Metric Spaces

Quote:

Originally Posted by ns1954

Let $(X,d\;)$ be a metric space. For arbitrary $A\subset X$ and $B\subset X$ put
$\displaystyle<br /> d\;(A,B)=\inf\limits_{x\in A, y\in B}d\;(x,y)<br />$

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

Prove that in conected metric spaces holds converse, naimly, if for every two closed and disjoint subsets... this distance is positive, than $(X,d\;)$ is compact.
Condition conected on metric space can't be omitted.

Hint: My proff is relativly long and is based on contrary asumptation that $(X,d\;)$ isn't compact, and then using knowing equivalent of compacity in terms of sequences, conscruct efectivly tho closed and disjoint subsets $A$ and $B$ for which is $d\;(A,B)=0$ . 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 $d\;(A,B) >0$ for suitable subsets, but space isn't compact.

Well, here's my attempt:

Assume $X$ is not compact then there exists a sequence $(x_n)$ with no accumulation points. Construct another sequence as follows $(y_n)$ such that $d(y_n,x_n)<\min \{ \frac{a_n}{2} , \frac{1}{n} \}$ where $a_n= \sup \{ r>0 : B_r(x_n) \cap (x_n)= \emptyset \}$ (we can guarantee that $y_n\neq x_n$ since $X$ is connected and so the singletons are not open). $(x_n)$ and $(y_n)$ have the same accum. points and so both are closed and disjoint but their distance is $0$ by construction.

For a counter-example take any infinite set with the discrete metric.