Normed Vector Space is Separable iff Weakly Separable

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 $\struct {X, \norm {\, \cdot \,} }$ is separable if and only if $\struct {X, w}$ is separable.

Proof

Necessary Condition

This follows from Separable Topological Space remains Separable in Coarser Topology.

$\blacksquare$

Sufficient Conditon

Suppose that $\struct {X, w}$ is separable.

Let $D$ be a countable dense subset of $\struct {X, w}$.

From Set Closure Preserves Set Inclusion, we have:

$\map {\cl_w} D \subseteq \map {\cl_w} {\map \span D}$

So that:

$\map {\cl_w} {\map \span D} = X$

From Mazur's Theorem, we then have:

$\map \cl {\map \span D} = X$

From Closed Linear Span of Countable Set in Topological Vector Space is Separable, we have:

$\struct {\map \cl {\map \span D}, \norm {\, \cdot \,} }$ is separable.

So $\struct {X, \norm {\, \cdot \,} }$ is separable.

$\blacksquare$

Sources

• 2001: Marián Fabian, Petr Habala, Petr Hájek, Vicente Montesinos Santalucía, Jan Pelant and Václav Zizler: Functional Analysis and Infinite-Dimensional Geometry ... (previous) ... (next): Proposition $3.26$