Weak Topology on Infinite Dimensional Normed Vector Space is not Metrizable

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$.

Let $w$ be the weak topology on $X$.

Then $w$ is not metrizable.

Proof
Let $X^\ast$ be the normed dual space of $X$.

From Metric Space is First-Countable, it suffices to show that $\struct {X, w}$ is not first-countable.

that $\struct {X, w}$ is first-countable.

Let $\sequence {U_n}_{n \mathop \in \N}$ be a local basis for $\mathbf 0_X$.

From Open Sets in Weak Topology of Topological Vector Space, for each $n \in \N$ we can pick $\epsilon_n > 0$ and a finite set $F_n \subseteq X$ such that:


 * $V_n = \set {x \in X : \cmod {\map f x} < \epsilon_n \text { for all } f \in F_n} \subseteq U_n$

Then $\sequence {V_n}_{n \mathop \in \N}$ is still a local basis for $\mathbf 0_X$.

Let $g \in X^\ast$ and consider:


 * $U = \set {x \in X : \cmod {\map g x} < 1}$

Then there exists $\map n g$ such that:


 * $U_{\map n g} \subseteq U$

Set $n = \map n g$.

Let:


 * $\ds x \in \bigcap_{f \mathop \in F_n} \ker f$

Then $\map f x = 0$ for each $f \in F_n$.

Then for each $\lambda \in \GF$ and $f \in F_n$ we have $\map f {\lambda x} = 0$.

So we have $\lambda x \in U$ for each $\lambda \in \GF$.

That is, $\cmod {\map g {\lambda x} } < 1$ for each $\lambda \in \GF$.

So for each $\lambda > 0$ we have:


 * $\ds \cmod {\map g x} < \frac 1 \lambda$

So we have $\map g x = 0$.

We conclude that:


 * $\ds \bigcap_{f \mathop \in F_n} \ker f \subseteq \ker g$

From Condition for Linear Dependence of Linear Functionals in terms of Kernel, we then have:


 * $g \in \map \span {F_n} = \map \span {F_{\map n g} }$

Since $g \in X^\ast$ was arbitrary, we have that:


 * $\ds X^\ast = \bigcup_{n \mathop = 1}^\infty \map \span {F_n}$

Since each $F_n \subseteq X$ is a finite set, $\map \span {F_n}$ is a finite-dimensional subspace of $X^\ast$.

So from Finite Dimensional Subspace of Normed Vector Space is Closed, $\map \span {F_n}$ is a closed set for each $n \in \N$.

Further, from Interior of Proper Subspace of Normed Vector Space is Empty, $\map \span {F_n}$ has empty interior for each $n \in \N$.

From Normed Dual Space is Banach Space, $X^\ast$ is complete.

But $X^\ast$ has then been written as the countable union of closed sets with empty interior, so $X^\ast$ is not a Baire space.

This contradicts the Baire Category Theorem.

So $\struct {X, w}$ is not first-countable and hence is not metrizable.