Normed Dual Space of Separable Normed Vector Space is Weak-* Separable

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Theorem

Let $X$ be a separable normed vector space.

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

Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.


Then $\struct {X^\ast, w^\ast}$ is separable.


Proof

Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$.

From Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Separable, $\struct {B^-_{X^\ast}, w^\ast}$ is separable space.

Let $S$ be a countable dense subset of $B^-_{X^\ast}$.

For $n \in \N$, we have:

$\map {\cl_{w^\ast} } {n S} = n B^-_{X^\ast}$

from Dilation of Closure of Set in Topological Vector Space is Closure of Dilation.

Now, note that:

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

from Countable Union of Countable Sets is Countable.

From Closure of Union contains Union of Closures, we therefore have:

\(\ds \map {\cl_{w^\ast} } {\bigcup_{n \mathop = 1}^\infty n S}\) \(\supseteq\) \(\ds \bigcup_{n \mathop = 1}^\infty \map {\cl_{w^\ast} } {n S}\)
\(\ds \) \(=\) \(\ds \bigcup_{n \mathop = 1}^\infty B^-_{X^\ast}\)
\(\ds \) \(=\) \(\ds X^\ast\)

So:

$\ds \map {\cl_{w^\ast} } {\bigcup_{n \mathop = 1}^\infty n S} = X^\ast$

So $\struct {X^\ast, w^\ast}$ is separable.

$\blacksquare$


Sources

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