The Timpson Lyceum · General Topology (Point Set Topology)

Compactness & Open Covers – Lecture 21 (Topology)

Overview

This lecture introduces compactness, the final topological property of the course and the least intuitive one. In familiar spaces like $\mathbb{R}^2$ or $\mathbb{R}^3$ , "compact" coincides with "closed and bounded" (the notion used in calculus), but the genuine topological definition is stated purely in terms of open covers and finite subcovers. We develop that definition, see why it is generally easier to disprove compactness than to prove it, and establish the core theorems: compactness is intrinsic (independent of ambient space), it is preserved by continuous images and finite unions, and — together with the Tube Lemma — it is preserved by finite products. We also see how compactness interacts with the Hausdorff condition (compact sets are closed) and with closed subsets.

Key Concepts

Open cover. A collection $\mathcal{O}$ of open sets whose union contains $X$ : $\bigcup_{U \in \mathcal{O}} U \supseteq X$ .
Finite subcover. A finite subcollection of a given cover that still covers the set.
Compact space. A topological space $X$ is compact if and only if every open cover of $X$ has a finite subcover. The quantifier is essential: exhibiting one finite cover (e.g. $\{X\}$ itself) proves nothing — you must handle every possible cover.

Examples on $\mathbb{R}$ (standard topology).

$\{(-\infty, 1),\,(-1, \infty)\}$ is a finite open cover of $\mathbb{R}$ , but this alone does not make $\mathbb{R}$ compact.
$\mathcal{O} = \{(n-1,\, n+1) : n \in \mathbb{Z}\}$ covers $\mathbb{R}$ , yet deleting even one interval $(n-1,n+1)$ leaves the integer $n$ uncovered. So $\mathcal{O}$ has no finite subcover, and $\mathbb{R}$ is not compact.
Asymmetry: to show a set is not compact, produce one bad cover; to show it is compact, give a procedure that extracts a finite subcover from an arbitrary cover.

Compactness is intrinsic (the subspace lemma). For $A \subseteq X$ , $A$ is compact in $X$ if and only if $A$ is compact in the subspace topology it inherits from $X$ . Recall that open sets of $A$ have the form $V = U \cap A$ with $U$ open in $X$ . The lemma lets us stop tracking whether a cover lives in $X$ or in $A$ — the two are logically equivalent.

Continuous images. If $f : X \to Y$ is continuous and $A$ is compact in $X$ , then $f(A)$ is compact in $Y$ . (Pull an open cover of $f(A)$ back through $f$ ; continuity makes the preimages open, compactness of $A$ gives a finite subcover, and pushing forward recovers a finite subcover of the original.) Compactness is a topological property, and continuous functions preserve it.

Closed subset of a compact set. If $D$ is compact in $X$ , $C \subseteq D$ , and $C$ is closed, then $C$ is compact. (Add the open set $C^{c}$ to any open cover of $C$ to get an open cover of $D$ ; extract a finite subcover, discard $C^{c}$ , and what remains covers $C$ .)

Compact sets in Hausdorff spaces are closed. If $X$ is Hausdorff and $A$ is compact in $X$ , then $A$ is closed. The Hausdorff condition — any two distinct points have disjoint neighborhoods — is exactly what powers the proof.

Unions and intersections.

A finite union $C_1 \cup \dots \cup C_n$ of compact sets is compact (concatenate the finite subcovers).
If $X$ is Hausdorff, an arbitrary intersection $\bigcap_\alpha C_\alpha$ of compact sets is compact. Compact sets thus behave much like closed sets.

Tube Lemma. Let $Y$ be compact. If $x \in X$ and $U$ is open in $X \times Y$ with $\{x\} \times Y \subseteq U$ , then there exists an open $W \subseteq X$ with $x \in W$ such that the whole "tube" satisfies $W \times Y \subseteq U$ . Compactness of $Y$ is what allows the defining intersection of the $W$ 's to be taken over finitely many sets, keeping $W$ open.

Products. If $X$ and $Y$ are compact, then $X \times Y$ is compact. (By induction this extends to any finite product $A_1 \times \dots \times A_n$ .) Each slice $\{x\} \times Y$ is compact, gets a finite subcover whose union is a tube; the Tube Lemma fattens it to $W_x \times Y$ ; the $W_x$ cover $X$ , so finitely many tubes — each with its own finite cover — tile $X \times Y$ with a finite cover.

Worked Example

Claim. The set $A = \{0\} \cup \left\{ \tfrac{1}{n} : n \in \mathbb{Z}^{+} \right\}$ is compact in $\mathbb{R}$ (standard topology).

The strategy is a general recipe: from an arbitrary open cover, capture all but finitely many points with a single set, then cover the finite remainder one set at a time.

\begin{aligned} &\text{Let } \mathcal{C} \text{ be any open cover of } A.\\ &\exists\, U_0 \in \mathcal{C} \text{ with } 0 \in U_0,\ \text{and } \exists\, a<0<b \text{ with } 0 \in (a,b) \subseteq U_0.\\ &\text{By the Archimedean property, } \exists\, N \in \mathbb{Z}^{+} \text{ with } \tfrac{1}{N} < b.\\ &\text{For all } n \ge N:\quad 0 < \tfrac{1}{n} \le \tfrac{1}{N} < b \ \Longrightarrow\ \tfrac{1}{n} \in (a,b) \subseteq U_0.\\ &\text{So } U_0 \text{ already contains } 0 \text{ and every } \tfrac{1}{n} \text{ with } n \ge N.\\ &\text{The only uncovered points are } \tfrac{1}{1}, \tfrac{1}{2}, \dots, \tfrac{1}{N-1} \ (\text{finitely many}).\\ &\text{For each } n < N \text{ choose } U_n \in \mathcal{C} \text{ with } \tfrac{1}{n} \in U_n.\\ &\{\,U_0, U_1, U_2, \dots, U_{N-1}\,\} \subseteq \mathcal{C} \text{ is a finite subcover of } A. \qquad \blacksquare \end{aligned}

Since an arbitrary open cover admits a finite subcover, $A$ is compact. (Geometrically: the interval around $0$ swallows the entire tail $\tfrac{1}{N}, \tfrac{1}{N+1}, \dots$ , leaving only finitely many early terms to mop up.)

Summary

$X$ is compact iff every open cover has a finite subcover — the "every" is the whole point; one lucky finite cover proves nothing.
Disproving compactness needs a single bad cover (e.g. $\{(n-1,n+1)\}$ on $\mathbb{R}$ ); proving it needs a procedure that works for all covers.
Compactness is intrinsic: a subset is compact in $X$ iff it is compact in its subspace topology, so we may ignore which space a cover lives in.
Continuous images of compact sets are compact, and a closed subset of a compact set is compact.
In a Hausdorff space, compact sets are closed; finite unions of compact sets are compact, and (Hausdorff) arbitrary intersections are compact too.
The Tube Lemma (requiring a compact factor) gives that a finite product of compact spaces is compact.