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If $X$ is a Hausdorff Space and there are 2 disjoint compact sets $A,B\subset X$, we want to prove $\exists V,U \subset X, V \bigcap U=\emptyset $ S.T. $A\subset U, B\subset V$.($U,V$ are open sets)

My idea is as follows:

We choose arbitray 2 points, $x_1\in A, y_1\in B$ and by definition of compactness, we can find a neighborhood of $x_1$ and $y_1$, denoted as $U_1,V_1$ S.T. $U_1\bigcap U_2=\emptyset$. By the same algorithm, we will have $U=\{U_1, U_2,...\},V=\{V_1,V_2,...\}$. So, what I want to ask is that are $U,V$ still disjoint? If it is then problem solved.

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There are a couple of unclear points with what you wrote. First of all, you probably meant to say that $U_1\cap V_1=\emptyset$, but didn't. Secondly, it looks like you're using $\{U_1,U_2\dots\}$ to denote something, but you probably did not mean a set of sets. Be sure to improve the notation in these so that all is clear. – rschwieb Apr 15 '13 at 17:06
Two steps. Do it for $A=\{x\}$ first. One open subcover application. Then for $A$ general compact. Second open subcover application. The trick in the end is that you take $V$ to be a finite intersection of open sets containing $B$. – 1015 Apr 15 '13 at 17:08
@rschwieb Yea, thanks, I've just upgraded. What I want to do here is the same as you mentioned, but just not sure if the statement I wrote above is correct or not? $U,V$ still disjoint? or you have a better way to express the idea? – Cancan Apr 15 '13 at 17:09
@Cancan They probably are not disjoint in general. I'm not sure if "what you do here" is referring to my comment or my solution. What you did in your post is not the same as what I did in my solution. – rschwieb Apr 15 '13 at 17:12
up vote 1 down vote accepted

It might make it a little easier to prove this in two pieces.

First show that you can separate any point $x\in A$ from the entire set $B$ with disjoint open sets.

Explicitly, show that there always exists $\mathcal{O}_x,\mathcal U_x$ disjoint and open such that $x\in \mathcal O_x$ and $B\subseteq \mathcal U_x$.

Then use the pairs of open sets generated this way to separate $A$ as a whole from $B$.

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The idea is very cool, but how can you surely say that $O_x,U_x$ are still disjoint if $x \in O_x, B\subseteq U_x$? I know that we can separate point in $A$ from the whole set $B$, but how can we be sure that we can also separate them from a neighborhood of $B$? – Cancan Apr 15 '13 at 17:16
My bad, I got your point. This explanation is very good! Thanks! – Cancan Apr 15 '13 at 17:20
@Cancan Thanks for the compliment, but if you could please edit out the profanity within the next 5 minutes, thta would be even better :) – rschwieb Apr 15 '13 at 17:22

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