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.
Best Answer
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 $y_i^{n}, 1\leq i \leq N_n$.