Let's say that a "complete resolution of GCH" is a definable class function $F: \operatorname{Ord}\longrightarrow \operatorname{ Ord}$ such that $2^{\aleph_\alpha} = \aleph_{F(\alpha)}$ for all ordinals $\alpha$. It is known of course that $F(\alpha) = \alpha+1$ is a complete resolution of GCH (in the positive) that is relatively consistent with ZFC. I read that it's an unpublished theorem of Woodin that $F(\alpha) = \alpha+2$ is a complete resolution of GCH that is relatively consistent with ZFC plus some large cardinal hypothesis. My questions are: (1) What's the weakest known complete resolution of GCH in consistency strength other than $F(\alpha) = \alpha+1$ and what large cardinal axiom is required for it? (2) What are some other complete resolutions of GCH that are known to be consistent relative to specific large cardinal hypotheses, what are their respective large cardinal hypotheses, and how do these consistency strengths relate to one another?

2$\begingroup$ You asked a good question, which has several sub questions, none of which is trivial. It is usually not the best idea to accept an answer after an hour or so. $\endgroup$ – Asaf Karagila Dec 20 '14 at 5:01

2$\begingroup$ Hello Jesse. In light of the comments by Asaf and Andres, maybe it is possible to deselect my answer so as to encourage further answers with richer technical details? There is much to say, both about the failure of SCH and the universal ("everywhere") failure of hypotheses that determine cardinal exponentiation. $\endgroup$ – Avshalom Dec 20 '14 at 5:38
One candidate answer scheme might be the following: if $F$ is any (sufficiently absolute) definable function on the class of regular alephs such that $\kappa < \lambda \Rightarrow F(\kappa) \leq F(\lambda)$ and $\operatorname{cf}(F(\kappa)) > \kappa$, then ZFC + $(\forall \kappa = \operatorname{cf}(\kappa))(2^\kappa = F(\kappa))$ + SCH is consistent, where SCH is the Singular Cardinals Hypothesis or, in an equivalent form, the Gimel Hypothesis, due to Solovay, asserting $(\forall \kappa > \operatorname{cf}(\kappa))( \kappa^{\operatorname{cf}(\kappa)} = \max(2^{\operatorname{cf}(\kappa)}, \kappa^+))$, and no large cardinals are required.
Knowledge of the gimel function $\gimel(\kappa) = \kappa^{\operatorname{cf}(\kappa)}$ suffices to determine cardinal exponentiation recursively (for example, see P. Komjath, V. Totik, (Problems and Theorems in Classical Set Theory): chapter 10, problem 26, sets this out). So it is natural to explore the gimel function in greater depth. Writing a singular $\kappa$ as the limit of an increasing sequence $a$ of smaller regular cardinals leads to the observation that the deeper problem concerns the cofinality $\operatorname{cf}(([\kappa]^{\leq \lambda}, \subseteq))$ of the partial order $([\kappa]^{\leq \lambda}, \subseteq)$ for regular $\lambda < \kappa$. In this direction, one comes eventually to pcf theory, which offers an analysis of the puppet master $\operatorname{pcf}(a)$ rather than his troupe of erratic marionettes $\langle 2^\lambda : \lambda \in Card \rangle$.

$\begingroup$ Ah, so the singular cardinals hypothesis is the one that's hard to contradict! What's the weakest complete resolution of GCH, then, that contradicts the singular cardinals hypothesis? What's the consistency strength of its negation? $\endgroup$ – Jesse Elliott Dec 19 '14 at 22:26

2$\begingroup$ Yes! Hard to contradict Solovay and the SCH. The failure of SCH implies strong settheoretic hypotheses, e.g. for all $x \subseteq \omega, x^\sharp$ exists. $\endgroup$ – Avshalom Dec 19 '14 at 22:33


2$\begingroup$ @JesseElliott There is a lot to say about this question. Too bad you accepted an answer already. $\endgroup$ – Andrés E. Caicedo Dec 20 '14 at 2:26

1$\begingroup$ In Solovay's theorem the function $F$ should be reasonably absolute. Otherwise, there is the silly function defined by setting $F(\kappa)=(2^\kappa)^+$ for which you certainly can't have $2^\kappa=F(\kappa)$. This silly $F$ is definable but it's not absolute and, when you try to match it with $2^\kappa$, it "runs away" so that you never succeed. $\endgroup$ – Andreas Blass Dec 20 '14 at 20:14
$\newcommand\Ord{\text{Ord}}$Easton's theorem allows us to control the continuum function on the infinite regular cardinals, and in particular, on the infinite successor cardinals, in a very flexible manner, without using any large cardinals.
For example, we can have $F(\alpha+1)=\alpha+5$ for all ordinals $\alpha$, with $F(\lambda)=\lambda+1$ for all limit ordinals, and many other possibilities. There are a huge number of possibilities.


$\begingroup$ No problem! It is good to have several answers, even if they overlap. $\endgroup$ – Joel David Hamkins Dec 19 '14 at 22:16


$\begingroup$ Oh, your answers and questions are outstanding! $\endgroup$ – Joel David Hamkins Dec 20 '14 at 2:29
Let me add more examples:
If we consider the global behavior of the power function, then we have for example:
(A) (ForemanWoodin): $F$ can be such that $F(\alpha)>\alpha+\omega,$ all $\alpha$ (modulo a supercompact and infinitely many inaccessibles above it). Note that by a result of Patai, there is no $\beta>\omega$ such that $F(\alpha)=\alpha+\beta,$ all $\alpha$.
Remark. In the above model, $F$ is not definable from the ground model, but we can go to intermediate submodel in which $F$ is definable.
(B) (Cummings): $F$ can be such that $F(\alpha)=\alpha+1,$ all successor $\alpha,$ and $F(\alpha)=\alpha+2,$ all limit $\alpha$ (modulo a $\kappa+3$strong cardinal $\kappa$. By work of GitikMitchell, we need more than a $\kappa+2$strong cardinal $\kappa$).
(C) (Merimovich): Let $2\leq n < \omega.$ Then $F$ can be taken to be $F(\alpha)=\alpha+n,$ all $\alpha$ (modulo a $\kappa+n+1$strong cardinal $\kappa$. By work of GitikMitchell, we need more than a $\kappa+n$strong cardinal $\kappa$).
(D) (FiredmanG): We can have (B) or (C) just by adding a single real to a model satisfying $GCH$. More precisely, the final model can be of the form $V[R],$ where $V\models GCH$ and $R$ is a real.
If we consider the local behavior of the power function, then we can say more:
(E) (GitikMerimovich): Let $2\leq m <\omega,$ and let $\phi: \omega\to \omega$ be such that $\phi$ is increasing and $\phi(n)>n,$ for all $n$. Then we can have $F(n)=\phi(n)$ and $F(\omega)=\omega+m$ (modulo a $\kappa+m$strong cardinal $\kappa$).
(F) (Gitik): We can have $F$ defined on $\omega_1$ such that both sets $\{ \alpha<\omega_1: F(\alpha)=\alpha+2\}$ and $\{ \alpha<\omega_1: F(\alpha)=\alpha+3\}$ are stationary in $\omega_1$ (modulo suitable large cardinals. Some similar results are also proved by GitikMerimovich).
If we avoid choice, then an Easton like theorem is valid for all cardinals:
Let $\theta(\kappa)=sup\{\nu:$ there exists a surjection $f: p(\kappa)\to \nu \}.$ It is easily seen that $\theta(\kappa)>\kappa^+$ is a cardinal and it is increasing. The next theorem shows that these are the only restrictions that $ZF$ imposes on $θ(κ)$:
(G) (FernengelKoepke, based on an earlier result of GitikKoepke) Let $M$ be a ground model of $ZFC + GCH +$Global Choice. In $M$, let $F$ be a function defined on the class of infinite cardinals such that
i. $F(κ)$ is a cardinal > $κ^+$;
ii. $κ < λ$ implies $F (κ)\leq F (λ)$.
Then there is an extension $N$ of $M$ which satisfies $ZF$, preserves cardinals and cofinalities, and such that $θ (κ) = F (κ)$ holds for all cardinals in $N$.

$\begingroup$ As for the final part, are you sure that these are all the restrictions? I think that with those you need to restrict $F$ to regular cardinals, or at least add some additional restriction for limit cardinals. I'm not sure that you can get every possible function like that. $\endgroup$ – Asaf Karagila Dec 22 '14 at 5:55

$\begingroup$ @AsafKaragila Yes, these are all the restrictions, see An Eastonlike Theorem for ZermeloFraenkel Set Theory Without Choice (Preliminary Report) $\endgroup$ – Mohammad Golshani Dec 22 '14 at 6:21

$\begingroup$ Interesting, thanks. I had the impression that there were some problems there, but maybe I was wrong. $\endgroup$ – Asaf Karagila Dec 22 '14 at 8:00

$\begingroup$ Does Martin's Maximum determine a complete resolution of GCH, or is there a natural resolution of GCH that is consistent with it? $\endgroup$ – Jesse Elliott Dec 30 '14 at 22:58

1$\begingroup$ $MM$ implies $SCH$, it also implies $2^{\aleph_0}=2^{\aleph_1}=\aleph_2.$ Also $MM$ is destructible under $\aleph_2$directed closed forcings, so we can force any pattern of the power function on regular cardinals above $\aleph_1$ consistent with Easton's conditions. $\endgroup$ – Mohammad Golshani Jan 1 '15 at 4:24