Sign up ×
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.

I am supposed to prove that ${\aleph_\omega}^{\aleph_1} = {\aleph_\omega}^{\aleph_0} \cdot {2}^{\aleph_1}$ , but I really have no idea how to start or what to do. I thought I could use the following fact: ${2}^{\aleph_1}= {\aleph_1}^{\aleph_1}$, because of the infiniteness of ${\aleph_1}$.

I hope someone will show me how this works. Thanks in advance!

share|cite|improve this question

marked as duplicate by Lord_Farin, Calvin Lin, Amzoti, TMM, Start wearing purple May 29 '13 at 15:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1 Answer 1

We have, since $\aleph_\omega$ is a limit, that \[ \aleph_\omega^{\aleph_1} = (\sup_n \aleph_n^{\aleph_1})^{\operatorname{cf}\aleph_\omega} \] By Hausdorff and induction \[ \aleph_{n+1}^{\aleph_1} = \aleph_n^{\aleph_1} \cdot \aleph_{n+1} = \aleph_1^{\aleph_1}\cdot \aleph_{n+1} = 2^{\aleph_1} \cdot \aleph_{n+1} \] Hence \[ \sup_n \aleph_n^{\aleph_1} = 2^{\aleph_1} \cdot \aleph_{\omega} \] As $\operatorname{cf}\aleph_\omega = \aleph_0$, finally \[ \aleph_\omega^{\aleph_1} = (2^{\aleph_1} \cdot \aleph_\omega)^{\aleph_0} = 2^{\aleph_1} \cdot \aleph_\omega^{\aleph_0}. \]

share|cite|improve this answer
Is $\lambda^\kappa = \left(\sup_{\mu<\lambda}\mu^\kappa\right)^{\mathrm{cf}(\lambda)}$ (for $\lambda$ a limit cardinal, $\kappa$ an arbitrary infinite cardinal) a theorem of ZFC? If not, how do you get the first line? – goblin Jun 30 at 16:15

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