The

First, the standard method of building B-valued models of set

Finally, let me remark that there

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.

1

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.

Finally, let me remark that 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.