TITLE: Weak-* sequential compactness and separability
QUESTION [12 upvotes]: Let $X$ be a Banach space, and let $B$ be the closed unit ball of $X^*$, equipped with the weak-* topology.  Alaoglu's theorem says that $B$ is compact.  If $X$ is separable, then $B$ is metrizable, and in particular it is sequentially compact.

What about the converse?  If $B$ is sequentially compact, must $X$ be separable?

This question was inspired by this one.

REPLY [4 votes]: For an explicit counterexample: let $H$ be a non-separable Hilbert space, and suppose we have a sequence of vectors $x_n$ in the unit ball of $H$.  Let $H_0$ be the closed linear span of the $x_n$, so that $H_0$ is a separable Hilbert space.  Then the $x_n$ have a subsequence converging weakly in $H_0$.  But because of the orthogonal decomposition $H = H_0 \oplus H_0^\perp$, it is easy to see that this subsequence also converges weakly in $H$.