4

Let $(X,d)$ be a metric space, $A \subseteq X$ and $x \notin A$. What are the condition which make true the statement

$d(x,A) = d(x,\partial A)$

If A is compact (not empty) I can think of a way to find a point $y \in A$ such that $d(x,y)=d(x,A)$. And if $X$ is a normed space that should be enough to prove $y \in \partial A$. Is there something beside this?

Markus Steiner
  • 313
  • One condition: if $A\neq\varnothing$ then $A$ should not be clopen. So this will not work in e.g. discrete metric space. – drhab Jan 03 '19 at 08:27
  • If $X$ is a normed space, then $A$ need only be weakly compact. All closed balls are weakly closed, so their intersection with weakly compact sets are weakly compact. If you consider the nested family of sets $B[x; d(x, A) + 1/n] \cap C$, they should have a non-empty intersection which lies on the boundary. – Theo Bendit Jan 03 '19 at 22:57

0 Answers0