# A kind of saturation property related to forcing notions

Forcing is typically done over well-founded models. There are lots of good reasons for this, but it can feel confining at times. Fortunately, we can equally well force over non-well-founded models! It takes a bit more work to show that everything remains nice, but basically the internal Boolean-valued model can be constructed in the usual way. See ??? for details.

Forcing over illfounded models has a number of serious applications, so let’s ignore those and do something silly instead. One thing we can do is make a “freshman’s dream” about forcing actually correct:

Freshman’s dream. If $\mathbb{P}$ is a homogeneous notion of forcing, then any two extensions by $\mathbb{P}$ are isomorphic.

This is of course false - indeed, any nontrivial forcing yields many pairwise nonisomorphic extensions, as long as our base model is well-founded and hence rigid (assuming enough generics exist in the first place, of course). However, in the context of nonstandard models, we can make the dream come true!

Of course, this takes a bit of work to make precise. Here’s what I mean:

Definition. (Within a ground model $V$.) A model $M$ of $ZFC$ has unique forcing extensions (UFE) if for all $\mathbb{P}\in M$, if $M\models$ "$\mathbb{P}$ is homogeneous" then $$V\models \mbox{ "\Vdash_{\mathbb{P}^2}M[G_0]\cong M[G_1]"},$$ where we conflate $\mathbb{P}$ (the element of $M$) with the $V$-partial order $(\{x: x\in M, M\models x\in\mathbb{P}\}, \{(x, y): M\models x\le_\mathbb{P} y\})$ to make sense of forcing over $V$ with $\mathbb{P}^2$.

We could of course replace “homogeneous” with, say, “almost homogeneous” here — I’m just using the stronger notion for simplicity.

It is easy (via a back-and-forth argument) to see that if $M$ is saturated, then $M$ has UFE (this also leads to a fun proof that “a homogeneous forcing extension of a saturated model of $ZFC$ is saturated” - where we force over $V$, and compute saturation of $M[G]$ in $V[G]$). Moreover, it is easy to see that saturation is strictly stronger than UFE: we can kill saturation by forcing in $V$, but we can’t kill UFE this way.

My question is whether this notion has been studied before. In particular, there are a few things about it which look particularly interesting:

Question 1. Do homogeneous forcings preserve UFE? Specifically, suppose $M$ has UFE, $\mathbb{P}\in M$ is a forcing which $M$ knows is homogeneous, and $G$ is $\mathbb{P}$-generic over $V$ (not just $M$!). Does $V[G]$ think that $M[G]$ has UFE? The obvious iteration doesn’t seem to necessarily be homogeneous, so I don’t know how to proceed here.

Question 2. My main motivation — there is a kind of Keisler-order-iness going on here, although I’m not sure how solid this vague connection really is: For $K$ a definable class of forcings, we can define $K$-UFE as above but restricting attention to forcings which $M$ think lie in $K$. Then we can ask how these notions compare for various $K$s. For example, if $M$ has $\{$c.c.c.$\}$-UFE, does $M$ have $\{$countably closed$\}$-UFE? I suspect that for the most part, questions like this reduce to just comparing the classes involved — e.g. since there are countably closed, non-c.c.c. forcings, the answer to the previous specific question should be “no” — but I don’t see how to prove this yet.

Question 3. How much weaker is UFE than full saturation, really? For example: suppose $\kappa$ is a cardinal such that no saturated models of ZFC exist with cardinality $\kappa$. Can there be a model of cardinality $\kappa$ which does have UFE? And relatedly, the observation that UFE is preserved by forcing shows that really, UFE is a version of “every constructible type over a small parameter set is realized” (where an $M$-type $p$ over a set $A\subset M$ is constructible if $V$ satisfies “whenever $N$ is a copy of $M$ with ordinal domain, there is a $q$ in $L[N]$ which pulls back to $p$”). Is UFE strictly weaker than “constructible saturation?” If so, there should be a good explicit-ish counterexample . . .

Smoryński proved long ago that any two countable computably saturated models of set theory with the same theory and the same standard system (the sets of natural numbers coded in the model) are isomorphic, and this result is used pervasively in the theory of models of arithmetic and set theory. You can find some details in lemma 2.4 of my paper

Since forcing does not change the natural number arithmetic of a model of set theory, it follows that it also does not change the standard system of an $\omega$-nonstandard model. And all extensions by almost homogeneous forcing have the same theory. So the only missing element is the following lemma, which I proved a few years ago, in part of a project with Andres Caicedo, and Dan Hathaway was also involved earlier-on.

Key Lemma. If $M$ is a countable computably saturated model of set theory with forcing extension $M[G]$, then $M[G]$ is also computably saturated.

It follows that if $M$ is any countable computably saturated model of ZFC and $M[G]$ and $M[H]$ are forcing extensions by any almost homogeneous forcing in $M$, then they are isomorphic. Thus:

Theorem. The freshman's dream is true, if one forces over a countable computably saturated model of set theory.

There is an interesting consequence, which is what got Andres and I interested in it:

Theorem. If ZFC is consistent, then there is a model $M\models\text{ZFC}$ that is isomorphic to all its extensions $M[c]$ to add a Cohen real.

The proof is to let $N$ be a countable computably saturated model and let $M=N[d]$ for a Cohen real $d$. Now, $M\cong M[c]$ because $M=N[d]\cong N[d*c]=M[c]$.

Andres and I have a draft written, but we haven't completed it yet.

• The project with Andres arose from an exchange on math.stackexchange, but I can't seem to find it now. – Joel David Hamkins Jun 26 '16 at 20:12
• Neat! I'm more interested in uncountable models right now, but this is really cool! – Noah Schweber Jun 26 '16 at 20:26
• Much of the argument goes through for uncountable saturated models. Unfortunately, though, we don't have the key lemma in that case. I'll send you the current draft of the article to which I referred, although it is not yet ready for general distribution. – Joel David Hamkins Jun 26 '16 at 21:55
• In the statement of Smorynski's result at the beginning of your answer, the model should be computably saturated. (And my browser refuses to put an acute accent on the n in Smorynski's name; the best I can do is Smory´nski.) – Andreas Blass Jun 26 '16 at 22:07
• Oh, dear, of course that is what I meant---I have now edited, and it seems I got the accent in as well. – Joel David Hamkins Jun 26 '16 at 22:10