7
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

12
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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