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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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    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}.2014-07-12

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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.

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    The best we know is how to make ${\aleph_\omega}^{\aleph_0}=\aleph_{\alpha+1}$, for any infinite \alpha<\omega_1 we want. It is open whether we can get past $\aleph_{\omega_1}$.2012-09-24