Weak Topology on Infinite Dimensional Normed Vector Space is not Metrizable

Theorem

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

Let $\struct {X, \norm {\, \cdot \,} }$ be an infinite dimensional 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.

Aiming for a contradiction, suppose 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:

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

Then:

$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is a generator for $X$.

We have that $X$ has infinite dimension.

Hence, from Normed Dual Space of Infinite-Dimensional Normed Vector Space is Infinite-Dimensional, $X^\ast$ also has infinite dimension.

Hence:

$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is infinite.

From Countable Union of Finite Sets is Countable, it follows that:

$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is countable.

From Generator of Vector Space Contains Basis, it follows that $X^\ast$ either has countable or finite dimension.

We have already established that it does not have finite dimension, hence it must have countable dimension.

From Normed Dual Space is Banach Space, $X^\ast$ is a Banach space of countable dimension.

This contradicts Infinite-Dimensional Banach Space has Uncountable Dimension.

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

$\blacksquare$