1. Feb 24, 2005

mansi

Let A be a compact subset of a metric space (X,d). Show that there exist a,b in A such that d(A) = d(a,b) where d(A) denotes the diameter of A.

I guess...we're supposed to use the fact that a compactness of A implies that it is closed and bounded or alternately...we could assume that there are no a,b in A such that d(A)=d(a,b) and arrive at a contradiction to the fact that A is compact.
Any suggestions??

2. Feb 24, 2005

master_coda

Find appropriate sequences (a_n) and (b_n) then apply Bolzano-Weierstrass.

3. Feb 24, 2005

matt grime

THis may be exactly the same hint, but it's also known that compact and seqentially compact are equivalent for metric spaces. But then there is another way of doing it: every continuous real valued function on a compact spaces is bounded and attains its bounds.

4. Feb 24, 2005

master_coda

That's what I meant by Bolzano-Weierstrass. But I probably should have been more explicit, since that's really a generalization of the theorem and not the theorem iteself.

5. Feb 25, 2005

mansi

thanks.... i did do the sequences thing but i feel that it's not an elegant way of doing it. why? that's because you're talking about distance in A (cross) A
that's what d(an,bn) means...
so i think it's definitely not the best way of doing it....
"every continuous real valued function on a compact spaces is bounded and attains its bounds. " can you please elaborate?

6. Feb 25, 2005

matt grime

A is compact, hence AxA is compact. d(?,?) is a continuous function on AxA hence there is a point in AxA such that d(?,?) is maximized, let that point be (a,b).

7. Feb 26, 2005

mansi

thanks a lot...i figured that out finally!