Is Every Subset of a Compact Set Compact?

A compact space can contain subsets that are not compact. The key distinction is whether the subset is closed in the compact space. Compactness is inherited by closed subsets, but not by arbitrary subsets.

What compactness means for a subset

A subset $A$ of a topological space $X$ is given the subspace topology: its open sets are sets of the form $A \cap U$, where $U$ is open in $X$. Compactness of $A$ means that every cover of $A$ by sets open in this subspace topology has a finite subcover.

The fact that $X$ is compact controls open covers of all of $X$. It does not ensure that every open cover of a smaller subset can be reduced to finitely many sets. A closedness condition supplies the missing link.

Why closed subsets of compact spaces are compact

Suppose $X$ is compact and $F$ is closed in $X$. Take any open cover of $F$ in its subspace topology. Each member of that cover can be written as $F \cap U$ for some open set $U$ in $X$. The corresponding open sets $U$ cover $F$.

Add the open set $X \setminus F$ to this collection. Since $F$ is closed, $X \setminus F$ is open, and the resulting collection covers all of $X$. Compactness of $X$ gives a finite subcover. Removing $X \setminus F$ if it was selected leaves finitely many sets that cover $F$. Thus $F$ is compact.

This proof works in any topological space; it does not require the space to be Hausdorff.

A concrete counterexample to the general claim

The interval $[0,1]$ is compact in the usual topology on the real line. Its subset $(0,1)$ is not compact.

For each integer $n \geq 2$, define the relatively open subset

$$U_n = (1/n,1)$$

of $(0,1)$. These sets cover $(0,1)$: for any $x$ with $0<x<1$, choose an integer $n$ large enough that $1/n<x$. But no finite collection of the $U_n$ covers $(0,1)$. In a finite collection, let $N$ be the largest index. Their union is just $(1/N,1)$, which misses points of $(0,1)$ smaller than or equal to $1/N$.

Therefore $(0,1)$ is a noncompact subset of the compact space $[0,1]$. It is not closed in $[0,1]$, which is why the closed-subset theorem does not apply.

When compact subsets are closed

The reverse implication needs a separation assumption. If $X$ is Hausdorff, every compact subset of $X$ is closed. Consequently, in a compact Hausdorff space, a subset is compact if and only if it is closed.

Without Hausdorffness, a compact subset need not be closed. For example, let $X=\{0,1\}$ have topology $\{\varnothing,\{1\},X\}$. Every subset is compact because $X$ is finite, but $\{1\}$ is not closed: its complement $\{0\}$ is not open.

So the precise general statement is: closed subsets of compact spaces are compact, while compact subsets of Hausdorff spaces are closed. Do not treat either direction as unconditional.

Common mistakes

• Assuming compactness passes to every subspace. It does not; the interval example gives a counterexample.
• Using “closed” without specifying the ambient space. The relevant condition is that $F$ is closed in the compact space $X$.
• Assuming compact always means closed. That implication requires the ambient space to be Hausdorff.
• Confusing a subset with its induced topology. Compactness here is evaluated using the subspace topology, not an independently chosen topology.