Compact Sets Are Closed in Hausdorff Spaces

Compactness Meets Separation

Compactness turns an arbitrary open cover into a finite subcover. A Hausdorff space, by contrast, provides a way to separate any two distinct points by disjoint open sets. The interaction between these properties is strong: if a set is compact, the finitely many separations needed for its points can be combined into one neighborhood separation. This is the key to proving that compact subsets of Hausdorff spaces are closed.

The metric spaces studied earlier are Hausdorff, but the result does not depend on having a metric. We will state and prove it for topological spaces, using only open sets, compactness, and the Hausdorff property.

This definition of a compact subset is equivalent to saying that \(K\), with the subspace topology, is compact. Indeed, a cover of \(K\) by open sets in \(X\) gives a cover by their intersections with \(K\). In the other direction, every open set in the subspace \(K\) is the intersection of \(K\) with an open set in \(X\); replacing each member of a subspace cover by such an ambient open set gives an open cover of \(K\) in \(X\).

Every metric space is Hausdorff. If \(x\ne y\), let \(r=d(x,y)/3\), which is positive. The open balls \(B_r(x)\) and \(B_r(y)\) are disjoint: if some \(z\) belonged to both, then

which is impossible. Thus the theorem we establish below applies, in particular, to compact subsets of metric spaces.

Separating a Compact Set from a Point

Fix a point outside a compact set. Hausdorffness gives a separate pair of open neighborhoods for that point and each individual point of the compact set. At first, the neighborhood of the outside point may depend on which point of the compact set is being separated. Compactness lets us reduce these choices to finitely many, so that their neighborhoods can be intersected.

Proof. If \(K\) is empty, take \(U=X\) and \(V=\varnothing\). Suppose instead that \(K\) is nonempty. For each \(y\in K\), we have \(y\ne x\). Since \(X\) is Hausdorff, choose open sets \(U_y,V_y\subseteq X\) such that

The sets \(V_y\), for \(y\in K\), form an open cover of \(K\). By compactness, there are finitely many points \(y_1,\ldots,y_m\in K\) such that

Set

The set \(U\) is open because it is a finite intersection of open sets, and it contains \(x\). The set \(V\) is open because it is a union of open sets, and it contains \(K\). For every \(i,j\), \(U\subseteq U_{y_i}\) and \(V_{y_j}\) contains \(V_{y_j}\), but these facts alone do not compare \(U_{y_i}\) and \(V_{y_j}\) when \(i\ne j\). Instead, observe that \(U\) is contained in every \(U_{y_j}\), so \(U\cap V_{y_j}\subseteq U_{y_j}\cap V_{y_j}=\varnothing\). This holds for each \(j\), and hence \(U\cap V=\varnothing\). The required neighborhoods have been constructed. \(\square\)

The finite intersection in this proof is crucial. A separate neighborhood of \(x\) for every \(y\in K\) would not by itself give one neighborhood of \(x\) disjoint from all of \(K\): an infinite intersection of open sets need not be open. Compactness supplies a finite subcover, making the intersection open.

Compact Subsets of Hausdorff Spaces Are Closed

Proof. Let \(K\) be a compact subset of a Hausdorff space \(X\). If \(K=X\), it is closed. Otherwise, take any \(x\in X\setminus K\). By the theorem on separating a point from a compact set, there is an open set \(U\) containing \(x\) that is disjoint from an open set containing \(K\). In particular, \(U\cap K=\varnothing\), so \(U\subseteq X\setminus K\).

We have shown that every point of \(X\setminus K\) has an open neighborhood contained in \(X\setminus K\). Therefore \(X\setminus K\) is open, and \(K\) is closed. \(\square\)

The argument establishes more than closedness: each point outside \(K\) has an open neighborhood that misses \(K\). In metric spaces, this theorem is a direct consequence of the fact that metric spaces are Hausdorff. It does not require a distance from the point to the set to be attained, nor does it require a minimum separation distance.

Worked Examples

A Related Consequence: Uniqueness of Limits

Hausdorffness also ensures that sequences cannot converge to two distinct points, even in a topological space where no metric is specified. Convergence here means that every open neighborhood of the proposed limit contains all terms from some point onward.

Proof. Suppose a sequence \((x_n)\) converges to both \(p\) and \(q\), and suppose for contradiction that \(p\ne q\). By Hausdorffness, there are disjoint open sets \(U,V\) with \(p\in U\) and \(q\in V\). Since \(x_n\to p\), there is an index \(N_1\) such that \(x_n\in U\) for every \(n\geq N_1\). Since \(x_n\to q\), there is an index \(N_2\) such that \(x_n\in V\) for every \(n\geq N_2\). For \(n\geq\max(N_1,N_2)\), the term \(x_n\) would belong to \(U\cap V\), which is empty. This contradiction proves \(p=q\). \(\square\)

For metric spaces, uniqueness of limits was established earlier. This result explains the topological source of that property: the Hausdorff condition supplies disjoint neighborhoods, while convergence would force sufficiently late terms into both neighborhoods.

Why the Hypotheses Matter

The theorem is useful whenever compactness is known but closedness is needed. In a metric space, one can often establish compactness by a sequence argument and then immediately conclude that the set is closed, without separately testing all points in its complement. The proof also clarifies which assumptions do the work: compactness reduces infinitely many pointwise separations to finitely many, and Hausdorffness provides those separations in the first place.

Do not reverse the theorem without additional hypotheses. A closed set need not be compact: for example, \(\mathbb{R}\) is closed in itself but is not compact, as the open cover \(\{(-n,n):n\geq1\}\) has no finite subcover. Nor should the conclusion be applied to compact sets in arbitrary topological spaces; the indiscrete example shows that Hausdorffness matters.