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.)



Can models of set theory contain extra ordinals?
I have two answers. 


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.


