Fix a countable transitive model $M$ of ZFC. In my answer to this question I indicated that there are forcing iterations $((Q_\alpha:\alpha\leq\omega),(\dot P_\alpha:\alpha<\omega))$ in $M$ and sequences $(G_\alpha:\alpha<\omega)$ of filters such that the following happens:

Each $G_\alpha$ is a filter in the evaluation of $\dot P_\alpha$ with respect to the filter $G_0*\dots*G_{\alpha-1}$ and $G_\alpha$ is generic over $M[G_0,\dots,G_{\alpha-1}]$ (call such a sequence $(G_\alpha:\alpha<\omega)$ a sequence of generics), but there is no $Q_\omega$-generic filter over $M$ whose $\alpha$-th projection is $G_\alpha$ for all $\alpha<\omega$.

An example can be obtained as follows:
Take the countable support iteration of Sacks forcing (or any other nontrivial $\omega^\omega$-bounding proper forcing notion) of length $\omega$ (i.e., the supports are actually everything, but this doesn't matter). This forcing adds no Cohen real.

Compare this to the finite support iteration of the same forcing notions. This iteration does add a Cohen real. The Cohen real is coded by the sequence of generics and hence this sequence of generics does not come from a generic filter for the countable support iteration mentioned before. This sequence of generics is not even contained in a forcing extension obtained using the countable support iteration.

Now here are two questions:

1) Is there an example of a sequence of generics (of length $\omega$) that cannot come from any iteration of the $\dot P_\alpha$?

I am asking here for iterations where the finite initial segments are as usual (just plain iteration) and we choose whatever ideal for the supports, including all finite subsets of the index set. But I am open to more general forms of iteration. For example take a large forcing notion $Q$ along with commuting complete embeddings of the $Q_\alpha$, $\alpha<\omega$. This would be an iteration of the $\dot P_\alpha$, too, the most general one that I can think of right now.

2) Is there an example of a sequence of generics over $M$ that is not contained in any countable transitive extension of $M$ with the same ordinals as $M$ that is a model of ZFC?

Obviously, a positive answer to 2) solves 1) as well.


2 Answers 2


There are a number of interesting things to say.

The answer to your first question is yes. Suppose that $M$ is a countable transitive model of set theory and we have a forcing iteration $P_\omega$ in $M$ of length $\omega$, forcing with, say, Cohen forcing $Q_n$ at stage $n$. Let $z$ be any real that cannot be added by forcing over $M$, such as a real that codes all the ordinals of $M$. This real cannot exist in any extension of $M$ to a model of ZFC with the same ordinals. Now, suppose that $G$ is any $M$-generic filter for the iteration, with $G_n$ being the stage $n$ generic filter. Let $H_n$ be the filter that results from $G_n$ by changing the first bit so as to agree with $z(n)$. That is, we change a single bit at each stage. The resulting sequence $\langle H_n | n\in\omega\rangle$ will be generic at every stage, since only finitely many bits are changed by a given stage, but the whole sequence computes $z$, which cannot be added by forcing.

Second, a similar phenomenon occurs even just with 2-step product forcing:

Theorem. If $M$ is a countable transitive model of ZFC, then there are two $M$-generic Cohen reals $c$ and $d$ such that $M[c]$ and $M[d]$ have no common extension to model of ZFC with the same ordinals.

The proof is to build $c$ and $d$ in stages. Fix a real $z$ which cannot exist in any extension of $M$ with the same ordinals, and enumerate the dense sets of $M$ by $D_0, D_1$ and so on. Build $c$ and $d$ in zig-zag fashion: first provide $c_0$ meeting $D_0$, and $d_0$ all $0$s of the same length as $c_0$, followed by the first digit of $z$. Now extend $d_0$ to $d_1$ meeting the dense set, adding all $0$s to $c_0$ making $c_1$ of the same length, and adding one more bit of $z$. And so on. The point is that we ensure that each of $c$ and $d$ is $M$-generic, but together, they reveal the coding points of $z$. So no model extending $M$ with the same ordinals can have both $c$ and $d$, for then it would have $z$.

Third, this is essentially the only kind of obstacle, for there is a positive result here. The following theorem is proved in a paper with G. Fuchs, myself and J. Reitz on the topic of set-theoretic geology:

Theorem. If $M$ is a countable transitive model of set theory, and $M[G_n]$ is a sequence of generic extensions of $M$ by forcing $G_n\subset P_n\in M$ of bounded size in $M$, such that the extensions are finitely amalgamble, in the sense that any finitely many of the $M[G_n]$ have a common forcing extension $M[H]$, then there is a single forcing extension $M[H]$ containing all $M[G_n]$.

I'll try to post a proof sketch later, but the main idea is to perform a very large combination of forcing, and then perform surgey so as to replace certain coordinate generics with $G_n$, in such a way that the resulting extension can see only finite fragments of the sequence $\langle G_n | n\lt\omega\rangle$, without being able to construct the whole sequence.

A special case of this theorem answers your second question, in a sense, for if we have a sequence of extensions $M\subset M[G_0]\subset M[G_1]\subset\cdots$, then these extensions are finitely amalgamable, and so there is indeed a common extension $M[H]$ containing every $M[G_n]$. This extension, however, is not an $\omega$-iteration of the forcing notions in your iteration, and in general we cannot expect that the sequence $\langle G_n | n\lt\omega\rangle$ is in $M[H]$, for the reasons described above.

  • $\begingroup$ I added a link to the geology paper. $\endgroup$ Jul 30, 2011 at 12:44

Take the steps $Q_\alpha$ to be Cohen forcing, and begin by taking a generic filter $G$ for the usual finite-support product. So $G$ codes an $\omega$-sequence of Cohen reals $x_n\in 2^\omega$. Fix some $z\in 2^\omega$ coding an ordinal larger than the height of $M$. Define $y_n\in 2^\omega$ to be exactly the same as $x_n$ except that $y_n(0)=z(n)$. Since only one entry in $x_n$ has been changed, it is clear that $y_n$ is Cohen-generic over $M$ and that, for each natural number $n$, $\langle y_k:k<n\rangle$ is generic for $P_0*\dots*P_{n-1}$; let $G_n$ be the corresponding generic filter in $P_0*\dots*P_{n-1}$. The sequence $\langle G_n:n\in\omega\rangle$ is not in any extension of $M$ with the same ordinals, because it encodes $z$.

  • $\begingroup$ While I was typing this, Joel gave the same construction (and more). $\endgroup$ Sep 14, 2010 at 17:52
  • $\begingroup$ Andreas, we gave the same answer simultaneously! $\endgroup$ Sep 14, 2010 at 17:53
  • $\begingroup$ I accept Joel's answer for being a hair earlier and more extensive. $\endgroup$ Sep 14, 2010 at 18:25

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