# Math4201 Topology I (Lecture 30)
## Compact and connected spaces
### Locally compact
#### 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:
1. $X$ is a subspace of $Y$.
2. $Y-X$ has one point (usually denoted by $\infty$).
3. $Y$ is compact and Hausdorff.
> $Y$ is called **one point compactification** of $X$.
Proof for existence of Y (forward direction)
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, $Y-U$ is a closed subspace of $X$. [Compact subspace of a Hausdorff space is closed](../Math4201_L25#proposition-of-compact-subspaces-with-hausdorff-property)
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.
Proof for properties from $Y$.(backward direction)
**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)
## 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 $B_n$.
Example for metric space satisfies the first countability axiom
Any metric space satisfies the first countability axiom.
Take $\{B_{\frac{1}{n}}(x)\}_{n=1}^\infty$.
#### 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.
Example for metric space satisfies the second countability axiom
$\mathbb{R}$ 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.
> [!WARNING]
>
> Not all topological spaces satisfy the second countability axiom is metrizable.