Compactness is the topological form of finite extraction. It lets infinitely many local pieces be controlled by finitely many of them.
Open covers and finite extraction
The definition is global, but it is used by pulling covers back, pushing them forward, and passing to closed subsets.
Definition. A space \(X\) is compact if every open cover \(\{U_\alpha\}_{\alpha\in A}\) of \(X\) has a finite subcover. A subset \(K\subseteq X\) is compact if it is compact in the subspace topology.
Proposition. A closed subset of a compact space is compact.
Proof. Let \(F\subseteq X\) be closed and let \(\{U_\alpha\}\) be an open cover of \(F\) in the subspace topology. Write \(U_\alpha=F\cap V_\alpha\) with \(V_\alpha\) open in \(X\). Then \(\{V_\alpha\}\cup\{X\setminus F\}\) covers \(X\). Compactness gives finitely many \(V_{\alpha_k}\) together with possibly \(X\setminus F\); intersecting with \(F\) gives a finite cover of \(F\).
Proposition. The continuous image of a compact space is compact.
Proof. If \(f:X\to Y\) is continuous and \(\{V_\alpha\}\) covers \(f(X)\), then \(\{f^{-1}(V_\alpha)\}\) is an open cover of \(X\). Choose finitely many inverse images covering \(X\); the corresponding \(V_\alpha\) cover \(f(X)\).
Finite intersection property
Closed-set language turns compactness into an intersection theorem.
Theorem. A space \(X\) is compact if and only if every family of closed sets with the finite intersection property has nonempty total intersection.
Proof. A family \(\{F_\alpha\}\) has the finite intersection property exactly when no finite subfamily of the open complements \(X\setminus F_\alpha\) covers \(X\). If \(\bigcap_\alpha F_\alpha=\varnothing\), then the complements form an open cover. Compactness gives a finite subcover, equivalent to a finite subfamily of the \(F_\alpha\) having empty intersection. The converse is the same argument in reverse.
Corollary. In a compact space, a decreasing sequence \(F_1\supseteq F_2\supseteq\cdots\) of nonempty closed sets has nonempty intersection.
Proof. Every finite subfamily has intersection equal to the smallest-indexed set among finitely many chosen tail sets, hence nonempty. Apply the finite intersection property criterion.
Compact subsets in Hausdorff spaces
Hausdorff separation makes compact sets behave like closed and bounded sets in metric analysis.
Theorem. If \(K\) is compact and \(Y\) is Hausdorff, then the graph of every continuous map \(f:K\to Y\) is closed in \(K\times Y\). In particular, compact subsets of Hausdorff spaces are closed.
Proof. For the graph, if \((x,y)\) is not on the graph, then \(y\ne f(x)\). Choose disjoint neighborhoods \(V\ni y\) and \(W\ni f(x)\). By continuity, some neighborhood \(U\ni x\) satisfies \(f(U)\subseteq W\), so \(U\times V\) avoids the graph. Thus the complement is open. The closedness of compact subsets follows by applying the point-versus-compact separation argument or by taking \(f\) to be the inclusion of a compact subset.
Metric compactness
In metric spaces, compactness admits sequential and total-bounded formulations, but these are not valid in arbitrary spaces.
Theorem. For a metric space, compactness is equivalent to sequential compactness, and also to completeness plus total boundedness.
Proof. Compactness gives sequential compactness by covering the space with finitely many balls of radius \(1\), then one ball of radius \(1/2\), and so on; a diagonal nested-ball argument extracts a Cauchy subsequence, whose cluster point exists by compactness. Sequential compactness implies total boundedness, otherwise one constructs a sequence with pairwise distances bounded below. It also implies completeness by applying sequential compactness to Cauchy sequences. Finally, complete and totally bounded spaces are compact because every sequence has a Cauchy subsequence with a limit, and sequential compactness in metric spaces implies the open-cover definition via the Lebesgue-number argument.
Remark. Outside metric spaces, sequences may miss compactness; nets and filters repair this defect.
Compactness beyond sequences
Theorem. For an arbitrary topological space \(X\), the following are equivalent: \(X\) is compact; every net has a convergent subnet; and every ultrafilter on \(X\) converges.
Proof outline. If a net has no convergent subnet, its tails generate a family of closed sets with the finite intersection property but empty intersection, contradicting compactness. Conversely, from an open cover with no finite subcover, direct the finite subcollections by inclusion and choose a point outside each union; no subnet can converge. Ultrafilters encode the same finite-intersection data: compactness gives a common point of the closures of their members, and maximality turns that cluster point into a limit.
Counterexample. The ordinal interval \([0,\omega_1)\) is countably compact but not compact. Every sequence is bounded below some countable ordinal and has a convergent subsequence, yet the open cover \[ \{[0,\alpha):\alpha<\omega_1\} \] has no finite subcover. Sequential criteria characterize compactness in metric and first-countable settings, not in all spaces.