$\inf_{x\in A}{\limsup_nd(x_n, x)} = \limsup_n[\inf_{x\in A}d(x_n, x)] $ for compact subset $A$.

Question

let $ (X, d) $ be a complete metric space, $ A\subset X $ be compact and take a sequence $ (x_n) \subset X $\ $ A $ as a bounded sequence.

Since infimum is independent from n , does the following hold?

$$ \inf_{x\in A}{\limsup_nd(x_n, x)} = \limsup_n[\inf_{x\in A}d(x_n, x)] $$

I failed to find counterexample.

If it is not true, please give a counterexample.

Answer 1

Consider $A=\{-1,1\}$, $X=\Bbb R$ and $x_n=(-1)^n$. LHS is $2$, while RHS is $0$.

After the additional request $x_n\notin A$: Consider the same $A$ and $$x_n=\begin{cases}4&\text{if }n\text{ is even}\\ -3&\text{if }n\text{ is odd} \end{cases}$$ $$d(x_n,1)=4,3,4,3,4,3,\ldots\\ d(x_n,-1)=2,5,2,5,2,5,\ldots$$ so LHS is $4$, while RHS is $3$.