Prove that if $A,B$ are closed sets such that $A\cap B\neq \emptyset$, there exists $U,V$ open sets such that $A\subset U$, $B\subset V$, and $U\cap V= \emptyset$.

I am thinking of going by contradiction, that is: $\forall\; U,V$ open sets such that $A\subset U$, $B\subset V$, and $U\cap V\neq \emptyset$.

Let $ U,V$ open. Then, $\exists\;r_1,r_2$ such that $B(x,r_1)\subset U$ and $B(x,r_2)\subset V.$ From here, I don't know how to get a contradiction. Can anyone help me out?

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    $\begingroup$ $A\subset U, B\subset V$ implies that $A\cap B\subset U\cap V$. $\endgroup$ – Tsemo Aristide Aug 9 '18 at 0:41
  • $\begingroup$ @Tsemo Aristide: But $A\cap B\neq \emptyset.$ $\endgroup$ – Omojola Micheal Aug 9 '18 at 0:42
  • $\begingroup$ @mfl: I'm sorry! But how would that help me? $\endgroup$ – Omojola Micheal Aug 9 '18 at 0:47
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    $\begingroup$ @ΘΣΦGenSan: You mean the question is not correct? $\endgroup$ – Omojola Micheal Aug 9 '18 at 0:50
  • $\begingroup$ are you interested in a proof if we assume $A, B$ are compact? $\endgroup$ – Matt A Pelto Aug 9 '18 at 1:36

This is not true in general.

Consider $\Bbb{R}$ with the usual topology, and set $A = [0,1]$ and $B = [1,2]$.

First, we have $A\cap B = \{1\}\ne\emptyset$. Any two open subsets $U\subset A$ and $V\subset B$ cannot contain $1$ since they would have to contain a small neighborhood around $1$, which neither $A$ nor $B$ do. Thus $1\notin U$ and $1\notin V$. Therefore, $U\cap V \subset A\cap B=\{1\}$, so $U\cap V =\emptyset$.


I just realized that your question asks for $U\cap V$ to be empty, and thus my counterexample is merely an example. Moreover, the question stipulates that $A\subset U$ and $B\subset V$, whereas I assumed the opposite inclusion.

If you assume that $A\subset U$ and $B\subset V$, then $A\cap B \ne\emptyset$ always implies that $U\cap V\ne\emptyset$ as Tsemo Atistide pointed out in the comments.

If you do in fact assume the opposite inclusion, where you are asking for open sets inside the closed sets, then $U\cap V = \emptyset$ can never hold for closed sets with empty interior, as they contain no open sets by definition.

Are you sure the question is correct as written?


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