What is the consistency strength of $$ZFC + G \neg CH$$

where $G \neg CH$ is Generalized anti-continuum hypothesis! Formally this is: $$\forall x [|x|> 1 \to \exists y \ ( |x|<|y|<|P(x)|)]$$

Also what is the least large cardinal property $\kappa$ such that $ZFC+G\neg CH$ cannot interpret $ZFC + \kappa $?

  • $\begingroup$ Ain't Easton's theorem gives as a model where $2^{\aleph_α}=\aleph_{α+2}$ for all $α$ without assuming large cardinal? $\endgroup$
    – ℋolo
    Apr 13, 2019 at 19:46
  • 3
    $\begingroup$ @Holo: That's only for regular cardinals. Singular cardinals are much harder to control. $\endgroup$ Apr 13, 2019 at 20:21
  • $\begingroup$ @EricWofsey I see, thanks for clarifying $\endgroup$
    – ℋolo
    Apr 13, 2019 at 20:27
  • $\begingroup$ Well, you violate SCH in every strong limit cardinal, so there are plenty of measurable cardinals. If my memory serves me right, some strongness is required. I don't know if there is an optimal large cardinal hypothesis for this. $\endgroup$
    – Asaf Karagila
    Apr 13, 2019 at 22:31
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    $\begingroup$ Your "in other words" isn't really accurate - the two questions aren't a priori the same (conceivably a principle could be equiconsistent with ZFC+a measurable without interpreting ZFC+a measurable, although I don't think this really happens with natural theories). $\endgroup$ Apr 13, 2019 at 22:40

1 Answer 1


The precise consistency strength of the global failure of the generalized continuum hypothesis is somewhat technical to state. As far as I know, it has not been published, but I think we have a decent understanding of what the correct statement should be. The most relevant paper towards this result is

MR2224051 (2007d:03082). Gitik, Moti Merimovich, Carmi. Power function on stationary classes. Ann. Pure Appl. Logic 140 (2006), no. 1-3, 75–103

Working in NBG (so that we can quantify over classes), the theorem should be that the statement "GCH fails everywhere" is equiconsistent with "for all $\alpha$ there are stationarily many $\kappa$ such that $o(\kappa)=\kappa^{++}+\alpha$". It is my understanding that Gitik and Merimovich have established this result or very close variants.

That a class $S$ is stationary means that if $C\subseteq\mathrm{ORD}$ is a club of ordinals, i.e., it is unbounded and contains all its limit points (that is, whenever $\alpha>0$ and $C\cap\alpha$ is unbounded in $\alpha$, then $\alpha\in C$), then $C\cap S\ne\emptyset$. Thus, the statement above requires quantification over classes. Working in ZFC, we would restrict attention to definable classes, but the statement above still needs to be stated as a schema, namely, for each formula $\phi(x,\vec y)$ in the language of set theory, we need a statement saying that, for any $\alpha$ and any $\vec b$, if $\{\beta:\phi(\beta,\vec b)\}$ is a club of ordinals, then there is a $\kappa$ such that $\phi(\kappa,\vec b)$ and $o(\kappa)=\kappa^{++}+\alpha$.

[ If one wants to avoid these technicalities, perhaps it is easier to work below an inaccessible $\Omega$ such that for all $\alpha<\Omega$, $S_{\Omega,\alpha}:=\{\kappa<\Omega:o(\kappa)=\kappa^{++}+\alpha\}$ is stationary in $\Omega$. The forcing construction of Gitik and Merimovich should give a model where $\Omega$ is still inaccessible and GCH fails everywhere below $\Omega$. Conversely, given an inaccessible $\Omega$ such that GCH fails everywhere below $\Omega$, one shows that there is an inner model where all the $S_{\Omega,\alpha}$ (from the point of view of the inner model) are stationary in $\Omega$. Thus, one gets an equiconsistency, but starting from the background theory ZFC+"there is an inaccessible" rather than just ZFC. This is akin to what the situation was in inner model theory for a while, where to develop the core model below a Woodin cardinal required the extra assumption that there was a measurable cardinal, and only recently was this assumption removed. ]

The argument makes essential use of techniques from Radin forcing to obtain the global failure of GCH from the given assumption. For the converse, starting with the global failure of GCH one argues that either, say, $0^{¶}$ exists, or else in the core model $K$ we have the required sets being stationary.

The assumption that $0^{¶}$ exists is a technical inner model-theoretic requirement that gives us an inner model with cardinals stronger than those mentioned in the second part of the dichotomy. The need for bringing it into the picture is that the existence of the core model $K$ requires some anti-large cardinal assumption (such as "$0^{¶}$ does not exist"). Thus, one argues that either we have an inner-model theoretic object that gives us inner models with way more large cardinals than we were even aiming for, or else, the object does not exist, which allows us to construct $K$ and verify it has nice properties that, when combined with the combinatorics of the global failure of GCH in the universe $V$, ensure that in $K$ we have the large cardinals we were looking for.

The core model is a generalization of Gödel's constructible universe $L$ that allows the existence of certain large cardinals ($L$ itself is not good enough in this respect, for instance, there are no measurable cardinals in $L$). It has the advantage over other models with large cardinals in that in possesses a very detailed well-understood structure, in particular a global well-ordering and means to deal with definability conditions. This fine structure makes it the ideal ambient structure to prove the existence of large cardinals from combinatorial hypotheses in the universe.

Here, $o(\cdot)$ is the Mitchell ordering on extenders. There are several ways of formalizing the concept of an extender. What matters here is that extenders are sets that code elementary embeddings of the whole universe of sets. The most basic example of an extender is a measure (a $\kappa$-complete ultrafilter on an uncountable cardinal $\kappa$). Any extender on $\kappa$ codes (via a generalization of the ultrapower construction) an embedding $j$ of the universe $V$ into a well-founded model $M$ with critical point $\mathrm{crit}(j)$ equal to $\kappa$.

Given extenders $E,F$ on $\kappa$, we say that $E\vartriangleleft F$ if and only if $E$ belongs to the (transitive collapse of the) ultrapower by $F$. If $\kappa$ is not a strong cardinal, $\vartriangleleft$ is a well-founded relation, and $o(\kappa)$ is its height.

That there is a strong cardinal (an assumption weaker in consistency strength than the existence of $0^{¶}$) can be expressed in terms of Mitchell's ordering by saying that there is a $\kappa$ with $o(\kappa)=+\infty$ (or, more precisely, $o(\kappa)=\mathrm{ORD}$).

That there is a measurable cardinal is to say that there is a $\kappa$ with $o(\kappa)>0$. The assumption $o(\kappa)\ge2$ is already much stronger than simply saying that there are 2 measurable cardinals. It is much stronger than saying that there is a measurable cardinal that is the limit of measurable cardinals.

The global failure of GCH implies in particular the failure of the singular cardinals hypothesis, SCH: Provably in ZFC there are strong limit singular cardinals $\kappa$, and SCH fails if for any of them we have $2^\kappa>\kappa^{+}$. The failure of SCH has been studied carefully and is well-understood. Gitik established about 30 years ago that it is equiconsistent with the existence of a $\kappa$ such that $o(\kappa)=\kappa^{++}$. This is a statement that can be expressed solely in terms of ultrafilters on $\kappa$. Once we are past $o(\kappa)=\kappa^{++}$ (as in the statement corresponding to the global failure of SCH), we definitely need the theory of extenders.

[I'll be happy to correct the statement of the equiconsistency or any other inaccuracies if I ended up misstating something.]

  • $\begingroup$ Can I infer from that, that ZFC + generalized failure of the continuum hypothesis can prove the consistency of ZFC + a measurable cardinal exists? $\endgroup$
    – Zuhair
    Apr 14, 2019 at 12:42
  • $\begingroup$ Yes, that there is a measurable cardinal can be expressed in terms of the Mitchell order by saying that there is a $\kappa$ such that $o(\kappa)>0$. The statement I mentioned gives in particular that for any $\alpha$ there are $\kappa$ with $o(\kappa)>\alpha$. All these $\kappa$ are measurable. So, the statement gives you the consistency of "there is a proper class of measurable cardinals", and significantly more. $\endgroup$ Apr 14, 2019 at 13:01
  • $\begingroup$ Perhaps it is worth pointing out that this is something we knew before we had any inkling on the right consistency strength of the global failure of GCH. The point is that the global failure of GCH in particular implies the failure of the singular cardinals hypothesis SCH: provably in ZFC, there are singular strong limit cardinals. SCH fails if $2^\kappa>\kappa^+$ for any of these cardinals $\kappa$. The failure of SCH has been understood for almost 30 years now: Gitik showed that it is equiconsistent with the existence of a $\kappa$ with $o(\kappa)=\kappa^{++}$. $\endgroup$ Apr 14, 2019 at 13:07
  • $\begingroup$ Now, already saying $o(\kappa)>1$ gives us that $\kappa$ is a measurable cardinal which itself is a limit of measurable cardinals (and much more), so it gives us (much more than) a set-sized model of ZFC+"there is a proper class of measurables". $\endgroup$ Apr 14, 2019 at 13:09
  • $\begingroup$ @Thanks Noah, it should be fixed now. $\endgroup$ May 14, 2019 at 0:04

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