Suppose $A$ and $B$ are two closed subsets of $\mathbb{R}$, and assume that : $$\text{dist}(A,B)=\inf_{(a,b)\in A\times B}|a-b|=0.$$
Is it true that $A\cap B\neq\varnothing$ ?
I found a counter-example using arithmetics : $A=\mathbb{N}^\star$ and $B=\pi\mathbb{N}^\star$ are two disjoint closed subets of $\mathbb{R}$, but from rational approximations, I showed that $\text{dist}(A,B)=0$.
Now, if both $A$ and $B$ are compact, by considering the function $\Phi:A\times B\to\mathbb{R}$ defined by $\Phi(a,b)=|a-b|$, since it is continuous on the compact $A\times B$, it is bounded and there is a $(a_\ast,b_\ast)\in A\times B$ such that $\Phi(a_\ast,b_\ast)=\inf\Phi=0$, and this means that $|a_\ast-b_\ast|=0$, i.e. $a_\ast=b_\ast$, whence $A\cap B\neq\varnothing$.
My question is : What if we assume that only $A$ is compact ? ($B$ is still supposed to be closed.)
I considered the exact same $\Phi$ as before. Now, by the sequential characterization of $\inf\Phi$, there exist sequences $(a_n)_{n\in\mathbb{N}}\in A^\mathbb{N}$ and $(b_n)_{n\in\mathbb{N}}\in B^\mathbb{N}$ such that : $$\Phi(a_n,b_n)\underset{n\to+\infty}{\longrightarrow}\inf\Phi=0.$$
From (sequential) compactness of $A$, there exists $\psi:\mathbb{N}\nearrow\!\!\!\nearrow\mathbb{N}$ such that : $$a_{\psi(n)}\underset{n\to+\infty}{\longrightarrow}a_\ast\in A.$$
In particular, we still have : $$|b_{\psi(n)}-a_{\psi(n)}|\underset{n\to+\infty}{\longrightarrow}0.$$
Now I wrote :
$\begin{align} |b_{\psi(n)}-b_{\psi(n+p)}|&\leqslant|b_{\psi(n)}-a_{\psi(n)}|\\ &+|a_{\psi(n)}-a_{\psi(n+p)}|\\ &+|a_{\psi(n+p)}-b_{\psi(n+p)}|. \end{align}$
$(a_{\psi(n)})_{n\in\mathbb{N}}$ is Cauchy since it converges, which implies that $(b_{\psi(n)})$ is also Cauchy. From completeness, it therefore converges to some $b_\ast\in\mathbb{R}$, but since $B$ is closed, we have $b_\ast\in B$, thus the same conclusion as before applies : $a_\ast=b_\ast$ and $A\cap B\neq\varnothing$.
Does this seem correct to you ? Is there a simpler method to handle with this case ? Is there a way to do this without using completeness of $\mathbb{R}$ ?
If you just assume that both sets are closed it does not follow that they intersect. Indeed, consider $A=\bigcup_{n=1}^\infty[2n,\text{ }2n+\frac{1}{2n}]$ and $B=\bigcup_{n=1}^\infty[2n+1,\text{ }2n+1+\frac{1}{2n}]$. As for the second question, note that the inequality $|b_{\psi(n)}-a_*|\leq |b_{\psi(n)}-a_{\psi(n)}|+|a_{\psi(n)}-a_*|$ already implies that $b_{\psi(n)}\rightarrow a_*$ so the completeness of $\mathbb{R}$ is not really needed.