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The problem says:

If every closed ball in a metric space $X$ is compact, show that $X$ is separable.

I'm trying to use an equivalence in metric spaces that tells us: let X be the matric space, the following are equivalent

X is 2nd countable

X is Lindeloff

X is separable

I also thought about taking balls of a "big" radius and that they are disjoint, but I do not see how to make the set of balls is countable.

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  • $\begingroup$ Wouldn't the entire space necessarily be compact as well ('ball of infinite radius')? $\endgroup$ – Justin Benfield Aug 11 '17 at 18:27
  • $\begingroup$ @JustinBenfield That's what the problem says, it's from a book $\endgroup$ – 447395 Aug 11 '17 at 18:44
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For a different approach, choose $x\in X$ so that $X = \cup_{n \in \mathbb{N}} \overline B_x(n).$ Now consider, for fixed $n\in \mathbb N,$ the fact that the compact ball $\overline B_x(n)$ is totally bounded. This means that, for each $m\in \mathbb N$, there is a $\textit{finite}$ set of points $x_{m,n}\in \overline B_x(n)$ such that if $y\in \overline B_x(n)$ then there is an $m\in \mathbb N$ and $x_{m,n}$ such that $d(y,x_{m,n})<1/m,$ which implies that $\left \{ x_{m,n} \right \}_{m\in \mathbb N}$ is a countable dense subset of $\overline B_x(n)$ and therefore that $\left \{ x_{m,n} \right \}_{m,n\in \mathbb N}$ is a countable dense subset of $X$. The result follows by the equivalence separable $\Leftrightarrow $ second countable.

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$X = \cup_{n \in \mathbb{N}} D(p, n)$, for any $p \in X$ (where $D(x,r)$ denotes the closed ball around $p$ of radius $r$).

So $X$ is $\sigma$-compact hence Lindelöf hence separable.

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  • $\begingroup$ At the end, after “Lindelöf hence separable”, you could have added “since $X$ is a metrice space”. $\endgroup$ – José Carlos Santos Aug 11 '17 at 18:55
  • $\begingroup$ @JoséCarlosSantos the OP already mentioned it. $\endgroup$ – Henno Brandsma Aug 11 '17 at 19:08
  • $\begingroup$ You're right. Sorry. $\endgroup$ – José Carlos Santos Aug 11 '17 at 19:42
  • $\begingroup$ @HennoBrandsma Where have you used that the closed balls are compact? Instead you used that X is compact $\endgroup$ – 447395 Aug 11 '17 at 19:46
  • $\begingroup$ @Ali I write $X$ as a countable union of closed balls (this is always true) and so a countable union of compact sets. $\endgroup$ – Henno Brandsma Aug 11 '17 at 19:52

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