# Is it known whether or not $\aleph_\alpha=\beth_\alpha$ can be proven by ZFC?

Given a cardinal number $\aleph_\alpha$, is it known whether or not $\aleph_\alpha=\beth_\alpha$ is independent of ZFC?

One could define $\mathrm{CH}(\aleph_\alpha)$ as $\aleph_\alpha=\beth_\alpha$. For which cardinals is it true that $\mathrm{ZFC}\models\mathrm{CH(\kappa)}$?

Clearly, $\mathrm{ZFC}\models\mathrm{CH}(\aleph_0)$, and $\mathrm{CH}(\aleph_1)$ is equivalent to $\mathrm{CH}$ (and is thus independent of ZFC). Because $\mathrm{GCH}$ is indepenedent of ZFC, $\neg\mathrm{CH}(\kappa)$ cannot be proven for any cardinal $\kappa$.

If $\mathrm{ZFC}\models\mathrm{CH}(\aleph_{\alpha+1})$, then $\aleph_{\alpha+1}=\beth_{\alpha+1}$. Thus, if we assume $\aleph_\alpha<\beth_\alpha$, we get a contradiction, because $\aleph_\alpha<\beth_\alpha$ and then $\aleph_{\alpha+1}<\beth_{\alpha+1}$. So, if $\mathrm{CH}(\kappa^+)$ is provable, then $\mathrm{CH}(\kappa)$ is also provable. Thus, for cardinals $\kappa<\aleph_\omega$, $\mathrm{CH}(\kappa)$ is independent of ZFC.

It is known that $\alpha\leq\aleph_\alpha$, and thus if $\beth_\kappa=\kappa$ then $\aleph_\kappa=\kappa$ (i.e. all $\beth$-fixed points are also $\aleph$-fixed points). Therefore, $\beth_\kappa=\aleph_\kappa=\kappa$, and $\beth_\kappa=\aleph_\kappa$. So, for all $\beth$-fixed points $\kappa$, $\mathrm{CH}(\kappa)$ is provable from ZFC.

Are there any other known $\kappa$ with $\mathrm{CH}(\kappa)$ independent of ZFC? Is $\exists\kappa\neq\aleph_0(\mathrm{CH}(\kappa))$ independent of ZFC? (If so, $\beth$ fixed points are independent of ZFC as well.)

Edit: Of course $\beth$-fixed points are not independent of ZFC, they do exist by $\alpha\mapsto\beth_\alpha$ being a normal function. So, ZFC actually proves the existence of cardinals larger than $\aleph_0$ which fulfill GCH.

• There are arbirarily high common fixed points to $\beth, \aleph$, so in fact you can prove $\exists \alpha\neq \omega, \aleph_\alpha=\beth_\alpha$ – Maxime Ramzi Oct 4 '17 at 6:15
• Yes, I realized that. I already edited that in. – Zetapology Oct 4 '17 at 6:32
• I must point out that your choice of notation is a bit unfortunate. The standard terminology uses $\operatorname{CH}(\kappa)$ for the statement that there are no sets of intermediate size between $\kappa$ and $2^\kappa$ (or equivalently, under choice, that $2^\kappa=\kappa^+$). – Andrés E. Caicedo Oct 4 '17 at 13:36

## 1 Answer

Claim. For any $\alpha >0 \colon$ $\mathrm{ZFC} \not \vdash \mathrm{CH}(\aleph_{\alpha})$

Proof. Starting with $L$, we may enlarge the continuum $\beth_1$ arbitrarily without changing cardinals. In particular, we may fix a generic $g$ such that $L$ and $L[g]$ have the same cardinals (thus $\aleph_{\alpha}^L = \aleph_{\alpha}^{L[g]}$) but $\beth_1^{L[g]} > \aleph_{\alpha}^{L[g]}$.

• So while $\mathrm{ZFC}$ proves that $\mathrm{CH}(\aleph_\alpha)$ occurs on a club class, it can't guarantee this for any particular $\alpha$. – Stefan Mesken Oct 4 '17 at 9:16
• The proof is quite clear, but the claim could easily be misunderstood. A background theory for claims of provability is often a rather weak theory, such as PA (Peano arithmetic, or even weaker theories), augmented by a necessary assumption (such as "Con(ZFC)"). But in PA and even its second order relatives you cannot talk about arbitrary ordinals $\alpha$. What is your background theory? -- I would not know how to state your claim syntactically at all; my version would say "For all ZFC-models M and all ordinals alpha in M there is..." – Goldstern Oct 4 '17 at 13:42
• Martin @Goldstern: I think that this can be mitigated by stating that ZFC does not prove that there is an $\alpha$ such that $\beth_1<\aleph_\alpha$. – Asaf Karagila Oct 4 '17 at 14:16
• This answer should be modified, because this doesn't hold for $\alpha=\beth_1+1$, for which $\aleph\alpha>\beth_1$. – Zetapology Oct 4 '17 at 14:32
• Zetapology, that's actually not an issue. It's just that $\beth_{1}$ represents different ordinals in different models. If we fix the ordinal $\alpha$ - not the definition $\alpha$ may satisfy - as I did, everything works. – Stefan Mesken Oct 4 '17 at 14:35