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http://planetmath.org/birkhoffkakutanitheorem

A topological group $(G,*,e)$ is metrizable if and only if $G$ is Hausdorff and the identity $e$ of $G$ has a countable neighborhood basis.

In particular, any topological vector space which is first countable would have a countable neighborhood basis at the identity (since it has one at every point).

Therefore, every first countable topological vector space is metrizable, and any topological vector space which is not metrizable is not first countable.

Is this correct? I think so, but I want to make sure.

Also, this might be a duplicate of these questions: The weak$^*$ topology on $X^*$ is not first countable if $X$ has uncountable dimension., A topological vector space with countable local base is metrizable, Countable neighborhood base at zero in a topological vector space, A locally convex space is metrizable if and only if it is first countable, A basic doubt on metrizable topological vector space but mine seems a lot simpler for some reason, so again, I want to be sure, yes/no.

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Chill2Macht
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1 Answers1

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Yes this is true. In fact we can even say more; this is actually an if and only if statement (i.e. any metrizable topological space is first countable). This follows because the topology induced by the metric is translation invariant, hence if a countable neighborhood basis exists at 0, then one must exist at every point of the space. See "Köthe, G.-- Topological Vector Spaces I, 15.11 (1), p. 163".

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