# Can we force $\aleph_\omega^\omega<2^{\aleph_\omega}$?

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.

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

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).