Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$, $B$ be non-intersecting compact subsets of a Hausdorff topological space $M$ . How to prove that there exist a pair of open subsets $V\supset A$, $W\supset B$ satisfying $V\cap W=\emptyset$.

share|cite|improve this question
Find "General Topology" by R.Engelking and you'll find the proof there in the chapter on compactness... – W_D Sep 27 '13 at 15:04
See Martin's comment here for an outline of the argument. – Cameron Buie Sep 27 '13 at 15:05
up vote 2 down vote accepted

For each point $a\in A$ and each point $b\in B$ there are disjoint open sets $U_{a,b}\ni a,\ V_{a,b}\ni b$. Fix $b\in B.$ Then the collection $\mathcal U_b=\{U_{a,b}\mid a\in A\}$ covers $A$. By compactness there are points $a_1,...,a_n\in A$ such that $A\subseteq U_b:=\bigcup_{i=1}^n U_{a_i,b}$. If you define $V_b=\bigcap_{i=1}^n V_{a_i,b}$, then $V_b$ is an open neighborhoood of $b$ which is disjoint from the open neighborhood $U_b$ of $A$. This means a point and a compact set can be separated by neighborhoods.
Now, $B$ is covered by $\{V_b\mid b\in B\}$, and again by compactness finitely many of these open sets suffice to cover $B$. By the same argument you get a union of finitely many $V_b$ that covers $B$ and an intersection of finitely many $U_{b_i}$ which is a neighborhood of $A$, and both sets are disjoint.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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