If $X^{\ast}$ separable there exists $(x_n) \subset S_E$ such that $x_n \rightharpoonup 0$.

Question

Let $X$ an infinite dimensional vector space. Show that if $X^{\ast}$ is separable, then there exists a sequence $(x_n) \subset S_E$, the unit sphere in $X$, such that $x_n \rightharpoonup 0$ in the $\sigma(X,X^{\ast})$ topology.

This is an exercise coming from Brezis book of Functional Analsyis. I don't see exactly why I have to use that $X^{\ast}$ is separable.

Since $X$ is infinite dimensional, the weak closure of $S_E$ is $B_E$, the unit closed ball. So by the definition of closure, since $0 \in B_E$ there exists a sequence $(x_n) \subset S_E$ such that $x_n \rightharpoonup 0.$

I see that one possible reason for using that $X^{\ast}$ is separable is by the fact that in this way $B_E$ is metrizable, and then it is second countable, so makes sense to talk about sequences. Is there other reason?

Thanks

Answer 1

You are correct about why we need separability of $X^*$. In general, your statement

Since $X$ is infinite dimensional, the weak closure of $S_E$ is $B_E$, the unit closed ball. So by the definition of closure, since $0\in B_E$ there exists a sequence $\{x_n\}$ in $S_E$ such that $x_n\to0$ weakly.

is false, without replacing sequences with nets.