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)
