It is often useful to allow taking direct limits in set theory. This happens all the time when taking about ultrapowers.

But let's not limit ourselves to ultrapowers. Suppose that we have a sequence of models of $\sf ZFC$, $M_n$ for $n<\omega$, such that all of them are transitive and even countable, with the same ordinals. What reasonable assumptions would guarantee that $\bigcup M_n$ is again a model of $\sf ZFC$? What happens if we consider longer sequences of models (where we don't require all the models to satisfy $\sf ZFC$, or we don't require the sequence to be continuous at all limits)?

I am particularly interested in the case where each model is a forcing extension of its predecessor in the sequence; but not necessarily just that.

For example, one could think that the requirement "eventual stabilization of the von Neumann hierarchy" is enough. But it is not. Consider the following counterexample:

Start with a [countable transitive] model of $\sf GCH$, $M_0$ and choose (externally) a cofinal sequence $\kappa_n$ for $n<\omega$ in the ordinals of $M_0$; without loss of generality, $M_0\models``\kappa_n\text{ is a strong limit cardinal}"$. Next define $M_{n+1}$ to be the forcing extension of $M_n$ by $\operatorname{Add}(\kappa_n^+,\kappa_n^{+3})^{M_n}$.

It is easily verified that the $M_n$'s are all countable with the same ordinals, and that $M_n\subseteq M_{n+1}$, as well as that the von Neumann hierarchy there stabilizes.

However, in $M=\bigcup M_n$, we can look at the function $\varphi(n,\alpha)$ stating that $n<\omega$ and $\alpha$ is the unique ordinal such that $\sf GCH$ is violated $n-1$ times below it. Easily, $\varphi$ actually defines the sequence $\kappa_n$, which is a contradiction to Replacement.

On the other hand, the stabilization is necessary for us to get a model of the Power Set axiom. Otherwise, there is some $\alpha$ such that $V_\alpha^{M_n}$ never stabilizes, and therefore never becomes an element of $V_{\alpha+1}$ in the limit model.


This doesn't exactly answer your question, but I find it to be in a similar spirit. Namely, one could naturally consider the version of your question where you ask merely that $\bigcup_n M_n$ is contained within a model of ZFC, rather than necessarily being a model of ZFC itself.

For this version of the question, particularly applied to forcing extensions, I have a characterization in my paper:

J. D. Hamkins, Upward closure and amalgamation in the generic multiverse of a countable model of set theory, RIMS Kyôkyûroku, pp. 17-31, 2016.

Follow the link for further links to slides of talks I've given on the topic. One theorem appearing in the paper is:

Theorem. An increasing chain of forcing extensions $$W\subset W[G_0]\subset W[G_1]\subset\cdots$$ of a countable model of ZFC has an upper bound $W[H]$ in a forcing extension of $W$ if and only if the forcing extensions $W[G_n]$ had uniformly bounded essential size in $W$.

The essential size of a forcing extension is the size of the smallest complete Boolean algebra by which the extension can be realized as a forcing extension.

In particular, any countable tower of extensions by adding a Cohen real has an upper bound, and furthermore in this case, you can find the upper bound which is also the extension by a single Cohen real, which is an attractive little argument, of the style of many computability theory constructions.

  • $\begingroup$ What happens if we take an Easton iteration or product of violating GCH on a proper class of cardinals, and externally cut the ordinals into an $\omega$-sequence to define the increasing sequence of models? $\endgroup$ – Asaf Karagila Sep 9 '17 at 1:11
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    $\begingroup$ The covering extension $W[H]$ hides whatever bad sequence you attempt to encode. This is actually how the proof goes: you make a big product of all possible forcing, and then use the fact that it has a certain chain condition to hide the models $W[G_n]$ amongst those factors, so that you can no longer see it. So $W[H]$ will not have the sequence $\langle G_n\mid n\in\omega\rangle$. $\endgroup$ – Joel David Hamkins Sep 9 '17 at 1:20
  • $\begingroup$ And this is why it is somewhat easier to arrange an upper bound, rather than the union as in your question. $\endgroup$ – Joel David Hamkins Sep 9 '17 at 1:21
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    $\begingroup$ If the essential size of the forcing extensions grows unbounded in the ordinals of $W$, then you cannot amalgamate them in a single set forcing extension $W[H]$, since essential size must grow as you move to further forcing extensions. So if the forcing is increasingly larger (unbounded in $W$), then there is no upper bound in the sense of my version of the question. $\endgroup$ – Joel David Hamkins Sep 9 '17 at 1:33
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    $\begingroup$ Well, even for the easy direction, the fact that the essential size of a forcing extension never goes down relies on the intermediate model theorem (that if $M\subset W\subset M[G]$, then $W$ is a forcing extension of $M$ by a subalgebra of the forcing for $M[G]$), which I think of as a fundamental result that is more profound than often considered. In particular, I don't think it should count as obvious. $\endgroup$ – Joel David Hamkins Sep 9 '17 at 1:50

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