Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 3 down vote accepted

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:


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.