10
$\begingroup$

The following question is about if it is compatible to add to $\sf ZF$ an axiom asserting the existence of a countable transitive model of $\sf ZF$ such that for every strictly increasing function $f$ on the ordinals, we have a transitive countable model of $\sf ZF $ satisfying: $$\forall\alpha>0:\beth_\alpha = \aleph_{f(\alpha)}$$

Formally:

$ \exists M: M \equiv \operatorname {CTM}(\mathsf {ZF}) \land \forall f \subseteq M \ \big{(}\\f: \operatorname {Ord}^M \to \operatorname {Ord}^M \land \forall \alpha \forall \beta \, ( \beta > \alpha \to f(\beta) > f(\alpha) ) \\ \implies \\ \exists N : N \equiv \operatorname {CTM}(\mathsf {ZF}) \land \operatorname {Ord}^N =\operatorname {Ord}^M \land (N \models \forall \alpha > 0 : \beth_\alpha=\aleph_{f(\alpha)})\big{)} $

Where "$\equiv\operatorname {CTM}(\mathsf {ZF})$" means "is a countable transitive model of ZF"

So this is to say that the generalized continuum hypothesis can fail everywhere and in everyway.

$\endgroup$
5
  • 1
    $\begingroup$ does ZF prove the cofinality restriction without choice? [I think so, but too many years have passed ...] $\endgroup$
    – NameNo
    Jan 14, 2022 at 11:09
  • 7
    $\begingroup$ The fact that for each ordinal $\alpha$, $\beth_\alpha$ is well defined as a well-ordered cardinal number, is by itself equivalent to AC. Thus, $N$ is really a model of ZFC. $\endgroup$ Jan 14, 2022 at 11:57
  • $\begingroup$ Ah I see thanks. $\endgroup$ Jan 14, 2022 at 13:10
  • 3
    $\begingroup$ My point (see mathoverflow.net/questions/78627/… ) is that also any other formalizaition of "fail everywhere in every way" is killed since restrictions do exist. $\endgroup$
    – NameNo
    Jan 14, 2022 at 14:14
  • $\begingroup$ @NameNo, thanks a lot! $\endgroup$ Jan 14, 2022 at 14:25

2 Answers 2

22
$\begingroup$

No. An early nontrivial constraint on the $\beth$ function comes from Kőnig's Theorem, that for all infinite $\kappa$, $\mathrm{cf}(2^\kappa)>\kappa$. This implies that we cannot have $\beth_\alpha = \aleph_{f(\alpha)}$ for all $\alpha$, when $f(1) = \omega$, nor when $f(\omega+1)$ is a cardinal below $\aleph_\omega$.

Another constraint is Silver's Theorem, that if GCH holds below a singular of uncountable cofinality, then it holds at that singular as well.

Other constraints come from Shelah's PCF theory. Shelah showed that if $\aleph_\omega$ is a strong limit, then $2^{\aleph_\omega} < \aleph_{\omega_4}$.

$\endgroup$
1
  • 4
    $\begingroup$ +1. To the OP, it's worth explicitly saying that, per Easton's theorem, Konig's theorem is the only real restriction on the behavior of the continuum function ... on regular cardinals; as Silver's theorem and PCF theory show, the situation on singular cardinals is much more intricate. $\endgroup$ Jan 14, 2022 at 17:05
12
$\begingroup$

When working in ZF, one can have more freedom. See An Easton-like Theorem for Zermelo-Fraenkel Set Theory with the Axiom of Dependent Choice and An Easton-like theorem for Zermelo-Fraenkel Set Theory without Choice.

$\endgroup$

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.