Since $\operatorname{cf}(\aleph_\omega)=\omega$, $\aleph_\omega<\aleph_\omega^\omega$. However, can we force $\aleph_\omega^\omega<2^{\aleph_\omega}$? I am especially interested in models for which $2^{\aleph_n}<\aleph_\omega$ for all $n<\omega$ in addition to our other condition.

Traditional ways of adding subsets of $\aleph_\omega$ seem unreliable, as they appear to collapse $\aleph_\omega$ to $\aleph_0$ or raise $2^{\aleph_n}$ at the same time.

  • $\begingroup$ Is it true in $\aleph_\omega^\omega$ that the subscript is the ordinal $\omega$, but the superscript is the cardinal $\aleph_0$ ? $\endgroup$ Feb 1 at 15:55
  • $\begingroup$ @GeraldEdgar Yes, here by $\aleph_\omega^\omega$ I mean $(\aleph_\omega)^{\aleph_0}$. $\endgroup$ Feb 1 at 17:31

1 Answer 1


By Theorem 5.16 in Jech, for limit cardinals $\kappa$, $2^{\kappa}=(2^{<\kappa})^{cf(\kappa)}$. Hence, if $\aleph_{\omega}$ is a strong limit, $2^{\aleph_{\omega}}=(2^{<\aleph_{\omega}})^{\omega}=\aleph_{\omega}^{\omega}$. However, if we do not require $\aleph_{\omega}$ to be a strong limit, we can simply start from a model of GCH and add many subsets to e.g. $\omega_1$ with countable conditions. This arbitrarily increases $2^{\aleph_{\omega}}$ while not changing $\aleph_{\omega}^{\omega}$ (it is unchanged as a set by the countable distributivity and by CH $Add(\omega_1)$ does not collapse any cardinals).


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