Is every limit cardinal $\alpha$ of the form $\alpha=card(X)+card(P(X))+cardP(P(X))+...$, for some set $X$?

Question

A limit cardinal is a cardinal number $\alpha$ such that if $\beta<\alpha,$ then there is a cardinal number $\gamma$ with $\beta<\gamma<\alpha.$ Now if $\alpha$ is a limit cardinal then can we find a set $X$ which satisfies $$\alpha= card X + card P(X) + card P(P(X))+...?$$

Answer 1

No. at least not without assuming GCH. Take $\aleph_ω$ and look at the model of ZFC where $2^{\aleph_0}=\aleph_{ω+1}$.

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Even assuming $GCH$, take $\aleph_{ω_1}$, and assume that your proposition is true, you get that $\mbox{cof}(\aleph_{ω_1})=\mbox{cof}(ω_1)=\aleph_0$, which is false.

Answer 2

No. An uncountable cardinal with that property is a strong limit cardinal of countable cofinality. An uncountable cardinal $a$ such that $\forall b<a\, (2^b<a)$ is called a strong limit cardinal.

GCH is consistent with ZFC (Kurt Godel, 1930's), and GCH implies that all limit cardinals are strong limits.

It is consistent with ZFC that $2^{\omega}>\omega_{\omega}$ (Paul Cohen, 1960's) so it is consistent with ZFC that the limit cardinal $\omega_{\omega}$ is not a strong limit.

In ZFC the recursively-defined beth function maps the class $On$ of ordinals into the class of cardinal ordinals: $\beth(0)=\omega,$ and $\beth(b+1)=2^{\beth(b)}$ , and if $0\ne b=\cup b$ then $\beth(b)=\cup_{c\in b}\beth(c).$ Consider $\beth (\omega_1)=\cup_{c\in \omega_1}\beth(c),$ which is a strong limit cardinal of cofinality $\omega_1.$