TITLE: Number of transitive models of set theory
QUESTION [14 upvotes]: 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.

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

Here are some specific versions:

Suppose we know there are two transitive models of set theory. Can we prove there are infinitely many?

Suppose we know there are two transitive models of set theory. Can we prove there are two with the same height?

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.

REPLY [0 votes]: If there is one model of set theory, transitive or otherwise, then there are class-many.  This follows from the upwards Lowenheim-Skolem theorem and Mostowski's transitive collapse theorem.
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