NoteNextra-origin/content/Math4201/Math4201_L25.md

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}

Example of compact spaces

(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)

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.

Proof

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.

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.

Proof

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.

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.

Proof

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.

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.

Proof

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

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.

Example distinguishing these two properties

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.

Proof

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.

Theorem of products of compact spaces

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

Incomplete Proof

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...