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Let $\tau$ be the topology induced from the metric space $(X, d)$. We know that the base of the open balls in the metric space is a base for this topology.

I was wondering that if we restrict $X$ to be a countable set we can find a countable basis for the base of the open balls in the metric space.

I thought of taking the open balls with a rational radius but I was stuck proving that each open ball can be written as a finite intersection of open balls with rational radius.

I also do not think it is true. So I think the claim above is not true but I did not find a counter example.

I would really appreciate any help. Thank you!

FAF
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  • I'm not sure I understood you correctly, but if X is countable, wouldn't collection of $B(x,\frac{1}{n})$ for all $x\in X, , n\in \mathbb{N}$ be countable basis for X? In this case, as x, n is countable, it would be countable union of countable sets, which would be countable. – kim May 13 '21 at 15:22
  • It should be a basis of the tha base of the topology induced of the metric space. Not for X. @kim – FAF May 13 '21 at 15:24
  • What do you mean by a basis for the base of the open balls? That is not a standard term. If you simply mean a base for the topology, then kim is correct; if you mean something else, you’ll have to explain what you mean. – Brian M. Scott May 13 '21 at 17:07

1 Answers1

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It is true that if $X$ has a countable dense set $D$ (so $\overline{D}=X$ and $|D| = \aleph_0$) then $\mathcal{B}=\{B(x,r)\mid x \in D, r \in \Bbb Q\}$ is a countable base for the topology of $(X,d)$.

It's proof can be found in the proof of 7 to 1 in this answer.

Henno Brandsma
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