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We know that every open set $A$ is in a metric space $(X,d)$ is the countable union of closed sets, and every open set $A$ is in a Euclid space $R^n$ is the countable union of closed balls.

My question is, in what spaces, every open set is the countable union of closed balls.

If every open set $A$ is in a metric space $(X,d)$ is the countable union of closed balls?

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No. Take the set $X=\mathbb R\times \mathbb [0,1]$ with metric $$d\bigl((x,y),(u,v)\bigr)=\begin{cases}1&\text{if }x\ne u\\|y-v|&\text{if }x=u\end{cases} $$ Then the set $$A=\mathbb R\times (0,1)=\bigcup_{x\in\mathbb R}B\bigl((x,\tfrac12),\tfrac12\bigr)$$ is open, but not the countable union of closed balls (nor is it the countable union of open balls).

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Yes, every countable open-ball-covering can be made into a countable closed-ball-covering.

Take a countable number of open balls that cover $A$. Each ball $B_i$ has a center $a_i$ and a radius $r_i$. Now for each $i$ look at the set of closed balls $\overline{B}_{i,n}$ centered at $a_i$ with radius $(1-1/n)r_i$ for $n \in \Bbb N$.

The full collection of these $\overline{B}_{i,n}$ is still countable, and cover exactly the same points as the original $B_i$-balls.

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