# $(\beth_{\omega})^\omega=\beth_{\omega+1}$

I'm trying to show that $(\beth_{\omega})^\omega=2^{\beth_\omega}$. This is an exercise in Kunen where he suggests to encode subsets of $\beth_\omega$ with functions from $\omega\rightarrow\beth_\omega$. Any help would be appreciated.

Thanks,

Cody

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Can you give the exact reference, not just "Kunen"? – Asaf Karagila Sep 24 '12 at 15:18
Oops, let me try that comment again. This is Exercise I.13.33 in the 2013 edition of Kunen's Set Theory. It actually asks you to show a little more: $(\beth_\omega)^\omega = \left|\prod_{n < \omega} \beth_n\right| = \beth_{\omega+1}$. – Nate Eldredge Jul 12 '14 at 16:16

First, note that $(A\cap\beth_n)_{n<\omega}$ is a function $f_A:\omega\to\bigcup_n\mathcal P(\beth_n)$, and that the assignment $A\mapsto f_A$ is 1-1.
Then, note that $\mathcal P(\beth_n)$ is in bijection with $\beth_{n+1}$. Fix bijections for each $n$, and use them to replace $f_A$ into a function that takes ordinal values.
I am slightly confused about SCH and this question, actually. Does that mean that SCH cannot fail at $\beth_\omega$? – Asaf Karagila Sep 24 '12 at 15:26
Asaf: SCH can fail at $\beth_\omega$; in fact, we can have GCH below $\aleph_\omega$, and $\mathcal P(\aleph_\omega)$ big. Remember, SCH is equivalent to asserting that, for all $\kappa$, we have $\kappa^{\mbox{cf}(\kappa)}=\kappa^{+}+2^{\mbox{cf}(\kappa)}$; it does not assert a separation between $\gimel(\kappa)$ and $2^\kappa$. – Andrés Caicedo Sep 24 '12 at 15:38
Yes, I knew that SCH could fail at $\aleph_\omega$ which is why it was a bit strange for me. Thanks for clarifying! – Asaf Karagila Sep 24 '12 at 15:49
By the way, regarding the failure of SCH at $\aleph_\omega$ do we know how far we can blow up $\aleph_\omega^\omega$ while we do that? (Obviously not above $\aleph_{\omega_4}$, but anything else?) – Asaf Karagila Sep 24 '12 at 16:00