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This question was a true /false question in my real analysis quiz today and I am unable to provide a reason for it.

Question is : A closed and bounded subset of complete metric space is compact.

I think it is not compact but only because any open cover may not have finite subcover but I am unable to provide a counterexample for the same.

amWhy
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  • the proposed duplicate shows that totally bounded and complete implies compact, but in contrast the answer here shows that a closed and bounded subset of a complete metric space is not necessarily compact – J. W. Tanner Jun 29 '20 at 04:11
  • @J.W Tanner so that means some people are wrongly associating it with different question? –  Jun 29 '20 at 09:07

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For a counterexample, you can consider $[0,1]$ with the discrete metric $d(x,y) = 1$ if $x \ne y$, $d(x,y) = 0$ if $x = y$. This is complete because any Cauchy sequence is eventually constant and hence convergent, but the open cover $\{B(x,\frac 12) : x \in [0,1]\}$ has no finite subcovers because each ball $B(x, \frac 12) = \{x\}$ is a singleton.

user6247850
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  • so you mean a ball of radius 1/2 around a point x , is a cover of x but it doesn't have a finite subcover? How? B( x, 1/2) is a single point in itself so why doesn't there exists a finite subcover? –  Jun 25 '20 at 17:42
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    @user You take one (singleton) ball for each point in $[0, 1]$, and you get a cover of $[0, 1]$ with (uncountably) infinitely many open sets. Such a cover has no proper subcover (removing any ${x}$ from the cover will leave $x$ uncovered), let alone finite subcovers. – user803264 Jun 25 '20 at 17:49