id stringlengths 17 29 | nl_statement stringlengths 43 377 | nl_proof stringlengths 0 3.81k | lean4_src_header stringclasses 14
values | lean4_formalization stringlengths 50 426 |
|---|---|---|---|---|
Ireland-Rosen|exercise_4_5 | Consider a prime $p$ of the form $4 t+3$. Show that $a$ is a primitive root modulo $p$ iff $-a$ has order $(p-1) / 2$. | \begin{proof}
Let $a$ a primitive root modulo $p$.
As $a^{p-1} \equiv 1(\bmod p), p \mid\left(a^{(p-1) / 2}-1\right)\left(a^{(p-1) / 2}+1\right)$, so $p \mid a^{(p-1) / 2}-1$ or $p \mid$ $a^{(p-1) / 2}+1$. As $a$ is a primitive root modulo $p, a^{(p-1) / 2} \not \equiv 1(\bmod p)$, so
$$
a^{(p-1) / 2} \equiv-1 ... | import Mathlib
open Real
open scoped BigOperators
| theorem exercise_4_5 {p t : ℕ} (hp0 : p.Prime) (hp1 : p = 4*t + 3)
(a : ZMod p) :
IsPrimitiveRoot a p ↔ ((-a) ^ ((p-1)/2) = 1 ∧ ∀ (k : ℕ), (0 < k ∧ k < (p-1)/2) → (-a)^k ≠ 1) := |
Ireland-Rosen|exercise_4_8 | Let $p$ be an odd prime. Show that $a$ is a primitive root modulo $p$ iff $a^{(p-1) / q} \not \equiv 1(p)$ for all prime divisors $q$ of $p-1$. | \begin{proof}
$\bullet$ If $a$ is a primitive root, then $a^k \not \equiv 1$ for all $k, 1\leq k < p-1$, so $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$.
$\bullet$ In the other direction, suppose $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$.
Let $\del... | import Mathlib
open Real
open scoped BigOperators
| theorem exercise_4_8 {p : ℕ} {a : ZMod p} (hp : Odd p) (hp' : p.Prime) :
IsPrimitiveRoot a p ↔ (∀ q : ℕ, q ∣ (p-1) → q.Prime → ¬ a^((p-1)/q) = 1) := |
Ireland-Rosen|exercise_5_13 | Show that any prime divisor of $x^{4}-x^{2}+1$ is congruent to 1 modulo 12 . | \begin{proof}
\newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
$\bullet$ As $a^6 +1 = (a^2+1)(a^4-a^2+1)$, $p\mid a^4 - a^2+1$ implies $p \mid a^6 + 1$, so $\legendre{-1}{p} = 1$ and $p\equiv 1 \pmod 4$.
$\bullet$ $p \mid 4a^4 - 4 a^2 +4 = (2a-1)^2 + 3$, so $\legendre{-3}{p} = 1$.
As $-3 \equiv 1 \pmod... | import Mathlib
open Real
open scoped BigOperators
| theorem exercise_5_13 {p x: ℤ} (hp : Prime p)
(hpx : p ∣ (x^4 - x^2 + 1)) : p ≡ 1 [ZMOD 12] := |
Ireland-Rosen|exercise_5_37 | Show that if $a$ is negative then $p \equiv q(4 a)$ together with $p\not | a$ imply $(a / p)=(a / q)$. | \begin{proof}
\newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
Write $a = -A, A>0$. As $p \equiv q \pmod {4a}$, we know from Prop. 5.3.3. (b) that $(A/p) = (A/q)$.
Moreover,
\begin{align*}
\legendre{a}{p}&= \legendre{-A}{p} = (-1)^{(p-1)/2} \legendre{A}{p}\\
\legendre{a}{q}&= \legendre{-A}{q} = (-1^{(q... | import Mathlib
open Real
open scoped BigOperators
| theorem exercise_5_37 {p q : ℕ} [Fact (p.Prime)] [Fact (q.Prime)] {a : ℤ}
(ha : a < 0) (h0 : p ≡ q [ZMOD 4*a]) (h1 : ¬ ((p : ℤ) ∣ a)) :
legendreSym p a = legendreSym q a := |
Ireland-Rosen|exercise_18_4 | Show that 1729 is the smallest positive integer expressible as the sum of two different integral cubes in two ways. | \begin{proof}
Let $n=a^3+b^3$, and suppose that $\operatorname{gcd}(a, b)=1$. If a prime $p \mid a^3+b^3$, then
$$
\left(a b^{-1}\right)^3 \equiv_p-1
$$
Thus $3 \mid \frac{p-1}{2}$, that is, $p \equiv_6 1$.
If we have $n=a^3+b^3=c^3+d^3$, then we can factor $n$ as
$$
\begin{aligned}
& n=(a+b)\left(a^2-a b+... | import Mathlib
open Real
open scoped BigOperators
| theorem exercise_18_4 : 1729 = sInf (fun (n : ℕ) => ∃ x y z w : ℤ,
x^3 + y^3 = n ∧ z^3 + w^3 = n ∧ x ≠ z ∧ x ≠ w ∧ y ≠ z ∧ y ≠ w) := |
Munkres|exercise_13_1 | Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x \in A$ there is an open set $U$ containing $x$ such that $U \subset A$. Show that $A$ is open in $X$. | \begin{proof}
Since, from the given hypothesis given any $x \in A$ there exists an open set containing $x$ say, $U_x$ such that $U_x \subset A$. Thus, we claim that
$$
A=\bigcup_{x \in A} U_x
$$
Observe that if we prove the above claim, then $A$ will be open, being a union of arbitrary open sets. Since, for ea... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_13_1 (X : Type*) [TopologicalSpace X] (A : Set X)
(h1 : ∀ x ∈ A, ∃ U : Set X, x ∈ U ∧ IsOpen U ∧ U ⊆ A) :
IsOpen A := |
Munkres|exercise_13_4a1 | If $\mathcal{T}_\alpha$ is a family of topologies on $X$, show that $\bigcap \mathcal{T}_\alpha$ is a topology on $X$. | \begin{proof}
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}_\a... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| def is_topology (X : Type*) (T : Set (Set X)) :=
univ ∈ T ∧
(∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧
(∀s, (∀t ∈ s, t ∈ T) → sUnion s ∈ T)
theorem 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) := |
Munkres|exercise_13_4b1 | 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 $\mathcal{T}_\alpha$. | \begin{proof}
(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 $\m... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| def is_topology (X : Type*) (T : Set (Set X)) :=
univ ∈ T ∧
(∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧
(∀s, (∀t ∈ s, t ∈ T) → sUnion s ∈ T)
theorem 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 ⊆... |
Munkres|exercise_13_5a | 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 $\mathcal{A}$. | \begin{proof}
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... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| def is_topology (X : Type*) (T : Set (Set X)) :=
univ ∈ T ∧
(∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧
(∀s, (∀t ∈ s, t ∈ T) → sUnion s ∈ T)
theorem exercise_13_5a {X : Type*}
[TopologicalSpace X] (A : Set (Set X)) (hA : IsTopologicalBasis A) :
generateFrom A = generateFrom (sInter {T | is_topology X T ∧ A ⊆ T}) := |
Munkres|exercise_13_6 | Show that the lower limit topology $\mathbb{R}_l$ and $K$-topology $\mathbb{R}_K$ are not comparable. | \begin{proof}
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... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| def lower_limit_topology (X : Type) [Preorder X] :=
generateFrom {S : Set X | ∃ a b, a < b ∧ S = Ico a b}
def Rl := lower_limit_topology ℝ
def K : Set ℝ := {r | ∃ n : ℕ, r = 1 / n}
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})
theorem ex... |
Munkres|exercise_13_8b | 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 $\mathbb{R}$. | \begin{proof}
(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 gr... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| def lower_limit_topology (X : Type) [Preorder X] :=
generateFrom {S : Set X | ∃ a b, a < b ∧ S = Ico a b}
theorem exercise_13_8b (T : Set (Set ℝ))
(hT : T = {S : Set ℝ | ∃ a b : ℚ, a < b ∧ S = Ioo ↑a ↑b}) :
IsTopologicalBasis T ∧
generateFrom T ≠
lower_limit_topology ℝ := |
Munkres|exercise_16_4 | 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. | \begin{proof}
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... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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 := |
Munkres|exercise_17_4 | 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 $X$. | \begin{proof}
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 $X$.
\end{proof} | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_17_4 {X : Type*} [TopologicalSpace X]
(U A : Set X) (hU : IsOpen U) (hA : IsClosed A) :
IsOpen (U \ A) ∧ IsClosed (A \ U) := |
Munkres|exercise_18_8b | 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. | \begin{proof}
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 fo... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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)) := |
Munkres|exercise_19_6a | 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 $\alpha$. | \begin{proof}
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 produ... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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)) := |
Munkres|exercise_21_6a | 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 $x \in[0,1]$. | \begin{proof}
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
\end{proof} | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_21_6a
(f : ℕ → Icc 0 1 → ℝ)
(h : ∀ x n, f n x = ↑x ^ n) :
∀ x, ∃ y, Tendsto (λ n => f n x) atTop (𝓝 y) := |
Munkres|exercise_21_8 | 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... | \begin{proof}
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 ... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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)) := |
Munkres|exercise_22_2b | 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. | \begin{proof}
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).
\end{proof} | import Mathlib
open Filter Set TopologicalSpace Topology
| theorem exercise_22_2b {X : Type*} [TopologicalSpace X]
{A : Set X} (r : X → A) (hr : Continuous r) (h : ∀ x : A, r x = x) :
QuotientMap r := |
Munkres|exercise_23_2 | 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. | \begin{proof}
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... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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) := |
Munkres|exercise_23_4 | Show that if $X$ is an infinite set, it is connected in the finite complement topology. | \begin{proof}
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.
\end{p... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
set_option checkBinderAnnotations false | theorem exercise_23_4 {X : Type*} [TopologicalSpace X] [CofiniteTopology X]
(s : Set X) : Infinite s → IsConnected s := |
Munkres|exercise_23_9 | 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. | \begin{proof}
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... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_23_9 {X Y : Type*}
[TopologicalSpace X] [TopologicalSpace Y]
(A₁ A₂ : Set X)
(B₁ B₂ : Set Y)
(hA : A₁ ⊂ A₂)
(hB : B₁ ⊂ B₂)
(hA : IsConnected A₂)
(hB : IsConnected B₂) :
IsConnected ({x | ∃ a b, x = (a, b) ∧ a ∈ A₂ ∧ b ∈ B₂} \
{x | ∃ a b, x = (a, b) ∧ a ∈ A₁ ∧ b ∈ B₁}) := |
Munkres|exercise_24_2 | Let $f: S^{1} \rightarrow \mathbb{R}$ be a continuous map. Show there exists a point $x$ of $S^{1}$ such that $f(x)=f(-x)$. | \begin{proof}
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 in... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_24_2 {f : (Metric.sphere 0 1 : Set ℝ) → ℝ}
(hf : Continuous f) : ∃ x, f x = f (-x) := |
Munkres|exercise_25_4 | Let $X$ be locally path connected. Show that every connected open set in $X$ is path connected. | \begin{proof}
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.
\end{proof} | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_25_4 {X : Type*} [TopologicalSpace X]
[LocPathConnectedSpace X] (U : Set X) (hU : IsOpen U)
(hcU : IsConnected U) : IsPathConnected U := |
Munkres|exercise_26_11 | 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. | \begin{proof}
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... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_26_11
{X : Type*} [TopologicalSpace X] [CompactSpace X] [T2Space X]
(A : Set (Set X)) (hA : ∀ (a b : Set X), a ∈ A → b ∈ A → a ⊆ b ∨ b ⊆ a)
(hA' : ∀ a ∈ A, IsClosed a) (hA'' : ∀ a ∈ A, IsConnected a) :
IsConnected (⋂₀ A) := |
Munkres|exercise_27_4 | Show that a connected metric space having more than one point is uncountable. | \begin{proof}
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}... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_27_4
{X : Type*} [MetricSpace X] [ConnectedSpace X] (hX : ∃ x y : X, x ≠ y) :
¬ Countable (univ : Set X) := |
Munkres|exercise_28_5 | 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. | \begin{proof}
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\}... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| def countably_compact (X : Type*) [TopologicalSpace X] :=
∀ U : ℕ → Set X,
(∀ i, IsOpen (U i)) ∧ ((univ : Set X) ⊆ ⋃ i, U i) →
(∃ t : Finset ℕ, (univ : Set X) ⊆ ⋃ i ∈ t, U i)
theorem exercise_28_5
(X : Type*) [TopologicalSpace X] :
countably_compact X ↔ ∀ (C : ℕ → Set X), (∀ n, IsClosed (C n)) ∧
(∀ n, C n ... |
Munkres|exercise_29_1 | Show that the rationals $\mathbb{Q}$ are not locally compact. | \begin{proof}
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}... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_29_1 : ¬ LocallyCompactSpace ℚ := |
Munkres|exercise_29_10 | 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 $\bar{V} \subset U$. | \begin{proof}
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 d... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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 := |
Munkres|exercise_30_13 | Show that if $X$ has a countable dense subset, every collection of disjoint open sets in $X$ is countable. | \begin{proof}
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 $U \in \mathcal{U}$ intesects $S$. 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_{... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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 := |
Munkres|exercise_31_2 | Show that if $X$ is normal, every pair of disjoint closed sets have neighborhoods whose closures are disjoint. | \begin{proof}
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 fo... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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 = ∅ := |
Munkres|exercise_32_1 | Show that a closed subspace of a normal space is normal. | \begin{proof}
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)... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_32_1 {X : Type*} [TopologicalSpace X]
(hX : NormalSpace X) (A : Set X) (hA : IsClosed A) :
NormalSpace {x // x ∈ A} := |
Munkres|exercise_32_2b | Show that if $\prod X_\alpha$ is regular, then so is $X_\alpha$. Assume that each $X_\alpha$ is nonempty. | \begin{proof}
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 $X_\alpha$ sat... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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) := |
Munkres|exercise_32_3 | Show that every locally compact Hausdorff space is regular. | \begin{proof}
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... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_32_3 {X : Type*} [TopologicalSpace X]
(hX : LocallyCompactSpace X) (hX' : T2Space X) :
RegularSpace X := |
Munkres|exercise_33_8 | 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 $f(B) = \{1\}$. | \begin{proof}
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... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem 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) :
∃ (f : X → Icc 0 1), Continuous f ∧ f '' A = {0} ∧ f '' B = {1} := |
Munkres|exercise_38_6 | Let $X$ be completely regular. Show that $X$ is connected if and only if the Stone-Čech compactification of $X$ is connected. | \begin{proof}
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}
\e... | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
| theorem exercise_38_6 {X : Type*} [TopologicalSpace X] [CompletelyRegularSpace X] :
IsConnected (univ : Set X) ↔ IsConnected (univ : Set (StoneCech X)) := |
Pugh|exercise_2_12a | 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. | \begin{proof}
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$ wh... | import Mathlib
open Filter Real Function
open scoped Topology
| theorem 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) := |
Pugh|exercise_2_29 | 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}$. | \begin{proof}
The bijection given by $x\mapsto X^C$ suffices.
\end{proof} | import Mathlib
open Filter Real Function
open scoped Topology
| theorem 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 := |
Pugh|exercise_2_41 | 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. | \begin{proof}
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 .
$$
Assuming (1) to be true, let us finish the problem. First let us show that $B$ is bounded... | import Mathlib
open Filter Real Function
open scoped Topology
| theorem exercise_2_41 (m : ℕ) [NormedSpace ℝ ((Fin m) → ℝ)] :
IsCompact (Metric.closedBall (0 : EuclideanSpace ℝ (Fin m)) 1) := |
Pugh|exercise_2_57 | Show that if $S$ is connected, it is not true in general that its interior is connected. | \begin{proof}
Consider $X=\mathbb{R}^2$ and
$$
A=([-2,0] \times[-2,0]) \cup([0,2] \times[0,2])
$$
which is connected, while $\operatorname{int}(A)$ is not connected.
To see this consider the continuous function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is defined by $f(x, y)=x+y$. Let $U=f^{-1}(0,+\infty)$ whi... | import Mathlib
open Filter Real Function
open scoped Topology
| theorem exercise_2_57 {X : Type*} [TopologicalSpace X]
: ∃ (S : Set X), IsConnected S ∧ ¬ IsConnected (interior S) := |
Pugh|exercise_2_126 | Suppose that $E$ is an uncountable subset of $\mathbb{R}$. Prove that there exists a point $p \in \mathbb{R}$ at which $E$ condenses. | \begin{proof}
I think this is the proof by contrapositive that you were getting at.
Suppose that $E$ has no limit points at all. Pick an arbitrary point $x \in E$. Then $x$ cannot be a limit point, so there must be some $\delta>0$ such that the ball of radius $\delta$ around $x$ contains no other points of $E$ :
... | import Mathlib
open Filter Real Function
open scoped Topology
| theorem exercise_2_126 {E : Set ℝ}
(hE : ¬ Set.Countable E) : ∃ (p : ℝ), ClusterPt p (𝓟 E) := |
Pugh|exercise_3_4 | Prove that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$. | \begin{proof}
$$
\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2 \sqrt{n}}
$$
\end{proof} | import Mathlib
open Filter Real Function
open scoped Topology
| theorem exercise_3_4 (n : ℕ) :
Tendsto (λ n => (sqrt (n + 1) - sqrt n)) atTop (𝓝 0) := |
Pugh|exercise_3_63b | Prove that $\sum 1/k(\log(k))^p$ diverges when $p \leq 1$. | \begin{proof}
Using the integral test, for a set $a$, we see
$$
\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}^... | import Mathlib
open Filter Real Function
open scoped Topology
| theorem exercise_3_63b (p : ℝ) (f : ℕ → ℝ) (hp : p ≤ 1)
(h : f = λ (k : ℕ) => (1 : ℝ) / (k * (log k) ^ p)) :
¬ ∃ l, Tendsto f atTop (𝓝 l) := |
Putnam|exercise_1998_a3 | Let $f$ be a real function on the real line with continuous third derivative. Prove that there exists a point $a$ such that
$f(a) \cdot f^{\prime}(a) \cdot f^{\prime \prime}(a) \cdot f^{\prime \prime \prime}(a) \geq 0$. | import Mathlib
open scoped BigOperators
| theorem 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 := | |
Putnam|exercise_2000_a2 | 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. | \begin{proof}
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.
\end{proof} | import Mathlib
open scoped BigOperators
| theorem exercise_2000_a2 :
∀ N : ℕ, ∃ n : ℕ, n > N ∧ ∃ i : Fin 6 → ℕ, n = (i 0)^2 + (i 1)^2 ∧
n + 1 = (i 2)^2 + (i 3)^2 ∧ n + 2 = (i 4)^2 + (i 5)^2 := |
Putnam|exercise_2010_a4 | Prove that for each positive integer $n$, the number $10^{10^{10^n}}+10^{10^n}+10^n-1$ is not prime. | \begin{proof}
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 h... | import Mathlib
open scoped BigOperators
| theorem exercise_2010_a4 (n : ℕ) (hn : n > 0) :
¬ Nat.Prime (10^10^10^n + 10^10^n + 10^n - 1) := |
Putnam|exercise_2017_b3 | 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. | \begin{proof}
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) = \... | import Mathlib
open scoped BigOperators
| theorem 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)) := |
Putnam|exercise_2018_b2 | 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 $\{z \in \mathbb{C}:|z| \leq 1\}$. | \begin{proof}
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 ... | import Mathlib
open scoped BigOperators
| theorem 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) := |
Putnam|exercise_2020_b5 | For $j \in\{1,2,3,4\}$, let $z_{j}$ be a complex number with $\left|z_{j}\right|=1$ and $z_{j} \neq 1$. Prove that $3-z_{1}-z_{2}-z_{3}-z_{4}+z_{1} z_{2} z_{3} z_{4} \neq 0 .$ | \begin{proof}
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_... | import Mathlib
open scoped BigOperators
| theorem 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 := |
Rudin|exercise_1_1a | If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $r+x$ is irrational. | \begin{proof}
If $r$ and $r+x$ were both rational, then $x=r+x-r$ would also be rational.
\end{proof} | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_1_1a
(x : ℝ) (y : ℚ) (hy : y ≠ 0) :
( Irrational x ) -> Irrational ( x + y ) := |
Rudin|exercise_1_2 | Prove that there is no rational number whose square is $12$. | \begin{proof}
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
$$
... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_1_2 : ¬ ∃ (x : ℚ), ( x ^ 2 = 12 ) := |
Rudin|exercise_1_5 | Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A=-\sup (-A)$. | \begin{proof}
We need to prove that $-\sup (-A)$ is the greatest lower bound of $A$. For brevity, let $\alpha=-\sup (-A)$. We need to show that $\alpha \leq x$ for all $x \in A$ and $\alpha \geq \beta$ if $\beta$ is any lower bound of $A$.
Suppose $x \in A$. Then, $-x \in-A$, and, hence $-x \leq \sup (-A)$. It ... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_1_5 (A minus_A : Set ℝ) (hA : A.Nonempty)
(hA_bdd_below : BddBelow A) (hminus_A : minus_A = {x | -x ∈ A}) :
sInf A = - sSup minus_A := |
Rudin|exercise_1_11a | If $z$ is a complex number, prove that there exists an $r\geq 0$ and a complex number $w$ with $| w | = 1$ such that $z = rw$. | \begin{proof}
If $z=0$, we take $r=0, w=1$. (In this case $w$ is not unique.) Otherwise we take $r=|z|$ and $w=z /|z|$, and these choices are unique, since if $z=r w$, we must have $r=r|w|=|r w|=|z|, z / r$
\end{proof} | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_1_11a (z : ℂ) :
∃ (r : ℝ) (w : ℂ), r ≥ 0 ∧ abs w = 1 ∧ z = r * w := |
Rudin|exercise_1_13 | If $x, y$ are complex, prove that $||x|-|y|| \leq |x-y|$. | \begin{proof}
Since $x=x-y+y$, the triangle inequality gives
$$
|x| \leq|x-y|+|y|
$$
so that $|x|-|y| \leq|x-y|$. Similarly $|y|-|x| \leq|x-y|$. Since $|x|-|y|$ is a real number we have either ||$x|-| y||=|x|-|y|$ or ||$x|-| y||=|y|-|x|$. In either case, we have shown that ||$x|-| y|| \leq|x-y|$.
\end{proof} | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_1_13 (x y : ℂ) :
|(abs x) - (abs y)| ≤ abs (x - y) := |
Rudin|exercise_1_16a | Suppose $k \geq 3, x, y \in \mathbb{R}^k, |x - y| = d > 0$, and $r > 0$. Prove that if $2r > d$, there are infinitely many $z \in \mathbb{R}^k$ such that $|z-x|=|z-y|=r$. | \begin{proof}
(a) Let w be any vector satisfying the following two equations:
$$
\begin{aligned}
\mathbf{w} \cdot(\mathbf{x}-\mathbf{y}) &=0, \\
|\mathbf{w}|^2 &=r^2-\frac{d^2}{4} .
\end{aligned}
$$
From linear algebra it is known that all but one of the components of a solution $\mathbf{w}$ of the first eq... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_1_16a
(n : ℕ)
(d r : ℝ)
(x y z : EuclideanSpace ℝ (Fin n)) -- R^n
(h₁ : n ≥ 3)
(h₂ : ‖x - y‖ = d)
(h₃ : d > 0)
(h₄ : r > 0)
(h₅ : 2 * r > d)
: Set.Infinite {z : EuclideanSpace ℝ (Fin n) | ‖z - x‖ = r ∧ ‖z - y‖ = r} := |
Rudin|exercise_1_18a | If $k \geq 2$ and $\mathbf{x} \in R^{k}$, prove that there exists $\mathbf{y} \in R^{k}$ such that $\mathbf{y} \neq 0$ but $\mathbf{x} \cdot \mathbf{y}=0$ | \begin{proof}
If $\mathbf{x}$ has any components equal to 0 , then $\mathbf{y}$ can be taken to have the corresponding components equal to 1 and all others equal to 0 . If all the components of $\mathbf{x}$ are nonzero, $\mathbf{y}$ can be taken as $\left(-x_2, x_1, 0, \ldots, 0\right)$. This is, of course, not tr... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_1_18a
(n : ℕ)
(h : n > 1)
(x : EuclideanSpace ℝ (Fin n)) -- R^n
: ∃ (y : EuclideanSpace ℝ (Fin n)), y ≠ 0 ∧ (inner x y) = (0 : ℝ) := |
Rudin|exercise_1_19 | Suppose $a, b \in R^k$. Find $c \in R^k$ and $r > 0$ such that $|x-a|=2|x-b|$ if and only if $| x - c | = r$. Prove that $3c = 4b - a$ and $3r = 2 |b - a|$. | \begin{proof}
Since the solution is given to us, all we have to do is verify it, i.e., we need to show that the equation
$$
|\mathrm{x}-\mathrm{a}|=2|\mathrm{x}-\mathrm{b}|
$$
is equivalent to $|\mathrm{x}-\mathbf{c}|=r$, which says
$$
\left|\mathbf{x}-\frac{4}{3} \mathbf{b}+\frac{1}{3} \mathbf{a}\right|=\fr... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_1_19
(n : ℕ)
(a b : EuclideanSpace ℝ (Fin n))
: (∃ c r, r > 0 ∧ (∀ x, ‖x - a‖ = 2 * ‖x - b‖ ↔ ‖x - c‖ = r)) ∧
(∀ c r, (r > 0 ∧ (∀ x, ‖x - a‖ = 2 * ‖x - b‖ ↔ ‖x - c‖ = r)) →
(3 • c = 4 • b - a ∧ 3 * r = 2 * ‖b - a‖)) := |
Rudin|exercise_2_24 | Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is separable. | \begin{proof}
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 m... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_2_24 {X : Type*} [MetricSpace X]
(hX : ∀ (A : Set X), Infinite A → ∃ (x : X), x ∈ closure A) :
SeparableSpace X := |
Rudin|exercise_2_27a | Suppose $E\subset\mathbb{R}^k$ is uncountable, and let $P$ be the set of condensation points of $E$. Prove that $P$ is perfect. | \begin{proof}
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.
\end{proof} | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem 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 := |
Rudin|exercise_2_28 | 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. | \begin{proof}
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 th... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem 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 := |
Rudin|exercise_3_1a | Prove that convergence of $\left\{s_{n}\right\}$ implies convergence of $\left\{\left|s_{n}\right|\right\}$. | \begin{proof}
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... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_3_1a
(f : ℕ → ℝ)
(h : ∃ (a : ℝ), Tendsto (λ (n : ℕ) => f n) atTop (𝓝 a))
: ∃ (a : ℝ), Tendsto (λ (n : ℕ) => |f n|) atTop (𝓝 a) := |
Rudin|exercise_3_3 | 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 $n=1,2,3, \ldots$. | \begin{proof}
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 inc... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| noncomputable def f : ℕ → ℝ
| 0 => sqrt 2
| (n + 1) => sqrt (2 + sqrt (f n))
theorem exercise_3_3
: ∃ (x : ℝ), Tendsto f atTop (𝓝 x) ∧ ∀ n, f n < 2 := |
Rudin|exercise_3_6a | Prove that $\lim_{n \rightarrow \infty} \sum_{i<n} a_i = \infty$, where $a_i = \sqrt{i + 1} -\sqrt{i}$. | \begin{proof}
(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 $\left(p=\frac{1}{2}\right)$.
\end{proof} | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| noncomputable def g (n : ℕ) : ℝ := sqrt (n + 1) - sqrt n
theorem exercise_3_6a
: Tendsto (λ (n : ℕ) => (∑ i in range n, g i)) atTop atTop := |
Rudin|exercise_3_8 | If $\Sigma a_{n}$ converges, and if $\left\{b_{n}\right\}$ is monotonic and bounded, prove that $\Sigma a_{n} b_{n}$ converges. | \begin{proof}
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 $... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_3_8
(a b : ℕ → ℝ)
(h1 : ∃ y, (Tendsto (λ n => (∑ i in (range n), a i)) atTop (𝓝 y)))
(h2 : Monotone b)
(h3 : Bornology.IsBounded (Set.range b)) :
∃ y, Tendsto (λ n => (∑ i in (range n), (a i) * (b i))) atTop (𝓝 y) := |
Rudin|exercise_3_20 | Suppose $\left\{p_{n}\right\}$ is a Cauchy sequence in a metric space $X$, and some sequence $\left\{p_{n l}\right\}$ converges to a point $p \in X$. Prove that the full sequence $\left\{p_{n}\right\}$ converges to $p$. | \begin{proof}
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)... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_3_20 {X : Type*} [MetricSpace X]
(p : ℕ → X) (l : ℕ+) (r : X)
(hp : CauchySeq p)
(hpl : Tendsto (λ n => p (l * n)) atTop (𝓝 r)) :
Tendsto p atTop (𝓝 r) := |
Rudin|exercise_3_22 | 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. | \begin{proof}
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$. S... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem 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 := |
Rudin|exercise_4_2a | 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$). | \begin{proof}
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(... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem 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) := |
Rudin|exercise_4_4a | 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)$. | \begin{proof}
To prove that $f(E)$ is dense in $f(X)$, simply use that $f(X)=f(\bar{E}) \subseteq \overline{f(E)}$.
\end{proof} | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem 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) := |
Rudin|exercise_4_5a | 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$. | \begin{proof}
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... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem 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 := |
Rudin|exercise_4_6 | 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. | \begin{proof}
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)=... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem 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 := |
Rudin|exercise_4_8b | Let $E$ be a bounded set in $R^{1}$. Prove that there exists a real function $f$ such that $f$ is uniformly continuous and is not bounded on $E$. | \begin{proof}
The function $f(x)=x$ is uniformly continuous on the entire line, but not bounded.
\end{proof} | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_4_8b
(E : Set ℝ) (h : Bornology.IsBounded E) :
∃ f : ℝ → ℝ, UniformContinuousOn f E ∧ ¬ Bornology.IsBounded (Set.image f E) := |
Rudin|exercise_4_12 | A uniformly continuous function of a uniformly continuous function is uniformly continuous. | \begin{proof}
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 ... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem 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) := |
Rudin|exercise_4_19 | 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. | \begin{proof}
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.
\end{proof} | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_4_19
{f : ℝ → ℝ} (hf : ∀ a b c, a < b → f a < c → c < f b → ∃ x, a < x ∧ x < b ∧ f x = c)
(hg : ∀ r : ℚ, IsClosed {x | f x = r}) : Continuous f := |
Rudin|exercise_4_24 | 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. | \begin{proof}
We shall prove that
$$
f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)
$$
for all "dyadic rational" numbers, i.e., all numbers of the form $\lambda=\frac{k}{2^n}$, where $k$ is a nonnegative integer not larger than $2^n$. We do this by induction on $n$. The case $n=0$ is trivial (si... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_4_24 {f : ℝ → ℝ}
(hf : Continuous f) (a b : ℝ) (hab : a < b)
(h : ∀ x y : ℝ, a < x → x < b → a < y → y < b → f ((x + y) / 2) ≤ (f x + f y) / 2) :
ConvexOn ℝ (Set.Ioo a b) f := |
Rudin|exercise_5_2 | Suppose $f^{\prime}(x)>0$ in $(a, b)$. Prove that $f$ is strictly increasing in $(a, b)$, and let $g$ be its inverse function. Prove that $g$ is differentiable, and that $g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \quad(a<x<b)$. | \begin{proof}
For any $c, d$ with $a<c<d<b$ there exists a point $p \in(c, d)$ such that $f(d)-f(c)=f^{\prime}(p)(d-c)>0$. Hence $f(c)<f(d)$
We know from Theorem $4.17$ that the inverse function $g$ is continuous. (Its restriction to each closed subinterval $[c, d]$ is continuous, and that is sufficient.) Now ob... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_5_2 {a b : ℝ}
{f g : ℝ → ℝ} (hf : ∀ x ∈ Set.Ioo a b, deriv f x > 0)
(hg : g = f⁻¹) :
StrictMonoOn f (Set.Ioo a b) ∧
DifferentiableOn ℝ g (Set.Ioo a b) ∧
∀ x ∈ Set.Ioo a b, deriv g (f x) = 1 / deriv f x := |
Rudin|exercise_5_4 | If $C_{0}+\frac{C_{1}}{2}+\cdots+\frac{C_{n-1}}{n}+\frac{C_{n}}{n+1}=0,$ where $C_{0}, \ldots, C_{n}$ are real constants, prove that the equation $C_{0}+C_{1} x+\cdots+C_{n-1} x^{n-1}+C_{n} x^{n}=0$ has at least one real root between 0 and 1. | \begin{proof}
Consider the polynomial
$$
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 .
$$
It is obvious that $p(0)=0$, and the hypothesis of the problem is that $p(1)=0$. Hence Rolle's th... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_5_4 {n : ℕ}
(C : ℕ → ℝ)
(hC : ∑ i in (range (n + 1)), (C i) / (i + 1) = 0) :
∃ x, x ∈ (Set.Icc (0 : ℝ) 1) ∧ ∑ i in range (n + 1), (C i) * (x^i) = 0 := |
Rudin|exercise_5_6 | Suppose (a) $f$ is continuous for $x \geq 0$, (b) $f^{\prime}(x)$ exists for $x>0$, (c) $f(0)=0$, (d) $f^{\prime}$ is monotonically increasing. Put $g(x)=\frac{f(x)}{x} \quad(x>0)$ and prove that $g$ is monotonically increasing. | \begin{proof}
Put
$$
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
$$
for some $c \in(0, x)$. Since $f^{\prime}$ is monotonically increasing, this result implies that $f(x)<x f^{\prime}(x)$. It therefore follow... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_5_6
{f : ℝ → ℝ}
(hf1 : ContinuousOn f (Set.Ici 0))
(hf2 : DifferentiableOn ℝ f (Set.Ioi 0))
(hf3 : f 0 = 0)
(hf4 : MonotoneOn (deriv f) (Set.Ioi 0)) :
MonotoneOn (λ x => f x / x) (Set.Ioi 0) := |
Rudin|exercise_5_15 | Suppose $a \in R^{1}, f$ is a twice-differentiable real function on $(a, \infty)$, and $M_{0}, M_{1}, M_{2}$ are the least upper bounds of $|f(x)|,\left|f^{\prime}(x)\right|,\left|f^{\prime \prime}(x)\right|$, respectively, on $(a, \infty)$. Prove that $M_{1}^{2} \leq 4 M_{0} M_{2} .$ | \begin{proof}
The inequality is obvious if $M_0=+\infty$ or $M_2=+\infty$, so we shall assume that $M_0$ and $M_2$ are both finite. We need to show that
$$
\left|f^{\prime}(x)\right| \leq 2 \sqrt{M_0 M_2}
$$
for all $x>a$. We note that this is obvious if $M_2=0$, since in that case $f^{\prime}(x)$ is constant,... | import Mathlib
open Topology Filter Real Complex TopologicalSpace Finset
open scoped BigOperators
| theorem exercise_5_15 {f : ℝ → ℝ} (a M0 M1 M2 : ℝ)
(hf' : DifferentiableOn ℝ f (Set.Ioi a))
(hf'' : DifferentiableOn ℝ (deriv f) (Set.Ioi a))
(hM0 : M0 = sSup {(|f x|) | x ∈ (Set.Ioi a)})
(hM1 : M1 = sSup {(|deriv f x|) | x ∈ (Set.Ioi a)})
(hM2 : M2 = sSup {(|deriv (deriv f) x|) | x ∈ (Set.Ioi a)}) :
(M1 ^ ... |
Shakarchi|exercise_1_13a | Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $\text{Re}(f)$ is constant, then $f$ is constant. | \begin{proof}
Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$.
Since $\operatorname{Re}(f)=$ constant,
$$
\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 .
$$
Thus, in $\Omega$,
$$
... | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
| theorem exercise_1_13a {f : ℂ → ℂ} (Ω : Set ℂ) (a b : Ω) (h : IsOpen Ω)
(hf : DifferentiableOn ℂ f Ω) (hc : ∃ (c : ℝ), ∀ z ∈ Ω, (f z).re = c) :
f a = f b := |
Shakarchi|exercise_1_13c | Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $|f|$ is constant, then $f$ is constant. | \begin{proof}
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... | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
| theorem exercise_1_13c {f : ℂ → ℂ} (Ω : Set ℂ) (a b : Ω) (h : IsOpen Ω)
(hf : DifferentiableOn ℂ f Ω) (hc : ∃ (c : ℝ), ∀ z ∈ Ω, abs (f z) = c) :
f a = f b := |
Shakarchi|exercise_1_19b | Prove that the power series $\sum z^n/n^2$ converges at every point of the unit circle. | \begin{proof}
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$.)
\end{proof} | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
| theorem exercise_1_19b (z : ℂ) (hz : abs z = 1) (s : ℕ → ℂ)
(h : s = (λ n => ∑ i in (range n), z ^ (i + 1) / (i + 1) ^ 2)) :
∃ y, Tendsto s atTop (𝓝 y) := |
Shakarchi|exercise_1_26 | Suppose $f$ is continuous in a region $\Omega$. Prove that any two primitives of $f$ (if they exist) differ by a constant. | \begin{proof}
Suppose $F_1$ adn $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.
\end{proof} | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
| theorem exercise_1_26
(f F₁ F₂ : ℂ → ℂ) (Ω : Set ℂ) (h0 : Nonempty Ω) (h1 : IsOpen Ω) (h2 : IsConnected Ω) (hf : ContinuousOn f Ω)
(hF₁ : DifferentiableOn ℂ F₁ Ω) (hF₂ : DifferentiableOn ℂ F₂ Ω)
(hdF₁ : ∀ x ∈ Ω, deriv F₁ x = f x) (hdF₂ : ∀ x ∈ Ω, deriv F₂ x = f x)
: ∃ c : ℂ, ∀ x, F₁ x = F₂ x + c := |
Shakarchi|exercise_2_9 | 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. | \begin{proof}
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\l... | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
| theorem exercise_2_9
{f : ℂ → ℂ} (Ω : Set ℂ) (b : Bornology.IsBounded Ω) (h : IsOpen Ω)
(hf : DifferentiableOn ℂ f Ω) (z : Ω) (hz : f z = z) (h'z : deriv f z = 1) :
∃ (f_lin : ℂ →L[ℂ] ℂ), ∀ x ∈ Ω, f x = f_lin x := |
Shakarchi|exercise_3_3 | Show that $ \int_{-\infty}^{\infty} \frac{\cos x}{x^2 + a^2} dx = \pi \frac{e^{-a}}{a}$ for $a > 0$. | \begin{proof}
$\cos x=\frac{e^{i x}+e^{-i x}}{2}$. 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 \ri... | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
| theorem 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))) := |
Shakarchi|exercise_3_9 | Show that $\int_0^1 \log(\sin \pi x) dx = - \log 2$. | \begin{proof}
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 (\si... | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
| theorem exercise_3_9 : ∫ x in (0 : ℝ)..(1 : ℝ), Real.log (Real.sin (Real.pi * x)) = - Real.log 2 := |
Shakarchi|exercise_3_22 | 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 $z \in \partial D$. | \begin{proof}
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.
\end{proof} | import Mathlib
open Complex Filter Function Metric Finset
open scoped BigOperators Topology
| theorem 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 := |
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