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This natural property is not true always, when I mean a bounded set, I mean a set that it's contained in some ball of radius $r$. It's not always true that if this is the case, I can cover fixing a radius $u < r$ cover with balls of radius $u$, my bounded set, using only finite or even numerable balls. An example is the discrete metric, and any no numerable set ($[0,1]$) . What conditions I can impose in my space, to have this nice property? In any bounded set, not only in compact sets (or at least in open balls, that's what I need now)

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    "I can impose in my space" should be "can I impose on my space" ("I can" is for a statement and "can I" is for a question, and "impose in my space" sounds like we already have a specific space and the conditions are on something else).2011-08-24

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If Countable Choice, then your property is equivalent to being a Lindelöf space.

Proof:


I assume we must have $0. $ \; $ (I am not sure if that is part of the definition of balls.)

Let $B_u(x)$ be the open ball of radius $u$ centered at $x$.

Let $\overline{B}_u(x)$ be the closed ball of radius $u$ centered at $x$.

Clearly, $\; B_u(x) \subseteq \overline{B}_u(x) \subseteq B_{\frac12 \cdot (u+r)}(x)$,
so it does not matter whether we cover with open or closed balls.

I also assume you are including finite sets as "numerable", and I include them as "countable".


part 1: Lindelöf spaces have your property

Let $C$ be the set of all open balls of radius $u$.
Since the space is Lindelöf, let C' be a countable subcover of $C$.
C' is a countable cover of every bounded subset, and this works for all $u$, which proves part 1.



part 2: If countable choice, then spaces with your property are Lindelöf


(CC) will mean a use of countable choice.
If the space is finite, then in space is compact, in which case the space is Lindelöf.
Assume the space is infinte. $ \; $ Let $x$ be a point in the space.
By your property, for all non-negative integers $n$,
there is a countable cover of $B_{n+1}(x)$ by balls of radius $\frac1{n+1}$.
(CC) $\;$ For all non-negative integers $n$, let $C_n$ be such a cover.
For all positive integers $n$, since $x_0\in B_n(x)$, $C_n$ is non-empty.

Define $\quad D \; = \; \displaystyle\bigcup_{n\in \{0,1,2,3,...\}} C_n$

(CC) $\;$ By the last property proven here, $D$ is countable.
Since the diameter of the balls in $C_n$ goes to zero as $n$ goes to infinity, and the space is infinite,
the number of balls in $C_n$ must go to infinity as $n$ goes to infinity. $ \; $This shows that $D$ is infinite. $\;\;$ Since $D$ is countably infinite, let $\; f : \{0,1,2,3,...\} \to D \;$ be some bijection.
(CC) $\;$ For all non-negative integers $n$, let $x_n$ be a member of $f(n)$.
That finishes the setup.
Now, let $U$ be a non-empty open subset of the space, and let $y$ be a member of $U$.
Let $r$ be a positive real number such that $\; B_r(y) \subseteq U \;$.
Since the diameter of the balls in $C_n$ goes to zero as $n$ goes to infinity, there is an $n$ such that the diameter of the balls in $C_n$ is less than $r$. $\;\;$ Let $m$ be such an $n$.
Let $B$ be a member of $C_m$ such that $y\in B$. $ \; $ By construction, $\; x_{f^{-1}(B)} \in B \subseteq B_r(y) \subseteq U \;$.
This works for all non-empty open subsets $U$ of the space, so the space is separable.
(CC) $\;$ Therefore, by Theorem 3.12, the space is Lindelof.


QED

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    @Brian: God forbid! :-) I use Arch Linux with awesome as a windows manager :-)2011-08-24