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...)

Answer 2

An answer for separability can be found here Why is $l^\infty$ not separable?. To make it clear: if $l_\infty$ is separable so is any of its subset (easily provable). So that consider the following set $$ G = \{ x\in l_\infty: x_n \in \{0,1\},\ \forall n\in \mathbb{N} \}. $$ Since we have a bijection $\iota: G \to 2^\mathbb{N}$ and the latter is uncountable, so is $G$. Consider $x\neq y\in G$, $\iota(x)\neq \iota(y)$, so that there is a coordinate $\tilde{n}\in\mathbb{N}$ such that $x_n = 1$ and $y_n=0$ (or viceversa). But then $\|x-y\|=\sup_n |x_n-y_n| \geq |x_{\tilde{n}}-y_{\tilde{n}}| = 1$. This is enough to prove that it cannot exists a dense countable subset. Indeed if there existed one, namely $R$, then for all $x\in G\subset l_\infty$ we could find $(r_n)\subset R$ such that $r_n\to x$. Choose $n=n_x\in\mathbb{N}$ such that $\|r_{n_x} -x\|<1/4$. This means that for $x\neq y$, $$ \|r_{n_x}-r_{n_y}\| \geq \|x-y\|-\|r_{n_x}-x\| - \|r_{n_y}-y\| \geq 1-1/4-1/4 = 1/2. $$ So that the cardinality of $r_{n_x} \in R$ equals the cardinality of $G$ which is uncountable. This contradiction shows that $l_\infty$ is not separable.