Originally Posted by

**ns1954** Let $\displaystyle (X,d\;)$ be a metric space. For arbitrary $\displaystyle A\subset X$ and $\displaystyle B\subset X$ put

$\displaystyle \displaystyle

d\;(A,B)=\inf\limits_{x\in A, y\in B}d\;(x,y)

$

It can be proved, if $\displaystyle (X,d)$ is compact space, then for every tho closed and disjoint subsets $\displaystyle A$ and $\displaystyle B$ is $\displaystyle d\;(A,B) > 0$

Prove that in conected metric spaces holds converse, naimly, if for every two closed and disjoint subsets... this distance is positive, than $\displaystyle (X,d\;)$ is compact.

Condition conected on metric space can't be omitted.

Hint: My proff is relativly long and is based on contrary asumptation that $\displaystyle (X,d\;)$ isn't compact, and then using knowing equivalent of compacity in terms of sequences, conscruct efectivly tho closed and disjoint subsets $\displaystyle A$ and $\displaystyle B$ for which is $\displaystyle d\;(A,B)=0$. But there are some other equivalents of compacity, maybe someone find another solution. I have the appropriate counterexemple of totaly disconected metric space in which is $\displaystyle d\;(A,B) >0$ for suitable subsets, but space isn't compact.