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Let $(M, d)$ be a non-separable metric space. Denote by $B = B(x, \epsilon)$ a generic open ball of $M$. Is there any condition on $M$ such that the following claim is true:

Claim: Given any open set $A$ of $M$, there is a countable family of open balls $B_i^A = B(x_i^A, \epsilon_i^A)$, $i \in \mathbb{N}$, such that $$ A = \bigcup_{i\in \mathbb{N}} B_i^A \qquad ? $$

Of course when the space is separable this is true, for separability is equivalent to second countability for metric spaces. That is why I am concerned with non-separable spaces.

Also, notice that I am not asking for a countable basis for the topology, for that would be asking for separability.

Instead, I am asking for a "locally countable basis", that is, conditions under which any open set can be covered by a countable family of open balls, the family changing according to the open set in question.

Brisão
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    Certainly, if $M$ is uncountable and $d$ is the discrete metric, it isn't true. – Daniel Schepler Feb 20 '20 at 00:01
  • Yes, naturally. I am aware this is not a universal property. I just want to know if there are any set of practial conditions which give me the property I need. I'm working with a not too exotic metric space, although it is non-separable, and I am concerned about borel measurable functions. The thing is that I am only able to check the measurability of open balls, so this question arose – Brisão Feb 20 '20 at 00:07
  • If the function is Borel measurable, all open sets are measurable by definition? – Henno Brandsma Feb 20 '20 at 08:29

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No, it follows from the theorem in this answer that for a metric space it is equivalent to be hereditarily Lindelöf (which implies your property) and separable (and second countable and many more properties).

So a metric space satisfies your property iff it is separable. (If $X$ is not separable, it has an uncountable pairwise disjoint family of open balls, and its union is an $A$ without this property.)

A concrete counterexample: $\Bbb R$ in the discrete metric ($d(x,y)=1$ for $x \neq y$) then the only open balls are $\Bbb R$ and all sets $\{x\}$ so we cannot write the open set $(0,1)$ as a countable union of open balls.

Henno Brandsma
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