Compactness contradiction physics

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Discussion Overview

The discussion revolves around proving that for a compact subset A of a metric space (X,d), there exist points a and b in A such that the diameter of A, denoted d(A), equals the distance d(a,b). The scope includes theoretical aspects of compactness in metric spaces and the application of related theorems.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using the closed and bounded nature of compact sets or assuming the absence of such points a and b to reach a contradiction regarding compactness.
  • Another participant proposes finding appropriate sequences (a_n) and (b_n) and applying the Bolzano-Weierstrass theorem.
  • A participant notes the equivalence of compactness and sequential compactness in metric spaces, indicating that every continuous real-valued function on compact spaces is bounded and attains its bounds.
  • One participant expresses concern about the elegance of using sequences to demonstrate the result, questioning the approach of discussing distances in A x A.
  • Another participant states that since A is compact, the product space A x A is also compact, and thus there exists a point in A x A where the distance function is maximized.

Areas of Agreement / Disagreement

Participants express differing views on the most elegant approach to the proof, with some favoring the use of sequences and others preferring a more direct application of properties of continuous functions on compact spaces. The discussion remains unresolved regarding the preferred method of proof.

Contextual Notes

Participants have not fully resolved the assumptions regarding the definitions of compactness and the implications of the distance function on the product space A x A.

mansi
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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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Find appropriate sequences (a_n) and (b_n) then apply Bolzano-Weierstrass.
 
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.
 
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.
 
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:
 
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).
 
thanks a lot...i figured that out finally!
 

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