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The book of Thoms Jech on set theory mentions some equalities involving $\aleph_{\omega}$:

  1. $\aleph_\omega^{\aleph_1} = \aleph_\omega^{\aleph_0} \cdot 2^{\aleph_1}$
  2. If $\alpha < \omega_1$, then $\aleph_\alpha^{\aleph_1} = \aleph_\alpha^{\aleph_0} \cdot 2^{\aleph_1}$.
  3. If $\alpha < \omega_2$, then $\aleph_\alpha^{\aleph_2} = \aleph_\alpha^{\aleph_1} \cdot 2^{\aleph_2}$

How can the above equalities be proved?

Asaf Karagila
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Davius
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    Did you read the chapter preceding these exercises and attempt to use any of the theorems proved there? – spaceisdarkgreen Apr 03 '21 at 17:41
  • Obviously, I read it, but I did not identify some valid clue. – Davius Apr 03 '21 at 20:19
  • If $2^{\aleph_1}\ge \aleph_\omega^{\aleph_0}$ then $2^{\aleph_1}= \aleph_{\omega}^{\aleph_1}$ (why?). If $2^{\aleph_1}<\aleph_\omega^{\aleph_0}$ then $2^{\aleph_1}<\aleph_\omega$ (why?) and so for any $n\in \omega,$ $\aleph_n^{\aleph_1} = 2^{\aleph_1}\cdot \aleph_n$ (why? hint: Hausdorff formula), which is less than $\aleph_\omega,$ so using theorem 5.20, $\aleph_{\omega}^{\aleph_1} = \gimel(\aleph_\omega)=\aleph_\omega^{\aleph_0}.$ For the other two, use induction on $\alpha$ along with thm 5.20 and Hausdorff formula. – spaceisdarkgreen Apr 03 '21 at 20:43

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