Here $X,Y$ are assumed to be vector spaces, and $Y$ a subset of the algebraic dual of $X$. The weak topology if of course generated by the family of seminorms $\{ |y(x)| < \epsilon, y \in Y, \epsilon>0\}$.

I need to check that $(X, \tau)$ (with the weak topology) is metrizable iff $Y$ has a countable dimension. I have already proved the right to left implciation. I did it using the fact that the weak topology on a vector space is metrizable (since it is a locally convex space) iff the topology can be generated by a countable family of seminorms(functions). So I picked the countable base of $Y$, and checked that the topology generated by that subset is the same as the original.

I would appreciate though some help with proving the other direction. Thanks.


1 Answer 1


Hints: Metrizabilty implies that there is a countable local base at $0$. So there exist weak neighborhoods $\{x: |y_i^{n}(x)| <r_i, 1 \leq i \leq N_n\}$ which form a local base at $0$. Claim: $\{y_i^{n}: 1 \leq i \leq N_n, n \geq 1\}$ span $Y$. For this take $y \in Y$ and consider the weak neighborhood $\{x: |y(x)|<1\}$. This contains one of the above neighborhoods, say $\{x: |y_i^{n}(x)| <r_i, 1 \leq i \leq N_n\}$. If $y_i^{n}(x)=0$ for each $i$ then $y(x)=0$. A standard Linear Algebra argument now shows that $y$ must be a linear combination of $y_i^{n}, 1\leq i \leq N_n$.

  • $\begingroup$ Hi! I have some questions: Does every vector space has a countable Hamel basis? If not, then aren't we assuming already that $Y$ has a countable dimension (which is what we should conclude)? Also, we need to assume a metric exists for $Y$, but why is it possible to define it from the beggining? $\endgroup$
    – user986747
    Commented Apr 16, 2022 at 13:03
  • $\begingroup$ I am assuming that $Y$ has countable dimension and proving that the weak topology is metrizable. $Y$ having countable dimension means that there is a countable Hamel basis for $Y$. I am using this basis to construct a metric for $X$ with the weak topology. @StorageOne $\endgroup$ Commented Apr 16, 2022 at 23:13
  • $\begingroup$ This is what I already managed to prove.. I was asking for some hints in the opposite direction. $\endgroup$
    – user986747
    Commented Apr 16, 2022 at 23:24
  • $\begingroup$ Sorry, I thought you wanted the right to left implication. I have now edited my answer. @StorageOne $\endgroup$ Commented Apr 16, 2022 at 23:47
  • $\begingroup$ May I ask how to show "If $y_i^n(x)=0$ for each $i$ then $y(x)=0$"? I don't understand how that follows from that neighborhood containment. Thanks! $\endgroup$
    – user760
    Commented May 24, 2023 at 20:29

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