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I'm currently trying to solve Exercise 5.27 of Jech's Set Theory (3rd Millennium ed.), viz:

If $2^{\aleph_1}=\aleph_2$, then $\aleph_{\omega}^{\aleph_0} \ne \aleph_{\omega_1}$.

The presumption is saying that there is no cardinal between $\aleph_1$ and $2^{\aleph_1}$ but I fail to see how this can be employed in this "negative computation" of $\kappa^{\operatorname{cf} \kappa}$ with $\kappa = \aleph_\omega$.

Any hints are much appreciated; I suspect that the right pointer will enable me to come up with a solution.

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Hint: Use the fact that for $\alpha<\omega_1$ $\aleph_{\alpha}^{\aleph_1}= \aleph_{\alpha}^{\aleph_0}\cdot 2^{\aleph_1}$; exercise $5.19$ of the book, and the assumption $2^{\aleph_1}=\aleph_2$to conclude $\aleph_{\omega}^{\aleph_0}=\aleph_{\omega}^{\aleph_1}$, then use König's Theorem.

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  • $\begingroup$ Thanks. It took me a few moments to figure out the last bit but I've got it now. I guess I was put off by the premise being very strong; we only actually used $2^{\aleph_1} < \aleph_\omega^{\aleph_0}$. $\endgroup$ – Lord_Farin Mar 28 '13 at 15:32
  • $\begingroup$ @Lord_Farin you're welcome $\endgroup$ – Camilo Arosemena-Serrato Mar 28 '13 at 15:34

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