This is a question I wonder a little about every now and then.


It is immediate, using forcing, that if there is a transitive set model of set theory, then there are continuum many.



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> Can one prove a weak version of this without using the forcing machinery?
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> (Perhaps in the presence of reasonable large cardinal assumptions?)
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> 
> 


Here are some specific versions:


1. Suppose we know there are two transitive models of set theory. Can we prove there are infinitely many?
2. Suppose we know there are two transitive models of set theory. Can we prove there are two with the same height?
3. Suppose we know there is an uncountable model. Are there continuum many?


I'm going to leave "reasonable" loose, but we do not want to assume much. For example, if there is a transitive model of "there is a measurable cardinal" then one can (easily) check that there are continuum many countable transitive models of set theory. There are also a few more or less obvious observations in the same spirit that follow from $\Sigma^1\_2$ absoluteness.


Also: if there is a countable transitive model $M$ of set theory, there is a comeager set of reals $C$ and a measure 1 set $R$ such that if $x$ is in $C\cup R$, then $M[x]$ is a model of set theory. Any weakening of this or something similar in spirit that can be established without forcing would also be welcome.


Now: I do not think I want something where we do forcing in disguise. So I am not sure presenting forcing as some variant of Bairwise compactness or that sort of thing would be appropriate here.


Of course, any references you think I should be aware of are more than welcome.



Number of transitive models of set theory
Suppose you could prove, in ZFC, without forcing, the statement 


(A) If there are two transitive models of ZFC, then there is a third.


Then you could also prove, in ZFC, without forcing, the statement 


(B) If there are two transitive models of ZFC, then there is a transitive model of ZFC + $V\neq L$.


[Proof: Work in ZFC and assume there are two transitive models of ZFC. By (A) there is a third. If one of them satisfies $V\neq L$ we're done, so assume all three satisfy $V=L$ and are therefore of the form $L\_\xi$ for some ordinals $\xi$. Let $\alpha<\beta<\gamma$ be the first three ordinals occurring as heights of transitive models of ZFC (+ $V=L$). Since $L\_\gamma$ sees both $L\_\alpha$ and $L\_\beta$, it must, by (A) again, see another transitive model of ZFC. By minimality of $\alpha<\beta<\gamma$, $L\_\gamma$ can't see another model of the form $L\_\xi$, so it must see a model of ZFC + $V\neq L$. (I've tacitly used that the notion of "transitive model of ZFC" is absolute between the real world and transitive models of ZFC.)] 


What does this have to do with the question? I claim that a non-forcing proof of (B) would be significant news --- it would say that people could have deduced the consistency of $V\neq L$ from highly plausible assumptions before Cohen. So I feel reasonably confident in saying that no ZFC proof of (B) without forcing is known. Therefore, by the argument above, no ZFC proof of (A) without forcing is known. 


[Concerning "highly plausible," note that the existence of lots of transitive models of ZFC follows from assumptions just slightly beyond ZFC itself. My favorite such extension of ZFC is to add a satisfaction predicate for formulas in the language of ZFC, add axioms saying this predicate obeys the usual recursive definition of satisfaction, and allow this new predicate in the replacement scheme of ZFC. With these assumptions, you can apply a L"owenheim-Skolem argument to get lots of elementary submodels of the universe.]


Note that the same discussion goes through if, in (A) we replace the conclusion (that there exists a third model) with the statement that there are two transitive models of ZFC of the same height. 


