From 4c715753b29d51ed5803517300533859faefad50 Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Fri, 14 Nov 2025 11:53:34 -0600
Subject: [PATCH] Update Math4201_L32.md

---
 content/Math4201/Math4201_L32.md | 137 ++++++++++++++++++++++++++++++-
 1 file changed, 136 insertions(+), 1 deletion(-)

diff --git a/content/Math4201/Math4201_L32.md b/content/Math4201/Math4201_L32.md
index 4055b42..6836ed6 100644
--- a/content/Math4201/Math4201_L32.md
+++ b/content/Math4201/Math4201_L32.md
@@ -4,7 +4,7 @@
 
 ### Locally compact
 
-#### Theorem of Homeomorphism over locally compact Hausdorff spaces
+#### Theorem of one point compactification
 
 $X$ is a locally compact Hausdorff space if and only if there exists topological space $Y$ satisfying the following properties:
 
@@ -12,3 +12,138 @@ $X$ is a locally compact Hausdorff space if and only if there exists topological
 2. $Y-X$ has one point (usually denoted by $\infty$).
 3. $Y$ is compact and Hausdorff.
 
+> $Y$ is called **one point compactification** of $X$.
+
+<details>
+<summary>Proof for existence of Y (forward direction)</summary>
+
+Let's defined the topology of $Y$ as follows:
+
+Let $U\subseteq Y$ is open if and only if either
+
+1. $U\subseteq X$ and $U$ is open in $X$. ($\infty\notin U$) (Type 1 open set)
+2. $\infty \in U$ and $Y-U\subseteq X$ with subspace topology from $X$ is compact. (Type 2 open set)
+
+---
+
+First, we prove that $X$ is a subspace of $Y$. (That is, every open set $U\subseteq X$ implies that $U\cap X$ is open in $X$.)
+
+Case 1: $U\subseteq X$ is open in $X$, then $U\cap X=U$ is open in $Y$.
+
+Case 2: $\infty\in U$, then $Y-U$ is a compact subspace of $X$, since $X$ is Hausdorff. So $Y-U$ is a closed subspace of $X$.
+
+So $X\cap U=X-(Y-U)$ is open in $X$.
+
+We also need to show any open $U\subseteq X$ can be written as the intersection of some open in $Y$ and $X$.
+
+Note that for an open set $U\subseteq X$, $U\cap X$ is open in $X$. So $U\cap X$ is open in $Y$.
+
+---
+
+The second part is trivial by observation.
+
+---
+
+**First we show that $Y$ is Hausdorff.**
+
+Let $x_1,x_2\in Y$, such that $x_1\neq x_2$.
+
+If one of $x$, without loss of generality, $x_1$ is $\infty$, then by the assumption on $X$, there is a compact set $K$ containing an open neighborhood $U$ of $x_2$.
+
+Note that $Y-K$ is an open subspace of Type 2 in $Y$. In particular, it contains $\infty$.
+
+This is disjoint from the open neighborhood $U$ of $x_2$.
+
+If $x_1,x_2$ are both in $X$, then by the assumption on $X$, then by Hausdorff property for $X$, there are disjoint open neightbors $U_1$ and $U_2$ such that $x_1\in U_1$ and $x_2\in U_2$. By Type 1 open sets, these are also open and disjoint in $Y$.
+
+**Then we show that $Y$ is compact.**
+
+Take an open cover $\{U_\alpha\}_{\alpha\in I}$ of $Y$.
+
+In particular, there is $\alpha_0\in I$ such that $\infty\in U_{\alpha_0}$
+
+Note that $Y-U_{\alpha_0}\subseteq X$ with subspace topology from $X$ is compact (by Type 2 set).
+
+So there exists a finite subcover $\{U_{\alpha_i}\}_{\alpha_i\in I}$ such that $Y-U_{\alpha_0}\subseteq \bigcup_{i=1}^n U_{\alpha_i}$.
+
+So $U_{\alpha_0},U_{\alpha_1},\dots,U_{\alpha_n}$ is a finite cover of $Y$.
+
+So $Y$ is compact.
+
+</details>
+
+<details>
+<summary>Proof for properties from $Y$.(backward direction)</summary>
+
+**Property 1**
+
+$X$ is Hausdorff because it's a subspace of Hausdorff space.
+
+**Property 2**
+
+By definition
+
+**Property 3**
+
+$X$ is locally compact.
+
+Let $x\in X$ since $Y$ is Hausdorff, there are disjoint open sets $U,V\subseteq Y$ such that $x\in U$ and $\infty\in B$.
+
+Let $K=Y-V$, $K$ is a subset of $X$ since $\infty\notin V$.
+
+To complete the proof, we need to show that $K$ is compact.
+
+Since $V$ is open in $Y$, then $K$ is closed in $Y$. Since $Y$ is compact, then $K$ is compact. (any closed subspace of compact space is compact)
+
+</details>
+
+## Countability axioms
+
+### First countability axiom
+
+#### Definition for first countability axiom
+
+Let $X$ be a topological space, then $X$ satisfies the first countability axiom if
+
+For any $x\in X$, there is a countable collection $\{B_n}_n$ of open neighborhoods of $x$ such that any open neighborhood $U$ of $x$ contains one of $B_n$.
+
+<details>
+<summary>Example for metric space satisfies the first countability axiom</summary>
+
+Any metric space satisfies the first countability axiom.
+
+Take $\{B_{\frac{1}{n}}(x)\}_{n=1}^\infty$.
+
+</details>
+
+#### Properties for topological spaces that satisfy the first countability axiom
+
+1. If $A\subseteq X$, then for any $x\in \overline{A}$, there is a sequence $\{x_n\}_{n=1}^\infty\subseteq A$ such that $x_n\to x$.
+2. If $f:X\to Y$ such that for any sequence $\{x_n\}_{n=1}^\infty\subseteq X$ such that $x_n\to x$, we have $f(x_n)\to f(x)$ in $Y$, then $f$ is continuous.
+
+### Second countability axiom
+
+#### Definition for second countability axiom
+
+Let $X$ be a topological space, then $X$ satisfies the second countability axiom if
+
+it has a countable basis.
+
+Clearly any second countable space also satisfies the first countability axiom.
+
+But the converse is not true.
+
+<details>
+<summary>Example for metric space satisfies the second countability axiom</summary>
+
+$\mathbb{R}$ satisfies the second countability axiom. Take $\{(a,b)|a,b\in\mathbb{Q}\}$ is a basis for $\mathbb{R}$.
+
+And $\mathbb{Q}$ is countable.
+
+More generally, $\mathbb{R}^n$ is also countable and satisfies the second countability axiom.
+
+</details>
+
+> [!WARNING]
+>
+> Not all topological spaces satisfy the second countability axiom is metrizable.