The distance between two disjoint compact subsets $A,B$ of a metric space $X$ is positive

Question

Please tell me whether my argument for the following result is true:

The distance between two disjoint compact subsets $A,B$ of a metric space $X$ is positive:

$d:X\times X\to \mathbb R$ is continuous$\implies d|_{A\times B}$ is continuous on $A\times B$

$A,B$ are compact $\implies A\times B$ is compact$\implies d|_{A\times B}$ assumes its minimum on $A\times B\implies\exists~a\in A,b\in B$ such that $$\inf_{x\in A,~y\in B} d|_{A\times B}(x,y)=d(a,b)>0.$$

There is no need to deal with the restriction on $d$, otherwise your proof is good. — copper.hat, Nov 07 '13 at 04:17
Related : https://math.stackexchange.com/questions/48714/a-and-b-disjoint-a-compact-and-b-closed-implies-there-is-positive-distance-bet — Arnaud D., Nov 28 '17 at 10:06

score 1 · Accepted Answer · answered Nov 07 '13 at 04:12

1

There is no need to deal with restrictions of $d$. All that is necessary is to note that $A \times B $ is compact.

Let $m = \min_{a\in A, b\in B} d(a,b)$. If $m = 0$, then $m=d(a,b)$ for some $a \in A, b \in B$, and since $d(a,b) = 0$, we have $A \cap B \ne \emptyset$, a contradiction.

answered Nov 07 '13 at 04:12

copper.hat

172,524

Sorry. Edited it. – Shantipriya Nov 07 '13 at 04:13
Me too. ${}{}{}{}$ – copper.hat Nov 07 '13 at 04:16
Could you please point where exactly did you use the fact that $A \times B$ is compact? – blahblah Apr 24 '19 at 18:12
1

@Tomasz: To conclude that there is a point $(a,b)$ at which the distance is a $\min$. – copper.hat Apr 24 '19 at 18:34
How do we conclude that? Because compact spaces imply that every sequence has a limit point? – blahblah Apr 24 '19 at 18:54
1

@math_beginner: A continuous function on a compact set achieves its $\max,\min$ value. One way of proving it is to take an accumulation point of an extremising sequence. – copper.hat Apr 24 '19 at 20:46

The distance between two disjoint compact subsets $A,B$ of a metric space $X$ is positive

1 Answers1