Compact metric implies uncountable w*-dense set I am reading a proof of the following: Let $X$ be a separable Banach space. The Szlenk index is countable iff $X*$ is separable.
In the proof of => it uses the following:
If $X^*$ is not separable, there is $\epsilon>0$ and uncountable $A\subset B_{X^*}$ with $\|x^*-y^*\|>\epsilon$ for all distinct $x^*,y^*\in A$.
Since $(B_{X^*},w^*)$ is a compact metric, 'there exists an uncountable $V\subset A$ which is $w^*$-dense in itself.'
My question is, why is this second line true? (the line which has quotation marks).
Thanks!
 A: I presume you are reading Ted Odell's survey paper on ordinal indices in Banach spaces :) . Anyway, here are some details that may be of assistance. 
Since $(B_{X^\ast},w^\ast)$ is compact metric it is second countable, and so is every topological subspace of $(B_{X^\ast},w^\ast)$. Besides this fact regarding second countability we need two results whose proof can be found in Section 8.5 of Semadeni's book Banach spaces of continuous functions (this book is very handy in general). The first of these results is Theorem 8.5.2 of Semadeni's book, which says that every topological space $A$ can be written as a union $A=V\cup S$, where $V\cap S=\emptyset$, $V$ is dense-in-itself and $S$ is scattered. Moreover, Proposition 8.5.5 of Semadeni's book asserts that a scattered subset of a second countable space is countable.
Once you know these facts, the claim of Odell's paper is now easy to verify. Let $A\subseteq (B_{X^\ast},w^\ast)$ be an uncountable $\epsilon$-separated set. Then we may write $A= V\cup S$, where $V\cap S=\emptyset$, $V$ is weak$^\ast$-dense-in-itself and $S$ is scattered. As $A$ is second countable, its scattered subset $S$ is countable. It follows that the $\epsilon$-separated, weak$^\ast$-dense-in-itself set $V$ satisfies $\vert V\vert=\vert A\vert \geq \aleph_1$.
