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Let $(X,d)$ be a complete metric space (I am not assuming that the metric is finite, there could be points in $X$ with infinite distance). Assume that each Ball in $X$ is pre-compact i.e. $\forall x \in X, \forall R > 0$ the closed Ball $B_{R}(x)$ is pre-compact (or if its closed then compact). Does it follow that $X$ is separeble? If not, does one need to make additional assumptions ?

Greetings Nina

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  • $\begingroup$ does one have an idea ? $\endgroup$ – Nina Jul 3 '13 at 14:13
  • $\begingroup$ This is not a MO question and should be probably closed. The answer readily follows from the standard fact that compact spaces are separable. $\endgroup$ – Valerio Capraro Jul 3 '13 at 14:29
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    $\begingroup$ how do i reach these points ? $\endgroup$ – Nina Jul 3 '13 at 14:33
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    $\begingroup$ in this case you should first reformulate the question using extended metrics instead of metrics. $\endgroup$ – Valerio Capraro Jul 3 '13 at 14:33
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    $\begingroup$ Take the trivial metric: the distance between two different points is infinite. This space is not separable as long as it contains uncountably many points and each ball of finite radius is compact (it's just a singleton) $\endgroup$ – Valerio Capraro Jul 3 '13 at 14:41
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If $d$ is an extended metric, taking values in $\Bbb R^+\cup\{\infty\}$, then the result is false, as Valerio Capraro noted in the comments. Let $X$ be any set, and for $x,y\in X$ define

$$d(x,y)=\begin{cases}0,&\text{if }x=y\\\infty,&\text{if }x\ne y\;.\end{cases}$$

Then $\langle X,d\rangle$ is complete, because every Cauchy sequence is eventually constant. However, $d$ induces the discrete topology on $X$, so the space is separable if and only if $X$ is countable. In particular, if $X$ is uncountable, then the space is not separable.

If $d$ is an ordinary metric, and you simply meant that the diameter of $X$ is not necessarily finite, then the result is true. Fix a point $p\in X$. For each $n\in\Bbb Z^+$ let $K_n=\operatorname{cl}B_n(p)$, the closure of the open ball of radius $n$ centred at $p$. By hypothesis each $K_n$ is compact, and a compact metric space is separable, so each $K_n$ has a countable dense subset $D_n$. Let $D=\bigcup_{n\in\Bbb Z^+}D_n$; clearly $D$ is countable. Finally, $X=\bigcup_{n\in\Bbb Z^+}K_n$, so $D$ is dense in $X$, and $X$ is therefore separable.

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  • $\begingroup$ Here it is anyways. I am not versed in ordinals and cardinals, so the answer there is not much of use now, but I kinda groke it. Of course, the question itself is about cardinals. But maybe you can help a little? Pedagogically? $\endgroup$ – Pedro Tamaroff Jul 4 '13 at 3:46

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