Let (X,d) be a metric space. Suppose that every epsilon neighborhood of every point has a compact closure. Prove that X is complete.
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1Well, compact metric spaces are complete, so your space is "locally complete". Now, since the notion of convergent and Cauchy sequences are local, you can easily conclude that $X$ is complete. – Crostul Dec 21 '15 at 11:44
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See http://math.stackexchange.com/questions/627667/every-compact-metric-space-is-complete – Crostul Dec 21 '15 at 11:44
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What are your thoughts on the problem? – Rory Daulton Mar 02 '16 at 08:47
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Hint: Every Cauchy sequence can be fit into the compact closure of some neighborhood.
Any sequence in a compact space has a convergent subsequence.
Suppose that $x_n$ is a Cauchy sequence. Select $N$ such that $d(x_m,x_n)<1$ for $m,n \geq N$.
All but finitely many elements of the sequence fall in a ball of radius $1$ about $x_N$, and therefore within its compact closure. Conclude that the Cauchy sequence has a convergent subsequence. Thus, $x_n$ converges.
So, every Cauchy sequence in $X$ converges. So, $X$ is complete.
Ben Grossmann
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Well, then you really need to show what you've done so far. Edit your question to indicate what you've tried. – Ben Grossmann Dec 21 '15 at 11:50
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And, if you haven't been able to do anything at all, then this hint is a good starting point. – Ben Grossmann Dec 21 '15 at 11:51
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i have an assignment due in an hour. I cant try anything right now. Please help? – James Hunt Dec 21 '15 at 11:56
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