0

For starters let's define the distance between a point $x$ and a set $A\subseteq X$ of a metric space $(X,d)$ as follows: $$\text{dist}(x,A)=\inf\{d(x,a):a \in A\}.$$

Now let's assume that $A,B$ are two closed non-empty subsets of $X$ and $A$$\cap$$B=$$\emptyset$

I want to prove that $U=${$x$$\epsilon$$X$$:dist(x,A)<dist(x,B)$}

and $V=${$x$$\epsilon$$X$$:dist(x,B)<dist(x,A)$} are

a) open

b) $A$$\subseteq$U , $B$$\subseteq$V

c) $U$$\cap$$V=$$\emptyset$

So here is my idea , for any random $x$$\epsilon$$U$ I choose a $\delta$ to satisfy $0<δ<dist(x,B)-dist(x,A)$ . For every $y$ in $B(x,$$\delta$$)$$\subseteq$$U$ I have that $dist(y,A)$$\le$$dist(x,A)+dist(x,y)<dist(x,B)$. So since , every $y$$\epsilon$$U$ is internal , $U$ is an open set . Is this correct ? Thanks

Stef M
  • 111

1 Answers1

1

The idea is the right one but the proof is a bit incomplete, since you want to show that $d(y,A) < d(y,B)$, not $< d(x,B)$. So consider the ball of radius $\delta/2$ instead.

orangeskid
  • 53,909