Does $(\ell^\infty)^*$ have the cardinality of the continuum?

Question

I have a sort of intuitive but non-rigorous understanding of the dual of $\ell^\infty(\Bbb N)$. It is the span of evaluation maps of bounded sequences and of evaluations "at infinity": consistent ways of picking out limit points of sequences.

This seems like a very big space. Is it still "small enough" to have the cardinality of the continuum?

Answer 1

Note that $\ell_\infty\cong C(\beta \mathbb{N})$. For each $p\in \beta \mathbb{N}$, point-evaluation at $p$, $\delta_p$, is a norm-one functional on $C(\beta \mathbb{N})$. By Pospišil's theorem, there are $2^{2^{\aleph_0}}$ elements of $\beta \mathbb{N}$. However, elements of $\ell_\infty^*$ are functions and so there are at most ${(2^{\mathbb{R}}})^{|\ell_\infty|} = 2^{2^{\aleph_0}}$ of them, so we conclude that $\ell_\infty^*$ has cardinality $2^{2^{\aleph_0}}$.

Answer 2

For ultrafilter $f$ on $\mathbb N$ and $x=(x_n)_n\in l_{\infty},$ define $f'(x)$ as the unique $r$ such that for every nbhd $U$ of $r$ we have $\{n:x_n\in U\}\in f.$

We may easily verify that $f'\in l_{\infty}^*$ with $\|f'\|=1$ and that if $f,g$ are distinct ultrafilters on $\mathbb N$ then $f'\ne g'.$ By Pospisil's Theorem ( see link in A by Tomek Kania) therefore $|l_{\infty}^*|\geq 2^{|\mathbb R|}.$

Since $|l_{\infty}|=|\mathbb R|$ we also have $|l_{\infty}^*|\leq |\mathbb R^{l_{\infty}}|=2^{|\mathbb R|}.$