Constructing a Banach space of cardinality $\beth_{\omega+1}$

Question

This is related to yesterday's question Constructing a vector space of dimension $\beth_\omega$; it's the next exercise (I.13.35 (a)) in Kunen's Set Theory.

Let $B_0 = \ell^1$ and let $B_{n+1} = B_n^{**}$ be the continuous second dual of $B_n$, so that we can consider $B_n$ as a subset of $B_{n+1}$ in the usual way. Let $B$ be the completion of $\bigcup_n B_n$. Show that $|B| = \beth_{\omega+1}$.

I have managed to solve part (b) of the exercise, which says that any Banach space $X$ with $|X| \ge \beth_\omega$ actually has $|X| \ge \beth_{\omega+1}$. So it will be enough to show $|B| \ge \beth_\omega$ (though given the ordering of the parts of the exercise, this may not be what Kunen had in mind).

Presumably we should try to show that $|B_{n+1}| \ge 2^{|B_n|}$ or something similar. But when working with continuous duals, I don't see how to do that. Additionally, each step of the induction is somehow going to have to use the fact that we started with $B_0 = \ell^1$, since if $B_0$ had been a reflexive space, this would never work. We also have to rule out the possibility that one of the later $B_n$ turns out to be reflexive.

Any suggestions are welcome.

Well, the completion of a metric space is usually of size $|X|^\omega$ (because we just add limits of countable sequences), so in the case of this question it's not very hard to show directly that $B$ has the wanted cardinality. Perhaps a sleeker argument (which you may or may not have used) is that it is a Baire space, so it cannot be the union of countably many sets with empty interior, so its cardinality cannot have a countable cofinality. — Asaf Karagila, Jul 12 '14 at 16:26
@AsafKaragila: I did discover the Baire category argument for part (b), but it only got me as far as $|X| > \beth_\omega$. To get $|X| \ge \beth_{\omega+1}$ I had to work harder. — Nate Eldredge, Jul 12 '14 at 16:28
D'oh, you're right. I think that in general you should be able to mimic the proof that an infinite dimensional Bananach space has dimension of at least $2^{\aleph_0}$ here. Although I don't quite remember it. The Baire category argument is used only to show that the dimension is not countable, and similarly here, it just shows it's not $\beth_\omega$. The gap to $\beth_{\omega+1}$ may be higher, of course. Thank you! — Asaf Karagila, Jul 12 '14 at 16:31
May be this can help http://math.stackexchange.com/questions/141535/cardinality-of-a-hamel-basis — Norbert, Jul 12 '14 at 18:21

score 5 · Answer 1 · edited Apr 13 '17 at 12:20

Let us try to estimate $|B|$ from below. Every space $B_{n-1}^*$ ($n\geqslant 1)$ is of the form $C(K_n)$ for some compact Hausdorff space. For example, $K_1 = \beta\mathbb{N}$ and $|K_1| = \beth_1$. In particular, by the Riesz–Markov–Kakutani representation theorem each space $B_n$ is isometric to the space $M(K_n)$ of Radon measures on $K_n$. Moreover, we can embed injectively $K_{n}$ ($n\geqslant 1$) into $B_n$ via $$x\mapsto \delta_x\;(x\in K_n).$$ The subspace $D_n:=\overline{\mbox{span}}\{\delta_x \colon x\in K_{n}\}\subset B_n$ is isometric to $\ell_1(K_{n})$.

We shall prove inductively that $|B_{n}|\geqslant \beth_{n}$ for each $n\geqslant 2$.

Suppose that $|D_n|\geqslant \beth_{n}$. It is clear that $|B_n|\geqslant |D_n|=|\ell_1(K_n)|=|K_n|$. In particular, $|K_n|\geqslant \beth_n$.

We claim that $|D_{n+1}|\geqslant \beth_{n+1}$. Indeed, $$|B_n^{**}|\geqslant |D_n^{**}|=|C(\beta |K_{n}|)^{*}|\geqslant |\beta |K_{n}||\geqslant 2^{\beth_{n}}=\beth_{n+1},$$ where the last inequality follows from Pospíšil's theorem.

We have thus proved that $|B|\geqslant \beth_\omega$. However, if $\kappa$ is the cardinality of a Banach space, then $\kappa^{\aleph_0}=\kappa$. Consequently, $|B|\geqslant (\beth_\omega)^{\aleph_0}=\beth_{\omega+1}$ (see this thread for the proof).

The opposite inequality is quite easy.

Thanks for your answer, I am working to absorb it. My first question is, how do we see that all the $B_n^*$ must be of the form $C(K_n)$? — Nate Eldredge, Jul 15 '14 at 06:14
Oh, that's easy. Fix $(\mu_i)$ a maximal family of pairwise singular probability measures in $B_0^{}$. Then you can write $B_0^{}$ as the $\ell_1$-sum of $L_1(\mu_i)$. In particular, $B_0^{***}$ is the $\ell_\infty$-sum of $L_\infty(\mu_i)$, hence of the form $C(K)$ by Gelfand-Naimark in the complex case or Kalton-Albiac in the real case. Now proceed inductively.
General philosophy: the second dual of a C-algebra is a C-algebra. Commutativity of the second dual follows from a w*-density argument. — Tomasz Kania, Jul 15 '14 at 08:13
http://www1.maths.leeds.ac.uk/~pmt6hgd/preprints/Dissertatmeasurealgebrafinal.pdf
is a great source of knowledge about those things. — Tomasz Kania, Jul 15 '14 at 09:18

Constructing a Banach space of cardinality $\beth_{\omega+1}$

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