It is known that not every locally compact Haussdorff space is normal, see for example


But it seems that the following is true, I just want to make sure I am not making any mistake:

Lemma Let $X$ be a locally compact Haussdorff space and let $K, W \subset X$ be compact with $K \cap W = \emptyset$. Then, there exists open sets $K \subset U, W \subset V$ such that $U \cap V =\emptyset$.


We have $K \subset (X \backslash W)$. Since $K$ is compact, $X \backslash W$ is open and $X$ is locally compact Haussdorff space, by Theorem 2.7 in Rudin, "Real and Complex Analysis" we can find an open set $U$ (with compact closure) such that $K \subset U \subset \bar{U} \subset X \backslash W$.

Let $V:= X \backslash \bar{U}$. Then $V$ is open and satisfies the above conditions.


Is this correct? I am a bit uneasy, especially since the proof only uses the fact that $W$ is closed, compactness is only needed for $K$.

  • $\begingroup$ Yes, this is true in any Hausdorff space; you don't even need local compactness. The proof is a nice exercise. With local compactness you get regularity and then indeed it's enough to have one set closed and the other compact. $\endgroup$ May 7, 2020 at 19:34
  • $\begingroup$ @NateEldredge Perfect, thank you. $\endgroup$
    – Nick S
    May 7, 2020 at 19:36
  • 1
    $\begingroup$ Roughly speaking, anywhere in the separation axioms that you see "point", you can upgrade it to "compact set" for free. $\endgroup$ May 7, 2020 at 19:38
  • $\begingroup$ @NateEldredge If I understand right, the reason for this is because for each of the points in the compact set we can find a pair of separating open sets. Then, we pick a finite subcover, and take the corresponding finite intersection of open sets, which is open since finite. The last step seems to also be the reason why these cannot be extended from compact to closed. $\endgroup$
    – Nick S
    May 7, 2020 at 20:10
  • $\begingroup$ Yes, that's it exactly. $\endgroup$ May 7, 2020 at 20:20

1 Answer 1


Yes, this is known and you don't need local compactness even as long as $X$ is completely regular (locally compact spaces are completely regular).

I think that the best way to prove is to go via the Čech–Stone functor, which is available precisely for completely regular spaces. Consider $\beta X$ and note that $K, W\subset X\subset \beta X$ remain compact in $\beta X$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.