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Compact implies Sequentially Compact

  • Thread starter boopbeep
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  • #1
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Homework Statement [/b]
I need help proving that if X is a metric space and E a subset of X is compact, then E is sequentially compact.


I know I need to consider a sequence x_n in E, and I want to say that there is a point a in E and a radius r > 0 so that Br(a) [the ball of radius r with center a] contains x_k for infinitely many k's. If I show this, then I think I can conclude that any subsequence of x_n converges to a. Can someone please help?
 

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  • #2
Hurkyl
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So... your conjecture is that every sequence in a compact metric space is actally convergent?
 
  • #3
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No, that is not one of the assumptions...my second paragraph in the problem description is a hint in the back of the textbook as to how to go about the problem.
 
  • #4
Office_Shredder
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If any subsequence of xn converges to a, then xn converges to a too. Rather, what you can conclude is a single subsequence of xn converges to a (which is what sequential compactness is about)
 
  • #5
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If any subsequence of xn converges to a, then xn converges to a too. Rather, what you can conclude is a single subsequence of xn converges to a (which is what sequential compactness is about)
This is not true- consider a_n=(-1)^n=1,-1,1,-1,1,-1,...

Then a_{2n}=1 is a subsequence that converges to 1, but a_n does not converge at all- nonetheless to 1.
 
  • #6
HallsofIvy
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If any subsequence of xn converges to a, then xn converges to a too. Rather, what you can conclude is a single subsequence of xn converges to a (which is what sequential compactness is about)
This is not true- consider a_n=(-1)^n=1,-1,1,-1,1,-1,...

Then a_{2n}=1 is a subsequence that converges to 1, but a_n does not converge at all- nonetheless to 1.

Office Shredder meant "if every subsequence". He chose the wrong word.
 

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