Definition 5.24 SupposeA ⊆X×Y. We say that B ⊆A uniformizes A if and only if
i)πX(A) =πX(B), and
ii) for allx∈πX(A) there is a uniqueb∈Y such that (x, b)∈B.
In other words, B is the graph of a function f : πX(A) → Y such that
(x, f(x))∈Afor allx∈πX(A).
The Axiom of Choice tells us that for every A ⊆ X×Y, there is B ⊆A uniformizingA. We will be interested in trying to understand how complicated B is relative toA.
Definition 5.25 We say that Γ has theuniformization property if for allA∈ Γ(N × N), there isB∈Γ a uniformization ofA.
We first show that uniformization can be difficult.
Proposition 5.26 There is a closed setC⊆ N ×N that can not be uniformized by aΣ1
Proof By 5.23 there areΠ1
1-setsA0, A1⊆ N such thatA0∩A1=∅but there is no Borel setB with A0⊆B andA1∩B=∅.
There are closed setsC0, C1⊆ N such that X\Ai ={x:∃y (x, y)∈Ci}.
Without loss of generality we can takeCi⊆ N ×Nhii. LetC=C0∪C1. Suppose
B ∈Σ1
1 uniformizes C. Then B is the graph of a function f : N → C, and, by 4.15f is Borel measurable. Let Bi = f−1(N ×Nhii). Then each Bi is a
Borel set andB0∩B1 =∅. If x ∈Ai, thenx ∈B1−i. Thus B1 is a Borel set
separatingA0 andA1, a contradiction.
While Borel sets can not be uniformized by Borel sets, or evenΣ11-sets, we will prove that anyΠ1
1-set can be uniformized by aΠ11-set. As a warm-up we first prove a uniformization theorem forΠ1
1-subsets ofN ×N.
Theorem 5.27 (Kriesel’s Uniformization Theorem) Every Π1
1 subset of N ×Ncan be uniformized by a Π11-set.
Proof LetA⊆X×NbeΠ1
1 and letf :X×N→T rbe continuous such that x∈Aif and only iff(x)∈WF. Let
B={(x, n)∈A:∀m∈Nρ(f((x, m))6< ρ(f(x, n)) and ∀m < n ρ(f(x, m))6≤ρ(f(x, n))}.
ThenB is Π1
1 and (x, n)∈B if and only if (x, n)∈A, ρ(x, n) = infmρ(x, m)
and for allm < n,ρ(x, m)> ρ(x, n). Clearly for allx∈π(A) there is a unique nsuch that (x, n)∈B. ThusB uniformizes A.
We can do even better ifπ(A) is Borel.
Corollary 5.28 (Selection) Suppose A ⊆ X ×N is Π1
1 and π(A) is Borel.
Then Ahas a Borel-uniformization.
Proof LetB be aΠ11-uniformization ofA. Then
(x, n)6∈B⇔ ∃m∈N(m6=n∧(x, m)∈B). This is aΠ1
1-definition of X\B. ThusB is Borel.
Theorem 5.29 (Kondo’s Theorem) Π1
1 has the uniformization property.
Proof LetA⊆ N × N byΠ1
1. There is a tree T onN<ω×N<ω×N<ω such that A={(x, y) :T(x, y)∈WF}. Fix σ0, σ1, . . .an enumeration of N<ω. We
may assume that such thatσ0=∅,|σi| ≤i, and ifσi⊂σj, theni < j.
We build a sequence ofΠ1
1-setsA=A0⊇A1⊇A2⊇. . .such that (x, y)∈ An+1 if and only if
i) (x, y)∈An;
iii) if (x, z)∈An andz(n) =y(n), thenρ(T(x, y)σn)≤ρ(T(x, z)σn).
In other words, we first findmx,nminimal such that there is azwith (x, z)∈
An and z(n) =mx,n, we then findαx,n minimal such that there is (x, z)∈An
withz(n) =mx,n and ρ(T(x, z)σn) =αx,n. Then (x, y)∈An+1 if and only if (x, y)∈An,y(n) =mx,nandρ(T(x, y)σn) =αx,n.
Let B =TnAn. We will show that B is a Π11-uniformization of A. Letπ be the projection (x, y)7→x. Ifx∈π(A), then, by induction,x∈π(An) for all
n. Defineyx∈ N byyx(n) =mx,n. Claim 1If (x, y)∈B, then y=yx.
If (x, y)∈An,y(n) =mx,n, thusy=yx.
We need to show that (x, yx)∈B for allx∈π(A). Fixx∈π(A). Claim 2Supposeσi, σj∈T(x, yx) andσi ⊂σj. Thenαx,j< αx,i.
Choose (x, z)∈Aj+1. Sincei < j, (x, z)∈Ai+1. Sincez|j+ 1 =y|j+ 1 and
|σj| ≤j,σj ∈T(x, z). Thus
αx,i=ρ(T(x, z)σi)> ρ(T(x, z)σj) =αx,j,
as desired.
Claim 3(x, yx)∈A.
If σi0 ⊂ σi1 ⊂ . . . is an infinite path through T(x, yx), then, by claim 2, αx,1> αx,2> . . .a contradiction. ThusT(x, yx) is well founded and (x, yx)∈A. Claim 4(x, yx)∈B.
An induction onT(x, yx) shows that ifσn ∈T(x, yx), thenρ(T(x, yx)σn)≤
αx,n. By choice ofαx,n, we have ρ(T(x, yx)σn) =αx,n and (x, yx)∈An for all
n.
ThusB is the graph of a function uniformizingA.
Claim 5B isΠ1 1. DefineR(x, y, n) by ∃z[(∀k < n(y(k) =z(k)∧ρ(T(x, y)σk) =ρ(T(x, z)σk))∧ (z(n)< y(n)∨(z(n) =y(n)∧ρ(T(x, z)σn)< ρ(T(x, y)σn). By 5.15RisΣ1
1. Suppose (x, y)∈An. If (x, y)6∈An+1, then eithery(n)6=mx,n
or ρ(T(x, y)σn) 6= αx,n. In either case R(x, y, n) holds. On the other hand,
supposeR(x, y, n) andzwitnesses the existential quantifier. Since (x, y)∈An,
(x, z)∈ An. It is easy to see that eithery(n)6=mx,n or ρ(T(x, y)σn)6=αx,n.
Thus (x, y)6∈An+1. Thus
(x, y)∈B↔(x, y)∈A∧ ∀n¬R(x, y, n) andB isΠ1
1.
Corollary 5.30 Σ1
2 has the uniformization property.
Proof Suppose A ⊆ N × N is Σ1
2. There is a Π11-set B ⊆ N3 such that A={(x, y) :∃z(x, y, z)∈B}. By Kondo’s Theorem, there is aΠ1
1-set Bb⊆B such that
and for all x ∈π(A) there is a unique pair (y, z) such that (x, y, z)∈ B. Letb
b
A={(x, y) :∃z(x, y, z)∈Bb}. Thenπ(A) =b π(A) and for allx∈π(A) there is a uniqueysuch that (x, y)∈Bb.
Exercise 5.31 a) Show thatΣ1
2 has the reduction property. b) Show thatΠ1
2 does not have the uniformization property. c) Suppose there is a ∆1
2 well-order of N. Show that Σ1n-uniformization
holds for alln≥2. (In particular this is true ifV=L). While Σ1
1 does not have the uniformizaton property, the next two exercises show that we can get close.
Exercise 5.32 Let ≤lex be the lexicographic order on N. Suppose F ⊆ N is closed and nonempty. Show that there is x ∈ F such that x ≤lex y for all y∈F.
Exercise 5.33[Von Neumann Uniformization] SupposeA⊆ N × N isΣ1 1. Let Cbe the smallestσ-algebra withΣ1
1⊂ C. There isB∈ CuniformizingA. [Hint: There is a continuousf : N → N × N withf(N) =A. Let π(x, y) =x. For a ∈ A, let Fx ={z ∈ N : π(f(z)) = x}. Let g : π(A) → N by g(x) is the
lexicographic least element ofFx. Show thatgis anC-measurable function. Let
B={(x, f(g(x))) :x∈π(A)}. Show thatB ∈ Cand B uniformizesA.] Conclude from 4.23 that every Σ1
1-set has a C-measurable uniformization, and hence a Lebesgue measurable uniformization.
Π
1 1-sets
Many of the proofs in this section work just as well for Π1
1-sets. Here are statements of the effective versions.
Theorem 5.34 i) IfA⊆X isΠ11, there is a computablef :X→T rsuch that x∈Aif and only iff(x)∈WFfor allx∈X.
ii) Π1
1 has the reduction property.
iii) Any two disjoint Σ1
1 sets can be separated by a∆11-set.
iv) Any Π1
1-subset ofN × N can be uniformized by a Π11-set.
v) IfA⊆ N ×NisΠ1
1 andπ(A) =N, thenAhas a ∆11-uniformization. Further analysis of Π1
1-sets will require looking at an effective version of “ordinals”.