I was reading about cardinal arithmetic and I came across the beth function $\beth_\alpha$. For some ordinal $\alpha$, $\beth_\alpha$ defined by recursion on $\alpha$ by $\beth_0 = \aleph_0$, $\enspace \beth_{\alpha + 1} = 2^{\beth_\alpha}$ , and $\enspace \beth_\eta = \sup\{\beth_\alpha : \alpha < \eta \} $, where $\eta$ is a limit ordinal.

After reading about various properties of $\beth_\alpha$, I was surprised to see that it isn't used much in most topics. I found an exercise in Kunen's set theory book that seemed interesting.

$(\beth_\omega)^{\aleph_0} = \prod_{n \in \omega}\beth_n = \beth_{\omega + 1}$

$ \prod_{n \in \omega}\beth_n$ is defined to be $|\mathcal{F}|$, where $\mathcal{F} = \{f \in {}^\omega (\beth_\omega) : \forall n f(n) \in \beth_n\}$.

Since the beth function is based on cardinality, I'm guessing I have to somehow compare the cardinality to get a equation throughout. So far, I'm thinking that since $n \in \omega$, $(\beth_\omega)^{\aleph_0} = \prod_{n \in \omega}\beth_\omega \geq \prod_{n \in \omega}\beth_n $ . Also, $(\beth_\omega)^{\aleph_0} \leq 2^{(\beth_\omega)^{\aleph_0}} = 2^{(\beth_\omega)\cdot{\aleph_0}} = 2^{\beth_{\omega}} = \beth_{\omega + 1}$. This leaves me with: $$ \prod_{n \in \omega}\beth_n \leq (\beth_\omega)^{\aleph_0} \leq \beth_{\omega + 1}.$$

I'm not sure where to go from here. I mean, I know that I have to somehow bring the $\leq$ back around to the product, but that is where I'm stuck. Could anyone help me out? Thanks in advance!

EDIT: I forgot to mention that Kunen has a hint for this exercise: Every subset of $\beth_\omega$ can be coded by a function from $\omega$ into $\beth_\omega$.


1 Answer 1


The standard trick is to split $\omega$ into countably many infinite sets $A_i$, $i\in\omega$, and note that for each $i$ we have $\prod_{n\in A_i}\beth_n\ge\sup_{m\in A_i}\beth_m=\beth_\omega$.

Combined with your inequality, this gives that the product $\prod_{n\in\omega}\beth_n$ equals $(\beth_\omega)^{\aleph_0}\le(\beth_{\omega})^{\beth_\omega}=2^{\beth_\omega}=\beth_{\omega+1}$.

To prove the equality, use that $\beth_{\omega+1}=|\mathcal P(\beth_\omega)|$: Take any subset $A$ of $\beth_\omega$ and note that it is entirely determined by $(A\cap\beth_n\mid n\in\omega)$, so this gives an injection of $\beth_{\omega+1}$ in $\prod_n 2^{\beth_n}=\prod_n\beth_{n+1}\le\prod_n\beth_n$.

  • $\begingroup$ Does $(\beth_\omega)^{\beth_\omega} = 2^{\beth_\omega}$ follow since $\beth_\omega = \sup\{\beth_{\alpha} : \alpha < \omega\}$ and every $\alpha < \omega$ is a successor ordinal? Then, $\alpha = \beta + 1$ for some $\beta < \alpha$, and $\beth_\alpha = \beth_{\beta + 1} = 2^{\beth_{\beta}}$, meaning that $(\beth_\omega)^{\beth_\omega} = 2^{\beth_{\beta} \cdot \beth_{\omega}} = 2^{\beth_\omega}$. $\endgroup$
    – Maria
    Sep 28, 2013 at 16:42
  • 2
    $\begingroup$ Oh, no, you don't need anything like that. It is a general fact about infinite cardinals $\kappa$: Using that $\kappa\cdot\kappa=\kappa$, we have that $2^\kappa\le\kappa^\kappa\le(2^\kappa)^\kappa=2^{\kappa\cdot\kappa}=2^\kappa$. $\endgroup$ Sep 28, 2013 at 17:22
  • $\begingroup$ Ahh I see! Thanks for clearing that up for me. $\endgroup$
    – Maria
    Sep 28, 2013 at 17:25
  • $\begingroup$ One more question. Why is $\prod_n\beth_{n+1}\le\prod_n\beth_n$? $\endgroup$
    – Maria
    Sep 28, 2013 at 18:46
  • 1
    $\begingroup$ Because $\prod_n\beth_{n+1}=\prod_{0<n\in\omega}\beth_n=1\cdot\prod_{0<n\in\omega}{\beth_n}\le\beth_0\cdot\prod_{0<n\in\omega}\beth_n=\prod_{n\in\omega}\beth_n$. $\endgroup$ Sep 28, 2013 at 19:13

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