# Compactness and Hausdorff Spaces

It's a nice fact that a continusous bijection from a compact space to a Hausdorff space is a homeomorphism. In other words, continuity of the inverse is implied by the setting. In reviewing some topology, I've found it nice to write up some notes that build up this fact. These are primarily adapted from Armstrong's Basic Topology.

Throughout, we'll let $$X$$ and $$Y$$ be topological spaces. In thinking about $$\mathbb{R}^n$$, it's nice when handling subspaces to see that they are closed and bounded. This notion is useful when dealing with surfaces, for example. So it would be nice to generalize this notion. We do so with the following definition:

Let $$\mathcal{F}$$ be a family of open subsets of $$X$$ such that $$\bigcup \mathcal{F} = X$$. We call $$\mathcal{F}$$ an open cover of $$X$$. If $$\mathcal{F}$$ is an open cover of $$X$$ and $$\mathcal{F}'$$ is a subfamily of $$\mathcal{F}$$ which also covers $$X$$, then $$\mathcal{F}'$$ is a subcover of $$\mathcal{F}$$. A space $$X$$ is compact if every open cover of $$X$$ has a finite subcover.

Compactness is a property of a topological space; that is, it's invariant under homeomorphism. But even stronger, it is preserved by any continuous surjection:

Lemma 1. The continuous image of a compact space is compact.

Proof. (click to open)

Let $$f:\:X\to Y$$ be a continuous surjection with $$X$$ compact. Pick an open cover of $$Y$$, call it $$\mathcal{F} := \{U_i\}$$. Then $$f^{-1}(U_i)$$ is open by the continuity of $$f$$, and so $$\bigcup f^{-1}(U_i)$$ is an open cover of $$X$$. By the compactness of $$X$$, we can pick a finite subcover $$\{f^{-1}(U_1),\dots,f^{-1}(U_k)\}$$ of $$\bigcup f^{-1}(U_i)$$. Since $$f$$ is surjective, we get \begin{align*} X &= f^{-1}(U_1) \cup\dots\cup f^{-1}(U_k)\\ f(X) &= f(f^{-1}(U_1) \cup\dots\cup f^{-1}(U_k))\\ Y &= U_1 \cup \dots \cup U_k. \end{align*} So $$Y$$ is compact. $$\Box$$

Lemma 2. A closed subset of a compact space is compact.

Proof.

Let $$X$$ be compact, $$C$$ a closed subset of $$X$$, and $$\mathcal{F}$$ an open cover of $$C$$. Since $$X\setminus C$$ is open, $$\mathcal{F} \cup (X\setminus C)$$ is an open cover of $$X$$. By the compactness of $$X$$, we can find a finite subcover $$U_1 \cup \dots \cup U_k \cup (X\setminus C)$$ with $$U_i \in \mathcal{F}$$. So then $$U_1 \cup \dots \cup U_k$$ is a finite subcover of $$\mathcal{F}$$. $$\Box$$

Lemma 3. If $$A$$ is a copmact subset of a Hausdorff space $$X$$, and if $$x \in X \setminus A$$, then there exist disjoint neighborhoods of $$x$$ and $$A$$. Therefore a compact subset of a Hausdorff space is closed.

Proof.

Take a point $$z\in A$$. Since $$X$$ is Hausdorff, we can find neighborhoods $$x \in U_z$$ and $$z \in V_z$$ which are disjoint. We now want to vary $$z$$ across all points of $$A$$. So for each $$z\in A$$, we get a neighborhood which we can put into a collection $$\mathcal{F} = \{V_z\:|\: z\in A\}$$. This is an open cover of $$A$$, so by the compactness of $$A$$, we can find a finite subcover $$\{V_{z_1},\dots,V_{z_k}\}$$. Note that $$V = V_{z_1}\cup\dots\cup V_{z_k}$$ is then a neighborhood of $$A$$. Let $$U = U_{z_1} \cap \dots \cap U_{z_k}$$ be an open neighborhood of $$x$$. Notice that $$U$$ and $$V$$ must be disjoint by construction, so these neighborhoods are disjoint. Since $$x \in X\setminus A$$, we see that $$A$$ must be closed. $$\Box$$

Theorem. A continuous bijection from a compact space $$X$$ to a Hausdorff space $$Y$$ is a homeomorphism, i.e., the continuity of $$f^{-1}$$ is automatic.

Proof. Let $$f\::\:X \to Y$$ be the continuous bijection and $$C$$ be a closed subset of $$X$$. Then $$C$$ is compact (Lemma 2) and so $$f(C)$$ is compact (Lemma 1). Thus $$f(C)$$ is closed (Lemma 3). This gives us that $$f$$ takes closed sets to closed sets, so $$f^{-1}$$ is continuous. $$\Box$$