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content/Math4201/Math4201_L26.md

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+# Math4201 Topology I (Lecture 26)
+
+## Continue on compact spaces
+
+### Compact spaces
+
+#### Tube lemma
+
+Let $X$ be a compact topological space and $y$ be a topological space. Let $N\subseteq X\times Y$ be an open set contains $X\times \{y_0\}$ for $y_0\in Y$. Then there exists an open set $W\subseteq Y$ is open containing $y_0$ such that $N$ contains $X\times W$.
+
+<details>
+<summary>Proof</summary>
+
+For any $x\in X$, there are open sets $x\in U_x\subseteq X$ and $y_0\in V_x\subseteq Y$ such that $U_x\times V_x\subseteq N$.
+
+In particular, $\{U_x\}_{x\in X}$ is an open cover of $X$. Since $\forall x\in X$, $x\in U_x$, so there exists a finite subcover $\bigcup_{i=1}^k U_{x_i}=X$.
+
+Take $W=\bigcap_{i=1}^k V_{x_i}$. This is intersection of finitely many open sets, so it is open.
+
+$y_0\in V_{x_i}$ for all $x_i\in X$, so $y_0\in W$.
+
+So $U_{x_i}\times W\subseteq U_{x_i}\times V_{x_i}\subseteq N$ for all $i=1, ..., k$.
+
+So $\bigcup_{i=1}^k U_{x_i}\times W\subseteq N$.
+
+</details>
+
+#### Product of compact space is compact
+
+Let $X$ and $Y$ be compact spaces, then $X\times Y$ is compact.
+
+<details>
+<summary>Proof</summary>
+
+Let $\{N_\alpha\}_{\alpha\in I}$ be an open covering of $X\times Y$.
+
+For any $y_0\in Y$, $X\times \{y_0\}$ is a compact subspace of $X\times Y$.
+
+So there are finitely many $N_\alpha$'s whose union $M_{y_0}$ is open containing $X\times \{y_0\}$.
+
+Using the tube lemma, $M_{y_0}$ contains $X\times W_{y_0}$ for some open neighborhood $W_{y_0}\subseteq Y$ of $y_0$.
+
+Now note that $\{W_{y_0}\}_{y_0\in Y}$ is an open cover of $Y$.
+
+In particular, there are finitely many $y_1, ..., y_n\in Y$ such that $Y=\bigcup_{i=1}^n W_{y_i}$ by compactness of $Y$.
+
+So $\forall i$, $X\times W_{y_i}\subseteq M_{y_i}$.
+
+This is a union of finitely many $N_\alpha$'s, so it is open.
+
+This implies that taking the union of all such $N_\alpha$'s, for all $1\leq i\leq k$, which is finite, covers $\bigcup_{i=1}^k X\times W_{y_i}=X\times Y$.
+
+</details>
+
+#### Closed intervals in real numbers are compact
+
+$[a,b]$ is compact in $\mathbb{R}$.
+
+<details>
+<summary>Proof</summary>
+
+Let $\{U_\alpha\}_{\alpha\in I}$ be an open cover of $[a,b]$.
+
+Define:
+
+$$
+C=\{c\in [a,b]\mid [a,c]\text{ is covered by finitely many } U_\alpha's\}
+$$
+
+Our goal is to show that $b\in C$.
+
+Clearly $a$ is covered by one $U_\alpha$, so $[a,a]\in C$.
+
+Take $y=\sup C\in [a,b]$.
+
+Since $y\in [a,b]$, there is $U_\alpha$ such that $y\in U_\alpha$.
+
+Since $U_\alpha$ is open, there exists an open interval $(y-\epsilon, y+\epsilon)\subseteq U_\alpha$. So there is some $z\in C$ such that $z\in (y-\epsilon, y+\epsilon)$. Otherwise $y-\epsilon$ is an upper bound of $C$, which contradicts the definition of $y$.
+
+So $[a,z]$ can be covered by finitely many $U_\alpha$'s. $U_{\alpha_1}, ..., U_{\alpha_k}$ and $\bigcup_{i=1}^k U_{\alpha_i}\supseteq [a,z]$.
+
+So $\bigcup_{i=1}^k U_{\alpha_i}\supseteq [a,z]\cup (y-\epsilon, y+\epsilon)=[a,y+\epsilon]$.
+
+If $y\neq b$, then there is an element $c\in [a,b]$ that belongs to $(y,y+\epsilon)$ and $[a,c]\subseteq [a,y+\epsilon)$ can be covered by finitely many $U_\alpha$'s, so $c\in C$. This contradicts the definition of $y$. $y<c\in C$.
+
+</details>
+
+#### Heine-Borel theorem
+
+A subset $K\subseteq \mathbb{R}^n$ is compact if and only if it is closed and bounded with respect to the standard metric on $\mathbb{R}^n$.
+
+#### Definition of bounded
+
+$A\subseteq \mathbb{R}^n$ is bounded if there exists $c\in \mathbb{R}^{>0}$ such that $d(x,y)<c$ for all $x,y\in A$.
+
+<details>
+<summary>Proof for Heine-Borel theorem</summary>
+
+Suppose $k\subseteq \mathbb{R}^n$ is compact.
+
+Since $\mathbb{R}^n$ is Hausdorff, $K\subseteq \mathbb{R}^n$ is compact, so $K$ is closed subspace of $\mathbb{R}^n$. by Proposition of compact subspaces with Hausdorff property.
+
+To show that $K$ is bounded, consider the open cover with the following balls:
+$$
+B_1(0), B_2(0), ..., B_n(0), ...
+$$
+
+Since $K$ is compact, there are $n_1, ..., n_k\in \mathbb{N}$ such that $K\subseteq \bigcup_{i=1}^k B_{n_i}(0)$. Note that $B_{n_i}(0)$ is bounded, so $K$ is bounded. $\forall x,y\in B_{n_i}(0)$, $d(x,y)<2n_i$. So $K$ is bounded.
+
+---
+
+Suppose $K\subseteq \mathbb{R}^n$ is closed and bounded.
+
+Continue next time.
+
+</details>

content/Math4201/_meta.js

@@ -29,4 +29,5 @@ export default {
     Math4201_L23: "Topology I (Lecture 23)",
     Math4201_L24: "Topology I (Lecture 24)",
     Math4201_L25: "Topology I (Lecture 25)",
+    Math4201_L26: "Topology I (Lecture 26)",
 }