Constructible Universe - L Is A Standard Inner Model of ZFC

L Is A Standard Inner Model of ZFC

L is a standard model, i.e. it is a transitive class and it uses the real element relationship, so it is well-founded. L is an inner model, i.e. it contains all the ordinal numbers of V and it has no "extra" sets beyond those in V, but it might be a proper subclass of V. L is a model of ZFC, which means that it satisfies the following axioms:

• Axiom of regularity: Every non-empty set x contains some element y such that x and y are disjoint sets.

(L,∈) is a substructure of (V,∈) which is well founded, so L is well founded. In particular, if x∈L, then by the transitivity of L, y∈L. If we use this same y as in V, then it is still disjoint from x because we are using the same element relation and no new sets were added.

• Axiom of extensionality: Two sets are the same if and only if they have the same elements.

If x and y are in L and they have the same elements in L, then by L's transitivity, they have the same elements (in V). So they are equal (in V and thus in L).

• Axiom of empty set: {} is a set.

{} = L₀ = {y | y∈L₀ and y=y} ∈ L₁. So {} ∈ L. Since the element relation is the same and no new elements were added, this is the empty set of L.

• Axiom of pairing: If x, y are sets, then {x,y} is a set.

If x∈L and y∈L, then there is some ordinal α such that x∈L_α and y∈L_α. Then {x,y} = {s | s∈L_α and (s=x or s=y)} ∈ L_α+1. Thus {x,y} ∈ L and it has the same meaning for L as for V.

• Axiom of union: For any set x there is a set y whose elements are precisely the elements of the elements of x.

If x ∈ L_α, then its elements are in L_α and their elements are also in L_α. So y is a subset of L_α. y = {s | s∈L_α and there exists z∈x such that s∈z} ∈ L_α+1. Thus y ∈ L.

• Axiom of infinity: There exists a set x such that {} is in x and whenever y is in x, so is the union y U {y}.

From transfinite induction, we get that each ordinal α ∈ L_α+1. In particular, ω ∈ L_ω+1 and thus ω ∈ L.

• Axiom of separation: Given any set S and any proposition P(x,z₁,...,z_n), {x|x∈S and P(x,z₁,...,z_n)} is a set.

By induction on subformulas of P, one can show that there is an α such that L_α contains S and z₁,...,z_n and (P is true in L_α if and only if P is true in L (this is called the "reflection principle")). So {x | x∈S and P(x,z₁,...,z_n) holds in L} = {x | x∈L_α and x∈S and P(x,z₁,...,z_n) holds in L_α} ∈ L_α+1. Thus the subset is in L.

• Axiom of replacement: Given any set S and any mapping (formally defined as a proposition P(x,y) where P(x,y) and P(x,z) implies y = z), {y | there exists x∈S such that P(x,y)} is a set.

Let Q(x,y) be the formula which relativizes P to L, i.e. all quantifiers in P are restricted to L. Q is a much more complex formula than P, but it is still a finite formula; and we can apply replacement in V to Q. So {y | y∈L and there exists x∈S such that P(x,y) holds in L} = {y | there exists x∈S such that Q(x,y)} is a set in V and a subclass of L. Again using the axiom of replacement in V, we can show that there must be an α such that this set is a subset of L_α ∈ L_α+1. Then one can use the axiom of separation in L to finish showing that it is an element of L.

• Axiom of power set: For any set x there exists a set y, such that the elements of y are precisely the subsets of x.

In general, some subsets of a set in L will not be in L. So the whole power set of a set in L will usually not be in L. What we need here is to show that the intersection of the power set with L is in L. Use replacement in V to show that there is an α such that the intersection is a subset of L_α. Then the intersection is {z | z∈L_α and z is a subset of x} ∈ L_α+1. Thus the required set is in L.

• Axiom of choice: Given a set x of mutually disjoint nonempty sets, there is a set y (a choice set for x) containing exactly one element from each member of x.

One can show that there is a definable well-ordering of L which definition works the same way in L itself. So one chooses the least element of each member of x to form y using the axioms of union and separation in L.

Notice that the proof that L is a model of ZFC only requires that V be a model of ZF, i.e. we do NOT assume that the axiom of choice holds in V.