we obtain that both the analytic sets and the co-analytic sets are closed under countable unions and intersections.
We have seen (Proposition 1) that the analytic sets are closed under continuous images. Taking continuous images of co-analytic sets, however, leads out of the co-analytic sets.
Using continuous images (or rather, the special case of projections), we define the projective hierarchy. Recall our notation ∃N for projection along N, with ∀N its dual. We denote by ∃NN and ∀NN projection along NN and its dual, respectively.
ΣΣ11 iff ΠΠ11 iff ΣΣ21 iff ΠΠ21 iff P(x)⇔∃αF(α,x)P(x)⇔∀αG(α,x)P(x)⇔∃α∀βG(α,β,x)P(x)⇔∀α∃βF(α,β,x)⋮ for a closed set F⊆NN×X, for an open set G⊆NN×X, for an open set G⊆NN×NN×X, for a closed set F⊆NN×NN×X,
These characterizations clearly indicate a relation between being projective and being definable in second order arithmetic using function quantifiers.
We have seen that the Borel sets of finite order correspond to the sets definable (from parameters) by formulas using only number quantifiers (arithmetical formulas). A similar relation holds between projective sets and sets definable by formulas using both number and function quantifiers. In fact, the way we defined the projective hierarchy makes this easy to see.
Historically, however, the topological approach and the definability approach happened separately, the former devised by the Russian school of Souslin, Lusin, and others, while the effective approach was pursued by Kleene. Kleene named the sets in his effective hierarchy analytical sets, which to this day is a source of much confusion.
As before, we can relativize this hierarchy with respect to a parameter γ∈NN, by requiring R to be computable only relative to γ. This gives rise to classes Σn1(γ), Πn1(γ), and Δn1(γ). Then the Fundamental Theorem can be extended to projective sets as follows:
Here are a few examples of projective sets that occur naturally in mathematics.
Analytic sets:
{K⊆X:K compact and uncountable} is a ΣΣ11 subset of the space K(X) of compact subsets of X.
{f∈C[0,1]:f continuously differentiable on [0,1]} is a ΣΣ11 subset of C[0,1].
Co-analytic sets:
{f∈C[0,1]:f differentiable on [0,1]} is a ΠΠ11 subset of C[0,1].
{f∈C[0,1]:f nowhere differentiable on [0,1]} is a ΠΠ11 subset of C[0,1].
WF={α∈2N:α codes a well-founded tree on N} is a ΠΠ11 subset of the space Tr of trees, which can be seen as a closed subspace of 2N<N, and hence is Polish. As we will see, the set WF is a prototypical ΠΠ11 set.
Higher levels:
{f∈C[0,1]:f satisfies the Mean Value Theorem [0,1]} is a ΠΠ21 subset of C[0,1].
(Here f satisfies the Mean Value Theorem if for all a<b∈[0,1] there exists c with a<c<b such that f′(c) exists and f(b)−f(a)=f′(c)(b−a).)
Some structural properties of the projective hierarchy¶
The quantifier manipulations mentioned above yield the following closure properties.
To show that the hierarchy is proper, we need the existence of universal sets.
Proof
By induction on n. We have seen (Theorem 3) that there exists a NN-universal set for ΣΣ11. Now note that if U∈ΣΣn1(NN×X) is NN-universal for ΣΣn1(X), then ¬U is NN-universal for ΠΠn1(X), and if U⊆NN×NN×X is NN-universal for ΠΠn1(NN×X), then
We have seen (Corollary 2 and Proposition 2) that all analytic sets are Lebesgue measurable and have the Baire property. Since these properties are closed under complements, it follows that the same holds for co-analytic (ΠΠ11) sets. Analytic sets also have the perfect-set property, but if you worked out the exercise, you will see that the proof does not carry over to complements of analytic sets. Can we find a different proof?
Similarly, it does not seem impossible to extend the regularity properties (LM) and (BP) to higher levels of the projective hierarchy. We will soon see that there are metamathematical limits that prevent us from doing so.
Without explicitly mentioning it, up to now we have been working in ZF, Zermelo-Fraenkel set theory, plus a weak form of Choice (ACω(NN)). If we add the full Axiom of Choice (AC), we saw that the regularity properties do not extend to all sets. Solovay’s model of ZF+DC shows that the use of a strong version of Choice is necessary for this.
On the other hand, the proofs gave us no direct indication how complex the non-regular sets we constructed are.
We will study a model of ZF in which exists a ΔΔ21 set which neither is Lebesgue measurable nor has the Baire property. This, together with the Solovay model, shows we cannot settle in ZF alone the question of whether the projective sets are measurable or have the Baire property. We would have to add additional axioms.
A key feature in the construction of a non-measurable ΔΔ21 set is the use of the well-ordering principle rather than the Axiom of Choice.
Lebesgue measure here refers to the product measure λ×λ, which is the unique translation invariant measure defined on the Borel σ-algebra generated by the rectangles I×J, where I and J are open intervals, and (λ×λ)(I×J)=λ(I)λ(J).
Proof
Since <W is of order type ω1, for every y∈R, the set Ay={x:x<Wy} is countable, and hence of Lebesgue measure zero.
Fubini’s Theorem implies that if A⊆R2 is measurable, then
(λ×λ)(A)=∫λ(Ay)dλ(y)=0.
So if A is measurable, then (λ×λ)(A)=0. The complement of A is ¬A={(x,y):x≥Wy}. As above, for any x∈R, (¬A)x={y:x≥Wy} is countable, and hence λ(¬A)x=0 for all x.
Again, by Fubini’s Theorem, (λ×λ)(¬A)=0, and thus (λ×λ)(R)=(λ×λ)(A∪¬A)=(λ×λ)(A)+(λ×λ)(¬A)=0, a contradiction.
We can apply a similar reasoning for Baire category, using the Lemma below. The sections Ay and ¬Ax are countable, and hence meager.
The following lemma provides a Baire category analogue to Fubini’s Theorem.
For a proof see Kechris (1995).
Therefore, if the Continuum hypothesis (CH) holds in a model and we can well-order R (or NN, 2N) within a certain complexity (as a subset of R2), we can find a non-regular set of the same complexity. The question now becomes how (hard it is) to define a well-ordering of R, and of course if CH holds.