# Math4201 Topology I (Lecture 25)

## Continue on compact spaces

### Compact spaces

#### Definition of compact spaces

A compact space $X$ is a topological space such that any open covering of $X$ has a finite subcovering.

$$
X=\bigcup_{\alpha\in A} U_\alpha\implies \exists \alpha_1, ..., \alpha_n\in A \text{ such that } X=\bigcup_{i=1}^n U_{\alpha_i}
$$

<details>
<summary>Example of compact spaces</summary>

$(0,1)$ is not compact, consider the open cover $\{(0,1/n):n\in \mathbb{N}\}$ which does not have a finite subcover.

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$\mathbb{R}$ is not compact, consider the open cover $\{(-n,n):n\in \mathbb{N}\}$ which does not have a finite subcover.

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Later we will see that $[0,1]$ is compact. (more generally, any closed and bounded interval is compact)

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> [!TIP]
>
> A property (or definition) is good for topologists if it is preserved by homeomorphism, or even better, by continuous maps.

#### Proposition of compact spaces preserved by continuous maps

Let $X$ be a compact space and $f:X\to Y$ be a continuous map. Then $f(X)$ is compact.

<details>
<summary>Proof</summary>

Consider an open covering of $f(X)$, So, there are open sets $\{f(x)\cap U_{\alpha}\}_{x\in X}$ such that $f(X)=\bigcup_{\alpha\in I} (f(x)\cap U_{\alpha})$.

This implies that $\{f^{-1}(f(x)\cap U_{\alpha})\}_{x\in X, \alpha\in I}$ consists of:

1. $f^{-1}(U_{\alpha})$ is open because $f$ is continuous.
2. $f^{-1}(f(x)\cap U_{\alpha})$ covers $X$ because $\forall x\in X, f(x)\in f(X)\subseteq \bigcup_{\alpha\in I} (f(x)\cap U_{\alpha})$ so $x\in f^{-1}( U_{\alpha})$.

Since $X$ is compact, there are finitely many $x_1, ..., x_n\in X$ such that $X=\bigcup_{i=1}^n f^{-1}(U_{\alpha_i})$.

So, $f(X)=\bigcup_{i=1}^n f(f^{-1}(U_{\alpha_i}))=\bigcup_{i=1}^n U_{\alpha_i}$.

This implies that $f(X)$ is compact.

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#### Corollary of compact spaces preserved by homeomorphism

If $f:X\to Y$ is homeomorphism and $X$ is compact, then $Y$ is compact.

#### Lemma of compact subspaces

Let $X$ be a topological space and $Y\subseteq X$ be a subspace with subspace topology from $X$.

Then $Y$ is compact if and only if for any open cover $\{U_\alpha\}_{\alpha\in I}$ of $Y$, there exists a finite subcover $\{U_{\alpha_1}, ..., U_{\alpha_n}\}$ of $Y$.

#### Proposition of closed compact sets

Every closed subspace $Y$ of a compact space $Y\subseteq X$ is compact.

<details>
<summary>Proof</summary>

Let $\{U_\alpha\}_{\alpha\in I}$ be an open cover of $Y$. Since $Y$ is closed, $X-Y$ is open. So, $(X-Y)\cup \bigcup_{\alpha\in I} U_\alpha$ is an open cover of $X$.

Since $X$ is compact, there are finitely many $\alpha_1, ..., \alpha_n\in I$ such that $X=\bigcup_{i=1}^n U_{\alpha_i}$ and possibly $X-Y\subseteq U_{\alpha_m}$.

So, $Y=\bigcup_{i=1}^n (U_{\alpha_i}\cap Y)=\bigcup_{i=1}^n U_{\alpha_i}$.

This implies that $Y$ is compact.

</details>

> [!WARNING]
>
> The converse of the proposition is almost true.

#### Proposition of compact subspaces with Hausdorff property

If $Y$ is compact subspace of a **Hausdorff space** $X$, then $Y$ is closed in $X$.

<details>
<summary>Proof</summary>

To show the claim, we need to show $x$ outside $y$, there is an open neighborhood of $x$ that is disjoint from $Y$.

For any $y\in Y$, there are disjoint open neighborhoods $U_y$ and $V_y$ of $x$ and $y$ respectively (by the Hausdorff property of $X$).

So $\bigcup_{y\in Y} V_y\supseteq Y$ and $Y$ is a compact subspace of $X$, so there are finitely many $y_1, ..., y_n\in Y$ such that $Y\subseteq \bigcup_{i=1}^n V_{y_i}$.

Since for each $y_i\in V_{y_i}$, there exists an open neighborhood $U_{y_i}$ of $x$ such that $U_{y_i}\cap V_{y_i}=\emptyset$, we have $U_{y_i}\cap Y=\emptyset$.

So $\bigcap_{i=1}^n U_{y_i}$ is disjoint from $\bigcup_{i=1}^n V_{y_i}\supseteq Y$, so disjoint from $Y$.

Furthermore, $x\in \bigcap_{i=1}^n U_{y_i}$, so $\bigcap_{i=1}^n U_{y_i}$ is open in $X$ because it is an finite intersection of open sets.

This holds for any $x\in X-Y$, so $X-Y$ is open in $X$, so $Y$ is closed in $X$.

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This the course of proving this proposition, we showed the following:

#### Proposition

If $X$ is Hausdorff and $Y\subseteq X$ is compact, and $x\in X-Y$, then there are disjoint open neighborhoods $U,V\subseteq X$ such that $x\in U$ and $Y\subseteq V$.

<details>
<summary>Proof</summary>

Use the proof from last proposition, take $U=\bigcap_{i=1}^n U_{y_i}$ and $V=\bigcup_{i=1}^n V_{y_i}$.

</details>

#### Theorem of closed maps from compact and Hausdorff spaces

If $f:X\to Y$ is continuous and $X$ is compact, $Y$ is Hausdorff, then $f$ is a closed map.

In particular, if $f:X\to Y$ is continuous and bijection with $X$ compact and $Y$ Hausdorff, then $f$ is a homeomorphism.

<details>
<summary>Example distinguishing these two properties</summary>

Consider the map $f:[0,2\pi)\to \mathbb{S}^1$ defined by $f(x)=(\cos x, \sin x)$. This is a continuous bijection.

$f$ is continuous bijection and $Y$ is Hausdorff, But $X$ is not compact.

Then $f$ is not a homeomorphism because $f^{-1}$ is not continuous.

</details>

<details>
<summary>Proof</summary>

Consider $Z\subseteq X$ is closed and $X$ is compact, so $Z$ is compact.

So $f(Z)$ is compact since $f$ is continuous. Note that $f(Z)\subseteq Y$ is Hausdorff, so $f(Z)$ is closed in $Y$.

So $f$ is a closed map.

</details>

#### Theorem of products of compact spaces

If $X,Y$ are compact spaces, then $X\times Y$ is compact. (More generalized version: Tychonoff's theorem)

<details>
<summary>Incomplete Proof</summary>

Let $\{U_\alpha\}_{\alpha\in I}$ be an open cover of $X\times Y$.

Step 1: For any $x\in X$, there are finitely many $\alpha_1, ..., \alpha_n\in I$ and open neighborhoods $x\in V\subseteq X$ such that $V\times Y\subseteq \bigcup_{i=1}^n U_{\alpha_i}\times Y$.

For any $y\in Y$, there is $U_\alpha$ and $x\in U_y\subseteq X$ and $y\in V_y\subseteq Y$ such that $(x,y)\in U_y\times V_y\subseteq U_\alpha$.

Continue next time...

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