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I am trying to solve the following exercise:

Let $(X,d)$ be a metric space that has the property that for any $x\in X$ and $r>0$, the closed ball $$\bar{B}(x,r):=\{y\in X:d(x,y)\leq r\}$$ is compact. Show that $X$ is complete.

I think I have a proof, but I am only using that the unit closed balls $\bar{B}(x,1)$ are compact. Maybe I am missing something?

Attempt: Let $(x_n)_{n=1}^\infty$ be a Cauchy sequence in $(X,d)$. Then, $\exists N\in\mathbb{N}$ such that $\forall n\geq N$, $d(x_n,x_N)<1$. Thus, $\forall n\geq N$, $x_n\in\bar{B}(x_N,1)$ which is compact by assumption, so $(x_n)_{n=N}^\infty$ is a sequence in this compact set and thus has a convergent subsequence. Hence, $(x_n)_{n=1}^\infty$ is a Cauchy sequence that has a convergent subsequence, so it converges in $(X,d)$. Therefore, $(X,d)$ is complete.

Spenser
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2 Answers2

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Your proof is right.

As a justification that it is not strange that you needed a little less than the hypothesis, let me show you what happens when your metric space is a normed space: since any two balls are homeomorphic, if one closed ball is compact than all are. Indeed, $$ \bar B(0,1)\simeq \bar B(y,r) $$ via the continuous function $x\mapsto r(x+y)$.

Martin Argerami
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  • Note that the Spenser asked about metric spaces in general and that kind of argument doesn't quite work then. – Martin Citoler Oct 24 '13 at 01:01
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    You are totally right. I seldom use metric spaces which are not normed spaces... – Martin Argerami Oct 24 '13 at 01:11
  • In time, another justification of why this is not strange might be the fact that every metric space admits a limited and topologically equivalent metric (for example $\tilde{d} = d/(1 + d)$ ), so the whole original space is a limited ball in this new equivalent metric, in the spirit of Neal's remark below. – ulilaka May 23 '18 at 18:15
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If there is an $r>0$ with $\overline{B(x,r)}$ compact for all $x$, then for any $r>\varepsilon>0$, $\overline{B(x,\varepsilon)}$ is compact. It's not so much stronger that you're just using compactness of $B(x,1)$ for all $x$; that implies compactness at all smaller scales.

Neal
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