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Let $X$ be a Hausdorff locally convex linear space, and denote by $\mathcal{N}_{0}$ the class of its (say, closed and absolutely convex) basis neighborhood of zero. Then, can we construct a sequence $U_{n}\subset \mathcal{N}_{0}$ in such way that given any $V\in\mathcal{N}_{0}$ there is $m\geq 1$ such that $U_{m}\subset V$?

Clearly, if $X$ is metrizable the above assert is true. Indeed, we can take $U_{n}:=1/nB$, $B$ being the unit ball. However, there is a "weaker" assumption on $X$ for this claim?

Many thanks in advance for your comments.

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If you could do that, this would mean that $X$ has a countable base of neighbourhoods at $0$, and this in turn implies (by a standard metrisation theorem (Birkhoff-Kakutani theorem)) that $X$ is metrisable. So this can only be done for metrisable vector spaces.

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  • $\begingroup$ Ok, thanks Henno Brandsman!! $\endgroup$ – user123043 Nov 8 '17 at 12:09

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