Theorem 11.11 (Doughrety-Jackson-Kechris) If E is a hyperfinite Borel equivalence relation, thenE≤BE0.
Corollary 11.12 IfE is a nontame hyperfinite Borel equivalence relation then
E≡B E0.
Proof By Theorem 9.17E0≤BE and by Theorem 11.11E≤BE0.
By Theorem 11.8 and Lemma 10.6 every hyperfinite Borel equivalence rela- tion is Borel-reducible toE(Z,C). Thus we may assume thatE=E(Z,C).
We say thatX ⊆ CZ is tame ifX isE-invaraint andE|X is tame.
Lemma 11.13 Suppose X ⊆ CZ is tame, and f : CZ \X → C is a Borel reduction ofE|Y toE0. Then E≤BE0.
Proof Letg:X → Cbe a Borel measurable function such that xEy⇔g(x) =g(y)
forx, y∈X. Since there is a perfect set ofE0-inequivalent elements, there is a continuousp:C → C such that p(x)E0p(y) for6 x =6 y. Let h,i:C2 → C be the ususal bijection
< x, yi= (x(0), y(0), x(1), y(1), . . .).
Finally let 0,1 ∈ C denote the infinte sequences that are constantly 0 and constantly 1, respectively. Definefb:CZ→ C by b f(x) = ½ hp(g(x)),0i ifx∈X hf(x),1i ifx6∈X .
Ifx∈X andy6∈X, thenf(x)6E0f(y), since the even part off(x) is 0 and the even part off(y) is 1.
Ifx, y∈X, then
b
f(x)E0f(y)b ⇔p(g(x))E0p(g(y))⇔g(x) =g(y)⇔xEy. Ifx, y6∈X, then
b
f(x)E0f(y)b ⇔f(x)E0f(y)⇔xEy. Thusfbis the desired reduction.
Lemma 11.14 If X0, X1, . . . ,⊆ CZ are tame, thenSXn is tame.
Proof Since eachXiis invariant we may assume that theXiare disjoint. Iffi:
Xi→ C is a Borel reduction andf :SXi→ C is the functionf(x) = 0i1bfi(x)
forx∈Xi, thenf is a Borel reducition ofE|X to ∆(C).
The two lemmas allow us to work “modulo tame sets”, i.e. ifX is tame we may ignore it and assume we are just working withE|(CZ\X).
Proof of Theorem 11.11 We will view each x ∈ CZ as a Z×N array of zeros and ones. The columns are. . . , x−2, x−1, x0, x1, x2, . . . where xi ∈ C. If
σ ∈ (2n)n, we view σ as (σ0, . . . , σ
n−1) where each σi ∈ 2n. We say that σ
occursinx atkifσi =xk+i|nfori= 0, . . . , n.
Fixσ∈(2n)n. LetY be the set of allx∈ CZ such that there is a largestk
such thatσoccurs inxatk. We will argue thatY is tame. Supposex∈X and kis maximal such that σ occurs inx at k. If n∈Z, then (nx)i(j) =xi−n(j).
Thusk−nis the largestisuch thatσoccurs innxati. ThusY isZ-invariant. Let s : X → X be the function s(x) = kx where k is maximal such that σ occurs inxatk. Thens(x) is the unique element of [x] where 0 is the largesti such thatσoccurs ati. ThussisE-invariant andY is tame.
Similarly, the set ofxsuch that there is a leastk such thatσoccurs inx at kis tame. By throwing out these tame Borel sets, we may restrict attention to a Borel setX that for allσ andx∈X, the set ofk such thatσoccurs inx at kis unbounded in both directions.
Ifσ∈(2n)n andm < nwe letσ|m= (σ0|m, . . . , σ
m−1|m). Fix<n a linear
order of (2n)n such that ifσ, τ ∈(2n)n andσ|m <
mτ|mfor somem < n, then
σ <nτ.
For x ∈ X let fn(x) be the <n-least element of (2n)n occuring inx. Our
assumptions on<n, insure thatfn(x)|m=fm(x) form < n. Definef :X → CN
by
f(x) = (y0, y1, . . .)
where fn(x) = (y0|n, y1|n, . . . , yn−1|n) for all n. Note that eachfn andf are
E-invariant.
We say thatg ∈ CN occursin x at kif xk+i =g(i) for all i∈N. LetY be
f(x) occurs inxatk. ThenY is Borel,E-invariant and the functions(x) =kx wherekis leastf(x) occurs atkis a Borel selector. ThusY is tame. LetW be the set of all x∈X such thatf(x) occurs inx at k for arbitrarily smallk. If x∈W, then the action ofZon [x] is periodic. Thus [x] is finite andW is tame. Throwing outY and W we may assume that f(x) does not occur inx for all x∈X.
Forx∈X andn∈Ndefine kx
0 = 0
k2xn+1 = the leastk such thatk > kx2n andf2n+1(x) occurs inx atk.
kx
2n+2 = the largestksuch thatk < k2xn andf2n+2(x) occurs inx atk.
Then
. . .≤kx
4 ≤k2x≤k0x< kx1 ≤k3x≤. . . . Sincef(x) does not occur inx,kx
2n→ ∞andk2xn+2→ −∞.
We make the usual identification between C and P(N) by identifying sets with their characteristic functions. Under this identification
AE0B⇔A4B is finite. Fix a bijection
p:N×(2<ω)<ω →N.
Forx∈X andn∈Nlet tx n =|knx+1−kxn|+ 1 and letrx n∈(2<ω)t x n be (σ0, . . . , σ tx n−1) where σi =xmin{kx n,kn+1}+1|n.
This looks more confusing then it is. Supposen is even. Then kx
n < knx+1 andrx
n is just the block of the matrixx where we look take rows 0, . . . , n−1
and columnskx
n tokxn+1. Let G(x) = {p(n, rx
n) :∈ N}. From G(x), and knowing k0x = 0, we can reconstruct the sequence (kx
i :i∈N) andx. ThusGis one-to-one.
SupposeG(x)4G(y) is finite. Then there is anmsuch that rx
n=ryn for all
n > m. It follows thaty is obtained by shiftingx. ThusxEy.
SupposexEy. There ism∈Z such thatxm+i =yi for alli∈N. Without
loss of generality assumem >0. Let n0 be least such that kx
2n0+1 > m. Since f(x) =f(y),
kx2n0+1=m+ky2n0+1. Thuskx
n =m+kyn for alln > n0. It follows thatG(x)4G(y) is finite.