# Taller models of ZFC

This question is somewhat related to a previous one, where I asked for new forms of infinite beyond the cardinal hierarchy.

Using forcing techniques, at least the ones I know of, one starts from a ground model and then enlarge it, by adding a generic set and closing up with respect to relative constructibility.

The new model shares the same ordinals and the same height of the original one.

On the other hand, if one postulates some suitably large cardinal axiom, one gets as a side effect new transitive models of ZFC, shorter than the one corresponding to the large cardinal's height.

Looks like one can fatten a model, keeping the same height, or make it shorter, but what about creating taller ones?

Question(s) :

1) Are there standard techniques to make a transitive model taller without adding new axioms to ZFC?

2) Does an axiom like this make sense (ie is neither a theorem of ZFC, nor blatantly inconsistent):

For any transitive model of ZFC $M$ of height $\eta_0=\text{ht}(M)$, there is always another transitive model $N$ such that $M=V_{\eta_0}^N$ of height $\eta_1=\text{ht}(N)$, where $\eta_0 < \eta_1$

NOTE: in the "axiom" above I have simply asked for the second ordinal to be higher than the height of the starting model, but I am also curious as to which extent one could strengthen that, by asking for the second height to be much higher (add your own brand of -much- ) than the first one

• In 1: “..without adding new axioms” – adding where? Can you clarify what does the sentence mean? – Emil Jeřábek Jan 28 '16 at 15:06
• In (2), are you referring to transitive set models (so that the question is interpreted within a single class model of ZFC, so to speak) or are you asking about class models (so that the question is about the relationship between different class models of ZFC)? In the former case, doesn't (2) follow from the axiom that every set is contained in a transitive set model of ZFC, which in turn follows just from the existence of arbitrarily large inaccessible cardinals? – Carl Mummert Jan 28 '16 at 16:26
• Concerning your question 1 which is stated quite vaguely, maybe iterated ultrapowers would meet your requirements. If you assume a say countable model $M$ with an $M$-ultrafilter $U$ you can form the ultrapower $Ult(M,U)$ and continue. At limit stages you take the direct limit. Performing this $\omega_1$-many times and assuming that the final model is wellfounded you end up with an $\omega_1$ sized model. – Stefan Hoffelner Jan 28 '16 at 17:07
• @Carl: Just one inaccessible cardinal is enough. Since an inaccessible cardinal is the limit of worldly cardinals (those that $V_\alpha$ is a model of ZFC), so if $\kappa$ is inaccessible $V_\kappa$ satisfies that "There is a proper class of $\alpha$'s such that $V_\alpha$ is a model of ZFC". You can probably even have something weaker like $\kappa$ is a worldly limit of worldly cardinals. – Asaf Karagila Jan 28 '16 at 18:14
• @Asaf Karagila: I think you are thinking about the consistency of the statement (you are saying it is consistent relative to the existence of one inaccessible), but I was thinking about the provability of the statement. Is it consistent with ZFC that the collection of wordly cardinals is nonempty and bounded? In my earlier comment, I was trying to suggest that the axiom would follow from very mild large cardinal assumptions. – Carl Mummert Jan 28 '16 at 18:18

Regarding Question 1:

It is certainly possible (i.e., consistent with ZFC) that there is only one transitive model of ZFC. For example, if there are transitive models $M$ and $N$ of ZFC such that $o(M)<o(N)$ (where $o(M)$ is the ordinal height of $M$), then there are ordinals $\alpha < \beta$ such that $L(\alpha)$ is the so-called Shepherdson-Cohen minimal model of set thery, which is the smallest transitive model of ZFC (where $L(\alpha)$ is the $\alpha$-th approximation to the constructible universe), and $L(\beta)$ is the next shortest transitive model of ZFC + V=L. In this situation $L(\beta)$ is a model of ZFC in which there is no transitive model of ZFC of height greater than $\alpha$.

So this shows that there is no universal technique within ZFC for making a transitive model of ZFC taller (where "$N$ is taller than $M$" is defined as $o(M)<o(N)$).

Regarding Question 2.

This question has already been answered by Carl Mummert's comment if "taller" is defined via: $N$ is taller than $M$ iff $o(M)<o(N)$ [since in the theory ZFC plus there are arbitrarily large inaccessibles, every transitive model of ZFC is shorter than some other transitive model].

On the other hand, if "taller'' is defined by the current version of the question which stipulates that $N$ is taller than $M$ if there is an ordinal $\alpha \in N$ such that $M = V_{\alpha}^{N}$, then the answer is in the negative since the minimal model of set theory $L(\alpha)$ (mentioned the answer above to Question 1) has no such taller extension. To see this, one needs to use the well-known fact that $L(\alpha)$ is pointwise definable [I will try to provide a reference in my next edit]. If $N$ is a model of ZFC such that $M = V_{\alpha}^{N}$, then $N$ would have to "notice" that its own $V_{\alpha}$ is countable (since $N$ can define define the satisfaction predicate for any set structure, a satisfaction predicate that will agree with the "real world" satisfaction predicate for $L(\alpha)$), which contradicts an easy theorem of ZFC that says that if $\alpha > \omega$ then $V_{\alpha}$ is uncountable.

• In your answer to Question 1, I think you meant to say only that $\alpha$ is the only possible height for a transitive model in $L(\beta)$, not that $L(\alpha)$ is the only possible model. Since $L(\beta)$ thinks $L(\alpha)$ is countable, it contains al sorts of forcing extensions of $L(\alpha)$. – Andreas Blass Jan 29 '16 at 19:50
• @Andreas Blass; Thanks Andreas, I meant to have written ZFC + V=L, I will fix it. – Ali Enayat Jan 29 '16 at 19:53