Compactness contradiction physics

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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??
 

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  • #2
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Find appropriate sequences (a_n) and (b_n) then apply Bolzano-Weierstrass.
 
  • #3
matt grime
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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.
 
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matt grime said:
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.
 
  • #5
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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? :confused:
 
  • #6
matt grime
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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
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thanks a lot...i figured that out finally!
 

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