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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:

  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.

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    $\begingroup$ Andres, I'm not quite clear on which construction methods you are allowing here; after all, the forcing arguments are not so difficult... $\endgroup$
    – Joel David Hamkins
    Commented Sep 14, 2010 at 19:58
  • $\begingroup$ Hi Joel. I think that is part of the question, I am not sure what methods we have for dealing with these problems, other than forcing. Of course, it is not that there are any issues with forcing, or any inherent difficulties to it (proof theoretic or otherwise). However, it is puzzling to me that I do not see how to show 1, for example, without some forcing construction used somewhere. I would like to tell you "descriptive set theoretic methods" or somesuch, but I don't want to overlook something valuable by adding this restriction. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 14, 2010 at 21:06
  • $\begingroup$ Without some kind of large cardinal assumption this seems unlikely as you could be in a model of ZFC+V=L where there are few transitive models of ZFC. Like if $\beta$ is the second countable ordinal where $L_\beta\models$ ZFC. Does this make sense? $\endgroup$
    – Dave Marker
    Commented Sep 14, 2010 at 22:43
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    $\begingroup$ Dave, that model has continuum many transitive models, since it can see the minimal model, and therefore knows that the minimal model is countable, and so has a perfect set of forcing extensions of it. $\endgroup$
    – Joel David Hamkins
    Commented Sep 15, 2010 at 0:07
  • $\begingroup$ Hi Dave. Yes, you are correct. We could assume $T$, the theory ZFC+V=L+there is only one $\alpha$ with $L_\alpha\models$ZFC. In $T$, I do not see any way of proving there are two transitive models of set theory other than by appealing to a forcing argument. So unless we come up with something very very ingenious, we would like some additional assumption that ensures the universe is not tiny. However, perhaps there is already a clever argument in $T$+there is a countable transitive model of $V\ne L$ that gives us infinitely many such models. (We know via forcing that there are continuum many.) $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 15, 2010 at 0:30

2 Answers 2

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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.

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  • $\begingroup$ Hi Andreas. This is a nice observation. I think, though, that the key question is really whether, knowing there are two transitive models of the same height, then we can produce a third. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 17, 2010 at 17:33
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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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    $\begingroup$ Amit, although this argument shows that there will be a proper class of models of ZFC, there is little reason to expect these models to be well-founded, and so you will not get transitive models this way. (In particular, you can only apply the Mostowski collapse to the well-founded models.) $\endgroup$
    – Joel David Hamkins
    Commented Sep 21, 2010 at 23:36
  • $\begingroup$ Sorry, I wasn't thinking. The model resulting from applying Lowenheim-Skolem will satisfy Foundation but you're right, there's nothing guaranteeing it's well-founded. $\endgroup$
    – Amit Kumar Gupta
    Commented Sep 22, 2010 at 0:34
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    $\begingroup$ Hi Amit. Yes, unfortunately, in many situations one can actually prove that no model obtained this way is well-founded. There is a tiny hope of doing something similar that actually gives a well-founded structure at the end, by applying Bairwise compactness in some appropriate setting, but at the moment I only see how to make this work starting with large cardinals around the level of measurability. $\endgroup$
    – Andrés E. Caicedo
    Commented Sep 22, 2010 at 3:34

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