In a complement of a closed set find a point that is closest to any point in the closed set

Question

Is it true that for $U$ closed and nonempty in a metric space $X$, let $a\in X\setminus U$, then there exists a $b\in U$ such that $d(a,b)\leq d(a,x) \forall x\in U$? I think it is correct because it asks for a "boundary" of a closed set. But how should I prove that such a "boundary" exist? Thanks

Answer 1

Consider $Q$ with the metric defined by $d(x,y)=\mid x-y\mid$. Let $C=\{x\in Q, x\leq \sqrt2\}$. It is a closed subset of $Q$. For $x>\sqrt2$ there does not exists $y\in C$ such that $d(x,C)=d(x,y)$.

Answer 2

If a given metric space satisfies Heine-Borel theorem then your question holds for that space. However if not it does not hold in general.

Consider the space $\Bbb{R}^\omega$ (i.e. set of all real sequences) with uniform metric $$d(f,g) = \min\{1,\, \sup\{|f(n)-g(n)| : n\in \Bbb{N}\}\}$$ with a subset $A$ of $\Bbb{R}^\omega$ given by $$A = \{(\underbrace{0,0,0,\cdots ,0}_{(n-1)\text{ times}},1+\tfrac{1}{n},0,0,\cdots): n\in\Bbb{N}\}$$ and take $b=(0,0,0,0,\cdots)$. You can check that $A$ is closed but no points in $A$ minimizes the distance from $b$ to $A$.

Answer 3

Let $p\not \in R.$ Let $S=R\cup \{p\}.$ Define a metric on $S$ by $d(x,y)=|x-y|/(1+|x-y|)$ for $x,y \in R,$ and $d(x,p)=2-d(x,0)$ for $x\in R.$ Then $R$ is closed in $S$ and $\inf_{x\in R}d(x,p)=1.$ But $d(x,p)>1$ for all $x\in R.$