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Show that every metric space $M$ can be written as a countable union of limited subsets.

Can someone give me some hints on this?

The only ideas that i had was that every limited subset is contained into an open ball and to show that the set $S_M$, which contains all these balls, is itself equal to $M$.

But i had no progress. Also i don't have a clue on how to show enumerability.

user2345678
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    Maybe I misunderstood, but what if you fix a point $x_0$ and consider the balls centered in $x_0$ with radius $n$ (natural)? – PhyM Aug 11 '17 at 22:33
  • That solved the case. Since the naturals numbers are not upper-limited, every point of $M$ must be inside one of these balls. And, consversely, every point of these balls are in $M$ (by definition). Since all these balls are limited, the result follows. – user2345678 Aug 11 '17 at 22:58
  • I'll post a CW answer for reference – PhyM Aug 11 '17 at 23:09
  • You're welcome :) – PhyM Aug 12 '17 at 09:36

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Fix a point $x_0$ and consider the countable family of balls centered in $x_0$ with radius $n\in \mathbb{N}$: every ball is limited and their union is $M$ (because $\forall y\in M, \ \exists n\in \mathbb{N} \ | \ d(y,x_0)<n$)

PhyM
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