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Is it true that in general complete metric space $(M,d)$, a closed ball of radius $r$ centered at $p\in M$ is always compact? That is, the ball is the set of all points $\left\{x:d(x,p)\leq r\right\}$.

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  • $\begingroup$ The closed unit ball in Hilbert space is not compact. Maybe you think that is more "natural" than a discrete space? $\endgroup$ – GEdgar Mar 19 '14 at 18:48
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No. If $(M,d)$ is an infinite set endowed with the discrete metric $d$, then $M$ is the closed ball around any one of its points of radius $1$. However, $M$ is not compact.

Given any metric space $(M,d)$, you can define a new metric space $(M,d')$ by $d'(x,y) = \min\{d(x,y),1\}$. These metric spaces have the same topology. But $M$ is now bounded, so if $M$ is not compact with the metric $d$ then it is still not compact with the metric $d'$.

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  • $\begingroup$ What if my space is complete? $\endgroup$ – user134070 Mar 19 '14 at 17:46
  • $\begingroup$ My $M$ is complete. Any Cauchy sequence in this space is eventually constant. $\endgroup$ – wckronholm Mar 19 '14 at 17:46
  • $\begingroup$ @wckronholm: I think you want to say that any Cauchy sequence is eventually constant (and therefore convergent). $\endgroup$ – user642796 Mar 19 '14 at 17:49
  • $\begingroup$ I did. And I edited my comment to that effect immediately before your comment. $\endgroup$ – wckronholm Mar 19 '14 at 17:50
  • $\begingroup$ Well thank you. But is what I am claim true if the metric is not discrete? $\endgroup$ – user134070 Mar 19 '14 at 17:51
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A subset $E$ of a metric space $X$ is said to be totally bounded if $E$ is contained in the union of finitely many open balls of radius $\varepsilon$, for every $\varepsilon>0$.

Let $X$ be a complete metric space.Then $E\subset X$ is compact if and only if it is closed and totally bounded.(Functional analysis by Walter Rudin appendix 4)

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For a Banach space $X$ the closed unit ball is compact iff dim $X<\infty$.

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Any metric space $(M,d)$ can be given a new metric which induces the same topology, but in which the entire space has finite diameter by setting $$d'(x,y)=\frac{d(x,y)}{1+d(x,y)}.$$ It follows that if $M$ is not compact, then all balls with radius larger than $1$ in this new metric $d'$ must also be non-compact.

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