Let $X$ be a compact metric space. Let $A$ be a closed subset of $X$ and let $x\in X$ be a point not in $A$. Show that there exists two disjoint open sets $U$ and $V$ such that one contains $A$ and the other contains $x$.

Now we consider $A$. Since $A$ is closed it is also compact. Because $A$ is closed, every sequence of points in $A$ has its limit point in $A$. Let $\{x_n\}$ be such a sequence. Let $\{O_n\}$ be a disc of radius $\epsilon_n$ centered at an $x_n.$ We can extract a finite subcover from this open cover that covers $A$. Let $O=\cup\{O_i\}_1^k.$

Now consider an $x\in (X\setminus O)$. This is a closed set. Again, we repeat the argument with open sets. None of these open sets will contain points of $O$. Hence another disjoint open set.

  • $\begingroup$ You can't select $x$, it is given. While irrelevant here, it may be the case that $X \setminus O$ is empty. $\endgroup$ – copper.hat Apr 28 '16 at 4:03
  • $\begingroup$ I'm not sure how to proceed then. $\endgroup$ – emka Apr 28 '16 at 4:14
  • $\begingroup$ There are a few ways. Here is one that follows an approach similar to the above: For $a \in A$, choose the open set $B(a, {1 \over 2} d(a,x))$. Note that the closure of this set does not contain $x$. $\endgroup$ – copper.hat Apr 28 '16 at 4:18
  • $\begingroup$ Could I do that with a sequence of points in A? $\endgroup$ – emka Apr 28 '16 at 4:20
  • $\begingroup$ @emka: How would that help? I think the most straightforward way to to use the definition of compactness directly. $\endgroup$ – copper.hat Apr 28 '16 at 4:28

First of all note that any two distinct points in $X$ can be separated by open sets. This is shown as follows -

Let $y,z\in X$ with $y\neq z$. Then $d(y,z)=:r>0$. Then the open balls $B(y,r/2)$ and $B(z.r/2)$ are disjoint open sets where the first one contains $y$ and the second contains $z$.

Now coming to your problem -

For each $a\in A$ there is a pair of disjoint open sets $U_a,V_a$ such that $a\in U_a$ and $x\in V_a$. the collection $\{U_a\}_{a\in A}$ forms an open cover of $A$ and by compactness there are a finite number that cover $A$. Call them $U_1,\cdots,U_n$. Let $V_1,\cdots,V_n$ be the corresponding disjoint open sets containing $x$. Let $U=\bigcup_{i=1}^{n}U_i$ and $V=\bigcap_{i=1}^nV_i$. Then $U$ and $V$ are disjoint open sets with $A\subseteq U$ and $x\in V$.

Remark - this proof works for any Hausdorff space $X$ and a compact subspace $A$ of $X$.

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    $\begingroup$ Note that we only need that $A$ is compact, not that $X$ is. $\endgroup$ – Henno Brandsma Apr 28 '16 at 18:32

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