Compact Sets Need Not Be Closed Without Hausdorffness

Why the Hausdorff Hypothesis Matters

In a Hausdorff space, every compact subset is closed. The Hausdorff assumption is essential: without it, compactness need not give enough separation to make a subset closed. An infinite indiscrete space provides a direct example. Its open-cover structure is so limited that every subset is compact, while its closed sets are limited to the empty set and the whole space.

Recall that a subset \(A\) of a topological space \((X,\mathcal{T})\) is given the subspace topology, consisting of all sets \(A\cap U\) with \(U\in\mathcal{T}\). The subset \(A\) is compact if every cover of \(A\) by open sets in this subspace topology has a finite subcover. Equivalently, every cover of \(A\) by open sets of \(X\) has a finite subcover: intersecting an ambient open cover with \(A\) gives a subspace-open cover, and each subspace-open set is an intersection of \(A\) with an ambient open set.

In an indiscrete space with at least two points, no two distinct points have disjoint open neighborhoods. Indeed, the only open neighborhood of any point is \(X\), and \(X\cap X=X\ne\varnothing\). Thus such a space is not Hausdorff. We will use both parts of the definition of this topology: the scarcity of open sets makes covers easy to reduce, and the scarcity of closed sets makes proper nonempty subsets fail to be closed.

Compactness of Every Subset

Proof. Let \((X,\mathcal{T})\) be indiscrete, and let \(A\subseteq X\). If \(A=\varnothing\), the empty finite family is a finite subcover of every cover of \(A\), so \(A\) is compact. Suppose \(A\ne\varnothing\). The subspace-open sets in \(A\) are \(A\cap\varnothing=\varnothing\) and \(A\cap X=A\), so the subspace topology on \(A\) is \(\{\varnothing,A\}\). Consider any open cover of \(A\) in this topology. Since its union is \(A\), at least one member of the cover must be \(A\): a union consisting only of empty sets would be empty. That single member is a finite subcover. Therefore \(A\) is compact. \(\square\)

The argument includes all cases. It does not depend on \(A\) being finite, on \(X\) being finite, or on \(A\) being a proper subset. In particular, the compactness of a subset here does not provide information about whether its complement is open.

Proof. A subset \(C\subseteq X\) is closed if and only if \(X\setminus C\) is open. In the indiscrete topology, \(X\setminus C\) must therefore equal either \(\varnothing\) or \(X\). In the first case \(C=X\), and in the second case \(C=\varnothing\). Both \(\varnothing\) and \(X\) are indeed closed because their complements, \(X\) and \(\varnothing\), are open. Thus these are exactly the closed subsets. Any proper nonempty subset is neither one, so it is not closed. \(\square\)

An Infinite Compact Subset That Is Not Closed

This example proves that the unrestricted statement “every compact subset is closed” is false. What fails is not compactness: the open-cover definition has been checked. What fails is the separation hypothesis that is available in a Hausdorff space. In an indiscrete space with at least two points, even two distinct points cannot be separated by disjoint open neighborhoods, because each has only \(X\) as an open neighborhood.

Reading the Example Alongside the Hausdorff Result

The earlier result on compact subsets of Hausdorff spaces says that if the ambient space is Hausdorff, then a compact subset of that space is closed. Its hypothesis concerns the ambient space. The example above has an ambient space that is not Hausdorff, so it does not contradict that result. Compactness of \(A\) alone is not a replacement for the Hausdorff assumption.

The examples separate three facts that should not be conflated:

• A set can be compact because every open cover has a finite subcover.
• A set is closed because its complement is open in the ambient space.
• The theorem that compact subsets are closed requires the ambient space to be Hausdorff.

In the infinite indiscrete example, every subset satisfies the first condition, while only the empty set and the entire space satisfy the second. The contrast is especially sharp because it applies not just to one chosen infinite subset, but to every proper nonempty subset.

A Common Mistake

Common pitfall. A tempting but incorrect calculation is: “\(A\) is compact, so \(A\) is closed.” For the specific set \(A=\{2,4,6,\dots\}\subseteq\mathbb{N}\) above, the compactness calculation is valid, but the conclusion is not. The closedness test requires checking whether \(X\setminus A=\{1,3,5,\dots\}\) is open. It is not: it is a nonempty proper subset of \(X\), whereas the only open sets are \(\varnothing\) and \(X\).

The correction is to include the hypothesis when using the compact-subset theorem: it applies to compact subsets of a Hausdorff space. Here \(X\) is not Hausdorff, since any two open neighborhoods of distinct points are both \(X\), and their intersection is \(X\), not \(\varnothing\). Thus the theorem cannot be applied to this example.