Source of most content: Gunhee Cho and Topology by Munkres.

Heine-Borel Theorem

\[\text{Let } X \text{ be a simply ordered set having the least upper bound property.} \nonumber\] \[\text{Then, any closed interval } [a, b] = \{x\in X : a \leq x \leq b \} \text{ is compact}. \nonumber\]

Proof

1. Take any $a,b\in X$ with $a < b$. Let $\mathcal{A}$ be an open cover of $[a, b]$. (*) If $x \in [a, b)$, then $\exists y \in (x ,b]$ s.t. $[x, y]$ can be covered by at most two elements of $\mathcal{A}$.
   • (*): Let $x \in [a,b)$. If $\exists y > x$ s.t. $(x, y) = \emptyset$, then $[x, y]$ consists of two points. Thus, it can be covered by at most two elements of $\mathcal{A}$.
   • (*): If not, choose an element $A$ of $\mathcal{A}$ contatining $x$. Since $x \ne b$, $\exists z \in (x, b]$ s.t. $(x, z) \subset A$. Since $x < z, \exists y$ s.t. $x < y < z $. Therefore, $[x, y] \subset A$.
2. Define $C := \{ y \in (a, b] : [a, y] \text{ can be covered by finite subcover of } \mathcal{A} \}$. Since $a\ne b$, by 1, $C \ne \emptyset$. Moreover, as $y \leq b$, $C$ is bounded aboved. By LUP, $\exists \sup C \in X$. THus, $a < \sup C \leq b$.
3. Since $\sup C \in (a, b], \, \exists A \in \mathcal{A}$ s.t. $\sup C \in A$. Let $z$ be a left end point of $A$. Since $z < \sup C, \, \exists y \in C$ s.t. $z < y \leq \sup C$. Thus, $[a, y]$ can be covered by finite subcover of $\mathcal{A}$ by def. of $C$. Also, $(y , \sup C] \subset A$. Thus, $[a, \sup C] = [a, y] \cup (y , \sup C]$, i.e., $[a, \sup C]$ can be covered by finite subcover of $\mathcal{A}$. Therefore, $ \sup C \in C$.
4. If $\sup C \ne b$, i.e., $\sup C < b$, then since $\sup C \in C$, we can take $A\in \mathcal{A}$ containing $\sup C$. Let $z$ be a right end point of $A$. Since $z \in (\sup C, b]$, $[\sup C, z]$ can be covered by finite subcover of $\mathcal{A}$. This means that $z \in C$, contradicting the definition of $\sup C$. Hence, $\sup C = b$.

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Cor1; Any closed interval of $\mathbb{R}$ is compact.
Cor2; In $\mathbb{R}^n, \, [a_1, b_1] \times \cdots [a_n, b_n] $ is compact.

Proposition

\[A \subseteq \mathbb{R}^n \text{ is compact iff } A \text{ is closed an bounded in the euclidean metric } d \text{ or the square metric }\rho. \nonumber\]

Proof

• $d : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \, (x,y) \mapsto d(x,y) := \sqrt{\sum_{i=1}^n (x_i - y_i)^2}$.
• $\rho: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \, (x,y) \mapsto \rho(x,y) := \max_{i}{\mid x_i - y_i\mid}$
• It suffices to consider only $\rho$ since $\rho(x,y) \leq d (x,y) \leq \sqrt{n} \rho(x,y)$ holds, i.e., if $A$ is bounded under $d$ iff it is bounded under $\rho$.
• $(\Rightarrow)$ Let $A$ be compact. Since $\mathbb{R}^n$ is Hausdorff, $A$ is closed by Prop2. Define $C := \{ B _\rho (O; M) : M \in \mathbb{R} _{+}\}$, the collection of open balls centered at origin with radius $M$. Then, $C$ is an open cover of $A$ since there exists $M$ s.t. $A \subset B _\rho (O; M)$. Since $A$ is compact, $\exists$ finite subcover of $C$. Take $N = \max \{ M _1, \ldots, M _k \}$, then $A \subset B _\rho (0; N)$. Thus, $\rho(x, O) \leq N$ for all $x\in X$, i.e., $A$ is bounded under $\rho$.
• $(\Leftarrow)$ Suppose $A$ is closed and bounded under $\rho$. Since $A$ is bounded, $\exists N \in \mathbb{R}$ s.t. $\forall x^1, x^2 \in A$, $\rho(x^1, x^2) \leq N$. Fix $a \in A$ and take $\rho(a, O) = M \in \mathbb{R}$. Then, $\forall y \in A, \, \rho(y, O) \leq \rho(y, a) + \rho(a, O) \leq N + M$ holds by triangle inequality. Take $P = M + N > 0$. Then, $A \subset [-P , P]^n$. Since closed interval is compact by Heine-Borel thm and product of compact spaces is compact, $[-P, P]^n$ is compact. The fact that $A$ is closed concludes the proof since closed subspace of compact set is compact

Extreme Value Theorem

\[\text{Let }(X,T_X) \text{ be C.S. and } Y \text{ be simply ordered set with } T_{order}. \text{ Let } f: X \to Y \text{ be cts. Then, } \exists \max f , \min f. \nonumber\]

Proof

Since $X$ is compact and $f$ is continuous, $\text{Im}f =: A$ is compact. Suppose $A$ has no largest element. Define $C := \{ (-\infty, a) : a \in A \}$. Since $A$ is compact, $\exists$ finite subcover of $A$ as $\{ (-\infty, a_1), \ldots, (-\infty, a_n) \}$. Take $M= \max \{ a_1, \ldots, a_n \} = a_k = f(x_k)$. Note that $A \subset (-\infty, M ) = (-\infty, f(x_k))$ holds. This implies $f(x_k) \in A$ and $f(x_k) \in (-\infty , f(x_k))$, which is a contradiction. Similarly, we can show that $\exists \min f$.

Uniformly Continuous

Def. Given metric spaces $(X, d_X), (Y, d_Y)$, we call $f:X \to Y$ uniformly continuous if $\forall \epsilon >0, \exists \delta >0$ s.t. if $d_X (x_1, x_2) < \delta$ for any $x_1, x_2 \in X$, then $d_Y (f(x_1), f(x_2)) < \epsilon$.

NB2; $f: \mathbb{R} \to \mathbb{R}$ is not uniformly continuous iff $\exists \epsilon >0$ s.t. $\forall n \in \mathbb{N}, \, \mid x_n - y_n \mid < \frac{1}{n} \implies \mid f(x_n) - f(y_n) \mid \geq \epsilon$ $\Leftrightarrow$ $\exists \epsilon >0$ and a seq. $(x_n), (y_n)$ s.t. $\mid x_n - y_n \mid < \frac{1}{n} \implies \mid f(x_n) - f(y_n) \mid \geq \epsilon$.

Def. Given metric space $(X, d)$, define diameter $dia A := \sup \{ d(a_1, a_2) : a_1, a_2 \in A \}$. $A$ is bounded if $dia A < \infty$.

Def. Given metric space $(X, d)$, let $A \subset X$. Define distance from $x$ to $A$ $d_A : X \to \mathbb{R}, \, x \mapsto \inf \{ d(x,a) : a \in A\}$, i.e., $d_A(x) = \inf_{a\in A} d (x,a)$.

NB3; $d_A : X \to \mathbb{R}$ is uniformly continuous.

Proof (NB3)

Fix $x \in X$. Then, $d_A (x) = \inf d(x,a) \stackrel{\forall a \in A}{\leq} d(x,a) \stackrel{\forall a \in A, \forall y \in X}{\leq} d(x,y) + d(y,a)$. Thus, $d_A (x) - d(x,y) \leq d(y,a)$ holds for all $a\in A, y\in X$. Fix $y\in X$. Then, $d_A(x) - d(x,y) \leq \inf d(y,a) = d_A (y)$, i.e., $d_A(x) - d_A(y) \leq d(x,y)$. Similarly, $d_A(y) - d_A(x) \leq d(y,x) = d(x,y)$. Therefore, $\mid d_A(x) - d_A(y) \mid \leq d (x,y)$, which concludes the proof.

Lebesgue number lemma

\[\text{Given metric space } (X, d) \text{ with compact } X. \text{ Let } \mathcal{A} \text{ be an open cover of } X. \nonumber\] \[\text{ Then, } \exists \delta > 0 \text{ s.t. } \forall A \subset X \text{ s.t. } dia A < \delta, \exists A_i \in \mathcal{A} \text{ satisfying } A \subset A_i. \nonumber\]

Proof

Since $X$ is compact, $\exists$ finite subcover of $\mathcal{A}$ called $\{ A _1, \ldots, A _n \}$, i.e., $X \subset \cup _{k=1}^n A _k$. Let $C _k := X \setminus A _k$ for all $1\leq k \leq n$. Define $f: X \to \mathbb{R}, \, x \mapsto f(x) := \sum _{k=1}^n d _{C _k}(x)$.

Then, $\forall x \in X$, $\exists$ ball $B$ s.t. $x \in B \subset A _j$ for some $j$ since ball is a basis of topology in metric space. Thus, $f(x) \geq \frac{1}{n}d _{C _j}(x) \geq \frac{1}{n}d _{B^c}(x) \geq \frac{1}{n}radius(B) > 0$ holds. Note that $f$ is continuous and $X$ is compact, we can apply extreme value theorem. Then, $\exists x _0 \in X$ s.t. $0 < f(x _0)= \min \text{Im} f$. Take $\delta = f(x _0) > 0$. For any $A \subseteq X$ with $dia A < \delta$, take $x \in A$. By definition, we have $\delta \leq f(x) = \frac{1}{n}\sum _{k=1}^n d _{C _k}(x) \leq d _{C _j}(x)$ where we take $j$ for maximum. Thus, $\delta \leq d _{C _j}(x)$, i.e., $B _\delta (x) \subset A _j$ holds. Since $dia A < \delta, \, A \subset B _\delta (x)$. Hence, $A \subset B _\delta (x) \subset A _j$.

Uniformly continuous theorem

\[\text{Let } (X,d_X), (Y, d_Y) \text{ be metric spaces for compact } X \text{ and } f:X \to Y \text{ be cts. Then, } f \text{ is uniformly continuous.} \nonumber\]

Proof

Given $\epsilon >0$, consider $C:= \{ B(y; \frac{\epsilon}{2}) : y \in Y \}$ that is an open cover of $Y$. Since $f$ is cts, $\{ f^{-1}(B(y; \frac{\epsilon}{2})) : y \in Y\}$ is open cover of $X$. By Lebesque number lemma, $\exists \epsilon$ s.t. $\forall A \subset X$ with $dia A < \delta$, $\exists A_i \in \{ f^{-1}(B(y; \frac{\epsilon}{2})) : y \in Y\}$ that contains $A$. Let $A = \{ x_1, x_2 \} \subset X$. Then, $dia A < \delta$ holds by definition. So $\exists y \in Y$ s.t. $A \subset f^{-1}(B(y; \frac{\epsilon}{2}))$ by Lemma. This implies that $f(A) = \{ f(x_1), f(x_2) \} \subset B(y; \frac{\epsilon}{2})$. Therefore, $d_Y(y, f(x_1)) < \frac{\epsilon}{2}$ and $d_Y(y, f(x_2)) < \frac{\epsilon}{2}$ holds. Applying triangle inequality concludes the proof.