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

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:

So:


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

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