Inspired by this article on Wikipedia:

Intuitively, forcing consists of expanding the set theoretical universe $V$ to a larger universe ${\displaystyle V^{*}}$. In this bigger universe, for example, one might have many new real numbers, identified with subsets of the set $\mathbb {N}$ of natural numbers, that were not there in the old universe, and thereby violate the continuum hypothesis.

All of the instances of this I can think of basically involve the principle of adding new natural numbers, as a first step, which thus adds new sets. For instance, if we use a non-principal ultrafilter to expand the universe, we clearly will get a nonstandard model of $\Bbb N$ in there. Thus, we add new sets of naturals because we have also added new naturals.

The question is: is it possible to somehow have a model of ZFC which adds new sets of natural numbers without actually adding new natural numbers?

Put another way: it is clearly possible for a model of ZFC with the true $\Bbb N$ to have less sets of naturals than the ambient theory, such as with $L$. But can the model somehow have more?

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    $\begingroup$ Well, yes, this is what forcing does. $\endgroup$ Feb 25 at 2:39
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    $\begingroup$ I think this question is based on a misunderstanding: forcing never adds ordinals, and in particular if $M$ is a forcing extension of $N$ then $M$ and $N$ have the same natural numbers. $\endgroup$ Feb 25 at 4:20
  • $\begingroup$ @NoahSchweber maybe I've misunderstood. I thought when you do forcing you take an ultrapower of the universe. Wouldn't that lead to nonstandard natural numbers? $\endgroup$ Feb 25 at 4:58
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    $\begingroup$ You definitely do not take an ultrapower when forcing. (Is there a particular text you're looking at?) $\endgroup$ Feb 25 at 5:03
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    $\begingroup$ Worth noting also that even when you do take an ultrapower of the universe, you don’t get nonstandard natural numbers (or any nonwellfoundedness at all), if the ultrafilter is countably complete. $\endgroup$ Feb 26 at 0:07

1 Answer 1


This question stems, I think, from conflating multiple different techniques in set theory. Let me quickly summarize what these techniques do (and don't) do:

  • The first technique, and in my opinion the really central one here, is classical (= poset-based) forcing. This is the direct "streamlining" of Cohen's original approach, and is what the vast majority of set theorists actually use; Kunen's book Set theory: an introduction to independence proofs contains a good treatment, as do these notes by Unger. There's nothing especially subtle here, and the three most important theorems can be stated snappily as follows:

    1. If $M$ is a countable transitive model (ctm) of $\mathsf{ZFC}$ and $\mathbb{P}$ is a poset in $M$, then there is (in $V$, not in $M$) a $\mathbb{P}$-generic-over-$M$ filter $G$. Such a filter determines a forcing extension $M[G]$ of $M$, which can be described explicitly (via names) or more abstractly as "the unique smallest ctm of $\mathsf{ZFC}$ containing $M$ as a submodel and $G$ as an element." Crucially, $M[G]$ is again a ctm and even has the same ordinals as $M$ itself.

    2. There is a hierarchy of finite subtheories $T_0\subset T_1\subset ...$ of $\mathsf{ZFC}$ such that the above bulletpoint holds with $T_i$ in place of $\mathsf{ZFC}$ for each $i$. This tweak lets us prove independence results merely assuming $Con(\mathsf{ZFC})$, not "$\mathsf{ZFC}$ has a ctm" (which is strictly stronger). It's really more of a technical improvement, but still worth noting; Kunen says more about it.

    3. Another way to fix the result of the first bulletpoint is to drop the assumption of transitivity: (1) holds with only the assumption that $M$ is a countable model of $\mathsf{ZFC}$, and by downward Lowenheim-Skolem the existence of such a model is equivalent to $Con(\mathsf{ZFC})$. In my opinion this is a much more natural tweak than (2); the added subtleties are actually substantive (which is both good and bad I suppose) and consist of $(i)$ the need to build $M[G]$ via "internal" recursion and $(ii)$ a less-trivial verification of Foundation in $M[G]$ (with $(i)$ being by far the most important point here). Embarrassingly, I don't really know a good source for this; it's just "folklore" and was presented to me as a good exercise back in grad school. Corazza has a paper on the topic, but I haven't read it so I can't vouch for it.

Before moving on to the other two points, let me just observe that the above - especially (3) - gives a very strong positive answer to your question: any countable model has "no-new-naturals, many-new-reals" expansions. Moreover, something like a countability hypothesis is obviously needed, since otherwise our "starting model" might already have all subsets of (what it thinks is) $\omega$. Note that transitivity/well-foundedness issues don't play a role here at all.

Now on to the other points:

  • The second technique worth mentioning is an alternative approach to forcing is via Boolean-valued models. I think this is what's tripping you up; it has nothing to do with ultrapowers, though. The idea is to get over the countability restriction in the poset-based approach. Of course we can't literally do this, since e.g. if $M$ is a transitive model of $\mathsf{ZFC}$ containing every real then it doesn't have any extensions by adding a single Cohen real, but we can fix this by broadening our notion of "model." Interestingly, the BVM-approach is also extremely useful in technical ways: the translation from a forcing poset $\mathbb{P}$ to the associated algebra of regular opens $\mathbb{B}_\mathbb{P}$ (nonstandard notation, sadly) is often quite helpful in answering concrete questions about forcing with $\mathbb{P}$ in the original sense. Jech's book Set theory (3rd Millenium edition) treats forcing in this way, but ultimately I don't recommend BVMs until you thoroughly understand the poset-based approach so I won't say more about that here.

  • Finally, there is the notion of Boolean-valued ultrapowers. But this is not the same as forcing (although they do interact in interesting ways)! It's a genuinely distinct topic that should be treated on its own after forcing is solidly understood, and to be honest I don't know very much about it. A good reference seems to be Hamkins/Seabold.

  • $\begingroup$ The two points of forcing over non-wellfounded models and boolean ultrapowers are connected. You can do the BWM stuff internally to the (non-necessariluy-wellfounded) model $M$, and then quotient by some ultrafilter on the boolean algebra to get a two-valued model $M^B/U$. The "forcing extension" $M^B/U$ has an inner model $\check M^U$ as its ground model, and $\check M^B$ is a boolean ultrapower of $M.$ The ultrapower embedding is an isomorphism iff $U$ is generic. (This is in the Hamkins/Seabold paper.) $\endgroup$ Feb 26 at 0:17
  • $\begingroup$ @spaceisdarkgreen It's definitely true that they're connected, but I stand by my statement that you should not attack Boolean ultrapowers until you solidly understand the basics of forcing itself. $\endgroup$ Feb 26 at 0:44
  • $\begingroup$ Another cool connection along the same lines is that the Boolean quotient $M^B/U$ is isomorphic to the (regular old non-boolean) ultrapower of $M$ by $U$ when $B$ is a powerset algebra (not super easy to see or to show). Forcing by a powerset algebra in the conventional way is trivial of course, since it's atomic, and this makes perfect sense here since the generic ultrafilters on a powerset algebra are exactly the principal ultrafilters, which renders the ultrapower trivial. $\endgroup$ Feb 26 at 0:46
  • $\begingroup$ Certainly no argument with that, just wanted to point out the connection and add some color on the non-wellfounded models. (As far as I know, building $M[G]$ by internal recursion is the same thing as taking $M^B/G$... or if you're using posets, quotienting the names by the relation $\exists p\in G\;(p\Vdash \sigma = \tau)^M$.) $\endgroup$ Feb 26 at 1:03
  • $\begingroup$ Hi @NoahSchweber and thanks for the response. Some of the forcing stuff is over my head; I will ask a new question about that... as far as it directly relates to my question though, I'm still trying to get in elementary terms what it means to add reals without adding naturals. (1/N) $\endgroup$ Mar 4 at 18:57

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