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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 $A\subseteq U, B\subseteq V$.

  2. Give an example which shows that $d(A,B) := \inf_{x\in A, y\in B}d(x,y)>0$ is not necessarly true.

A hint is given for 1.

Cover $A$ and $B$ with open balls for a certain radius.

But how do I find this radius?

For 2. I was thinking of the following:

Let $$A=\left\{\frac{1}{2n}: n\in \mathbb{N} \right\}\qquad B=\left\{\frac{1}{2n+1}: n\in \mathbb{N} \right\}$$

However, these sets are not closed. Since $\mathbb{R}\setminus A$ is not open (I can't find an neighbourhood of the point $0$ as it is a limit point)

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dietervdf
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1 Answers1

3

This is another proof for 1,

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 . And they are disjoint. Also if $x\in A$, then $d(x,A)=0$ and $d(x,B)>0$ by closedness of $B$ ,so $x\in U$. Hence, $A\subset U$. Similarly $B \subset V$.

For 2, consider $M=\mathbb{R}^2$, $A=\{(x,y)|y=0\}$, the x-axis, and $B$ be the graph of $y=\frac{1}{x}$, say $B=\{(x,y)|y=\frac{1}{x},x>0\}$. Their distance is $0$.

I think the point of part 2 is that in part 1, we can't cover $A$ by $U=\bigcup_{x\in A}B_{\frac{d(A,B)}{3}}(x)$, since $B_{\frac{d(A,B)}{3}}(x)$ may be empty.

John
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  • Could you elaborate on your last remark? Where is that $\dfrac{1}{3}\cdot d(A,B)$ coming from? – dietervdf Apr 27 '15 at 20:21
  • @dietervdf You can check if $d(A,B)>0$, then $U=\bigcup_{x\in A}B_{\frac{d(A,B)}{3}}(x)$, and $V=\bigcup_{x\in B}B_{\frac{d(A,B)}{3}}(x)$, is a disjoint open cover for $A,B$. But part 2 says in general $d(A,B)$ can be $0$, so the proof is not valid. – John Apr 28 '15 at 00:59