This question is somewhat related to a [previous one](https://mathoverflow.net/questions/100981/ultrainfinitism-or-a-step-beyond-the-transfinite), 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



Taller models of ZFC

> 
> 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.


