Proof Compact Hausdorff Subsets Are Closed

The Separation Argument Behind Closedness

The previous tutorial established that, in a Hausdorff space, a point outside a compact set can be separated from the whole set by disjoint open neighborhoods. This is stronger than separating the point from each individual point of the set: compactness turns those individual separations into one separation from the entire set. Here we use that result to explain the proof that compact subsets of Hausdorff spaces are closed, and then apply the same ideas to a further separation property of compact Hausdorff spaces.

The key observation is that a subset is closed exactly when every point outside it has an open neighborhood that misses it. Thus, for a compact set \(K\subseteq X\), the point-versus-compact separation theorem supplies precisely the neighborhoods needed to show that \(X\setminus K\) is open. The closedness conclusion is not a consequence of compactness alone: the separation step depends on Hausdorffness.

From Separated Neighborhoods to a Closed Set

The point-versus-compact theorem gives a neighborhood of each exterior point that does not meet \(K\). The argument below records how that local information gives the desired global conclusion. It is an application of the earlier separation theorem, rather than a new proof of that theorem.

Proof. If \(K=\varnothing\), then \(K\) is closed. Otherwise, take any \(x\in X\setminus K\). By the Point-versus-Compact Separation Theorem, there are disjoint open sets \(U,V\subseteq X\) such that \(x\in U\) and \(K\subseteq V\). Since \(U\cap V=\varnothing\) and \(K\subseteq V\), we have \(U\cap K=\varnothing\). Therefore \(U\subseteq X\setminus K\). Every point of \(X\setminus K\) consequently has an open neighborhood contained in \(X\setminus K\), so \(X\setminus K\) is open. Hence \(K\) is closed. \(\square\)

This proof has a useful dependency structure. Hausdorffness provides separations between distinct points; compactness lets the point-versus-compact theorem combine finitely many of those separations; and the definition of a closed set turns the resulting neighborhoods into closedness. In the point-versus-compact proof, the finite subcover matters because the neighborhood of the fixed point is a finite intersection of open sets, which is open. An arbitrary intersection of open sets need not be open.

Why the Hypotheses Matter

The conclusion does not follow from compactness alone. For example, let \(X=\{p,q\}\) have the indiscrete topology \(\{\varnothing,X\}\). The singleton \(K=\{p\}\) is compact: every open cover of \(K\) must include \(X\), the only open set containing \(p\), and that one set is a finite subcover. But \(K\) is not closed, since its complement \(\{q\}\) is not open. The space is not Hausdorff, because the only open neighborhood of either point is \(X\), so the two points cannot have disjoint open neighborhoods.

This example identifies the role of the separation assumption. Compactness controls open covers; it does not by itself provide neighborhoods that distinguish a point outside a set from points inside it. Hausdorffness supplies pointwise separation, and compactness is what makes the pointwise separations effective for an entire compact set.

A Further Consequence: Compact Hausdorff Spaces Are Normal

The same separation ideas yield a stronger ambient-space consequence. A topological space is called normal if every pair of disjoint closed subsets has disjoint open neighborhoods: whenever \(A\) and \(B\) are closed and \(A\cap B=\varnothing\), there are disjoint open sets \(U,V\) with \(A\subseteq U\) and \(B\subseteq V\). We prove this property for compact Hausdorff spaces. The proof uses two earlier results: closed subsets of compact spaces are compact, and disjoint compact subsets of a Hausdorff space can be separated by disjoint open neighborhoods.

Proof. Let \(X\) be compact and Hausdorff, and let \(A,B\subseteq X\) be disjoint closed sets. Since \(A\) and \(B\) are closed subsets of the compact space \(X\), each is compact, by the theorem on closed subsets of compact spaces. Now apply the theorem on separating disjoint compact sets in a Hausdorff space. It gives disjoint open sets \(U,V\subseteq X\) such that \(A\subseteq U\) and \(B\subseteq V\), as required.

This reasoning also covers empty sets. If either closed set is empty, the theorem on separating disjoint compact sets provides the required open neighborhoods; alternatively, one can take the empty set as the neighborhood of the empty set and \(X\) as the neighborhood of the other set. Thus every pair of disjoint closed sets has the required separation, and \(X\) is normal. \(\square\)

Normality is a statement about separating two sets, while the compact-closed corollary is a statement about separating one point from a set. The proof of normality works because, inside a compact space, closed sets are compact; that converts a separation problem for closed sets into one already handled for compact sets. Notice that compactness of the ambient space is essential to this argument: in a general Hausdorff space, closed subsets need not be compact.

Using the Proof Without Losing Its Scope

When applying the compact-closed result, check first that compactness refers to the subspace topology on \(K\), and that Hausdorffness holds in the ambient space \(X\). Then take an arbitrary point outside \(K\), not an arbitrary point inside it. The separation theorem gives an open neighborhood of that exterior point disjoint from \(K\); this is exactly what is needed to prove the complement open.

For normality, there is one additional step: closed subsets of a compact space are compact. Only after establishing that fact can the disjoint-compact-set separation theorem be applied. Keeping these dependencies explicit prevents a common error: assuming that any two disjoint closed subsets of any Hausdorff space can be separated by disjoint open neighborhoods. Hausdorffness alone does not guarantee that broader conclusion.