Let$(X,d)$ be a metric space, $A$ nonempty, compact subset of $X$. Let $$\mathrm{dist}(A,B)=\inf_{\substack{a\in A\\b\in B}}d(a,b).$$ I want to show that for every compact subset $B$ of $X$ disjoint from $A$ $\mathrm{dist}(A,B)>0$. I tried to show it using the extreme value theorem. Since $A,B$ are compact, then so is $A\times B$, $D:A\times B\to (0,\infty)$ defines a metric on $A\times B$, so if $f:A\times B\to (0,\infty)$, $\inf f = \mathrm{dist}(A,B)$ showing that $f$ attains it's minimum, that is $\inf f = \min f$ would complete the proof, but I'm not sure how to show $f$ is continuous. Any hints? Edit: I'm familiar with that linked proof, but I was trying to use a different method, using extreme value theorem, but I'm not quite sure how to prove continuity of $f$ for $D((a,b),(c,d))=d_1(a,c)+d_1(b,d)$.
Asked
Active
Viewed 166 times
1 Answers
0
Assume that $\mathrm{dist}(A,B)=0$. Then there exists $a_n\in A$ and $b_n\in B$ such that $$d(a_n,b_n)\to0$$
But since B is compact there is subsequence $b_{n_k}$ converging to some $b\in B$
Hence, $$ \lim_{k\to \infty} d(a_{n_k},b) \le d(a_{n_k},b_{n_k})+d(b_{n_k},b) = 0$$
This means that $a_{n_k} \to b$ with $a_{n_k}\in A$ and A is closed then $b\in A$. finally $$b\in B\cap A =\emptyset$$ Contradiction
Guy Fsone
- 23,903