[Math] How to conclude that $\ell_\infty$ is not separable from this exercise

Question

I have done an exercise that goes like this:

Consider the operator $\Phi: \ell_1\to(\ell_\infty)'$ that associates each $x=(x_j)_j\in\ell_1$ to $\Phi (x)\in (\ell_\infty)'$ given by $\Phi(x)(y)=\sum x_j y_j$, for all $y=(y_j)_j\in\ell_\infty$. Show that $\Phi$ is well defined, is linear and bounded. Construct an element $g\in (\ell_\infty)'\backslash \ \Phi(\ell_1)$. Is $\ell_\infty$ separable?

So… I went through all this exercise, but not the final. How can I conclude that $\ell_\infty$ is not separable from this? Or is this question not related to the exercise (what doesn't make sense)?

Answer 1

It seems a bit of a stretch, but it's possible to show that $\ell_\infty$ is not separable in a way that's somewhat related to the rest of the exercise.

Recall the Banach-Alaoglu theorem:

If $X$ is a Banach space the closed unit ball of $X'$ is weak-* compact.

And recall this:

If $X$ is a separable Banach space then the closed unit ball of $X'$ is metrizable in the weak-* topology (and hence is weak-* sequentially compact).

Now it's easy to find a sequence $(e_n)$ in the closed unit ball of $\ell_1$ such that the sequence $(\Phi(e_n))$ has no weak-* convergent subsequence; hence $\ell_\infty$ is not separable.

This seems a little silly, since we really don't need $\Phi$ to show that the unit ball of $\ell_\infty'$ is not weak-* sequentially compact. But it does give some connection between the last bit of the exercise and the rest of it. (If this is an exercise for a class: If you've covered that result about weak-* metrizability then it seems likely that this is more or less the "intended" solution...)