# Unit ball of a Separable Banach Spaces is metrizable

Please help me to understand why the unit ball of a separable banach space is metrizable, when it is given the induced topology from the weak topology. Specifically, my problem is I think I'd be able to do it if I knew the duel was separable, but not with the current given information. (If I had this additional information, I'd use the infinite sum metric tricks that one uses all too often.)

The underlying field can be real or complex.

• A weak* topology is defined on a dual. You mean: the unit ball of the dual is metrizable when $X$ is separable. By Banach-Alaoglu, it is a nice compact metric space. See here. – Julien Apr 5 '13 at 20:53
• No I'm sorry. I meant that supposedly one can metrize the unit ball of the weak topology of the original space? – Jeff Apr 5 '13 at 21:41
• Then you need to edit your question. And I think your assumption is most likely that the dual is separable. See here in this case. – Julien Apr 5 '13 at 22:10
• If the dual space is separable then yup's answer (or Banach-Alaoglu) does the trick: the inclusion $J \colon X \to X^{\ast\ast}$ is a homeomorphism from $X$ with the weak topology to $X^{\ast\ast}$ with the weak*-topology and the unit ball in $X^{\ast\ast}$ is metrizable. – Martin Apr 5 '13 at 22:21
• The unit ball in $\ell^1$ is not metrizable in its weak topology: the unit sphere is norm-closed and the weak closure of the unit sphere is the entire unit ball; yet the weak and norm-topologies have the same convergent sequences by the Schur property of $\ell^1$. – Martin Apr 5 '13 at 22:28

Choose a dense sequence $\langle x_n : n \in \mathbb{N}\rangle$ in the unit sphere of $X$, this exists by separability of $X$. Let $B'$ be the unit ball in the dual space. By the definition of the weak*-topology we have a continuous map $f_n \colon B' \to D$ given by $f_n(\varphi) = \varphi(x_n)$ where $D = \{z \in \mathbb{C} : |z| \leq 1\}$ is the closed unit disc in $\mathbb{C}$. These assemble to a continuous map $f \colon B' \to D^{\mathbb{N}}$ given by $$f(\varphi) = \langle f_n(\varphi) : n \in \mathbb{N}\rangle = \langle \varphi(x_n) : n \in \mathbb{N}\rangle$$ by the definition of the product topology on $D^\mathbb{N}$. I claim that this map is injective. Indeed, if $\varphi_1 \neq \varphi_2$ then there is $x \in X$ such that $\varphi_1(x) \neq \varphi_2(x)$. Normalizing $x$ we can assume hat $\|x\| = 1$. By choosing $x_n$ close enough to $x$ we will have $\varphi_1(x_n) \neq \varphi_2(x_n)$. This implies that $f(\varphi_1) \neq f(\varphi_2)$ and we're done.
To be a little more explicit, the standard metric on $D^\mathbb{N}$ is $d(a,b) = \sum_{n=1}^\infty 2^{-n} |a_n - b_n|$ and since $f$ is injective and continuous, $\delta(\varphi_1,\varphi_2) = d(f(\varphi_1),f(\varphi_2))$ is a continuous metric on $B'$ with the weak*-topology and this shows metrizability of $B'$.
1. show that convergence with respect to $\delta$ is equivalent to weak*-convergence.
2. show that $B'$ is compact with respect to $\delta$.