If $B_n$ is compact and $B_n\subseteq B_{n+1}$, is there a $n_0$ with $d(x,B_n)\ge d(x,\bigcap_nB_n)/2$ for all $n\ge n_0$?

Question

Let $(E,d)$ be a metric space, $x\in E$ and $$d(x,B):=\inf_{y\in B}d(x,y)\;\;\;\text{for }B\subseteq E.$$

Let $(B_n)_{n\in\mathbb N}$ be nonincreasing and $B:=\bigcap_{n\in\mathbb N}B_n$. It seems like that in the proof on page 9 in this paper, it s claimed that if $x\not\in B$ (hence $\delta:=d(x,B)>0$), there is a $n_0\in\mathbb N$ with $d(x,B_n)\ge\frac{d(x,B)}2$ for all $n\ge n_0$.

As the example in this answer shows, this is not true in general. In the paper, every $B_n$ (hence $B$) is compact, but I don't see how this makes any difference. So, is there anything else I'm missing or is the proof in the paper wrong?

Answer 1

There exists $b_n \in B_n$ such that $d(x,B_n)+\frac 1 n >d(x,b_n)$ $\, \, (1)$. There is a subsequence $(b_{n_k})$ of $(b_n)$ converging to some $b$. Let $N \geq 1$. Since $b_{n_k} \in B_{n_k} \subseteq B_N$ for all $k$ sufficiently large it follows that $b \in B_N$. This is true for all $N$, so $x \in B$. Now $d(x,B) \leq d(x,b)=\lim d(x,b_{n_k})\leq \lim d(x,B_n)$ by (1). Of course, $d(x,B_n) \leq d(x,B)$ for all $n$. Conclusion: $d(x,B_n) \to d(x,B)$ as $ n \to \infty$.