$\alpha<\beth_\omega\implies 2^\alpha<\beth_\omega$

Question

"Let $\alpha$ be a cardinal such that $\alpha<\beth_\omega$. Then $2^\alpha<\beth_\omega$".

I'm stuck, any suggestions or bibliographies of the $\beth$ function? :(

Answer 1

This is really just an application of the definition of the $\beth$ function:

• $\beth_0=\aleph_0$;
• $\beth_{\alpha+1}=2^{\beth_\alpha}$; and
• if $\alpha$ is a limit ordinal, then $\beth_\alpha=\sup\{\beth_\beta\mid\beta<\alpha\}$.

So, if $\alpha<\beth_\omega$, by definition, there is some $n<\omega$ such that $\alpha\leq\beth_n$. But that means that $2^\alpha\leq\beth_{n+1}<\beth_{n+2}\leq\beth_\omega$.