Compactness contradiction physics

[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??

 

[master_coda]

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

 

[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.

 

[master_coda]

THis may be exactly the same hint, but it's also known that compact and seqentially compact are equivalent for metric spaces.

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.

 

[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?

 

[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).

 

[mansi]

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