Hausdorff and Compactness.

Question

I have a lot of problems this week that start with a statement like let $X$ be a compact topological space (every open cover has a finite subcover). Then if $X$ is Hausdorff .....

My question is what does being Hausdorff have to do with compactness?

Answer 1

Almost by definition you can see that a point $\left\{x\right\}$ is a compact subset of $X$. But is it closed? That depends on the topology.

Now suppose $X$ is Hausdorff. Fix a point $x\in X$. Then for every other point $y$ there are opens $U_y$ and $V_y$ such that $x\in U_y$, $y\in V_y$ and $U_y\cap V_y=\emptyset$. So in particular $x\notin V_y $ for any $y\neq x$. Thus $\bigcup_{y\neq x}V_y$ is an open cover of $X\setminus \left\{x\right\}$, but not of $X$. So in fact I just showed that $\left\{x\right\}$ is closed.

If we adjoin one open $U_y$ to this collection, we get an open cover of $X$. So if we assume $X$ to be compact as well, there is a finite subcover say $U_{y_1}\cup\bigcup_{i=1}^nV_{y_i}$ of $X$. Now this cover has peculiar form, we know that $x\notin \bigcup_{i=1}^nV_{y_i}$. So being Hausdorff and compact allows you to consider particular finite covers that otherwise might not exist.

As a non-Hausdorff example. Consider the set $X=\left\{1,2,3, \dots n\right\}$ and define a topology by declaring $U\subset X$ to be open iff $1\in U$ (except for $U=\emptyset$). Check that this defines a topology. Clearly, this is not a Hausdorff topology as we cannot separate any point from $1$. Notice that $\left\{1\right\}$ is not closed as its complement is not open, but it is compact.

Answer 2

It's convenient, because compact Hausdorff spaces are normal, which is often used in proofs (to apply Urysohn's lemma etc.). This is mostly an analysis convenience. It also ensures compact sets are closed, among other things, which is often used.