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.


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.