I wrote a test yesterday in which one of the questions asked us to prove that if $A$ and $B$ are disjoint closed subsets of a metric space $(X, d)$, then there exist disjoint open subsets $U$ and $V$ of $X$ such that $A\subseteq U$ and $B\subseteq V$.

This is not the question which I am asking here, because this question has already been answered here on StackExchange.

Instead, I have a question related to the approach that I attempted to take the during the test (but was not able to complete).

If we define $$U_n = \bigcup \left\{B\left(a;\frac{1}{n}\right)\mid a\in A\right\}$$ and $$V_n = \bigcup \left\{B\left(b;\frac{1}{n}\right)\mid b\in B\right\}$$ where $B(x;r)$ is the open ball centred at $x$ of radius $r$, then $U_n$ and $V_n$ are open sets containing $A$ and $B$ respectively for any $n > 0$. So if we could show that there is some $n > 0$ such that $U_n\cap V_n\neq\emptyset$, then we would be done.

If this is not the case, then for each $n$, we have a $x_n\in U_n\cap V_n$.

But then we would have that $$x_n\in B\left(a_n, \frac{1}{n}\right)$$ and $$x_n\in B\left(b_n; \frac{1}{n}\right)$$ for some $a_n\in A$ and $b_n\in B$. This gives us that $$d(a_n,b_n)\leq d(a_n,x_n)+d(x_n,b_n)<\frac{2}{n}$$

Thus we have sequences $(a_n)$ and $(b_n)$ in $A$ and $B$ respectively such that $d(a_n, b_n)\to 0$ as $n\to\infty$. At this point, the problem would be solved if the existence of these two sequences implies that $A$ and $B$ have non-empty intersection. Will this always be the case? It seems reasonable, but I was not able to prove it.

If we make some additional assumptions about $X$, then this becomes easier. For example, if $X$ is compact, then $(a_n)$ has a convergent subsequence $(a_{n_k})$ which converges to a point $a\in A$. We see that $b_{n_k}$ also converges to $a$, and so $a\in A\cap B$.

Thus the result holds in compact metric spaces. (And so any counter-example must be in a metric space which is not compact)

  • $\begingroup$ In response to the title - I am pretty sure this holds only if, say, $A$ is compact and $B$ is closed. $\endgroup$
    – Eoin
    Jun 2, 2015 at 23:37
  • 1
    $\begingroup$ In $\Bbb R$, take $A=\{ n\mid n\in\Bbb N, n>2 \}$ and $B=\{ n+1/n\mid n\in\Bbb N\}$. $\endgroup$ Jun 2, 2015 at 23:44

2 Answers 2


Consider the subsets of the plane given by the $x$ axis and the graph of the exponential function.


Take a graph of a continuos positive function that approach to zero as x approach infinity and take the x axis.


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