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It is well known that if $X$ is a Banach space and $X^*$ contains separating sequence (i.e. sequence $(f_n) \subset X^{*}$ such that $f_n x = 0$, $n = 1,2, \dots$, implies that $x=0$) and $K \subset X$ is weakly compact then $K$ is weakly metrizable (i.e. metrizable in weak topology).

Using this we can easily prove, for example, that closed unit ball $B$ in $\ell^p$, $1 < p < \infty$, is weakly metrizable. Indeed, $\ell^p$ is reflexive and hence $B$ is weakly compact, which follows from Banach-Alaoglu theorem (it is even equivalent; closed unit ball is weakly compact if and only if $X$ is reflexive). Moreover $\ell^p$ contains separating sequence, thus $B$ is metrizable.

Can we prove that closed unit ball $B_{c_0}$ in $c_0$ is weakly metrizable? Main problem here is that $c_0$ is not reflexive and we can't mimic previous argument.

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$c_o$? Could you explain this notation? – MathOverview Nov 5 '12 at 13:33
@Elias: This is the space of real sequences converging to zero. – xen Nov 10 '12 at 22:09

1 Answer 1

up vote 2 down vote accepted

You can extend your statement as follows: Let $X$ be a seperable normed space, then every (norm-)bounded subset of $X^*$ is metrizable in the relative weak*-topology. Now apply this to $X=l^1$. Then $B_{c_0}$ as a bounded subset of $l^\infty$ is metrizable in the relative weak*-topology $\sigma(l^\infty, l^1)$ which is equal to $\sigma(c_0,l^1)$.

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