Let " alt="(X,d\" /> be a metric space. For arbitrary and put
It can be proved, if is compact space, then for every tho closed and disjoint subsets and is
Prove that in conected metric spaces holds converse, naimly, if for every two closed and disjoint subsets... this distance is positive, than " alt="(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 " alt="(X,d\" /> isn't compact, and then using knowing equivalent of compacity in terms of sequences, conscruct efectivly tho closed and disjoint subsets and for which is . 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 for suitable subsets, but space isn't compact.