# The weak topology on a dual pair $(X,Y)$ is metrizable iff the dimension of $Y$ is at most countable.

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.

Hints: Metrizabilty implies that there is a countable local base at $$0$$. So there exist weak neighborhoods $$\{x: |y_i^{n}(x)| which form a local base at $$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)| . 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 $$y_i^{n}, 1\leq i \leq N_n$$.
• 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?
• 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 Commented Apr 16, 2022 at 23:13
• 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! Commented May 24, 2023 at 20:29