# The Unapologetic Mathematician

## Some compact subspaces

Let’s say we have a compact space $X$. A subset $C\subseteq X$ may not be itself compact, but there’s one useful case in which it will be. If $C$ is closed, then $C$ is compact.

Let’s take an open cover $\{F_i\}_{i\in\mathcal{I}}$ of $C$. The sets $F_i$ are open subsets of $C$, but they may not be open as subsets of $X$. But by the definition of the subspace topology, each one must be the intersection of $C$ with an open subset of $X$. Let’s just say that each $F_i$ is an open subset of $X$ to begin with.

Now, we have one more open set floating around. The complement of $C$ is open, since $C$ is closed! So between the collection $\{F_i\}$ and the extra set $X\setminus C$ we’ve got an open cover of $X$. By compactness of $X$, this open cover has a finite subcover. We can throw out $X\setminus C$ from the subcover if it’s in there, and we’re left with a finite open cover of $C$, and so $C$ is compact.

In fact, if we restrict to Hausdorff spaces, $C$ must be closed to be compact. Indeed, we proved that if $C$ is compact and $X$ is Hausdorff then any point $x\in X\setminus C$ can be separated from $C$ by a neighborhood $U\subseteq X\setminus C$. Since there is such an open neighborhood, $x$ must be an interior point of $X\setminus C$. And since $x$ was arbitrary, every point of $X\setminus C$ is an interior point, and so $X\setminus C$ must be open.

Putting these two sides together, we can see that if $X$ is compact Hausdorff, then a subset $C\subseteq X$ is compact exactly when it’s closed.

January 15, 2008