New solutions to exercises from Thomas Jech’s book ‘‘Set Theory’’:

The exercises from this chapter was a good opportunity to play a bit more with the forcing method. Exercise 15.15 seemed a straightforward generalization of Easton’s forcing but turned out to be a bit technical. I realized that the forcing notion used in that exercise provides a result in ZFC (a bit like Exercises 15.31 and 15.32 allow to prove some theorems on Boolean Algebras by Forcing).

Remember that $\beth_{0}=\aleph_{0},\beth_{1}=2^{{\beth_{0}}},\beth_{2}=2^{{\beth_{1}}}...,% \beth_{\omega}=\sup\beth_{n},...,\beth_{{\alpha+1}}=2^{{\beth_{{\alpha}}}},...$ is the normal sequence built by application of the continuum function at successor step. One may wonder: is $\beth_{\alpha}$ regular?

First consider the case where $\alpha$ is limit. The case $\alpha=0$ is clear ($\beth_{0}=\aleph_{0}$ is regular) so assume $\alpha>0$. If $\alpha$ is an inacessible cardinal, it is easy to prove by induction that for all $\beta<\alpha$ we have $\beth_{\beta}<\alpha$: at step $\beta=0$ we use that $\alpha$ is uncountable, at successor step that it is strong limit and at limit step that it is regular. Hence $\beth_{\alpha}=\alpha$ and so is regular. If $\alpha$ is not a cardinal then $\operatorname{cf}(\beth_{\alpha})=\operatorname{cf}(\alpha)\leq|\alpha|<\alpha% \leq\beth_{\alpha}$ so $\beth_{\alpha}$ is singular. If $\alpha$ is a cardinal but not strong limit then there is $\beta<\alpha$ such that $2^{\beta}\geq\alpha$. Since $\beta<\alpha\leq\beth_{\alpha}$ there is $\gamma<\alpha$ such that $\beth_{\gamma}>\beta$. Then $\beth_{\alpha}\geq\beth_{{\gamma+1}}=2^{{\beth_{\gamma}}}>2^{\beta}\geq\alpha$. So $\operatorname{cf}(\beth_{\alpha})=\operatorname{cf}(\alpha)\leq\alpha<\beth_{\alpha}$ and $\beth_{\alpha}$ is singular. Finally, if $\alpha$ is a singular cardinal, then again $\operatorname{cf}(\beth_{\alpha})=\operatorname{cf}(\alpha)<\alpha\leq\beth_{\alpha}$ and $\beth_{\alpha}$ is singular.

What about the successor case i.e. $\beth_{{\alpha+1}}$? By Corollary 5.3 from Thomas Jech’s book any $\alpha$, we can show that $\aleph_{{\alpha+1}}$ is a regular cardinal. The Generalized Continuum Hypothesis says that $\forall\alpha,\aleph_{\alpha}=\beth_{\alpha}$. Since it holds in $L$ we can not prove in ZFC that for some $\alpha$, $\beth_{{\alpha+1}}$ is singular.

The generic extension $V[G]\supseteq V$ constructed in exercise 15.15 satisfies GCH and so it’s another way to show that $\beth_{{\alpha+1}}$ can not be proved to be singular for some $\alpha$. However, it provides a better result: by construction, $V[G]\models\beth_{{\alpha+1}}^{V}={(\beth_{{\alpha}}^{V})}^{+}$ and so $V[G]\models[\beth_{{\alpha+1}}^{V}\text{ is a regular cardinal}]$. Since ‘‘regular cardinal’’ is a $\Pi_{1}$ notion we deduce that $\beth_{{\alpha+1}}$ is a regular cardinal in $V$.

Now the question is: is there any ‘‘elementary’’ proof of the fact that $\beth_{{\alpha+1}}$ is regular i.e. without using the forcing method?

--update: of course, I forgot to mention that by König’s theorem, $2^{{\beth_{{\alpha}}}}=\beth_{{\alpha+1}}\geq\operatorname{cf}(\beth_{{\alpha+% 1}})=\operatorname{cf}(2^{{\beth_{{\alpha}}}})\geq{(\beth_{{\alpha}})}^{+}$ so the singularity of $\beth_{{\alpha+1}}$ would imply the failure of the continuum hypothesis for the cardinal $\beth_{\alpha}$ and this is not provable in ZFC.