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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?

s.harp
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  • Presumably you know that $\ell^\infty(\Bbb{N})$ is canonically isometrically isomorphic to $C(\beta\Bbb N)$, the space of continuous functions on the Stone Cech compactification of $\Bbb N$, and so its dual is isometrically isomorphic to the space of Radon measures on $\beta\Bbb N$. But I don't know the answer to your question... – Olivier Bégassat Jan 30 '17 at 16:45
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    @Max $\ell^\infty$ has cardinal $\mathfrak c$. How do you prove that assertion of yours? – Olivier Bégassat Jan 30 '17 at 16:46
  • @Max: If I recall correctly, the dual of $\ell^\infty$ is the BA space, which is also the space of all finitely additive measures on $\Bbb N$. So one just has to count how many of these measures we can construct. Probably a lot of them. – Asaf Karagila Jan 30 '17 at 16:55
  • @Max that under AC $(\ell^\infty)^*$ is isomorphic to $\ell^\infty$. – Olivier Bégassat Jan 30 '17 at 17:08
  • Uhm sorry I made a mistake, I don't know the answer to your question. I'll delete my comments – Maxime Ramzi Jan 30 '17 at 17:17
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    An exercise in Diestel's Sequences and Series in Banach Spaces, using Rosenthal's $\ell_1$-Theorem, shows that if $X$ contains $\ell_1$ isomorphically, then the cardinality of $X^{**}$ exceeds that of the continuum. (I don't know if that's overkill here.) – David Mitra Jan 31 '17 at 09:26
  • In the book he first wants you to show that $\ell^\infty$ contains an isometric copy of $\ell^1(2^{\Bbb N})$. From here one can see that $(\ell^\infty)^*$ must have cardinality greater than continuum with Hahn-Banach. So that statement must be the application of Rosenthal's theorem, but I don't see how to do it. – s.harp Jan 31 '17 at 09:52
  • (Maybe for each subset $A\subset\Bbb N$ one constructs a bounded sequence $x_{A,n}$ that has no weakly Cauchy subseq. so that $\overline{\mathrm{span}(x_{A,n})}\cap \overline{\mathrm{span}({x_{B,n}\mid B\subset \Bbb N, B\neq A})}={0}$ ?) – s.harp Jan 31 '17 at 09:55
  • I don't think it's needed, actually. Wojtaszczyk gives this hint: Every separable Banach space is a quotient of $\ell_1$ and $\ell_1(2^{\Bbb N})$ is a subspace of $C[0,1]^*$. – David Mitra Jan 31 '17 at 10:07
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    Kalton and Albiac also has this problem. In fact they offer: for an infinite set $\Gamma$, $\ell_1(\cal P \Gamma)$ is isometric to a subspace of $\ell_\infty(\Gamma)$, and the hint: for each $\gamma\in\Gamma$ define $\phi_\gamma\in\ell_\infty(\cal P\Gamma)$ by $\phi_\gamma=1$ when $\gamma\in A$ and $-1$ when $\gamma\not\in A$. ($\cal P \Gamma$ is the power set of $\Gamma$.) – David Mitra Jan 31 '17 at 10:12
  • I cannot find this exercise in the book by Kalton and Albiac, in what part is it? (I'm afraid I can't see the construction from your hint) – s.harp Jan 31 '17 at 11:35

2 Answers2

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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 $2^{2^{\aleph_0}}$.

Community
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Tomasz Kania
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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 $|l_{\infty}^*|\leq |\mathbb R^{l_{\infty}}|=2^{|\mathbb R|}.$

DanielWainfleet
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