0

I have a quick question about an old question on the website want to prove

Let $A, B$ be two disjoint closed subsets of a certain metric space $(M,d)$.

  1. Show that there exist disjoint open subsets $U, V \subseteq M$ such that $A\subseteq U, B\subseteq V$.

In the answer, the author claims

Let $U = \{x | d(x,A) < d(x,B)\},V = \{x|d(x, A) > d(x, B)\}$. Since $d(x, A), d(x, B)$ are continuous functions, U,V are open sets

I am trying to figure out why the continuity of the distance functions tells us that these sets are open.

I tried considering complements but to no avail. Then I tried open balls but I quickly got worried for things with bad regularity.

Question: Could someone elaborate on how the sets $U$ and $V$ are open?

Maths Wizzard
  • 2,259
  • If $f\colon X \to Y$ is a map between topological spaces, then by definition (or by characterization, depends on the approach), $f$ is continuous if and only if $f^{-1}(U)$ is open in $X$ whenever $U$ is open in $Y$. Apply this with $X=M$, $Y=\Bbb R$ and $f(x) = d(x,A) - d(x,B)$. – Didier May 11 '23 at 14:40
  • Hi, thanks for the comment. Just to clarify, here I would take the preimage of the negative axis (excluding zero) and of the positive axis (excluding zero) to be my two sets (U) and (V) correct? – Maths Wizzard May 11 '23 at 14:44
  • Yes, this is exactly this. Heuristically, the set of points that are strictly closer to $A$ than to $B$ is open because, because "strict" is an open condition – Didier May 11 '23 at 15:01

0 Answers0