Compactness

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 $|J| < \infty$.

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.

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. $\square$

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.

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.

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.

Lemma 5.5 (Tube Lemma). Let $X$ be compact and $Y$ any space. If $N \subseteq X \times Y$ is an open set containing the slice $\{x_0\} \times Y$, then there exists a neighbourhood $U$ of $x_0$ in $X$ such that $U \times Y \subseteq N$.

The tube lemma is the key ingredient in proving Tychonoff’s theorem for finite products and shows how compactness of the first factor allows local control to be extended globally.

In metric spaces, compactness admits several equivalent characterisations:

Theorem 5.6. For a metric space $(X, d)$, the following are equivalent:

1. $X$ is compact.
2. $X$ is sequentially compact.
3. $X$ is complete and totally bounded (every $\varepsilon$-net is finite).

Definition. A space is locally compact if every point has a compact neighbourhood. $\mathbb{R}^n$ is locally compact but not compact. Locally compact Hausdorff spaces admit a one-point compactification (the Alexandroff compactification).

Example 5.3 (Cantor Set). The Cantor set $C$ is a compact subset of $[0, 1]$ that is uncountable, nowhere dense, and has Lebesgue measure zero. It is homeomorphic to $\{0, 1\}^{\mathbb{N}}$ with the product topology.

Problem. Show that the set of continuous functions $F = \{f \in C([0, 1]) : \|f\|_\infty \leq 1\}$ is not compact in the sup-norm topology.

Definition. A space $X$ is paracompact if every open cover has a locally finite open refinement. Every compact space is paracompact, and every metric space is paracompact. Paracompactness is essential for the existence of partitions of unity, which enable the construction of global objects from local data.

Definition. A collection $\{A_\alpha\}$ of subsets of $X$ has the finite intersection property (FIP) if every finite subcollection has nonempty intersection.

Theorem 5.7 (FIP Characterisation). $X$ is compact iff for every collection $\{F_\alpha\}$ of closed subsets of $X$ with the FIP, $\bigcap_\alpha F_\alpha \neq \emptyset$.

This characterisation is often more convenient for proofs. For example, it immediately shows that the continuous image of a compact space is compact, and that a compact Hausdorff space is normal.

Problem. Show that $\mathbb{R}^n$ is homeomorphic to $S^n \setminus \{p\}$ (the sphere minus a point), and describe the one-point compactification of $\mathbb{R}^n$.

Problem. Prove that the closed unit ball $B = \{f \in C([0,1]) : \|f\|_\infty \leq 1\}$ is not compact using the finite intersection property.