If $2^{\aleph_{0}} \ge \aleph_{\omega_1}$, show that $\beth_{\aleph_{\omega}} = 2^{\aleph_{0}}$ , and that $\beth_{\aleph_{\omega_1}} = 2^{\aleph_{1}}$

I don´t know how to start, can you give me a little hint for start?, please.

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what's a potence? – mbsq Feb 26 '14 at 22:33
I'm not sure what is the "beth of aleph omega [one]". – Asaf Karagila Feb 26 '14 at 22:35
the functional beth aplied to the cardinal aleph sub omega one. – Julio Feb 26 '14 at 22:36
with omega one the first cardinal non-contable – Julio Feb 26 '14 at 22:36
a potence (of two) is the cardinality of the potence set that has cardinality the exponent – Julio Feb 26 '14 at 22:37

This is impossible. Recall the definition of the $\beth$ function:

1. $\beth_0=\aleph_0$.
2. $\beth_{\alpha+1}=2^{\beth_\alpha}$.
3. For a limit $\delta$, $\beth_\delta=\sup\{\beth_\alpha\mid\alpha<\delta\}$.

It follows that regardless to its size, $2^{\aleph_0}=\beth_1$ and it is much, much, so very much, smaller than $\beth_{\aleph_\omega}$.

It seems that you might have meant $\gimel$ (Gimel) instead. The $\gimel$ function is defined by $\gimel(\kappa)=\kappa^{\operatorname{cf}(\kappa)}$. In which case one can note that:

$$\aleph_\omega<2^{\aleph_0}\leq\aleph_\omega^{\aleph_0}\leq(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}.$$

For the case with $\aleph_{\omega_1}$ it works the same way.

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