# 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 $X$.

<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, $Y-U$ 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.

</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.
