Compact Spaces and Open Covers

Compactness is one of the most powerful and pervasive concepts in topology and analysis. It generalizes the properties of closed and bounded subsets of Euclidean space to arbitrary topological spaces, providing finiteness conditions that enable existence proofs, convergence arguments, and structural theorems.

Definition

Definition5.1Open Cover

Let $X$ be a topological space. An open cover of $X$ is a collection $\{U_\alpha\}_{\alpha \in A}$ of open subsets of $X$ such that $X = \bigcup_{\alpha \in A} U_\alpha$ . A subcover is a subcollection that still covers $X$ .

Definition5.2Compact Space

A topological space $X$ is compact if every open cover of $X$ has a finite subcover. That is, for every open cover $\{U_\alpha\}_{\alpha \in A}$ , there exist $\alpha_1, \ldots, \alpha_n \in A$ such that $X = U_{\alpha_1} \cup \cdots \cup U_{\alpha_n}$ .

Definition5.3Compact Subset

A subset $K$ of a topological space $X$ is compact if $K$ is compact in the subspace topology. Equivalently, every cover of $K$ by open sets of $X$ has a finite subcover.

Basic Properties

Theorem5.1Fundamental Properties of Compactness

A closed subset of a compact space is compact.
A compact subset of a Hausdorff space is closed.
A finite union of compact sets is compact.
The continuous image of a compact space is compact.
A finite topological space is compact.

Proof

(1): Let $C$ be closed in the compact space $X$ , and let $\{V_\alpha\}$ be an open cover of $C$ (open in $X$ ). Then $\{V_\alpha\} \cup \{X \setminus C\}$ is an open cover of $X$ . Extract a finite subcover; removing $X \setminus C$ (if present) gives a finite subcover of $C$ .

(2): Let $K$ be compact in the Hausdorff space $X$ . We show $X \setminus K$ is open. For $x \notin K$ and each $k \in K$ , choose disjoint open sets $U_k \ni x$ and $V_k \ni k$ (Hausdorff). Then $\{V_k\}_{k \in K}$ covers $K$ ; extract a finite subcover $V_{k_1}, \ldots, V_{k_n}$ . Set $U = U_{k_1} \cap \cdots \cap U_{k_n}$ . Then $U$ is an open neighborhood of $x$ disjoint from $V_{k_1} \cup \cdots \cup V_{k_n} \supseteq K$ , so $U \subseteq X \setminus K$ .

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Examples

ExampleCompact and Non-Compact Spaces

Compact: Finite sets (in any topology), $[a, b] \subseteq \mathbb{R}$ , $S^n$ , any finite product of compact spaces.
Not compact: $\mathbb{R}$ (covered by $\{(-n, n)\}_{n \geq 1}$ with no finite subcover), $(0, 1)$ (covered by $\{(1/n, 1)\}_{n \geq 2}$ ), any infinite discrete space.
Compact but not Hausdorff: The cofinite topology on an infinite set $X$ . Every open cover has a finite subcover: if $U_1$ is in the cover, then $X \setminus U_1$ is finite, and finitely many additional sets cover it.
Compact and Hausdorff: $[0, 1]$ , $S^n$ , the Cantor set.

The Finite Intersection Property

Definition5.4Finite Intersection Property

A collection $\{C_\alpha\}_{\alpha \in A}$ of sets has the finite intersection property (FIP) if every finite subcollection has nonempty intersection: $C_{\alpha_1} \cap \cdots \cap C_{\alpha_n} \neq \emptyset \quad \text{for all finite } \{\alpha_1, \ldots, \alpha_n\} \subseteq A.$

Theorem5.2Compactness via FIP

A topological space $X$ is compact if and only if every collection of closed sets with the finite intersection property has nonempty total intersection.

Proof

This is the contrapositive of the open cover definition, via De Morgan's laws.

$X$ is compact $\iff$ every open cover has a finite subcover $\iff$ if $\{U_\alpha\}$ are open with $\bigcup U_\alpha = X$ , then $\bigcup_{i=1}^n U_{\alpha_i} = X$ for some finite subcollection $\iff$ if $\{C_\alpha\}$ are closed with $\bigcap C_\alpha = \emptyset$ , then $\bigcap_{i=1}^n C_{\alpha_i} = \emptyset$ for some finite subcollection $\iff$ if all finite intersections of the $C_\alpha$ are nonempty, then $\bigcap C_\alpha \neq \emptyset$ .

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ExampleNested Interval Theorem

The Cantor intersection theorem is a consequence of the FIP characterization: if $C_1 \supseteq C_2 \supseteq \cdots$ is a decreasing sequence of nonempty closed subsets of a compact space, then $\bigcap_{n=1}^\infty C_n \neq \emptyset$ . The nested sequence automatically has the FIP.

Applied to $[0, 1]$ : if $[a_1, b_1] \supseteq [a_2, b_2] \supseteq \cdots$ is a nested sequence of closed intervals, their intersection is nonempty.

RemarkCompactness and Finiteness

Compactness is a topological analogue of finiteness. Just as every function on a finite set attains its maximum, every continuous real-valued function on a compact space attains its maximum (extreme value theorem). Many arguments that work for finite sets extend to compact spaces.