# Can models of set theory contain extra ordinals?

In the paper "Complete topoi representing models of set theory" by Blass and Scedrov, they consider a general notion of Boolean-valued model of set theory, and one of the conditions they impose is that the model contain "no extra ordinals after those of V", i.e. that for all z in the model we have

$$\Vert z \text{ is an ordinal} \Vert = \bigvee_{\alpha \text{ is an ordinal of } V} \Vert z=\check{\alpha}\Vert$$

where $\Vert-\Vert$ denotes the truth function of the model valued in some complete Boolean algebra.

My question is: do there exist models which do contain "extra ordinals" in this sense? I presume so, or they wouldn't have needed to impose this condition. What do such models look like?

(By way of clarification, certainly if the starting model V is a set model in some larger universe, then one can find other set models in that larger universe which contain more ordinals. But I'm interested in just starting with a single model V and building models from it, which can be sets or proper classes.)

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Please clarify the difference between "$z$ is an ordinal" and "$\alpha$ is an ordinal of $V$", if there is one. I can guess that "$z$ is an ordinal" is just the suitable internal statement about $z$. What does "$\alpha$ is an ordinal of $V$" mean then? Is $V$ your model or the meta-universe? –  Andrej Bauer Feb 23 '10 at 7:29
When you say that Blass and Scedrov have a class of B-valued models of set theory, do you mean that they are just using the usual V^B, but for many different B, or is B fixed and they are entertaining various nonstandard B-valued constructions? –  Joel David Hamkins Feb 23 '10 at 14:07
@Andrej, V is the model we started from, I guess what you call the "meta-universe." It would be equivalent to take the join over "all ordinals" since that join is happening outside of M, but I thought this way was clearer. Silly me. –  Mike Shulman Feb 23 '10 at 14:54
@Joel, they define a general notion of "model" which includes all the usual V^B, but also permutation and symmetric models. They aren't constructing models, but performing a construction on models which are assumed to be constructed in some other (usually standard) way. –  Mike Shulman Feb 23 '10 at 14:55
Are the models somehow uniformly presented as a class? Otherwise, I don't see that you would have a "class" of models from the perspective of V, and you are really living in a context with more ordinals than V already, in order to have this collection of models at all. –  Joel David Hamkins Feb 23 '10 at 15:02

First, the standard method of building B-valued models of set theory, where B is any complete Boolean algebra, always satisfies your condition.

Suppose that B is any complete Boolean algebra, and denote the original set-theoretic universe by V. One constructs the B-valued universe VB by building up the collection of B-names by recursion, so that τ is a B-name, if it consists of pairs ⟨σ,b⟩ where σ is a previously constructed name and b ∈ B. One may impose a B-valued structure on the class VB of all B-names, by first defining it for atomic formulas by induction on names and then extending to all formulas by induction on formulas. This is the usual way to do forcing with Boolean-valued models, and VB is the Boolean-valued structure that results.

The remarkable thing, providing the power of forcing, is that every ZFC axiom gets Boolean value 1 in VB. In particular, the assertion that any two ordinals are comparable will have Boolean value 1.

Suppose that z is any B-name. If β is an ordinal above the Levy rank of z, that is, the place in the Vβ hierarchy where the name z first exists, then it is not difficult to see that [[ β ∈ z ]] = [[ β = z ]] = 0. It follows that [[ z ∈ β ]] = 1. But this latter Boolean value is the same as Vα<β [[ z = α ]]. Thus, we have proved your identity

• [[ z is an ordinal ]] = Vα ∈ ORD[[ z = α ]],

since all of the terms in this join are 0 beyond β and below β it is the expression we already observed.

There is a subtle point about whether your condition actually expresses "no new ordinals" or not. Suppose that VB is the B-valued model we have constructed and let U be any ultrafilter on B. One may form the quotient model VB/U, and there is a Los theorem, showing that the quotient satisfies φ if and only if [[ φ ]] ∈ U. If U is not V-generic, for example, if U is in V and B is not atomic, then there will be names z such that [[ z is an ordinal ]] = 1, but [[ z = α ]] ∉ U for any ordinal α in V. One way of thinking about this is that VB knows that z is definitely an ordinal, and by your property, Vα ∈ ORD [[ z = α ]] = 1, but VB doesn't know that z is any particular ordinal α. Thus, the ultrafilter U is able to squeeze between these two requirements, and in the quotient, z is a new ordinal. But this doesn't contradict your property.

Now, second, I can give a negative example. It is implicit in your question that the B-valued model somehow includes V, since you refer to the V ordinals α inside the Boolean brackets. Suppose that V is the univese of all sets, and let j:V to M be any elementary embedding that is not an isomorphism. For example, perhaps M is the ultrapower of V by an ultrafilter (M may or may not be well-founded). In particular, not every ordinal of M has the form j(α) for an ordinal α of V. Since M is a model of ZFC, we may regard it as a 2-valued Boolean model, or as a B-valued model for any B, since 2 = {0, 1} is a subalgebra of B. But M has ordinals not of the form j(α) for any V ordinal α. If one identifies V with its image in M, then this would provide a counterexample to the desired property.

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Thanks! I did have some expectation that this condition would allow names of ordinals whose "ordinal-ness" is "spread out over more than one ordinal" and thus not assertably equal to any particular one, but I think your last example is the one I was really looking for. I didn't state all of Blass & Scedrov's other requirements, but I'm guessing that that example also violates their condition defining the inclusion of V in M, namely $[[z \in \check{x}]] = \bigvee_{y\in x} [[z = \check{y}]]$ for all x in V. Is that right? –  Mike Shulman Feb 23 '10 at 15:06
Ah, looks like maybe you already answered this question in the comments to Francois -- I was going to ask, in your last example, is it still true that the ordinals of the form j(a) are cofinal in the ordinals of M? –  Mike Shulman Feb 23 '10 at 15:08
Yes, they are cofinal. –  Joel David Hamkins Feb 23 '10 at 15:10
Mike, yes, that condition is exactly what it takes to run the argument I gave above. Your condition on ordinals is just a special case of it, unless z is an ordinal completely on top of the V ordinals. –  Joel David Hamkins Feb 23 '10 at 15:25

As Joel pointed out, that requirement is true for internal Boolean valued models. It is also true for symmetric models, permutation models (Blass & Scedrov allow atoms), and a variety of mixed constructions. Blass and Scedrov were simply isolating the key common features of these constructions.

That said, there are ways to violate this condition. By an earlier requirement by Blass and Scedrov on canonical names, this would require the Boolean valued model to be an end-extension of V. For example, an old result of Keisler and Morley says that every countable models of ZF has an elementary end-extension. There is no internal way to construct such things in plain ZFC since that would easily violate Gödel's Theorem, but it is possible to have decent approximations assuming large cardinals.

There is an interesting construction by Sy Friedman where he essentially forces with a poset of size Ord+. (You have to jump through several hoops to do this, see Chapter 5 of Sy Friedman's book Fine Structure and Class Forcing.) Let loose, this forcing would naturally violate the no new ordinals condition. However, Friedman is careful to cut down the model so that no new ordinals appear.

Here are a few additional remarks on a potential construction that comes very close to what you want.

Other than the fact that we just ran out of ordinals, there is no real reason to stop the usual construction of L at Ord. We can construct LOrd+1 in the same way except that we can't replace definitions for classes by actual sets. Nevertheless, we can define elements of LOrd+1 to be (Gödel codes for) one-variable formulas with ordinal parameters, identifying formulas that define the same class. This is tricky to do internally since we have no truth definition, but we can work around that using 0#, assuming it exists in V. Once we have LOrd+1, we can similarly construct LOrd+2. In fact, with large enough cardinals, we can keep going like this for quite a while. There is no reason we will ever hit a model of ZF in this way, but I don't think it's impossible. To accomplish this, you need to have L live inside a model V where all the relevant information is packed into a set. This is where Friedman (building on seminal work of Jensen and others) picks up. His methods in Chapter 5 suggest that you could pack all of the relevant information into a single real number. Unfortunately, this is also where I stop, opening Friedman's book at a random page should explain why...

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Francois, I give an example in my last paragraph, where the new ordinals come in between V ordinals, so it is not necessary that they come only on top. –  Joel David Hamkins Feb 23 '10 at 14:52
Yes, however Blass and Scedrov also require that canonical names evaluate correctly. –  François G. Dorais Feb 23 '10 at 14:57
What does that mean? Even the usual V^B names don't evaluate correctly in all quotients (unless the ultrafilter is V-generic). –  Joel David Hamkins Feb 23 '10 at 15:00
Blass and Scedrov don't pass to a generic extension, they work with V^B directly as a topos. The requirement has an obvious typo in the paper, but it should say that $\|z \in \check{x}\| = \sup_{y \in x} \|z = \check{y}\|$. This prevents new ordinals to squeeze in between old ones. –  François G. Dorais Feb 23 '10 at 15:06
Thanks! Talking about countable models is the sort of thing I wanted to avoid with my "clarification" comment, but the large cardinal approximations and Friedman's construction sound interesting. –  Mike Shulman Feb 23 '10 at 15:07