Compactness

5.1 Open Covers

Definition. Let $X$ be a topological space. An open cover of $X$ is a collection $\{U_\alpha\}_{\alpha \in I}$ of open sets such that $\bigcup_{\alpha \in I} U_\alpha = X$.

A subcover is a subcollection $\{U_\alpha\}_{\alpha \in J}$ ($J \subseteq I$) that still covers $X$. A finite subcover has $|J| < \infty$.

5.2 Compact Spaces

Definition. A topological space $X$ is compact if every open cover of $X$ has a finite subcover.

Theorem 5.1 (Heine–Borel). A subset of $\mathbb{R}^n$ (with the standard topology) is compact if and only if it is closed and bounded.

Example 5.1. $[0, 1]$ is compact in $\mathbb{R}$. $(0, 1)$ is not compact: the open cover $\{(1/n, 1)\}_{n=2}^\infty$ has no finite subcover.

Example 5.2. $\mathbb{R}$ is not compact (it is not bounded). Any finite set is compact.

Proposition 5.1. Every closed subset of a compact space is compact.

Proposition 5.2. Every compact subset of a Hausdorff space is closed.

5.3 Compactness in $\mathbb{R}$

Theorem 5.2. The closed interval $[a, b]$ is compact (with the standard topology on $\mathbb{R}$).

Proof. Let $\mathcal{U} = \{U_\alpha\}$ be an open cover of $[a, b]$. Let

$S = \{x \in [a, b] : [a, x] \text{ has a finite subcover from } \mathcal{U}\}.$

Then $a \in S$ (since $a \in$ some $U_\alpha$), so $S \neq \emptyset$. Let $s = \sup S$. One shows $s \in S$ and $s = b$, completing the proof. $\square$

5.4 Products: Tychonoff”s Theorem

Theorem 5.3 (Tychonoff). The product of any collection of compact spaces is compact.

For finite products, Tychonoff’s theorem follows from the tube lemma and is accessible without the axiom of choice. For arbitrary products, the full axiom of choice is required.

5.5 Sequential Compactness

Definition. A space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.

Theorem 5.4. In metric spaces, compactness and sequential compactness are equivalent.

Proposition 5.3. For $\mathbb{R}^n$: compact $\Leftrightarrow$ sequentially compact $\Leftrightarrow$ closed and bounded.

5.6 Compactness and Continuity

Theorem 5.5 (Extreme Value Theorem, generalised). If $X$ is compact and $f : X \to \mathbb{R}$ is continuous, then $f$ attains its maximum and minimum on $X$.

Proof. Since $f$ is continuous, $f(X)$ is compact in $\mathbb{R}$, hence closed and bounded. A closed bounded subset of $\mathbb{R}$ contains its supremum and infimum. $\square$

Proposition 5.4. The continuous image of a compact space is compact.