Let (X,d) be a metric space and let A be a compact subset of X. Let x be a point in X.
I want to show that there exists a point a* in A such that d(x,a*) is less than or equal to d(x,a) for all a in A.
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Attempt:
Let
Then there is a sequence in A such that as
Because A is compact there is a subsequence in A which converges to a point in A, call it a*
Then as
Then... something... to give d(x,a*) = D which is equivalent to what I was trying to show
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I'm not sure... it's all a little cloudy to me... I feel like I'm on the right track but I may not be appreciating some of the subtleties...
Also, the question came with a hint to construct a sequence in A satisfying less than or equal to D + 1/n, then use the compactnes of A... That's even cloudier to me :S
Thanks for any help