# Is the intersection of a closed set and a compact set always compact?

I am going through Rudin's Principles of Mathematical Analysis in preparation for the masters exam, and I am seeking clarification on a corollary.

Theorem 2.34 states that compact sets in metric spaces are closed. Theorem 2.35 states that closed subsets of compact spaces are compact. As a corollary, Rudin then states that if $L$ is closed and $K$ is compact, then their intersection $L \cap K$ is compact, citing 2.34 and 2.24(b) (intersections of closed sets are closed) to argue that $L \cap K$ is closed, and then using 2.35 to show that $L \cap K$ is compact as a closed subset of a compact set.

Am I correct in believing that this corollary holds for metric spaces, and not in general topological spaces?

In any Hausdorff space compact sets are automatically closed, and so the above argument works as written. It is true that in non-Hausdorff spaces, a compact set need not be closed.

On the other hand, it is true in general that a closed subset of a compact topological space is compact (whether or not the compact space is Hausdorff); this is easily proved directly in terms of the open cover characterization of compact topological spaces.

So if $K$ is compact in an arbitrary topological space (which is just to say that it is a compact topological space when given its induced topology) and $L$ is closed then $K \cap L$ is a closed subset of $K$ in its induced topology, and hence is compact with its induced topology, i.e. is again a compact subset of the ambient topological space (even though it need not be closed).

Summary: The corollary does hold for arbitrary (not necessarily Hausdorff) topological space, but you need to rewrite the proof slightly.

• This makes a lot of sense (and is a bit of a d'oh moment for me as well). Thank you! – Michael Chen Apr 25 '11 at 16:58

As usual, nets provide an elegant proof. Let $L$ be closed and $K$ compact. Suppose $(x_i)_i$ is a net in $L\cap K$. As $K$ is compact, there is a subnet $(x_{i_k})_k$ converging to some $x\in K$. As $L$ is closed, we must have $x\in L$. Hence $x\in L\cap K$ and consequently $L\cap K$ is compact.

• It came to my attention a few months back that many people don't know about nets. It is a nice and powerful tool, although confusing at times. – Asaf Karagila Apr 25 '11 at 17:40
• This proof also works with filters instead of nets. – Chris Eagle Apr 25 '11 at 17:49
• Analysts should know about nets, filters seem to be less common in analysis but I do use them for the sake of being a bit recalcitrant ;-). – Jonas Teuwen Apr 25 '11 at 18:10
• I'm afraid I don't know much about nets -- I'm reading through the Wikipedia link at the moment. But this looks interesting. Thank you for this answer. – Michael Chen Apr 25 '11 at 23:37