b_\alpha\right)$ for rationals $a_\alpha, b_\alpha$. Then $r \in\left[a_\alpha, b_\alpha\right)$ for some $\alpha$. Since $r$ is irrational we must have $a_\alpha<r$, but then $a_\alpha \notin[r, s)$, a contradiction. It follows that the topology generated by $\mathcal{C}$ is strictly coarser than the lower limit topology on $\mathbb{R}$.


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 $f\left(\bigcup_\alpha W_\alpha\right)=\bigcup_\alpha f\left(W_\alpha\right)$, it follows that $\pi_1$ and $\pi_2$ are open maps.


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 $X$.


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 pasting lemma that
$$
h: X \rightarrow Y, \quad h(x)=\min \{f(x), g(x)\}= \begin{cases}f^{\prime}(x) & \text { if } x \in A \\ g^{\prime}(x) & \text { if } x \in B\end{cases}
$$
is continuous


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 $\prod_\alpha X_\alpha$. Fix an index $\beta$ and let $U$ be a neighbourhood of $\pi_\beta(\mathbf{x})=x_\beta$. Let $V=\prod_\alpha U_\alpha$, where $U_\alpha=X_\alpha$ for each $\alpha \neq \beta$ and $U_\beta=U$. Then $V$ is a neighbourhood of $\mathbf{x}$, so there exists $N \in \mathbb{Z}_{+}$such that $\mathbf{x}_n \in V$ for all $n \geq N$. Therefore $\pi_\beta\left(\mathbf{x}_n\right)=x_n^\beta \in U$ for all $n \geq N$. Since $U$ was arbitrary, it follows that $\pi_\beta\left(\mathbf{x}_1\right), \pi_\beta\left(\mathbf{x}_2\right), \ldots$ converges to $\pi_\beta(\mathbf{x})$. Since $\beta$ was arbitrary, this holds for all indices $\alpha$.


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


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 to $f(x)$.

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 all $n \geq N_1$, so that $d\left(f_n\left(x_n\right), f\left(x_n\right)\right)<\varepsilon / 2$ for all $n \geq N_1$. Moreover, $f$ is continuous, so there exists $N_2 \in \mathbb{Z}_{+}$such that $d\left(f\left(x_n\right), f(x)\right)<\varepsilon / 2$ for all $n \geq N_2$. Thus, if $N>\max \left\{N_1, N_2\right\}$, then
$$
d\left(f_n\left(x_n\right), f(x)\right) \leq d\left(f_n\left(x_n\right), f\left(x_n\right)\right)+d\left(f\left(x_n\right), f(x)\right)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
$$
for all $n \geq N$, so $f_n\left(x_n\right) \in V$ for all $n \geq N$. It follows that $\left(f_n\left(x_n\right)\right)_n$ converges to $f(x)$.


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).


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 $A_n \subset B$ then $A_{n+1} \subset B$, for if $A_{n+1} \subset C$ then $A_n \cap A_{n+1} \subset B \cap C=\emptyset$, in contradiction with the assumption. By induction, $A_n \subset B$ for all $n \in \mathbb{Z}_{+}$, so that $\bigcup_n A_n \subset B$. It follows that $\bigcup_n A_n$ is connected.


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.


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
  sorry

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 $U=\bigcup_{x \in X \backslash A} U_x$ is connected because it is the union of the connected spaces $U_x$ which have the point $c \times d$ in common. Similarly, for each $y \in Y \backslash B$ the set
$$
V_y=(X \times\{y\}) \cup(\{c\} \times Y)
$$
is connected, so $V=\bigcup_{y \in Y \backslash B} V_y$ is connected. Thus $(X \times Y) \backslash(A \times B)=U \cup V$ is connected since $c \times d$ is a common point of $U$ and $V$.


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 theorem, since $S^1$ is connected and $g(-x)<0<g(x)$, there exists $y \in S^1$ such that $g(y)=0$. i.e, $f(y)=f(-y)$. Similarly, if $f(x)<f(-x)$, then $g(x)<0<g(-x)$ and again the intermediate value theorem gives the result.


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.


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 containing $C$ and $D$, respectively. We show that
$$
\bigcap_{A \in \mathcal{A}}(A \backslash(U \cup V))
$$
is not empty. Let $\left\{A_1, \ldots, A_n\right\}$ be a finite subcollection of elements of $\mathcal{A}$. We may assume that $A_i \subsetneq A_{i+1}$ for all $i=1, \ldots, n-1$. Then
$$
\bigcap_{i=1}^n\left(A_i \backslash(U \cup V)\right)=A_1 \backslash(U \cup V) \text {. }
$$
Suppose that $A_1 \backslash(U \cup V)=\emptyset$. Then $A_1 \subset U \cup V$. Since $A_1$ is connected and $U \cap V=\emptyset, A_1$ lies within either $U$ or $V$, say $A_1 \subset U$. Then $Y \subset A_1 \subset U$, so that $C=Y \cap C \subset Y \cap V=\emptyset$, contradicting the fact that $C$ and $D$ form a separation of $Y$. Hence, $\bigcap_{i=1}^n\left(A_i \backslash(U \cup V)\right)$ is non-empty. Therefore, the collection $\{A \backslash(U \cup V) \mid A \in \mathcal{A}\}$ has the finite intersection property, so
$$
\bigcap_{A \in \mathcal{A}}(A \backslash(U \cup V))=\left(\bigcap_{A \in \mathcal{A}} A\right) \backslash(U \cup V)=Y \backslash(U \cup V)
$$
is non-empty. So there exists $y \in Y$ such that $y \notin U \cup V \supset C \cup D$, contradicting the fact that $C$ and $D$ form a separation of $Y$. We conclude that there is no such separation, so that $Y$ is connected.


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 since $d_x(x)=0$. Thus, if $y \in X$ and $y \neq x$, then $d_x(X)$ contains the set $[0, \delta]$, where $\delta=d_x(y)>0$. Therefore $X$ must be uncountable.


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 $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$is a countable collection of open sets with no finite subcollection covering $X$, for if $U_{i_1} \cup \cdots \cup U_{1_n}$ covers $X$, then $C_{i_1} \cap \cdots \cap C_{i_n}$ is empty, contrary to the assumption. Hence $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$does not cover $X$, so there exist $x \in X \backslash \bigcup_{n \in \mathbb{Z}_{+}} U_n=\bigcap_{n \in Z_{+}}\left(X \backslash U_n\right)=\bigcap_{n \in Z_{+}} C_n$.

Conversely, assume that every nested sequence $C_1 \supset C_2 \supset \cdots$ of closed non-empty sets of $X$ has a non-empty intersection and let $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$be a countable open covering of $X$. For each $n$, let $V_n=U_1 \cup \cdots \cup U_n$ and $C_n=X \backslash V_n$. Suppose that no finite subcollection of $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$. Then each $C_n$ is non-empty, so $C_1 \supset C_2 \supset \cdots$ is a nested sequence of non-empty closed sets and $\bigcap_{n \in \mathbb{Z}_{+}} C_n$ is non-empty by assumption. Then there exists $x \in \bigcap_{n \in \mathbb{Z}_{+}} C_n$, so that $x \notin V_n$ for all $n$, contradicting the fact that $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$. It follows that there exists $N \in \mathbb{Z}_{+}$such that $C_N=\emptyset$, so that $X=V_N$ and hence some finite subcollection of $\left\{U_n\right\}_{n \in \mathbb{Z}_{+}}$covers $X$. We deduce that $X$ is countable compact.


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 $U_i=\mathbb{Q} \cap\left[a, q_i\right)$ for each $i$, is an open covering of $\mathbb{Q} \cap[a, b]$ with no finite subcovering. Now let $x \in \mathbb{Q}$ and suppose that $\mathbb{Q}$ is locally compact at $x$. Then there exists a compact set $C$ containing a neighbourhood $U$ of $x$. Then $U$ contains a set $\mathbb{Q} \cap[a, b]$ where $a, b$ are irrational numbers. Since this set is closed and contained in the compact $C$, it follows $\mathbb{Q} \cap[a, b]$ is compact, a contradiction. Therefore, $\mathbb{Q}$ is not locally compact.


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 $V_1$ and $V_2$ of $X$ containing $x$ and $C \backslash(U \cap W)$ respectively. Let $V=V_1 \cap U \cap W$. Since $\bar{V}$ is closed in $C$, it is compact. Furthermore, $\bar{V}$ is disjoint from $C \backslash(U \cap W) \supset C \backslash U$, so $\bar{V} \subset U$.


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 $U_1 \cap U_2=\emptyset$.
Thus, the function $\mathcal{U} \rightarrow S$ given by $U \mapsto x_U$ is injective and therefore, since $S$ is countable, it follows that $\mathcal{U}$ is countable.


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 $\bar{U}$ instead of $A$ and $B$ respectively and you will have open set $V^{\prime}$ which contains $B$ and its closure doesn't intersect with $\bar{U}$.


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 $X$ is a $T_4$-space.
Let $F, G \subseteq Y$ be disjoint closed sets. Since $F$ and $G$ are closed in $Y$ and $Y$ is closed in $X$, it follows that $F$ and $G$ are closed in $X$.

Since $X$ is normal, $X$ is also a $T_4$-space and therefore there exist disjoint open sets $U, V \subseteq$ $X$ such that $F \subseteq U$ and $G \subseteq V$.
However, then $U \cap Y$ and $V \cap Y$ are open disjoint sets in $Y$ (in relative topology) which separate $F$ and $G$.


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 $T_1$ axiom.

Let now $F \subseteq X_\alpha$ be a closed set and $x \in X_\alpha \backslash F$ a point.
Then $\prod_\beta F_\beta$, where $F_\alpha=F$ and $F_\beta=X_\beta$ for $\beta \neq \alpha$, is a closed set in $X$ since $\left(\prod_\beta F_\beta\right)^c=\prod_\beta U_\beta$, where $U_\alpha=F^c$ and $U_\beta=X_\beta$ for $\beta \neq \alpha$, which is an open set because it is a base element for the product topology.
Since all $X_\beta$ are nonempty, there exists a point $\mathbf{x} \in X$ such that $x_\alpha=x$. Then $\mathbf{x} \notin \prod_\beta F_\beta$.
Now since $X$ is regular (and therefore satisfies the $T_3$ axiom), there exist disjoint open sets $U, V \subseteq X$ such that $\mathbf{x} \in U$ and $\prod_\beta F_\beta \subseteq V$.

Now for every $\beta \neq \alpha$ we have that $x_\beta \in X_\beta=\pi_\beta(V)$. However, since $x_\beta \in \pi_\beta(U)$, it follows that $\pi_\beta(U) \cap \pi_\beta(V) \neq \emptyset$.
Then $U \cap V=\emptyset$ implies that $\pi_\alpha(U) \cap \pi_\alpha(V)=\emptyset$.. Also, $x \in \pi_\alpha(U)$ and $F \subseteq \pi_\alpha(V)$ and $\pi_\alpha(U), \pi_\alpha(V)$ are open sets since $\pi_\alpha$ is an open map.
Therefore, $X_\alpha$ satisfies the $T_3$ axiom.


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 $T_1$ axiom.
Then it follows that $X$ is regular.


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 $A$, so since $A$ is compact we have a finite subcover $\left\{U_{a_i} \mid 1 \leq i \leq m\right\}$. For each $1 \leq i \leq m$ define
$$
\begin{aligned}
\tilde{f}_{a_i}: X & \rightarrow[0,1] \\
x & \mapsto \frac{\max \left(f_{a_i}(x), \epsilon_{a_i}\right)-\epsilon_{a_i}}{1-\epsilon_{a_i}}
\end{aligned}
$$
so that $\forall x \in U_{a_i}: \tilde{f}_{a_i}(x)=0$ and $\forall x \in B, \forall 1 \leq i \leq m: \tilde{f}_{a_i}(x)=1$, and define $f:=$ $\prod_{i=1}^m \tilde{f}_{a_i}$. Then since $A \subset \cup_{i=1}^m U_{a_i}$ we have that $f(A)=\{0\}$ and also we have $f(B)=\{1\}$.


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 that $\{0,1\}$ is compact and Hausdorff we can extend $f$ to a surjective map $\bar{f}: \beta(X) \rightarrow\{0,1\}$ such that $\bar{f}^{-1}(\{0\})$ and $\bar{f}^{-1}(\{1\})$ are disjoint open sets that cover $\beta(X)$, which makes this space not-connected.


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 $p_{f(n)} \rightarrow L$, as required.


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.


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 how we call the Euclidean distance in $\mathbb{R}^m$. Indeed, given $x \in B,\|x\|_E \leq \frac{1}{C_1}\|x\| \leq \frac{1}{C_1}$. Hence $B \subset\left\{x \in \mathbb{R}^m: d_E(x, 0)<\frac{1}{C_1}+1\right\}$, which means $B$ is bounded w.r.t $d_E$.
Now let us show that $B$ is closed w.r.t. $d_E$. Let $x_n \rightarrow x$ w.r.t. $d_E$, where $x_n \in B$. Notice that this implies that $x_n \rightarrow x$ w.r.t. $d(x, y)=\|x-y\|$, the distance coming from $\|\cdot\|$, since by (1) we have
$$
d\left(x_n, x\right)=\left\|x_n-x\right\| \leq C_2\left\|x_n-x\right\|_E \rightarrow 0 .
$$
Also, notice that
$$
\|x\| \leq\left\|x_n-x\right\|+\left\|x_n\right\| \leq\left\|x_n-x\right\|+1,
$$
hence passing to the limit we obtain that $\|x\| \leq 1$, therefore $x \in B$ and so $B$ is closed w.r.t. $d_E$. Since $B$ is closed and bounded w.r.t. $d_E$, it must be compact. Now we claim that the identity function, $i d:\left(\mathbb{R}^m, d_E\right) \rightarrow\left(\mathbb{R}^m, d\right)$ where $\left(\mathbb{R}^m, d_E\right)$ means we are using the distance $d_E$ in $\mathbb{R}^m$ and $\left(\mathbb{R}^m, d\right)$ means we are using the distance $d$ in $\mathbb{R}^m$, is a homeomorphism. This follows by (1), since $i d$ is always a bijection, and it is continuous and its inverse is continuous by (1) (if $x_n \rightarrow x$ w.r.t. $d_E$, then $x_n \rightarrow x$ w.r.t. $d$ and vice-versa, by (1)). By a result we saw in class, since $B$ is compact in $\left(\mathbb{R}^m, d_E\right)$ and $i d$ is a homeomorphism, then $i d(B)=B$ is compact w.r.t. $d$.

We are left with proving (1). Notice that it suffices to prove that $C_1 \leq\|x\| \leq$ $C_2, \forall x \in \mathbb{R}^m$ with $\|x\|_E=1$. Indeed, if this is true, given $x \in \mathbb{R}^m$, either $\|x\|_E=0$ (which implies $x=0$ and (1) holds in this case), or $x /\|x\|_E=y$ is such that $\|y\|_E=1$, so $C_1 \leq\|y\| \leq C_2$, which implies $C_1\|x\|_E \leq\|x\| \leq C_2\|x\|_E$.
We want to show now that $\|\cdot\|$ is continuous w.r.t. $d_E$, that is, given $\varepsilon>0$ and $x \in \mathbb{R}^m$, there exists $\delta>0$ such that if $d_E(x, y)<\delta$, then $\|\mid x\|-\|y\| \|<\varepsilon$.

By the triangle inequality, $\|x\|-\|y\| \leq\|x-y\|$, and $\|y\|-\|x\| \leq\|x-y\|$, therefore
$$
|\|x||-\| y|\|\leq\| x-y \| .
$$
Writing now $x=\sum_{i=1}^m a_i e_i, y=\sum_{i=1}^m b_i e_i$, where $e_i=(0, \ldots, 1,0, \ldots, 0)$ (with 1 in the i-th component), we obtain by the triangle inequality,
$$
\begin{aligned}
\|x-y\| & =\left\|\sum_{i=1}^m\left(a_i-b_i\right) e_i\right\| \leq \sum_{i=1}^m\left|a_i-b_i\left\|\left|\left\|e_i\right\| \leq \max _{i=1, \ldots, m}\left\|e_i\right\| \sum_{i=1}^m\right| a_i-b_i \mid\right.\right. \\
& =\max _{i=1, \ldots, m}\left\|e_i\right\| d_{s u m}(x, y) \leq \max _{i=1, \ldots, m}\left\|e_i\right\| m d_{\max }(x, y) \\
& \leq \max _{i=1, \ldots, m}\left\|e_i\right\| m d_E(x, y) .
\end{aligned}
$$
Let $\delta=\frac{\varepsilon}{m \max _{i=1, \ldots, m}\left\|e_i\right\|}$. Then if $d_E(x, y)<\delta,\|x\|-\|y\|||<\varepsilon$.
Since $\|\cdot\|$ is continuous w.r.t. $d_E$ and $K=\left\{x \in \mathbb{R}^m:\|x\|_E=1\right\}$ is compact w.r.t. $d_E$, then the function $\|\cdot\|$ achieves a maximum and a minimum value on $K$. Call $C_1=\min _{x \in K}\|x\|, C_2=\max _{x \in K}\|x\|$. Then
$$
C_1 \leq\|x\| \leq C_2, \forall x \in \mathbb{R}^m \text { such that }\|x\|_E=1,
$$
which is what we needed.


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 $\mathbb{R}^2$ and so $U \cap \operatorname{int}(A)$ is open in $\operatorname{int}(A)$. Also, since $(0,0) \notin \operatorname{int}(A)$, so for all $(x, y) \in \operatorname{int}(A), f(x, y) \neq 0$ and $U \cap \operatorname{int}(A)=f^{-1}[0,+\infty) \cap \operatorname{int}(A)$ is closed in $\operatorname{int}(A)$. Furthermore, $(1,1)=f^{-1}(2) \in U \cap \operatorname{int}(A)$ shows that $U \cap \operatorname{int}(A) \neq \emptyset$ while $(-1,-1) \in \operatorname{int}(A)$ and $(-1,-1) \notin U$ shows that $U \cap \operatorname{int}(A) \neq \operatorname{int}(A)$.


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 $E$ :
$$
B_\delta(x) \cap E=\{x\}
$$
Call this "point 1 ". For the next point, take the closest element to $x$ and on its left; that is, choose the point
$$
\max [E \cap(-\infty, x)]
$$
if it exists (that is important - if not, skip to the next step). Note that by the argument above, this supremum, should it exist, cannot equal $x$ and is therefore a new point in $E$.

Call this "point 2 ". Now take the first point to the right of $x$ for "point 3 ". Take the first point to the left of point 2 for "point 4 ". And so on, ad infinitum.

This gives a countable list of unique points; we must show that it exhausts the entire set $E$. Suppose not. Suppose there is some element $a<x$ which is never included in the list (picking $a$ on the negative side of $x$ is arbitrary, and the same argument would work for the second case). Then the element closest and to the right of $a$ in $E$ (which exists, by the no-limit-points argument at the beginning) is also not in the list; if it was, $a$ would have been in one of the next two spots. And same with that point (call it $a_1$ ); there is a closest $a_2>a_1 \in E$ such that $a_2$ is not in the list. Repeating, we generate an infinite monotone-increasing sequence $\left\{a_i\right\}$ of elements in $E$ and not in the list, which is clearly bounded above by $x$. By the Monotone
Convergence Theorem this sequence has a limit. But that means the sequence $\left\{a_i\right\} \subset E$ converges to a limit, and hence $E$ has a limit point, contradicting the assumption. Therefore our list exhausts $E$, and we have enumerated all its elements.


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


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,
$$
\sum_{n=2}^{\infty} \frac{1}{n \log (n)^c}
$$
converges if and only if $c>1$.


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 replacing $f(x)$
by $f(-x)$ if necessary, we may assume $f'''(x)>0$.  (Notice that these
substitutions do not change the sign of $f(x) f'(x) f''(x) f'''(x)$.)
Now $f''(x)>0$ implies that $f'(x)$ is increasing, and $f'''(x)>0$
implies that $f'(x)$ is convex, so that $f'(x+a)>f'(x)+a f''(x)$
for all $x$ and $a$.  By
letting $a$ increase in the latter inequality, we see that $f'(x+a)$
must be positive for sufficiently large $a$; it follows that
$f'(x)>0$
for all $x$.  Similarly, $f'(x)>0$ and $f''(x)>0$ imply
that $f(x)>0$ for all $x$.  Therefore $f(x) f'(x) f''(x) f'''(x)>0$ for
all $x$, and we are done.


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.

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.


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 similarly $10^{10^n}$ is divisible by $2^{10^n}$ and hence by $2^{m+1}$. It follows that
\[
N \equiv 1 + 1 + (-1) + (-1) \equiv 0 \pmod{10^{2^m} + 1}.
\]
Since $N \geq 10^{10^n} > 10^n + 1 \geq 10^{2^m} + 1$, it follows that $N$ is composite.


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
\[
f(x) = \sum_{i=0}^{n-1} c_i x^i + \frac{x^n}{1-x^m} \sum_{i=0}^{m-1} c_{n+i} x^i.
\]
Evaluating at $x = 2/3$, we may equate $f(2/3) = 3/2$ with
\[
\frac{1}{3^{n-1}} \sum_{i=0}^{n-1} c_i 2^i 3^{n-i-1} + \frac{2^n 3^m}{3^{n+m-1}(3^m-2^m)} \sum_{i=0}^{m-1} c_{n+i} 2^i 3^{m-1-i};
\]
since all terms on the right-hand side have odd denominator, the same must be true of the sum, a contradiction.


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 only happens for $z=1$.
Thus there can be no root of $f_n$ with $|z| \leq 1$.


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
\[
f(\alpha, \beta, \gamma)=|3-z_1-z_2-z_3|^2-|1-z_1z_2z_3|^2.
\]
 A routine calculation shows that
\begin{align*}
f(\alpha, \beta, \gamma)&=
10 - 6\cos(\alpha) - 6\cos(\beta) - 6\cos(\gamma) \\
&\quad + 2\cos(\alpha + \beta + \gamma) + 2\cos(\alpha - \beta) \\
&\quad + 2\cos(\beta - \gamma) + 2\cos(\gamma - \alpha).
\end{align*}
Since the function $f$ is continuously differentiable, and periodic in each variable, $f$ has a maximum and a minimum and it attains these values only at points where $\nabla f=(0,0,0)$.  A routine calculation now shows that
\begin{align*}
\frac{\partial f}{\partial \alpha} + \frac{\partial f}{\partial \beta} + \frac{\partial f}{\partial \gamma} &=
6(\sin(\alpha) +\sin(\beta)+\sin(\gamma)-  \sin(\alpha + \beta + \gamma)) \\
&=
24\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\beta+\gamma}{2}\right)
\sin\left(\frac{\gamma+\alpha}{2}\right).
\end{align*}
Hence every critical point of $f$ must satisfy one of $z_1z_2=1$, $z_2z_3=1$, or $z_3z_1=1$. By symmetry, let us assume that $z_1z_2=1$. Then
\[
f = |3-2\mathrm{Re}(z_1)-z_3|^2-|1-z_3|^2;
\]
since $3-2\mathrm{Re}(z_1)\ge 1$, $f$ is nonnegative and can be zero only if the real part of $z_1$, and hence also $z_1$ itself, is equal to $1$.


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.


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
$$
4\left(s^2+s-3 t^2-3 t\right)=2 .
$$
But this is absurd, since 2 cannot be a multiple of 4 .


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 $x \geq-\sup (-A)$, i.e., $\alpha \leq x$. Thus $\alpha$ is a lower bound of $A$.

Now let $\beta$ be any lower bound of $A$. This means $\beta \leq x$ for all $x$ in $A$. Hence $-x \leq-\beta$ for all $x \in A$, which says $y \leq-\beta$ for all $y \in-A$. This means $-\beta$ is an upper bound of $-A$. Hence $-\beta \geq \sup (-A)$ by definition of sup, i.e., $\beta \leq-\sup (-A)$, and so $-\sup (-A)$ is the greatest lower bound of $A$.


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 $r=r|w|=|r w|=|z|, z / r$


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 ||$x|-| y|| \leq|x-y|$.


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. The remaining component is then uniquely determined. Also, if $w$ is any non-zero solution of the first equation, there is a unique positive number $t$ such that $t$ w satisfies both equations. (For example, if $x_1 \neq y_1$, the first equation is satisfied whenever
$$
z_1=\frac{z_2\left(x_2-y_2\right)+\cdots+z_k\left(x_k-y_k\right)}{y_1-x_1} .
$$
If $\left(z_1, z_2, \ldots, z_k\right)$ satisfies this equation, so does $\left(t z_1, t z_2, \ldots, t z_k\right)$ for any real number $t$.) Since at least two of these components can vary independently, we can find a solution with these components having any prescribed ratio. This ratio does not change when we multiply by the positive number $t$ to obtain a solution of both equations. Since there are infinitely many ratios, there are infinitely many distinct solutions. For each such solution $\mathbf{w}$ the vector $\mathbf{z}=$ $\frac{1}{2} \mathrm{x}+\frac{1}{2} \mathrm{y}+\mathrm{w}$ is a solution of the required equation. For
$$
\begin{aligned}
|\mathrm{z}-\mathbf{x}|^2 &=\left|\frac{\mathbf{y}-\mathbf{x}}{2}+\mathbf{w}\right|^2 \\
&=\left|\frac{\mathbf{y}-\mathbf{x}}{2}\right|^2+2 \mathbf{w} \cdot \frac{\mathbf{x}-\mathbf{y}}{2}+|\mathbf{w}|^2 \\
&=\frac{d^2}{4}+0+r^2-\frac{d^2}{4} \\
&=r^2
\end{aligned}
$$
and a similar relation holds for $|z-y|^2$.


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 $k=1$, since the product of two nonzero real numbers is nonzero.


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 the infinite set, hence that no point could be a limit point of this set, contrary to hypothesis. Hence $X$ is totally bounded, i.e., for each $\delta>0$ there is a finite set $x_1, \ldots, x_{N\delta}$such that $X=\bigcup_{j / 1}^{N\delta} B_\delta\left(x_j\right)$

Let $x_{n_1}, \ldots, x_{n N_n}$ be such that $X=\bigcup_{j / 1}^{N_n} B_{\frac{1}{n}}\left(x_{n j}\right), n=1,2, \ldots$ We claim that $\left\{x_{n j}: 1 \leq j \leq N_n ; n=1,2, \ldots\right\}$ is a countable dense subset of $X$. Indeed
25
if $x \in X$ and $\delta>0$, then $x \in B_{\frac{1}{n}}\left(x_{n j}\right)$ for some $x_{n j}$ for some $n>\frac{1}{\delta}$, and hence $d\left(x, x_{n j}\right)<\delta$. By definition, this means that $\left\{x_{n j}\right\}$ is dense in $X$.


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.


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 the real numbers, the set $P$ must be empty if $E$ is countable. The at-most countable set $E \backslash P$ cannot be perfect, hence must have isolated points if it is nonempty.


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 $\left\{\left|s_n\right|\right\}$ is also a Cauchy sequence, and therefore must converge.


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 that is bounded above (by 2 ) and so converges. Since the limit $s$ satisfies $s^2-s-2=0$, it follows that the limit is 2.


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 $\left(p=\frac{1}{2}\right)$.


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 $M$ be an upper bound for both $\left|b_n\right|$ and $\left|S_n\right|$, and let $S=\sum a_n$ and $b=\lim b_n$. Choose $N$ so large that the following three inequalities hold for all $m>N$ and $n>N$ :
$$
\left|b_n S_n-b S\right|<\frac{\varepsilon}{3} ; \quad\left|b_m S_m-b S\right|<\frac{\varepsilon}{3} ; \quad\left|b_m-b_n\right|<\frac{\varepsilon}{3 M} .
$$
Then if $n>m>N$, we have, from the formula for summation by parts,
$$
\sum_{k=m+1}^n a_n b_n=b_n S_n-b_m S_m+\sum_{k=m}^{n-1}\left(b_k-b_{k+1}\right) S_k
$$
Our assumptions yield immediately that $\left|b_n S_n-b_m S_m\right|<\frac{2 \varepsilon}{3}$, and
$$
\left|\sum_{k=m}^{n-1}\left(b_k-b_{k+1}\right) S_k\right| \leq M \sum_{k=m}^{n-1}\left|b_k-b_{k+1}\right| .
$$
Since the sequence $\left\{b_n\right\}$ is monotonic, we have
$$
\sum_{k=m}^{n-1}\left|b_k-b_{k+1}\right|=\left|\sum_{k=m}^{n-1}\left(b_k-b_{k+1}\right)\right|=\left|b_m-b_n\right|<\frac{\varepsilon}{3 M},
$$
from which the desired inequality follows.


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
$$
d\left(p_n, p\right) \leq d\left(p_n, p_{n_{N+1}}\right)+d\left(p_{n_{N+1}}, p\right)<\varepsilon
$$
For the first term on the right is less than $\frac{\varepsilon}{2}$ since $n>N_1$ and $n_{N+1}>N+1>$ $N_1$. The second term is less than $\frac{\varepsilon}{2}$ by the choice of $N$.


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 $U \backslash F_1$ is open, there exists $r_1>0$ such that $B_1$, defined as the open ball of radius $r_1$ about $x_1$, is contained in $U \backslash F_1$. Let $E_1$ be the open ball of radius $\frac{r_1}{2}$ about $x_1$, so that the closure of $E_1$ is contained in $B_1$. Now $F_2$ does not contain $E_1$, and so we can find a point $x_2 \in E_1 \backslash F_2$. Since $E_1 \backslash F_2$ is an open set, there exists a positive number $r_2$ such that $B_2$, the open ball of radius $R_2$ about $x_2$, is contained in $E_1 \backslash F_2$, which in turn is contained in $U \backslash\left(F_1 \cup F_2\right)$. We let $E_2$ be the open ball of radius $\frac{r_2}{2}$ about $x_2$, so that $\bar{E}_2 \subseteq B_2$. Proceeding in this way, we construct a sequence of open balls $E_j$, such that $E_j \supseteq \bar{E}_{j+1}$, and the diameter of $E_j$ tends to zero. By the previous exercise, there is a point $x$ belonging to all the sets $\bar{E}_j$, hence to all the sets $U \backslash\left(F_1 \cup F_2 \cup \cdots \cup F_n\right)$. Thus the point $x$ belongs to $U \cap\left(\cap_1^{\infty} G_n\right)$.


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 any neighborhood of $f(x)$, it follows that $f(x) \in \overline{f(E)}$


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 $f(X)=f(\bar{E}) \subseteq \overline{f(E)}$.


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 $b$ (if these last two points exist). Define $g(x)$ to be $f(b)$ for $x>b$ and $f(a)$ for $x<a$ if $a$ and $b$ exist. On the interval $\left(a_k, b_k\right)$ let
$$
g(x)=f\left(a_k\right)+\frac{x-a_k}{b_k-a_k}\left(f\left(b_k\right)-f\left(a_k\right)\right) .
$$
Of course we let $g(x)=f(x)$ for $x \in E$. It is now fairly clear that $g(x)$ is continuous. A rigorous proof proceeds as follows. Let $\varepsilon>0$. To choose $\delta>0$ such that $|x-u|<\delta$ implies $|g(x)-g(u)|<\varepsilon$, we consider three cases.
i. If $x>b$, let $\delta=x-b$. Then if $|x-u|<\delta$, it follows that $u>b$ also, so that $g(u)=f(b)=g(x)$, and $|g(u)-g(x)|=0<\varepsilon$. Similarly if $x<a$, let $\delta=a-x$
ii. If $a_k<x<b_k$ and $f\left(a_k\right)=f\left(b_k\right)$, let $\delta=\min \left(x-a_k, b_k-x\right)$. Since $|x-u|<\delta$ implies $a_k<u<b_k$, so that $g(u)=f\left(a_k\right)=f\left(b_k\right)=g(x)$, we again have $|g(x)-g(u)|=0<\varepsilon$. If $a_k<x<b_k$ and $f\left(a_k\right) \neq f\left(b_k\right)$, let $\delta=\min \left(x-a_k, b_k-x, \frac{\left(b_k-a_k\right) \varepsilon}{\left|f\left(b_k\right)-f\left(a_k\right)\right|}\right)$. Then if $|x-u|<\delta$, we again have $a_k<u<b_k$ and so
$$
|g(x)-g(u)|=\frac{|x-u|}{b_k-a_k}\left|f\left(b_k\right)-f\left(a_k\right)\right|<\varepsilon .
$$
iii. If $x \in E$, let $\delta_1$ be such that $|f(u)-f(x)|<\varepsilon$ if $u \in E$ and $|x-u|<\delta_1$. (Subcase a). If there are points $x_1 \in E \cap\left(x-\delta_1, x\right)$ and $x_2 \in E \cap\left(x, x+\delta_1\right)$, let $\delta=\min \left(x-x_1, x_2-x\right)$. If $|u-x|<\delta$ and $u \in E$, then $|f(u)-f(x)|<\varepsilon$ by definition of $\delta_1$. if $u \notin E$, then, since $x_1, x$, and $x_2$ are all in $E$, it follows that $u \in\left(a_k, b_k\right)$, where $a_k \in E, b_k \in E$, and $\left|a_k-x\right|<\delta$ and $\left|b_k-x\right|<\delta$, so that $\left|f\left(a_k\right)-f(x)\right|<\varepsilon$ and $\left|f\left(b_k\right)-f(x)\right|<\varepsilon$. If $f\left(a_k\right)=f\left(b_k\right)$, then $f(u)=f\left(a_k\right)$ also, and we have $|f(u)-f(x)|<\varepsilon$. If $f\left(a_k\right) \neq f\left(b_k\right)$, then
$$
\begin{aligned}
|f(u)-f(x)| & =\left|f\left(a_k\right)-f(x)+\frac{u-a_k}{b_k-a_k}\left(f\left(b_k\right)-f\left(a_k\right)\right)\right| \\
& =\left|\frac{b_k-u}{b_k-a_k}\left(f\left(a_k\right)-f(x)\right)+\frac{u-a_k}{b_k-a_k}\left(f\left(b_k\right)-f(x)\right)\right| \\
& <\frac{b_k-u}{b_k-a_k} \varepsilon+\frac{u-a_k}{b_k-a_k} \varepsilon \\
& =\varepsilon
\end{aligned}
$$
(Subcase b). Suppose $x_2$ does not exist, i.e., either $x=a_k$ or $x=a_k$ and $b_k>a_k+\delta_1$. Let us consider the second of these cases and show how to get $|f(u)-f(x)|<\varepsilon$ for $x<u<x+\delta$. If $f\left(a_k\right)=f\left(b_k\right)$, let $\delta=\delta_1$. If $u>x$ we have $a_k<u<b_k$ and $f(u)=f\left(a_k\right)=f(x)$. If $f\left(a_k\right) \neq f\left(b_k\right)$, let $\delta=$ $\min \left(\delta_1, \frac{\left(b_k-a_k\right) \varepsilon}{\left|f\left(b_k\right)-f\left(a_k\right)\right|}\right)$. Then, just as in Subcase a, we have $|f(u)-f(x)|<\varepsilon$.
The case when $x=b_k$ for some $k$ and $a_k<x-\delta_1$ is handled in exactly the same way.


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 a mapping of $E$ into $E \times Y$.

We claim that the mapping $\varphi$ is continuous if $f$ is continuous. Indeed, let $x \in X$ and $\varepsilon>0$ be given. Choose $\eta>0$ so that $d_Y(f(x), f(u))<\frac{\varepsilon}{2}$ if $d_E(x, y)<\eta$. Then let $\delta=\min \left(\eta, \frac{\varepsilon}{2}\right)$. It is easy to see that $\rho(\varphi(x), \varphi(u))<\varepsilon$ if $d_E(x, u)<\delta$. Conversely if $\varphi$ is continuous, it is obvious from the inequality $\rho(\varphi(x), \varphi(u)) \geq d_Y(f(x), f(u))$ that $f$ is continuous.

From these facts we deduce immediately that the graph of a continuous function $f$ on a compact set $E$ is compact, being the image of $E$ under the continuous mapping $\varphi$. Conversely, if $f$ is not continuous at some point $x$, there is a sequence of points $x_n$ converging to $x$ such that $f\left(x_n\right)$ does not converge to $f(x)$. If no subsequence of $f\left(x_n\right)$ converges, then the sequence $\left\{\left(x_n, f\left(x_n\right)\right\}_{n=1}^{\infty}\right.$ has no convergent subsequence, and so the graph is not compact. If some subsequence of $f\left(x_n\right)$ converges, say $f\left(x_{n_k}\right) \rightarrow z$, but $z \neq f(x)$, then the graph of $f$ fails to contain the limit point $(x, z)$, and hence is not closed. A fortiori it is not compact.


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.


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 $d_Z(g(u), g(v))<\varepsilon$ if $d_Y(u, v)<\eta$. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d_Y(f(x), f(y))<\eta$ if $d_X(x, y)<\delta$

It is then obvious that $d_Z(g(f(x)), g(f(y)))<\varepsilon$ if $d_X(x, y)<\delta$, so that $g \circ f$ is uniformly continuous.


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.


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 $\lambda=1$ ). In the case $n=1$ we have $\lambda=0$ or $\lambda=1$ or $\lambda=\frac{1}{2}$. The first two cases are again trivial, and the third is precisely the hypothesis of the theorem. Suppose the result is proved for $n \leq r$, and consider $\lambda=\frac{k}{2^{r+1}}$. If $k$ is even, say $k=2 l$, then $\frac{k}{2^{r+1}}=\frac{l}{2^r}$, and we can appeal to the induction hypothesis. Now suppose $k$ is odd. Then $1 \leq k \leq 2^{r+1}-1$, and so the numbers $l=\frac{k-1}{2}$ and $m=\frac{k+1}{2}$ are integers with $0 \leq l<m \leq 2^r$. We can now write
$$
\lambda=\frac{s+t}{2},
$$
where $s=\frac{k-1}{2^{r+1}}=\frac{l}{2^r}$ and $t=\frac{k+1}{2^{r+1}}=\frac{m}{2^r}$. We then have
$$
\lambda x+(1-\lambda) y=\frac{[s x+(1-s) y]+[t x+(1-t) y]}{2}
$$
Hence by the hypothesis of the theorem and the induction hypothesis we have
$$
\begin{aligned}
f(\lambda x+(1-\lambda) y) & \leq \frac{f(s x+(1-s) y)+f(t x+(1-t) y)}{2} \\
& \leq \frac{s f(x)+(1-s) f(y)+t f(x)+(1-t) f(y)}{2} \\
&=\left(\frac{s+t}{2}\right) f(x)+\left(1-\frac{s+t}{2}\right) f(y) \\
&=\lambda f(x)+(1-\lambda) f(y)
\end{aligned}
$$
This completes the induction.
Now for each fixed $x$ and $y$ both sides of the inequality
$$
f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)
$$
are continuous functions of $\lambda$. Hence the set on which this inequality holds (the inverse image of the closed set $[0, \infty)$ under the mapping $\lambda \mapsto \lambda f(x)+(1-$ $\lambda) f(y)-f(\lambda x+(1-\lambda) y))$ is a closed set. Since it contains all the points $\frac{k}{2^n}$, $0 \leq k \leq n, n=1,2, \ldots$, it must contain the closure of this set of points, i.e., it must contain all of $[0,1]$. Thus $f$ is convex.


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 $f(x)=y$ and $f(x+h)=y+k$, we have
$$
\frac{g(y+k)-g(y)}{k}-\frac{1}{f^{\prime}(x)}=\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac{1}{f^{\prime}(x)}
$$
Since we know $\lim \frac{1}{\varphi(t)}=\frac{1}{\lim \varphi(t)}$ provided $\lim \varphi(t) \neq 0$, it follows that for any $\varepsilon>0$ there exists $\eta>0$ such that
$$
\left|\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac{1}{f^{\prime}(x)}\right|<\varepsilon
$$
if $0<|h|<\eta$. Since $h=g(y+k)-g(y)$, there exists $\delta>0$ such that $0<|h|<\eta$ if $0<|k|<\delta$. The proof is now complete.


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 $p^{\prime}(x)=0$ for some $x$ between 0 and 1 .


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
$$
g^{\prime}(x)=\frac{x f^{\prime}(x)-f(x)}{x^2}>0,
$$
so that $g$ is also monotonically increasing.


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 ∈ (Set.Ioi a)})
    (hM0 : M0 = sSup {(|f x|) | x ∈ (Set.Ioi a)})
    (hM1 : M1 = sSup {(|deriv f x|) | x ∈ (Set.Ioi a)})
    (hM2 : M2 = sSup {(|deriv (deriv f) x|) | x ∈ (Set.Ioi a)}) :
    (M1 ^ 2) ≤ 4 * M0 * M2 := by
  sorry

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 function, and the only bounded linear function is a constant, whose derivative is zero. Hence we shall assume from now on that $0<M_2<+\infty$ and $0<M_0<+\infty$.
Following the hint, we need only choose $h=\sqrt{\frac{M_0}{M_2}}$, and we obtain
$$
\left|f^{\prime}(x)\right| \leq 2 \sqrt{M_0 M_2},
$$
which is precisely the desired inequality.
The case of equality follows, since the example proposed satisfies
$$
f(x)=1-\frac{2}{x^2+1}
$$
for $x \geq 0$. We see easily that $|f(x)| \leq 1$ for all $x>-1$. Now $f^{\prime}(x)=\frac{4 x}{\left(x^2+1\right)^2}$ for $x>0$ and $f^{\prime}(x)=4 x$ for $x<0$. It thus follows from Exercise 9 above that $f^{\prime}(0)=0$, and that $f^{\prime}(x)$ is continuous. Likewise $f^{\prime \prime}(x)=4$ for $x<0$ and $f^{\prime \prime}(x)=\frac{4-4 x^2}{\left(x^2+1\right)^3}=-4 \frac{x^2-1}{\left(x^2+1\right)^3}$. This shows that $\left|f^{\prime \prime}(x)\right|<4$ for $x>0$ and also that $\lim _{x \rightarrow 0} f^{\prime \prime}(x)=4$. Hence Exercise 9 again implies that $f^{\prime \prime}(x)$ is continuous and $f^{\prime \prime}(0)=4$.

On $n$-dimensional space let $\mathbf{f}(x)=\left(f_1(x), \ldots, f_n(x)\right), M_0=\sup |\mathbf{f}(x)|$, $M_1=\sup \left|\mathbf{f}^{\prime}(x)\right|$, and $M_2=\sup \left|\mathbf{f}^{\prime \prime}(x)\right|$. Just as in the numerical case, there is nothing to prove if $M_2=0$ or $M_0=+\infty$ or $M_2=+\infty$, and so we assume $0<M_0<+\infty$ and $0<M_2<\infty$. Let $a$ be any positive number less than $M_1$, let $x_0$ be such that $\left|\mathbf{f}^{\prime}\left(x_0\right)\right|>a$, and let $\mathbf{u}=\frac{1}{\left|\mathbf{f}^{\prime}\left(x_0\right)\right|} \mathbf{f}^{\prime}\left(x_0\right)$. Consider the real-valued function $\varphi(x)=\mathrm{u} \cdot \mathrm{f}(x)$. Let $N_0, N_1$, and $N_2$ be the suprema of $|\varphi(x)|,\left|\varphi^{\prime}(x)\right|$, and $\left|\varphi^{\prime \prime}(x)\right|$ respectively. By the Schwarz inequality we have (since $|\mathbf{u}|=1) N_0 \leq M_0$ and $N_2 \leq M_2$, while $N_1 \geq \varphi\left(x_0\right)=\left|\mathbf{f}^{\prime}\left(x_0\right)\right|>a$. We therefore have $a^2<4 N_0 N_2 \leq 4 M_0 M_2$. Since $a$ was any positive number less than $M_1$, we have $M_1^2 \leq 4 M_0 M_2$, i.e., the result holds also for vector-valued functions.

Equality can hold on any $R^n$, as we see by taking $\mathbf{f}(x)=(f(x), 0, \ldots, 0)$ or $\mathbf{f}(x)=(f(x), f(x), \ldots, f(x))$, where $f(x)$ is a real-valued function for which equality holds.


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 $\Omega$,
$$
f^{\prime}(z)=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=0+0=0 .
$$
3
Thus $f(z)$ is constant.


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,
$$
\left\{\begin{array}{l}
0=\frac{\partial\left(u^2+v^2\right)}{\partial x}=2 u \frac{\partial u}{\partial x}+2 v \frac{\partial v}{\partial x} . \\
0=\frac{\partial\left(u^2+v^2\right)}{\partial y}=2 u \frac{\partial u}{\partial y}+2 v \frac{\partial v}{\partial y} .
\end{array}\right.
$$
Plug in the Cauchy-Riemann equations and we get
$$
\begin{gathered}
u \frac{\partial v}{\partial y}+v \frac{\partial v}{\partial x}=0 \\
-u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=0 \\
(1.14) \Rightarrow \frac{\partial v}{\partial x}=\frac{v}{u} \frac{\partial v}{\partial y}
\end{gathered}
$$
Plug (1.15) into (1.13) and we get
$$
\frac{u^2+v^2}{u} \frac{\partial v}{\partial y}=0 .
$$
So $u^2+v^2=0$ or $\frac{\partial v}{\partial y}=0$.

If $u^2+v^2=0$, then, since $u, v$ are real, $u=v=0$, and thus $f=0$ which is constant.

Thus we may assume $u^2+v^2$ equals a non-zero constant, and we may divide by it. We multiply both sides by $u$ and find $\frac{\partial v}{\partial y}=0$, then by (1.15), $\frac{\partial v}{\partial x}=0$, and by Cauchy-Riemann, $\frac{\partial u}{\partial x}=0$.
$$
f^{\prime}=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=0 .
$$
Thus $f$ is constant.
Why is the above only mostly a proof? The problem is we have a division by $u$, and need to make sure everything is well-defined. Specifically, we need to know that $u$ is never zero. We do have $f^{\prime}=0$ except at points where $u=0$, but we would need to investigate that a bit more.
Let's return to
$$
\left\{\begin{array}{l}
0=\frac{\partial\left(u^2+v^2\right)}{\partial x}=2 u \frac{\partial u}{\partial x}+2 v \frac{\partial v}{\partial x} . \\
0=\frac{\partial\left(u^2+v^2\right)}{\partial y}=2 u \frac{\partial u}{\partial y}+2 v \frac{\partial v}{\partial y} .
\end{array}\right.
$$
Plug in the Cauchy-Riemann equations and we get
$$
\begin{array}{r}
u \frac{\partial v}{\partial y}+v \frac{\partial v}{\partial x}=0 \\
-u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=0 .
\end{array}
$$
We multiply the first equation $u$ and the second by $v$, and obtain
$$
\begin{aligned}
u^2 \frac{\partial v}{\partial y}+u v \frac{\partial v}{\partial x} & =0 \\
-u v \frac{\partial v}{\partial x}+v^2 \frac{\partial v}{\partial y} & =0 .
\end{aligned}
$$
Adding the two yields
$$
u^2 \frac{\partial v}{\partial y}+v^2 \frac{\partial v}{\partial y}=0,
$$
or equivalently
$$
\left(u^2+v^2\right) \frac{\partial v}{\partial y}=0 .
$$
We now argue in a similar manner as before, except now we don't have the annoying $u$ in the denominator. If $u^2+v^2=0$ then $u=v=0$, else we can divide by $u^2+v^2$ and find $\partial v / \partial y=0$. Arguing along these lines finishes the proof.


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 $p=2$.)


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.


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 $k \geq 1$,
$$
\varphi^k(z)=z+k a_n\left(z-z_0\right)^n+O\left(\left(z-z_0\right)^{n+1}\right) .
$$
Let $r>0$ be such that when $\left|z-z_0\right| \leq r$, then $z \in \Omega$. Then by (1),
$$
k a_n=\frac{1}{2 \pi i} \int_{\left|z-z_0\right|=r} \frac{\varphi^k(z)}{\left(z-z_0\right)^{n+1}} d z .
$$
Since $\varphi^k(\Omega) \subset \Omega$ and since $\Omega$ is bounded, there exists $M>0$, independent of $k$, such that $\left|\varphi^k\right| \leq M$ on $\Omega$. Then by (2),
$$
k\left|a_n\right| \leq M r^{-n} .
$$
Since $k$ is arbitrary, $a_n=0$, a contradiction.


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 $\lim _{x \rightarrow i a} \frac{e^{i x}}{x^2+a^2}(x-i a)=\frac{e^{-a}}{2 i a}$, which multiplied by $2 \pi i$ gives the answer.


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
$$
\begin{aligned}
\int_0^1 f(z) d z & =\log (-2 i)+\frac{i \pi}{2}+\int_0^1 \log (\sin (\pi z)) d z \\
& =\int_0^1 \log (\sin (\pi z)) d z+\log (-2 i)+\log (i) \\
& =\log (2)+\int_0^1 \log (\sin (\pi z)) d z
\end{aligned}
$$
Now it suffices to show that $\int_0^1 f(z) d z=0$. Consider the contour $C(\epsilon, R)$ (which is the contour given in your question) given by the following.
1. $C_1(\epsilon, R)$ : The vertical line along the imaginary axis from $i R$ to $i \epsilon$.
2. $C_2(\epsilon)$ : The quarter turn of radius $\epsilon$ about 0 .
3. $C_3(\epsilon)$ : Along the real axis from $(\epsilon, 1-\epsilon)$.
4. $C_4(\epsilon)$ : The quarter turn of radius $\epsilon$ about 1 .
5. $C_5(\epsilon, R)$ : The vertical line from $1+i \epsilon$ to $1+i R$.
6. $C_6(R)$ : The horizontal line from $1+i R$ to $i R$.
$f(z)$ is analytic inside the contour $C$ and hence $\oint_C f(z)=0$. This gives us
$$
\begin{aligned}
\int_{C_1(\epsilon, R)} f d z+\int_{C_2(\epsilon)} f d z+\int_{C_3(\epsilon)} f d z+\int_{C_4(\epsilon)} f d z+\int_{C_5(\epsilon, R)} f d z+\int_{C_6(R)} f d z \\
=0
\end{aligned}
$$
Now the integral along 1 cancels with the integral along 5 due to symmetry. Integrals along 2 and 4 scale as $\epsilon \log (\epsilon)$. Integral along 6 goes to 0 as $R \rightarrow \infty$. This gives us
$$
\lim _{\epsilon \rightarrow 0} \int_{C_3(\epsilon)} f d z=0
$$
which is what we need.


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.


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


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


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 $a=b$.


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 $(v+F u) \cap W_j(j \neq i)$ contains at most one vector since otherwise $W_j$ would contain $u$. Hence
$$
|v+F u|=|F| \leq n-1 .
$$
Corollary: Avoidance lemma for vector spaces.
Let $E$ 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.


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 one group of order $q$ and again we are done.


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 $G \rightarrow S_7$ which must be injective if $G$ is simple (why?).

But this cannot be since then we would embed $G$ into $S_7$, which is impossible since $|G| \nmid 7 !=\left|S_7\right|$ (why?)


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


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 is in $IJ$ because $k \in J$, and the second term is in $IJ$ because $k \in I$, so $k \in IJ$ as desired.


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 $z=i b$. 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 $I$.
3. 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,
$$
(a+b i)^2=a^2-b^2+2 a b i \in I
$$
Furthermore, since $-2 a \in \mathbb{Z}[i]$, and $z \in I$ and $I$ is an ideal, $-2 a z \in I$; that is,
$$
-2 a z=-2 a(a+b i)=-2 a^2-2 a b i \in I
$$
Since $I$ is closed under addition,
$$
\left(a^2-b^2+2 a b i\right)+\left(-2 a^2-2 a b i\right) \in I \Longrightarrow-a^2-b^2 \in I
$$
Notice 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 $I$.


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 $y=0$.


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.


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 $x \in \mathbb{F}_31$.


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 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.