In this section, we are going to show that if ZF is consistent, so are ZF+AC and ZF+GCH.
The usual way to do this is to exhibit a model ZF in which the additional axioms holds, too, assuming a model of ZF exists. The universe of a model is supposed to be a set, and we will work with such set models when we will construct a model of ZF in which CH does not hold.
In this section, we will work with class models instead, in particular, L. The satisfaction relation is not formalizable for arbitrary classes, so we have to argue syntactically.
In the previous section, we showed that L is an inner model for ZF. What the “model” part here means is simply that we can prove in ZF that every axiom of ZF holds relative to L, or, using the standard notation for provability,
For suppose ZF+τ is inconsistent. Then there exists a proof of θ∧¬θ from ZF+τ, for some formula θ. Every formal proof uses only finitely many steps, so there exists a finitely manyσ1,…,σn∈ZF+τ such that
This means (σ1∧⋯∧σn)→(θ∧¬θ) is a validity and derivable by purely logical arguments (not assuming any additional axioms). But any such validity will remain valid when relativized (recall that classes are always defined via a formula φ):
We can add to ZF the axiom that all sets are constructible, i.e.
(V=L)∀x∃y(y is an ordinal ∧x∈Ly).
This axiom is usually denoted by V=L. We may be tempted to think that L is then trivially a model of ZF+V=L. But this is not at all clear, since this has to hold relative to L, i.e. (V=L)L.
This means that
∀x∈L∃y∈L(y is an ordinal ∧(x∈Ly)L).
To verify this, we need to make sure that insideL, L “means the same as” L. This is, of course, an absoluteness property, and we therefore revisit the complexity of the formulas defining the constructible universe.
We have seen that the map a↦PDef(a) is Σ1. This has important implications for the map α↦Lα.
Proof
We first show that the mapping is Σ1. The mapping is obtained by ordinal recursion over the functions a↦PDef(a) and a↦⋃a.
In general, if a function G:V→V is Σ1 and F:Ord→V is obtained by recursion from G, i.e. F(α)=G(F↾α), then F is also Σ1. This is because
y=F(α)↔α∈Ord∧∃f(f function ∧dom(f)=α∧∀β<α(f(β)=G(f↾β))∧y=G(f)).
Applying some of the various prefix transformations for Σ1-formulas, and using that being an ordinal, being an function, being the domain of a function, etc., are all Δ0 properties, the above formula can be shown to be Σ1, too.
In our case, G is a function that applies either PDef or ⋃ (both at most Δ1), depending on whether the input is a function defined on a successor ordinal or a limit ordinal (or applies the identity if neither is the case). Fortunately, this case distinction is also Δ0, and hence we obtain that F:α↦Lα is Σ1.
Finally, as in Theorem 2, observe that if F is a Σ1 function with a Δ1 domain (Ord), then F is actually Δ1, since we have
so the complement of the graph of F is Σ1-definable, too.
We can relativize the definition of L to other classes M. If M is is an inner model, then the development of L can be done relative to M. Since M is a ZF-model, it has to contain all the sets LαM (as we developed definability and proved facts about it insideZF). As M is transitive, the mapping F:α↦Lα is absolute for M and we obtain, for all α,
Every well-ordering on a transitive set X can be extended to a well-ordering of PDef(X).
Note that every element of PDef(X) is determined by a pair (ψ,a), where ψ is a set-theoretic formula, and a=(a1,…,an)∈X<ω is a finite sequence of parameters.
For each z∈PDef(X) there may exist more than one such pair (i.e. z can have more than one definition), but by well-ordering the pairs (ψ,a), we can assign each z∈PDef(X) its least definition, and subsequently order PDef(X) by comparing least definitions. Elements already in X will form an initial segment.
Such an order on the pairs (ψ,a) can be obtained in a definable way: First use the order on X to order X<ω length-lexicographically, order the formulas by their Gödel numbers, and finally put
Based on this, we can define an order <L all levels of L so that the following hold:
(1) <L↾Vω is a standard well-ordering of Vω (as for example given by a canonical isomorphism (Vω,∈)↔(N,E), see Ackermann (1937))
(2) <L↾Lα+1 is the order on PDef(Lα) induced by <L∣Lα.
(3) <L↾Lλ=⋃α<λ<L↾Lα for a limit ordinal λ>ω.
It is straightforward to verify that this is indeed a well-ordering on L. Moreover, the relation <L is Δ1. (To verify this, we have to spell out all the details of the above definition. This is a little involved, so we skip this here and refer to Jech (2003).)
We now show that V=L implies the Continuum Hypothesis. The proof works by showing that under V=L, every subset of a cardinal κ will be constructed by stage κ+. This is made possible by a “condensation” argument: If any subset x of κ is in L, then it must show up at some stage Lλ. κ and x generate an elementary substructure M of Lλ of cardinality κ. If we could show that this M itself must be an Lβ, we can use the fact following fact. Essentially, it tells us that the cardinality of the Lα evolves “tamely” along the ordinals (as opposed to (Vα)).
Proof
We know that α⊆Lα. Hence ∣α∣≤∣Lα∣. To show ∣α∣≥∣Lα∣, we work by induction on α.
If α=β+1, then by Proposition 1(4), ∣Lα∣=∣Lβ∣=∣β∣≤∣α∣.
If α is limit, then Lα is a union of ∣α∣ many sets of cardinality ≤∣α∣ (by inductive hypothesis), hence of cardinality ≤∣α∣.
But why would an elementary substructure of an Lλ have to be itself an Lβ? This is where the absoluteness of the construction of L strikes yet again!
Proof
Let T be the axioms of ZF (including Pairing, Union, Set Existence) used to prove that all the theorems leading up to the fact that for all α, Lα exists and that α↦Lα is Δ1 (and hence absolute). Any proof is finite, so we have used only finitely many (instances of) axioms of ZF to prove these facts. In particular, T is finite. Let φV=L be the sentence we obtain by taking the conjunction (∧) of all axioms in T together with the axiom V=L.
Suppose for a transitive set M, M⊨φV=L. Let λ be the least ordinal not in M. We must have that OrdM=λ, by the absoluteness of
ordinals. Moreover, λ must be a limit ordinal since for each α∈M, α∪{α} is in M since M satisfies Pairing and Union.
N is not necessarily transitive, but since it is well-founded we can take its Mostowski collapse (Theorem 1) and obtain a transitive set M
together with an isomorphismπ:(N,∈)→(M,∈).
Since κ is contained in both M and N, and is already transitive, it is straightforward to show via induction that π(α)=α for all α∈κ. Since x⊆κ, this also yields π(x)=x. This implies in turn that x∈M.
As (M,∈) is isomorphic to (N,∈) and N⪯Lλ, M satisfies the same sentences as (Lλ,∈). In particular, M⊨φV=L. By the condensation lemma, M=Lβ for some β.