Countable basis in a metric space is equivalent to being a separable metric space

Asked Sep 14 '19 at 13:38

Active Sep 14 '19 at 13:40

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Say we have a metric space space $(\Omega, d)$. Is the following true:

$(\Omega, d)$ has a countable basis in a metric space $\iff$ $(\Omega, d)$ separable metric space

To me the equivalence seems intuitive (in terms of open balls) but I have not seen it actually stated anywhere

edited Sep 14 '19 at 13:40

Gabriel Romon

asked Sep 14 '19 at 13:38

MinaThuma

https://math.stackexchange.com/questions/145303/equivalence-of-three-properties-of-a-metric-space this might be of interest – Riquelme Sep 14 '19 at 13:47
https://math.stackexchange.com/questions/1810581/every-separable-metric-space-has-a-countable-base This too. – Arnaud Mortier Sep 14 '19 at 13:48
This answer of mine generalises this observation considerably. – Henno Brandsma Sep 14 '19 at 14:24

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