# The Unapologetic Mathematician

## Compact Spaces

An amazingly useful property for a space $X$ is that it be “compact”. We define this term by saying that if $\{U_i\}_{i\in\mathcal{I}}$ is any collection of open subsets of $X$ indexed by any (possibly infinite) set $\mathcal{I}$ so that their union $\bigcup\limits_{i\in\mathcal{I}}U_i$ is the whole of $X$ — the sexy words are “open cover” — then there is some finite collection of the index set $\mathcal{A}\subseteq\mathcal{I}$ so that the union of this finite number of open sets $\bigcup\limits_{i\in\mathcal{A}}U_i$ still contains all of $X$ — the sexy words are “has a finite subcover”.

So why does this matter? Well, let’s consider a Hausdorff space $X$, a point $x\in X$, and a finite collection of points $A\subseteq X$. Given any point $a\in A$, we can separate $x$ and $a$ by open neighborhoods $x\in U_a$ and $a\in V_a$, precisely because $X$ is Hausdorff. Then we can take the intersection $U=\bigcap\limits_{a\in A}U_a$ and the union $V=\bigcup\limits_{a\in A}V_a$. The set $U$ is a neighborhood of $X$, since it’s a finite intersection of neighborhoods, while the set $V$ is a neighborhood of $A$. These two sets can’t intersect, and so we have separated $x$ and $A$ by neighborhoods.

But what if $A$ is an infinite set? Then the infinite intersection $\bigcap\limits_{a\in A}U_a$ may not be a neighborhood of $x$! Infinite operations sometimes cause problems in topology, but compactness can make them finite. If $A$ is a compact subset of $X$, then we can proceed as before. For each $a\in A$ we have open neighborhoods $x\in U_a$ and $a\in V_a$, and so $A\subseteq\bigcup\limits_{a\in A}V_a$ — the open sets $V_a$ form a cover of $A$. Then compactness tells us that we can pick a finite collection $A'\subseteq A$ so that the union $V=\bigcup\limits_{a\in A'}V_a$ of that finite collection of sets still covers $A$ — we only need a finite number of the $V_a$ to cover $A$. The finite intersection $U=\bigcap\limits_{a\in A'}U_a$ will then be a neighborhood of $x$ which doesn’t touch $V$, and so we can separate any point $x\in X$ and any compact set $A\subseteq X$ by neighborhoods.

As an exercise, do the exact same thing again to show that in a Hausdorff space $X$ we can separate any two compact sets $A\subseteq X$ and $B\subseteq X$ by neighborhoods.

In a sense, this shows that while compact spaces may be infinite, they sometimes behave as nicely as finite sets. This can make a lot of things simpler in the long run. And just like we saw for connectivity, we are often interested in things behaving nicely near a point. We thus define a space to be “locally compact” if every point has a neighborhood which is compact (in the subspace topology).

There’s an equivalent definition in terms of closed sets, which is dual to this one. Let’s say we have a collection $\{F_i\}_{i\in\mathcal{I}}$ of closed subsets of $X$ so that the intersection of any finite collection of the $F_i$ is nonempty. Then I assert that the intersection of all of the $F_i$ will be nonempty as well if $X$ is compact. To see this, assume that the intersection is empty:
$\bigcap\limits_{i\in\mathcal{I}}F_i=\varnothing$
Then the complement of this intersection is all of $X$. We can rewrite this as the union of the complements of the $F_i$:
$X=\bigcup\limits_{i\in\mathcal{I}}F_i^c$
Since we’re assuming $X$ to be compact, we can find some finite subcollection $\mathcal{A}\subseteq\mathcal{I}$ so that
$X=\bigcup\limits_{i\in\mathcal{A}}F_i^c$
which, taking complements again, implies that
$\bigcap\limits_{i\in\mathcal{A}}F_i=\varnothing$
but we assumed that all of the finite intersections were nonempty!

Now turn this around and show that if we assume this “finite intersection property” — that if all finite intersections of a collection of closed sets $F_i$ are nonempty, then the intersection of all the $F_i$ are nonempty — then we can derive the first definition of compactness from it.

January 14, 2008