# Unit ball in dual space is weak*separable

Let $X$ be a separable Banach space, $X'$ the dual space of $X$ endowed with the weak$*$-topology. How to prove that the unit ball $B$ in $X'$ is weak*-separable ?

I only know Banach-Alaoglu, which states that $B$ is compact in the weak*-topology, but not sure if it helps.

Let $x_n$ be a countable dense subset of the unit ball $B_X = \{x \in X: \|x\| \le 1\}$ of $X$. Then if $f \in B_{X'}$, the unit ball of $X'$ in the weak$^\ast$ topology, we define $F(f) = (f(x_n))_n \in D^\mathbb{N}$ where $D = \{x \in \mathbb{F}: \|x\| \le 1\}$, where $\mathbb{F}$ is the scalar field (reals or complex numbers), so $D$ is compact. Then in this answer I show that $F$ is 1-1 and so as $B_{X'}$ is compact, $F$ is an embedding and so $B_{X'}$ is metrisable and being compact separable as well.

• -1 It's not necessary to use compactness or metrizability. – fourierwho Jan 29 '18 at 17:34
• @fourierwho why not? He already knows about the compactness and metrisability is required background for functional analysis anyway. – Henno Brandsma Jan 29 '18 at 17:44
• I think another standard proof is to use functionals of norm $1$ hat are $1$ on $x_n$ and take their rational multiples that lie inside the ball. These are dense aren’t they? @fourierwho – Henno Brandsma Jan 29 '18 at 19:18
• @fourierwho I don't think downvoting an answer that doesn't follow the same approach you would provide is sufficient reason to downvote. – Aweygan Jan 29 '18 at 19:51

Use this question to show that

Theorem. If X is separable, then $B$ endowed with the weak*-topology is metrizable.

Since you also know that $B$ is compact, all you need now is contained in that question:

Theorem. If $X$ is a compact metric space, then $X$ is separable.

• -1 The question can be answered without using metrizability. Also, this answer leaves too little for the OP to do in my opinion, particularly since he hasn't given us any idea what he's tried or not yet. – fourierwho Jan 29 '18 at 15:12