$A$ closed, $B$ compact, both non-empty - counterexample in R to show that $d(A,B)$ NOT = $d(a,b)$?

Question

Defining $d(A,B) = \inf d(a,b)$ for $a \in A$ and $b \in B$, give a counterexample to show that there does NOT exist $a \in A$ and $b \in B$ so that $d(A,B) = d(a,b)$. I can think of $X = \mathbb{R}\setminus\{0\}$ where $B = [-10,-1]$ and $A = (0,10]$ but cannot think of something in $\mathbb{R}$.

Answer 1

If $A,B$ are closed,non-empty, and $B$ is compact : For $n\in \mathbb N, $ take $a_n\in A$ and $b_n\in B$ such that $$d(A,B)\leq |a_n-b_n|<d(A,B)+1/n.$$ Since $B$ is compact, take a strictly increasing $f:\mathbb N \to \mathbb N$ such that $$\lim_{n\to \infty}b_{f(n)}= b\in B.$$ For clarity of notation let $b_{f(n)}=b'_n$ and $a_{f(n)}=a'_n.$ We have $\lim_{n\to \infty}|a'_n-b'_n|=d(A,B).$

For all but finitely many $n$ we have $|a'_n-b'_n|<d(A,B)+1.$ And for all but finitely many $n$ we have |$b'_n-b|<1.$ So $(a'_n)_{n\in \mathbb N}$ is a bounded sequence.

So take a strictly increasing $g:\mathbb N\to \mathbb N$ such that $\lim_{n\to \infty} a'_{g(n)}=a.$ Since $A$ is closed we have $a\in A.$ Now we have $$d(A,B)\leq |a-b|\leq |a-a'_{g(n)}|+|a'_{g(n)}-b'_{g(n)}|+|b'_{g(n)}-b|.$$ The RHS above converges to $d(A,B)$ as $n\to \infty.$ $$ \text {So } d(A,B)=|a-b|.$$

Remark: If it it not obvious to you that $(a'_n)_n$ is a bounded sequence: Take $n_0$ such that $n>n_0\implies |a'_n-b'_n|<d(A,B)+1$ and take $n_1$ such that $n>n_1\implies |b'_n-b|<1.$ Let $n_2=\max (n_0,n_1).$ Then $$n>n_2\implies |a'_n|\leq |a'_n-b'_n|+|b'_n-b|+|b|<(d(A,B)+1)+1+b=K.$$ So for all $n$ we have $|a'_n|\leq \max (K, \max \{|a'_j|:j\leq n_2\}).$