JOURNAL OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 10, Number 1, January 1997, Pages 215–242
S 0894-0347(97)00221-X
THE CLASSIFICATION OF HYPERSMOOTH
BOREL EQUIVALENCE RELATIONS
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
This paper is a contribution to the study of Borel equivalence relations in standard Borel spaces, i.e., Polish spaces equipped with their Borel structure. A class of
such equivalence relations which has received particular attention is the class of hyperfinite Borel equivalence relations. These can be defined as the increasing unions
of sequences of Borel equivalence relations all of whose equivalence classes are finite
or, as it turns out, equivalently those induced by the orbits of a single Borel automorphism. Hyperfinite equivalence relations have been classified in [DJK], under
two notions of equivalence, Borel bi-reducibility, and Borel isomorphism.
An equivalence relation E on X is Borel reducible to an equivalence relation
F on Y if there is a Borel map f : X → Y with xEy ⇔ f (x)F f (y). We write
then E ≤ F . If E ≤ F and F ≤ E we say that E, F are Borel bi-reducible, in
symbols E ≈∗ F . When E ≈∗ F the quotient spaces X/E, Y /F have the same
“effective” or “definable” cardinality. We say that E, F are Borel isomorphic if
there exists a Borel bijection f : X → Y with xEy ⇔ f (x)F f (y). Below we denote
by E0 , Et the equivalence relations on the Cantor space 2N given by: xE0 y ⇔
∃n∀m ≥ n(xm = ym ), xEt y ⇔ ∃n∃k∀m(xn+m = yk+m ). We denote by ∆X the
equality relation on X, and finally we call E smooth if E ≤ ∆2N . This just means
that elements of X can be classified up to E-equivalence by concrete invariants
which are members of some Polish space.
It is shown now in [DJK] that up to Borel bi-reducibility there is exactly one
non-smooth hyperfinite Borel E, namely E0 , and up to Borel isomorphism there
are exactly countably many non-smooth hyperfinite aperiodic (i.e., having no finite
equivalence classes) Borel E, namely Et , E0 × ∆n (1 ≤ n ≤ ℵ0 ), E0 × ∆2N (where
∆n = ∆X , with card(X) = n, if 1 ≤ n ≤ ℵ0 ).
In this paper we investigate and classify the class of Borel equivalence relations which are the “continuous” analogs of the hyperfinite ones. We
S call a Borel
equivalence relation E hypersmooth if it can be written as E = n En , where
E0 ⊆ E1 ⊆ · · · is an increasing sequence of smooth Borel equivalence relations.
These have been also studied (in a measure theoretic context) in the Russian literature under the name tame equivalence relations. They include many interesting examples such as: The increasing union of a sequence of closed or even Gδ
equivalence relations (like for example the coset equivalence relation of a Polish
group modulo a subgroup, which is the increasing union of a sequence of closed
subgroups), the hyperfinite equivalence relations, the “tail” equivalence relations
Received by the editors September 1, 1994 and, in revised form, June 11, 1996.
1991 Mathematics Subject Classification. Primary 04A15, 03E15.
Key words and phrases. Borel equivalence relations, hypersmooth, dichotomy theorems.
The first author’s research was partially supported by NSF Grant DMS-9317509.
c
1997
American Mathematical Society
215
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216
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
E0 (U ), Et (U ) of a Borel map U : X → X given by xE0 (U )y ⇔ ∃n(U n (x) = U n (y))
and xEt (U )y ⇔ ∃n∃m(U n (x) = U m (y)), the equivalence relations induced by the
orbits of a Borel action of a Polish locally compact group which is compactly generated of polynomial growth (e.g., Rn ), the equivalence relation induced by the
composants of an indecomposable continuum, etc.
Denote by E1 the equivalence relation on (2N )N given by xE1 y ⇔ ∃n∀m ≥ n
(xm = ym ). This is the “continuous” analog of E0 and is clearly hypersmooth. It
is well-known that E0 < E1 (i.e., E0 ≤ E1 , but E1 6≤ E0 ) and it is easy to see
that E ≤ E1 for any Borel hypersmooth E. The main result in this paper is now
the following dichotomy, which was motivated by results in the measure theoretic
context, see [V], [VF], [VG].
Theorem 1. If E is a hypersmooth Borel equivalence relation, then exactly one of
the following holds:
(I) E ≤ E0 ;
(II) E1 ≤ E.
(Actually in (II) the reducing function can be taken to be injective, i.e., an
embedding.)
From this it follows that up to Borel bi-reducibility there are exactly two nonsmooth hypersmooth Borel equivalence relations, namely E0 and E1 . With some
further work one can obtain also results on classification up to Borel isomorphism.
For example, up to Borel isomorphism there are only two non-smooth hypersmooth
Borel E, satisfying some mild natural conditions, that have equivalence classes of
size 2ℵ0 , namely E0 × I2N and E1 (where I2N = 2N × 2N ).
Despite the fact that our main result involves only notions of classical descriptive
set theory, the proof makes heavy use of effective descriptive set theory, as was
the case with the proof of the Glimm-Effros type dichotomy for Borel equivalence
relations proved in [HKL].
Although the dichotomy expressed in Theorem 1 is of a “local” nature, as it refers
only to hypersmooth Borel equivalence relations, it turns out surprisingly to have
also global consequences concerning the structure of arbitrary Borel equivalence
relations. Consider the partial (pre-)order ≤ on Borel equivalence relations. A
node is a Borel equivalence relation E such that for any Borel F, E ≤ F or F ≤ E,
i.e., E is comparable to any Borel equivalence relation. It is trivial that each ∆n
(n = 1, 2, . . . ) is a node and by Silver’s Theorem in [S], which implies that for any
Borel E either E ≤ ∆ℵ0 or ∆2N ≤ E, we have that ∆ℵ0 , ∆2N are also nodes. We
now have:
Theorem 2. The only nodes in the partial order ≤ on Borel equivalence relations
are ∆n (1 ≤ n ≤ ℵ0 ), ∆2N , and E0 .
This has the following immediate implication. Say that a pair of Borel equivalence relations (E, E ∗ ) with E < E ∗ has the dichotomy property if for any
Borel equivalence relation F we have F ≤ E or E ∗ ≤ F . Clearly (∆n , ∆n+1 ),
n = 1, 2, . . . , have this property. By Silver’s Theorem so does (∆ℵ0 , ∆2N ), and by
the result in [HKL] the same holds for (∆2N , E0 ). It follows from Theorem 2 that
these are the only such pairs, i.e., except for the trivial case of (∆n , ∆n+1 ), the only
global dichotomy theorems for Borel equivalence relations are Silver’s Theorem and
the general Glimm-Effros Dichotomy established in [HKL].
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
217
The paper is organized as follows: Section 0 contains preliminaries on descriptive set theory and equivalence relations. Section 1 discusses the basic properties
of hypersmooth relations and several examples. In Section 2 we prove the main
theorem. Section 3 contains consequences concerning isomorphism classifications.
In Section 4 we discuss results and examples relating to the possibility of reducing
E1 to other Borel equivalence relations. Finally, Section 5 contains the “global”
consequences of our main results mentioned above.
0. Preliminaries
A) A standard Borel space is a set X equipped with a σ-algebra S such that
for some Polish (i.e., separable completely metrizable) topology τ on X, S in the
class of Borel sets of τ . We call the members of S the Borel sets in X. Every
uncountable standard Borel space is Borel isomorphic to the Baire space N = NN
and to the Cantor space C = 2N .
We use the customary notation and terminology concerning descriptive set theory, see, e.g., [Mo]. In particular Σ11 denotes the class of analytic sets, Π11 the class
of co-analytic sets and ∆11 the class of bi-analytic sets, i.e., these which are both
analytic and co-analytic. By Souslin’s Theorem the bi-analytic sets are exactly the
Borel sets.
The use of effective descriptive set theory is crucial for the proof of our main
result. Again we use standard terminology and notation as in [Mo]. Thus Σ11 , Π11 , ∆11
denote resp. the classes of effectively analytic, co-analytic and bi-analytic sets. We
denote by ω1x the first ordinal not recursive in x and by ω1CK the first non-recursive
ordinal.
The results from (both classical and effective) descriptive set theory that we
will use can be found in [Mo], and in [HKL] in regards to the Gandy-Harrington
topology, with the exception of two reflection theorems that we will now state.
Their proofs can be found in [HMS], [K3].
0.1. First Reflection Theorem. Let Φ ⊆ P(N ) (= the power set of N ) be Π11
on Σ11 , i.e., for B ⊆ N × N in Σ11 , {y : Φ(By )} is in Π11 . Then if Φ(A) holds for
A ∈ Σ11 , there is A0 ⊇ A, A0 ∈ ∆11 such that Φ(A0 ) holds.
0.2. Burgess Reflection Theorem. Let R ⊆ N N × N n (n ∈ N) be Π11 and let
Φ ⊆ P(N ) be given by
Φ(A) ⇔ ∀x ∈ N N ∀y ∈ N n {∀n(xn ) ∈ A) &
∀i < n(yi 6∈ A) ⇒ R(x, y)}.
If A ⊆ N is Σ11 and Φ(A) holds, then there is A0 ⊇ A, A0 ∈ ∆11 such that Φ(A0 )
holds.
B) By a Polish group we mean a topological group whose topology is Polish.
If X is a standard Borel space, a Borel action of G on X is an action (g, x) 7→ g · x
of G on X which is Borel as a function from G × X into X.
C) If X is a set and E an equivalence relation on X, we denote by [x]E the
equivalence class of x, by X/E = {[x]E : x ∈ X} the quotient space of X by
E, and by [A]E = {x : ∃y ∈ A(xEy)} the E-saturation of A ⊆ X. If [A]E = A we
say that A is E-invariant.
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218
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
A transversal for E is a subset T ⊆ X which meets every equivalence class
in exactly one point. A selector for E is a map s : X → X with xEy ⇒ s(x) =
s(y)Ey.
We denote by ∆X , IX respectively the smallest and largest equivalence relations
on X, i.e., ∆X is equality on X and IX = X 2 .
If A ⊆ X, we denote by E|A the restriction of E to A, i.e., E|A = E ∩ A2 . If
F is also an equivalence relation on X, E ⊆ F means that E is a subequivalence
relation of F , i.e., xEy ⇒ xF y.
Suppose now E, F are equivalence relations on X, Y resp. A reduction of E
into F is a map f : X → Y with xEy ⇔ f (x)F f (y). Note that this induces an
injection f ∗ : X/E → Y /F given by f ∗ ([x]E ) = [f (x)]F . If f is 1-1 we call this an
embedding. If f is 1-1 and onto it is called an isomorphism of E, F . If f is
an embedding and f [X] = B is F -invariant, then we say that it is an invariant
embedding. It is clearly an isomorphism of E with F |B. Invariant embeddings of
E into F and F into E give rise, via the standard Schroeder-Bernstein argument,
to an isomorphism of E and F .
The product of E, F is the equivalence relation E × F on X × Y defined by
(x, y)E × F (x0 , y 0 ) ⇔ xEx0 & yF y 0 .
D) Assume now E, F are equivalence relations on standard Borel spaces X, Y .
We write
E ≤ F ⇔ ∃ a Borel reduction of E into F ,
E < F ⇔ E ≤ F & F 6≤ E;
E ≈∗ F ⇔ E ≤ F & F ≤ E;
E v F ⇔ ∃ a Borel embedding of E into F ;
E ≈ F ⇔ E v F & F v E;
E vi F ⇔ ∃ a Borel invariant embedding of E into F ;
E∼
= F ⇔ ∃ a Borel isomorphism of E, F .
Note that
E∼
= F ⇔ E vi F & F vi E.
Now let E be a Borel equivalence relation on a standard Borel space X. We call
E smooth if E has a countable Borel separating family, i.e., a sequence (An )
of Borel sets in X with
xEy ⇔ ∀n(x ∈ An ⇔ y ∈ An ).
This is easily equivalent to saying that E ≤ ∆X , for some standard Borel space X.
If E admits a Borel transversal (equivalently a Borel selector), then E is smooth.
The converse is in general false (see, e.g., [K3, 18.D]), but holds for most natural
examples.
The following dichotomy result was proved in [HKL]. Let E0 be the equivalence
relation on 2N given by
xE0 y ⇔ ∃n∀m ≥ n(xm = ym ).
Then for any Borel E, exactly one of the following holds: E is smooth or E0 v E.
In fact the following effective version is proved in [HKL]: If E is a ∆11 equivalence
relation on N , then exactly one of the following holds: E ≤ ∆2N via a ∆11 reduction
or E0 v E.
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
219
E) A Borel equivalence relation E on X is called finite, resp. countable, if
everySequivalence class [x]E is finite, resp. countable. It is called hyperfinite if
E = n En , with E0 ⊆ E1 ⊆ · · · an increasing sequence of finite Borel equivalence
relations. Clearly hyperfinite equivalence relations are countable. For more about
their structure, see [DJK]. For example, they can be characterized as those that
are induced by the orbits of a Borel action of Z on X, i.e., which are of the form
E = {(x, T n (x)) : n ∈ Z} with T a Borel automorphism
of X. Also they turn out
S
to be exactly those that can be written as E = n En , with E0 ⊆ E1 ⊆ · · · an
increasing sequence of smooth countable Borel equivalence relations.
1. Basic facts and examples
Let X be a standard Borel S
space and E a Borel equivalence relation on X. We
call E hypersmooth if E = n Fn , where F0 ⊆ F1 ⊆ F2 ⊆ · · · is an increasing
sequence of smooth Borel equivalence relations. Such equivalence relations are
called tame in the Russian literature; see [V], [VF], [VG].
Let us note some simple closure properties of hypersmooth relations.
Proposition 1.1. (i) If F is hypersmooth and E ≤ F , then E is hypersmooth;
(ii) If E is hypersmooth and A is Borel, E|A is hypersmooth;
(iii) If E, F are hypersmooth, so is E × F .
The proofs are straightforward.
The following is a basic open problem.
S
Problem 1.2. If E = n Fn , where F0 ⊆ F1 ⊆ · · · is an increasing sequence of
Borel hypersmooth equivalence relations, is E hypersmooth?
We next discuss examples:
0) It is well-known (see, e.g., [K1, 2.2]) that ever closed equivalence relation
is
S
smooth, and in [HKL] this is extended to Gδ equivalence relations. So if E = n En ,
E0 ⊆ E1 ⊆ · · · an increasing sequence of closed or even Gδ equivalence relations,
then E is hypersmooth. Conversely, it follows from [K3, 13.11] that if E is Borel
hypersmooth on the standard Borel space S
X, there is a Polish topology τ giving
the Borel structure of X, such that E = n En , with E0 ⊆ E1 ⊆ · · · closed in
(X 2 , τ 2 ) equivalence relations.
1) Every Borel hyperfinite equivalence relation (see [DJK]) is hypersmooth. In
fact, we view hypersmooth relations as “continuous” analogs of the hyperfinite ones.
2) For any standard Borel space Ω, let E0 (Ω), Et (Ω) be the following equivalence
relations on X = ΩN :
xE0 (Ω)y ⇔ ∃n∀m ≥ n(xm = ym ),
xEt (Ω)y ⇔ ∃n∃m∀k(xn+k = ym+k ).
It is clear that E0 (Ω) is hypersmooth, and it is shown in [DJK] that so is Et (Ω).
Put
E0 = E0 (2),
E1 = E0 (2N ).
3) We can generalize the examples in 2) as follows:
Let X be a standard Borel space and U : X → X a Borel map. Put
xE0 (U )y ⇔ ∃n(U n (x) = U n (y)),
xEt (U )y ⇔ ∃n∃m(U n (x) = U m (y)).
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220
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
Then E0 (U ), Et (U ) are hypersmooth (see [DJK]). If we take X = ΩN and U ((xn )) =
(xn+1 ), the shift on ΩN , we obtain the examples in 2).
4) Let G be a Polish group and H ⊆ G a subgroup. Let G/H = {xH : x ∈ G}
be the (left) coset space of H in G and put
xEH y ⇔ xH = yH
for the associated equivalence relation. If H is closed, then it is well-known that
EH is smooth, in fact has a Borel transversal. Conversely (see [Mi]), if H is Borel
and EH is smooth,
S then H is closed.
If now H = n Hn , with H0 ⊆ H1 ⊆ · · · an increasing sequence of closed
subgroups of G, then EH is clearly hypersmooth. Both E0 , E1 are of this form.
n
N
For E0 , we take G = ZN
2 , Hn = Z2 (viewed as a subgroup of Z2 by identifying
N
(x1 , . . . , xn ) ∈ Z2 with (x1 , x2 , . . . , xn , 0, 0, . . . )). For E1 let G = TN (T the unit
circle), Hn = Tn . (This does not give literally E1 , which lives on 2N , but a Borel
isomorphic copy of it.)
5) If G is a Polish locally compact group and (g, x) 7→ g · x a Borel action of G
on X, we denote by EG the (Borel) equivalence relation induced by the orbits of
this action, i.e.,
xEG y ⇔ ∃g ∈ G(g · x = y).
It is shown in [W] and [K1] that ER ≤ E0 , so ER is hypersmooth. Thus the
orbit equivalence relation of a flow (i.e., an R-action) is hypersmooth. This was
extended in [JKL] to show that EG ≤ E0 for any G which is compactly generated
of polynomial growth (e.g., Rn ); thus all such EG are hypersmooth.
6) The following interesting example was discovered recently by Solecki: Let X
be a continuum (i.e., a compact connected metric space). It is called indecomposable if it is not the union of two proper subcontinua. For any indecomposable
continuum X and x ∈ X, the composant of x is the union of all proper subcontinua containing x. The composants form a partition of X (into 2ℵ0 pieces), and let
us denote by EX the corresponding equivalenceSrelation. By a result of Rogers [R],
EX is Fσ . Solecki has in fact shown that E = n En , E0 ⊆ E1 ⊆ · · · an increasing
sequence of closed equivalence relations, so E is hypersmooth.
The equivalence relation E1 is universal among hypersmooth Borel equivalence
relations.
Proposition 1.3. Let E be a hypersmooth Borel equivalence relation. Then E v
E1 .
S
Proof. Let E = n Fn , with Fn an increasing sequence of smooth Borel equivalence
relations on X. Let fn : X → 2N be Borel with xFn y ⇔ fn (x) = fn (y) and assume
that F0 = ∆X , so that f0 is injective. Define f : X → (2N )N by
f (x) = (f0 (x), f1 (x), · · · ).
Then f is Borel injective and xEy ⇔ f (x)E1 f (y), so E v E1 .
The universal relation E1 also has the following important property which has
been known for some time (see, e.g., [FR], [K1, §5]).
Proposition 1.4. If F is a countable Borel equivalence relation, then E1 6≤ F . In
particular, E0 < E1 .
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
221
In fact 1.4 is also a consequence of the following stronger result, which also has
other applications.
Theorem 1.5. Let X be a standard Borel space, F a countable Borel equivalence
relation on X and f : (2N )N → X a Borel map such that xE1 y ⇒ f (x)F f (y). Then
there are (xn ), (yn ) ∈ (2N )N with n 7→ xn , n 7→ yn injective, xn 6= ym for all n, m
and f ((xn )) = f ((yn )).
Proof. We can identify (2N )N with 2N×N . It has the usual product topology, whose
basic nbhds are given by Np = {x ∈ 2N×N : x|(m × n) = p}, where p ∈ 2m×n ,
m, n ∈ N. Similarly identity (2N )m with 2m×N with the product topology, whose
(m)
basic nbhds are Np = {x ∈ 2m×N : x|(m × n) = p}, for p ∈ 2m×n , n ∈ N. We call
m×n
p∈2
(m, n ∈ N) conditions.
We use below the following general notation:
∀∗ xP (x) means “P (x) holds on a comeager set”,
∀+ xP (x) means “P (x) holds on a non-meager set”,
and for U open,
∀∗ x ∈ U P (x) means “P holds on a comeager in U set”,
∀+ x ∈ U P (x) means “P holds on a non-meager in U set”.
In this notation, the Kuratowski-Ulam Theorem asserts that if P has the property of Baire, then
∀∗ (x, y)P (x, y) ⇔ ∀∗ x∀∗ yP (x, y) ⇔ ∀∗ y∀∗ xP (x, y).
Assume now f is as in the theorem.
Then f is Baire measurable, so f is conT
tinuous on a dense Gδ set G = n Gn ⊆ (2N )N , where the Gn are open, dense and
decreasing.
We will construct inductively for n ∈ N :
1) conditions pn , qn ∈ 2(ln ×kn ) , with ln , kn strictly increasing;
2) xi , yi ∈ 2N for i < ln , such that i 7→ xi , i 7→ yi are injective, xi 6= yj , ∀j ≤ i,
yi 6= xj , ∀j ≤ i and pn = (x0 , . . . , xln −1 )|kn , qn = (y0 , . . . , yln −1 )|kn ,
which moreover satisfy:
(a) Npn , Nqn ⊆ Gn ;
(b) ∀∗ α ∈ (2N )N ((x0 , . . . , xln −1 )b α ∈ G & (y0 , . . . , yln −1 )bα ∈ G);
(c) ∀+ α ∈ (2N )N (f ((x0 , . . . , xln −1 )bα) = f ((y0 , . . . , yln −1 )bα)).
We will write below xk for (x0 , . . . , xk ) and similarly for the y’s.
Assuming this can be done, by (b), (c) we can find {αn } such that
xln −1 bαn , y ln−1 bαn ∈ G
and
f (xln −1 bαn ) = f (yln −1 bαn ).
If x = (xn ), y = (yn ), by (a) we have x, y ∈ G and since xln −1 bαn → x,
yln −1 bαn → y and xln −1 bαn , y ln −1 bαn ∈ G, we have f (x) = f (y) by continuity.
But also n 7→ xn , n 7→ yn are injective and xn 6= ym for all n, m, so we are done.
To show that this construction is possible, we use the following lemma:
Lemma. Let p ∈ 2m×n , ϕ a Borel function defined on a comeager in Np set, such
that on its domain
xE1 y ⇒ ϕ(x)F ϕ(y).
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222
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
0
Then we can find q ∈ 2m×n with q ⊇ p and a condition r such that
∀∗ δ ∈ Nq(m) ∀∗ ε ∈ Nq(m) ∀∗ x ∈ Nr (ϕ(δbx) = ϕ(εbx)).
Proof. Fix x ∈ (2N )N . Define a partial function ξx : (2N )m → X by ξx (δ) = ϕ(δbx),
for δ ∈ (2N )m such that ϕ(δbx) is defined. By the Kuratowski-Ulam Theorem ξx is
(m)
defined on a comeager in Np set of δ’s, for a comeager in (2N )N set of x’s.
Since for δ, ε ∈ (2N )m , δbxE1 εbx, the image of ξx is contained in some F equivalence class, so is countable. Thus we can find some qx ⊇ p, qx ∈ 2m×nx ,
(m)
such that ξx is constant on a comeager set in Nqx . Then find conditions r and
m×n0
q∈2
with q ⊇ p such that on a comeager in Nr set of x’s, qx = q. Then we
have
∀∗ x ∈ Nr ∀∗ δ ∈ Nq(m) ∀∗ ε ∈ Nq(m) (ϕ(δbx) = ϕ(εbx)),
and so, by Kuratowski-Ulam,
∀∗ δ ∈ N1
(m) ∗
∀ ε ∈ Nq(m) ∀∗ x ∈ Nr (ϕ(δbx) = ϕ(εbx)).
We now construct the pn , qn , xn , yn . Assume the construction has been completed up to n. By (b), (c), find a condition p0 such that
∀∗ α ∈ Np0 [f (xln −1 bα) = f (yln −1 bα) & xln −1 bα, y ln −1 bα ∈ G].
0
Fix such an α. As xln −1 bα, yln −1 bα ∈ G, we can find conditions p0n+1 , qn+1
∈
0
l0n+1 ×kn+1
0
0
0
0
0
0
2
with ln+1 > ln , kn+1 > kn and pn+1 ⊇ pn t p , qn+1 ⊇ qn t p (where
pn t p0 |(ln × kn ) = pn and pn t p0 (ln + i, j) = p0 (i, j) and similarly for qn t p0 ),
0
0
p0n+1 ⊆ xln −1 bα, qn+1
⊆ yln −1 bα, Np0n+1 , Nqn+1
⊆ Gn+1 . Notice that
0
0
0
0
0
0
0
p0n+1 |([ln , ln+1
) × kn+1
) = qn+1
|([ln , ln+1
) × kn+1
) = α|([0, ln+1
− ln ) × kn+1
).
0
0
Define p ∈ 2(ln+1 −ln )×kn+1 by p(i, j) = p0n+1 (ln + i, j). Thus p ⊇ p0 .
Use the lemma for this p and ϕ(x) = f (xln −1 bx) = f (yln −1 bx), to obtain q and
r as in the lemma.
0
We have then, if m = ln+1
− ln :
∀∗ δ ∈ Nq(m) ∀∗ ε ∈ Nq(m) ∀∗ x ∈ Nr
[f (xln −1 bεbx) = f (yln −1 bδbx) = f (xln −1 bδbx) = f (yln −1 bεbx)],
and by (b)
∀∗ δ ∈ Nq(m) ∀∗ ε ∈ Nq(m) ∀∗ x ∈ (2N )N
(xln −1 bεbx ∈ G & xln −1 bδbx ∈ G
& y ln −1 bεbx ∈ G & y ln −1 bδbx ∈ G).
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
(m)
Since the set of all (δ, ε) ∈ Nq
conditions:
δi = xj
δi = δj
εi = yj
εi = εj
δi = εj
ε i = δj
δi = yj
εi = xj
(m)
(m)
×Nq
223
which satisfy at least one of the following
for some i < m, j < ln ,
for some j 6= i < m,
for some i < m, j < ln ,
for some j 6= i < m,
for some j ≤ i < m,
for some j ≤ i < m,
for some i < m, j < ln ,
for some i < m, j < ln ,
(m)
is meager in Nq × Nq , we can find xln , . . . , xln +m−1 , yln , . . . , yln +m−1 with
i 7→ xi , i 7→ yi injective for i < ln + m and xi 6= yj , yi 6= xj for j ≤ i < ln + m and
such that for ln+1 = ln + m
∀∗ α ∈ Nr [f (xln+1 −1 bα) = f (yln+1 −1 bα)]
and
∀∗ α ∈ (2N )N [xln+1 −1 bα ∈ G & y ln+1 −1 bα ∈ G].
Finally, choose kn+1 large enough.
2. The main theorem
Our main result is that up to Borel bireducibility, E0 and E1 are the only nonsmooth Borel hypersmooth equivalence relations. More precisely we have the following:
Theorem 2.1. Let E be a hypersmooth Borel equivalence relation. Then exactly
one of the following holds:
(I) E ≤ E0 ;
(II) E1 v E.
Since by [HKL], if E is non-smooth Borel, then E0 v E, it follows that for any
hypersmooth Borel E, exactly one of the following holds:
(i) E is smooth;
(ii) E ≈∗ E0 ;
(iii) E ≈ E1 .
The proof of 2.1 uses the methods of effective descriptive set theory. In fact we
prove the following effective result.
Theorem 2.2. Let {Fn } be a sequence of equivalence relations on the BaireSspace
N such that F0 ⊆ F1 ⊆ · · · and each Fn is Π01 , uniformly on n. Let E = n Fn .
Then exactly one of the following holds:
(I) E ≤ E0 via a ∆11 map;
(II) E1 v E via a continuous embedding.
Before we prove 2.2, let us argue that it, and its obvious relativization, implies
2.1. Indeed by the relativized version of 2.2, if F0 ⊆ F1S⊆ · · · is an increasing
sequence of closed equivalence relations in N and E = n Fn , then either E ≤
E0 or E1 v E (via a continuous function). Assume now that E is an arbitrary
hypersmooth Borel equivalence relation on the standard Borel space X. Then (see
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224
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
[K3, 13.11]) there is a Polish topology on X generating its
S Borel structure and
closed relations in this topology F0 ⊆ F1 ⊆ · · · , with E = n Fn . Let C ⊆ N be
closed and π : C → X a continuous injective map from C onto X.
Define Fn0 on N by
xFn0 y ⇔ (x = y) or [x, y ∈ C & π(x)Fn π(y)].
Then S
F00 ⊆ F10 ⊆ · · · and each Fn0 is a closed equivalence relation on N . Let
E 0 = n Fn0 . Then E 0 ≤ E0 or E1 v E 0 (via a continuous function). If E 0 ≤ E0
via f , then E ≤ E0 via f ◦ π −1 . If E1 v E 0 via a continuous embedding g, then
g[(2N )N ] ⊆ C, so π ◦ g is a continuous embedding of (2N )N into X, which witnesses
that E1 v E.
Proof of 2.2. For each n < m, put
[
Yn,m = {A ∈ Σ11 : A2 ∩ Fm ⊆ Fn }.
By the First Reflection Theorem, if A ∈ Σ11 and A2 ∩ Fm ⊆ Fn , there is B ∈ ∆11 ,
B ⊇ A with B 2 ∩ Fm ⊆ Fn , so
[
Yn,m = {A ∈ ∆11 : A2 ∩ Fm ⊆ Fn }
and in particular Yn,m is Π11 , uniformly in n, m. Put
\ [
Xn,m = N \Yn,m ,
X∗ =
Xn,m .
n m>n
Thus X ∗ ∈ Σ11 .
Case I. X ∗ = ∅.
S T
We show then that (I) holds. Since N = n m>n Yn,m , by effective reduction
we
disjoint sequence {Sn } of ∆11 sets, uniformly in n, such that
S can find a pairwise T
n Sn = N and Sn ⊆
m>n Yn,m , i.e.,
∀x ∈ Sn ∀m > n∃A ∈ ∆11 (x ∈ A & A2 ∩ Fm ⊆ Fn ).
For equivalence relations R ⊆ S, we say that S/R is countable, if every Sequivalence class contains only countably many R-equivalence classes. We claim
now that (E|Sn )/(Fn |Sn ) is countable: It is clearly enough to show that
(Fm |Sn )/(Fn |Sn ) is countable for any m > n. But if C is an Fm |Sn -equivalence
class and D ⊆ C an Fn |Sn -equivalence class, then there is a ∆11 nonempty set A
such that A ∩ C ⊆ D, so clearly there are only countably many such D in C.
Now define a new equivalence relation F00 on N by
xF00 y ⇔ ∃n(x, y ∈ Sn & xFn y).
Clearly F00 ⊆ E, F00 is ∆11 and smooth. Moreover, E/F00 is countable. Put also,
for n > 0,
[
xFn0 y ⇔ (x, y ∈
Sn0 & xFn y) or
n0 ≤n
∃m > n(x, y ∈ Sm & xFm y).
Then Fn0 is smooth ∆11 , uniformly on n, F00 ⊆ F10 ⊆ · · · and E =
Now let ϕ : N → 2N be ∆11 such that
xF00 y ⇔ ϕ(x) = ϕ(y).
Put ϕ[N ] = A ⊆ 2N , so that A ∈ Σ11 .
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S
n
Fn0 .
HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
(n)
225
(n)
Let {Ck }n,k∈N be a uniformly ∆11 family of sets such that for each n, {Ck }k∈N
is a separating family for Fn0 . Define the equivalence relation F n on A and subsets
(n)
C k of A by
⇔ ∃x∃y[ϕ(x) = α & ϕ(y) = β & xFn0 y],
αF n β
(n)
(n)
α ∈ C k ⇔ ∃x[ϕ(x) = α & x ∈ Ck ].
Since F00 ⊆ Fn0 and each Ck
(n)
is Fn0 -invariant, we also have for α, β ∈ A :
⇔ ∀x∀y[ϕ(x) = α & ϕ(y) = β ⇒ xFn0 y],
αF n β
(n)
(n)
α ∈ C k ⇔ ∀x[ϕ(x) = α ⇒ x ∈ Ck ].
(n)
(n)
Thus F n , {C k }Sare uniformly ∆11 on A. Clearly {C k } is a separating family for
F n . Also if E = n F n , then E is a countable ∆11 equivalence relation on A.
(n)
(n)
(n)
Now let {C k } be uniformly ∆11 such that C k = A ∩ C k . Consider then the
e ⊆ 2N and Fe = {Fen }n∈N , Fe ⊆ N × 2N × 2N :
statements (1)–(6) below, in variables A
e
(1) Fen is an equivalence relation on A;
1
e
e
e
(2) ∀x ∈ A∀y ∈ A(xFn y ⇒ y ∈ ∆1 (x));
(3) Fen ⊆ Fen+1 ;
(n)
e is Fen -invariant];
(4) ∀n∀k[C k ∩ A
(n)
(n)
e&y∈A
e & ¬xFen y ⇒ ∃k(x ∈ C k & y 6∈ C k )];
(5) ∀x∀y[x ∈ A
e
e
e ⇒ xF n y].
(6) ∀x∀y[xFn y & x ∈ A & y ∈ A
These are clearly satisfied by A and F = {F n }n∈N . They also have the form for
applying the Burgess Reflection Theorem. (It is understood here that in (6) we use
a Π11 definition for F n .) So we can find ∆11 sets A∗ ⊇ A, F ∗ = {Fn }, Fn∗ ⊇ F n , still
satisfying (1)–(6). By (1) and (3), Fn∗ are increasing
relations on A∗ ,
S equivalence
∗
∗
∗
while each Fn is countable by (2), thus so is E = n Fn . Also (4), (5) imply that
(n)
{C k ∩ A}k∈N is a separating family for Fn∗ , so Fn∗ is smooth. Finally, (6) shows
that E ∗ |A = E.
Since E ∗ is hypersmooth and countable, by [DJK] E ∗ can be reduced by a ∆11
function to E0 . Since ϕ reduces E to E ∗ as well, we have that ψ ◦ ϕ is a ∆11
reduction of E to E0 .
This completes Case I.
Case II. X ∗ 6= ∅.
We will show then that (II) holds. Since X ∗ is nonempty Σ11 , the set
X = X ∗ ∩ {x ∈ N : ω1x = ω1CK }
is also Σ11 and nonempty.
Note that the Gandy-Harrington topology when restricted to {x ∈ N : ω1x =
CK
ω1 } has a clopen basis (since the intersection of a Π11 set with this set is a countable
union of ∆11 sets), so regular; thus by the Choquet Criterion (see, e.g., [K3, 8.18]) it
is Polish. Denote the Gandy-Harrington topology restricted to X by τ . Fix also a
complete metric d for τ on X. We can of course assume that d ≥ δ, where δ is the
ordinary metric on N . We will embed E1 into E|X (continuously for the ordinary
topology on N .)
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226
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
Fix the canonical bijection h i of N2 with N given by the Cantor diagonal enumeration, i.e.,
(n + k)(n + k + 1)
hn, ki =
+ k.
2
For s ∈ 2p , where p = hn, ki, and j ∈ N we let sj (i) = s(hj, ii), provided hj, ii < p.
This associates to s a sequence hsj : j ∈ Ni of finite sequences, which are eventually
∅. Put
L(p) = min{j : sj = ∅} = min{j : hj, 0i ≥ p}.
Define also for s, t ∈ 2p , j ≤ L(p),
s ∼j t ⇔ ∀j 0 ≥ j(sj0 = tj0 ).
Then ∼j is an equivalence relation on 2p and ∼0 ⊆∼1 ⊆ · · · ⊆∼L(p) . Moreover, ∼0
is equality and ∼L(p) = 2p × 2p .
For α ∈ 2N , let also αm (k) = α(hm, ki). Then, identifying α ∈ 2N with
{αm }m∈N ∈ (2N )N , we have that
αE1 β ⇔ ∃n∀m ≥ n(αm = βm )
⇔ ∃n∀p(n ≤ L(p) ⇒ α|p ∼n β|p).
For an equivalence relation E (on some set S) and sets A, B (⊆ S) let
AEB ⇔ ∀x ∈ A∃y ∈ B(xEy) & ∀y ∈ B∃x ∈ A(xEy).
Note that AEB ⇔ [A]E = [B]E , and this is an equivalence relation too.
We first claim that in order to embed E1 into E|X (continuously), it is enough
to build a family {Us }s∈2<N and a strictly increasing function N : N → N satisfying:
τ
(i) Us is a nonempty Σ11 subset of X, t % s → U t ⊂ Us , Usb0 ∩ Usb1 = ∅ and
d(Usbi ) ≤ 2−lh(s) .
(ii) If s, t ∈ 2p , p = hn, ki and s ∼n t, then (Usb0 × Utb1 ) ∩ Fn+1 = ∅.
(iii) If s, t ∈ 2p , p = hn, ki, j ≤ L(p) and s ∼j t, then Us FN (j) Ut .
T
Indeed, assume this can be done. For α ∈ 2N , define f (α) by {f (α)} = n Uα|n .
This is clearly well-defined and 1-1 by (i). It is also continuous for the ordinary
topology on N as d ≥ δ. We argue that f embeds E1 into E.
Suppose that αE1 β, say αm = βm for m ≥ n. Then for p such that n ≤ L(p),
α|p ∼n β|p. By (iii) Uα|p FN (n) Uβ|p , so there are αp ∈ Uα|p , βp ∈ Uβ|p with
αp FN (n) βp . Since αp → f (α), βp → f (β) in the ordinary topology on N and FN (n)
is closed in that topology, f (α)FN (n) f (β), so f (α)Ef (β).
Assume now (α, β) 6∈ E1 . Let p = hn, ki be such that α(p) 6= β(p). Let n0 be
smallest with α|(p + 1) ∼n0 +1 β|(p + 1). Clearly n0 ≥ n. Let k0 be least with
α(p0 ) 6= β(p0 ) for p0 = hn0 , k0 i. Thus p0 ≤ p. Now α|p0 ∼n0 β|p0 , so by (ii)
(Uα|(p0 +1) × Uβ|(p0 +1) ) ∩ Fn0 +1 = ∅ and thus (Uα|(p0 +1) × Uβ|(p0 +1) ) ∩ Fn = ∅, so
¬f (α)Fn f (β). Since this happens for infinitely many n, ¬f (α)Ef (β).
For the rest of the proof, let us introduce the following terminology:
Given n < m and ∅ 6= A ∈ Σ11 , A ⊆ X, we will say that Fn is meager in Fm
on A if Fn is meager in Fm on A2 with the τ × τ -topology.
Since Fn , Fm are both closed in the product of the ordinary topology and thus in
the (τ × τ )-topology, this means that there are no nonempty Σ11 subsets C, D ⊆ A
with ∅ 6= (C × D) ∩ Fm ⊆ Fn . We claim that this is equivalent to saying that there
is no nonempty Σ11 set B ⊆ A with B 2 ∩ Fm ⊆ Fn . Indeed, if such C, D exist, we
can assume first that CFm D by replacing them by C ∩ [D]Fm , D ∩ [C]Fm . We claim
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
227
then that C 2 ∩ Fm ⊆ Fn . Indeed, if x, y ∈ C and xFm y, find z ∈ D with xFm z,
hence xFn z. Also yFm z, hence yFn z and thus xFn y.
We can use this argument to prove immediately the following basic lemma.
Lemma 2.3. Let A ⊆ X be a nonempty Σ11 set and x1 , . . . , xk ∈ A. For any n ∈ N
there is m > n and a Σ11 set A∗ ⊆ A such that x1 , . . . , xk ∈ A∗ and Fn is meager
in Fm on A∗ .
T S
S
Proof. Recall that X ⊆ X ∗ = n m>n Xm . So A ⊆ m>n Xm,n . So find m > n
with x1 , . . . , xk ∈ Xm,n . This can be done as {Xn,m }m>n is increasing. Put
A∗ = A ∩ Xn,m . If Fn is not meager in Fm on A∗ , then by the preceding argument,
there is a nonempty Σ11 set B ⊆ A∗ with B 2 ∩ Fm ⊆ Fn , so by the definition of
Yn,m , B ⊆ Yn,m , so B = ∅, a contradiction.
In order to construct the family {Us } and the function N satisfying (i)–(iii)
above, we will impose the following requirements:
• R(0): U∅ will be a nonempty Σ11 subset of X and N (0) > 0 will be such that
F0 is meager in FN (0) on U∅ .
τ
• R(1) (as (i) before): For s ∈ 2p , i = 0 or 1, U sbi ⊆ Us , Usb0 ∩ Usb1 = ∅ and
−p
d(Usbi ) ≤ 2 .
• R(2) (as (ii) before): For s, t ∈ 2p , p = hn, ki, if s ∼n t, then (Usb0 × Utb1 ) ∩
Fn+1 = ∅.
S
• R(3): For j ≤ L(p + 1), Fj is meager in FN (j) on s∈2p+1 Us .
• R(4): (a) For s, t ∈ 2p , j ≤ L(p), i = 0 or 1:
s ∼j t ⇒ Usbi FN (j) Utbi ;
(b) for s ∈ 2p , p = hn, ki,
Usb0 FN (n+1) Usb1 .
We claim that these are enough, i.e., they imply (i)–(iii).
Clearly R(0)–(2)⇒(i), (ii). We will verify that R(4) ⇒(iii):
Assume R(4). We have to show that for all p, s, t ∈ 2p , j ≤ L(p), if s ∼j t, then
Us FN (j) Ut . This is clear for p = 0. Suppose it holds for p = hn, ki and consider
p + 1, s, t ∈ 2p+1 , say s = s bi, t = tbi0 with s, t ∈ 2p , j ≤ L(p + 1). First let
i = i0 : If j ≤ L(p), then s bi ∼j tbi implies s ∼j t, so by R(4)(a) Us FN (j) Ut . If
j = L(p+1), since s ∼L(p) t, by R(4)(a) again we have Us FN (L(p)) Ut , so Us FN (j) Ut ,
since L(p + 1) ≥ L(p) and N is increasing. Consider now the case i 6= i0 , say
i = 0, i0 = 1. If s b0 ∼j tb1, then s b1 ∼j tb1, so, by the first case (i = i0 ), we
have Us b1 FN (j) Utb1 . By R(4)(b) also, Us b0 FN (n+1) Us b1 . But s b0, tb1 differ at
p = hn, ki, so clearly j ≥ n + 1 and, since N is increasing, N (j) ≥ N (n + 1), so
Us b0 FN (j) Us b1 . By transitivity Us b0 FN (j) Utb1 , and we are done.
We construct now, by induction on p ∈ N, {Us }s∈2p and N (j), for j ≤ L(p),
satisfying R(0)–(4).
For p = 0, we choose a nonempty Σ11 set U∅ ⊆ X and N (0) > 0, so that F0 is
meager in FN (0) on U∅ . This can be done by Lemma 2.3.
For the inductive step we will need some new concepts and a few combinatorial
lemmas.
A tree is a finite undirected graph which is connected and has no loops. A
labelled tree is a tree T together with an assignment (s, t) 7→ n(s, t) which gives
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228
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
n
for each edge (s, t) of T a natural number n(s, t) (its label). We usually write s — t
if n(s, t) = n.
By a tree structure we mean a triple (T, U, M ), where
(i) T is a labelled tree;
(ii) U is a map assigning to each vertex s of T a nonempty Σ11 set U (s) = Us ⊆ X;
(iii) M is a mapping from the set of labels of T into N.
A tree structure (T, U, M ) is good if moreover
n
(iv) s — t ⇒ Us FM(n) Ut .
A tree structure (T, U 0 , M ) refines (T, U, M ) if Us0 ⊆ Us for every vertex s of T .
Lemma 2.4. (i) Let U, V be Σ11 nonempty sets and x ∈ U , y ∈ V . If F is a Σ11
equivalence relation and xF y, then there are nonempty Σ11 sets U 0 ⊆ U , V ⊆ V
with x ∈ U 0 , y ∈ V 0 and U 0 F V 0 .
(ii) If U, V are nonempty Σ11 sets, F a Σ11 equivalence relation and U F V , then
for any nonempty Σ11 set A ⊆ U , we can find a nonempty Σ11 set B ⊆ V with AF B.
Moreover, if x ∈ A, y ∈ V and xF y then y ∈ B.
Proof. (i) Let U 0 = U ∩ [V ]F and V 0 = V ∩ [U ]F .
(ii) Let B = V ∩ [A]F .
Lemma 2.5. Let (T, U, M ) be a good tree structure. Let s0 be a vertex of T and A
a nonempty Σ11 subset of Us0 . Then there is a refinement (T, U 0 , M ) of (T, U, M )
which is good, such that Us0 0 = A. Moreover, if xs ∈ Us for all vertices and
n
s — t ⇒ xs FM(n) xt , then if xs0 ∈ A we can insure that xs ∈ Us0 , for all s.
Proof. Let l(s, t) be the distance function of T , i.e., the length of the unique path
from s to t. Let l(s) = l(s, s0 ). We will define Us0 by induction on l(s). For
l(s) = 0, i.e., s = s0 , we have Us0 0 = A. Assume now l(s) > 0 and let t be the vertex
following s on the unique path from s to s0 , so that l(t) = l(s) − 1. Thus Ut0 has
n
been defined. If s — t, then Us FM(n) Ut and, since Ut0 ⊆ Ut , we can find Us0 ⊆ Us so
0
0
that Us FM(n) Ut , by 2.4(ii).
Lemma 2.6. Let (T, U, M ) be a tree structure and xs ∈ Us for every vertex s of
T . If
n
s — t ⇒ xs FM(n) xt ,
then there is a refinement (T, U 0 , M ) of (T, U, M ) which is good and xs ∈ Us0 for all
s.
Proof. By induction on the cardinality of the set of vertices of T . Let s be a terminal
vertex of T , i.e., one which belongs to a unique edge. Let t be the other vertex of
this edge. If we delete s and this unique edge (but not t), we obtain a new tree T ∗
with one fewer vertex. Let (U ∗ , M ∗ ) be (U, M )|T ∗ . By the induction hypothesis,
0
0
there is a good refinement (T ∗ , U ∗ , M ∗ ) of (T ∗ , U ∗ , M ∗ ) satisfying xs∗ ∈ Us∗∗ for
0
n
s∗ 6= s. Applying 2.4(i) to FM(n) , where s — t, Us , Ut∗ and the points xs , xt , we
0
can find Us0 ⊆ Us , Ut0 ⊆ Ut∗ with xs ∈ Us0 , xt ∈ Ut0 and Us0 FM(n) Ut0 . Now apply 2.5
to (T ∗ , U ∗ , M ∗ ) with s0 = t, A = Ut0 , to define Us0 ∗ for all vertices s∗ of T ∗ . This
defines U 0 for all vertices of T .
Lemma 2.7. Let (T, U, M ) be a good tree structure. Let n ∈ N. Then there is
m > n and
(T, U 0 , M ) of (T, U, M ) which is good and Fn is meager in
S a refinement
0
Fm on s∈V Us , where V = set of vertices of T .
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
229
Proof. Since (T, U, M ) is good, if s0 is a fixed vertex of T , we can define by induction
n
on l(s, s0 ) a sequence
of points xs ∈ Us with s — t ⇒ xs FM(n) xt . By Lemma 2.3
S
applied to A = s∈V Us and the points {xs }s∈V , we can find m > n and A∗ ⊆ A
Σ11 such that Fn is meager in Fm on A∗ and xs ∈ A∗ , ∀s ∈ V . Put Us∗ = A∗ ∩ Us .
Then xs ∈ Us∗ and, by 2.6 appliedSto (T, U ∗ , M ), we can find a good refinement
(T, U 0 , M ) with xs ∈ Us0 . If A0 = s∈V Us0 , then Fn is meager in Fm on A0 , since
A0 ⊆ A∗ .
We now come to the final and key lemma. First we need a definition.
Let T be a labelled tree. Given n ∈ N, we say that two vertices s, t of T are
n-connected if all the labels in the path from s to t are ≤ n.
Lemma 2.8. Let (T, U, M ) be a good tree structure with M monotone. Let L be
the
in FN on
S largest label of T and n ≤ L. Let N be such that Fn+1 is meager
0
s∈V Us (V = the set of vertices of T ), where M (n) ≤ N ≤ M (n ) for any label
n0 > n (if such exists.) Then there are two refinements (T, U 0 , M ), (T, U 1 , M ) of
(T, U, M ) which are good and
(i) Us0 FN Us1 , for any s ∈ V ;
(ii) (Us0 × Ut1 ) ∩ Fn+1 = ∅, if s, t are n-connected.
Proof. Clearly n-connectedness is an equivalence relation on V , dividing it into
components which are subtrees of T . Enumerate these as C1 , . . . , CK .
We will consider first the case K = 1, i.e., n = L, in which case the requirement N ≤ M (n0 ), ∀n0 > n is vacuous. So we must have Us0 FN Us1 , ∀s ∈ V and
(Us0 × Ut1 ) ∩ Fn+1 = ∅, ∀s, t ∈ V .
Enumerate in a sequence (s1 , t1 ), . . . , (sp , tp ) the set V × V . We will define by
induction on 0 ≤ j ≤ p good tree structures (T, U i,j , M ) for i = 0, 1 such that
A) U i,0 = U , i ∈ {0, 1};
and for j + 1 ≤ p :
B) (T, U i,j+1 , M ) refines (T, U i,j , M );
C) (U 0,j+1 (sj+1 ) × U 1,j+1 (tj+1 )) ∩ Fn+1 = ∅;
D) U 0,j+1 (sj+1 )FN U 1,j+1 (tj+1 ).
If this can be done, put Us0 = U 0,p (s), Us1 = U 1,p (s). By C), if (s, t) =
(sj+1 , tj+1 ), we have Us0 × Ut1 = U 0,p (s) × U 1,p (t) ⊆ U 0,j+1 (sj+1 ) × U 1,j+1 (tj+1 ),
which is disjoint from Fn+1 . So (ii) is satisfied. For (i), notice that we have
Us0p FN Ut1p . Given any s ∈ V , there is a path from s to sp with labels ≤ L = n, so
by transitivity and the fact that M is monotone and M (n) ≤ N , we have Us0 FN Us0p .
Similarly, Us1 FN Ut1p , so Us0 FN Us1 .
For the inductive construction of U i,j , note that U i,0 is given. Assume U i,j is
given for both i = 0, 1 in order to construct U i,j+1 . Since Fn+1 is meager in FN
on (U 0,j (sj+1 ) ∪ U 1,j (tj+1 )), we can shrink U 0,j (sj+1 ), U 1,j (tj+1 ) to nonempty Σ11
sets A, B resp., so that (A × B) ∩ Fn+1 = ∅ but AFN B. (Notice that, by the
induction hypothesis, U 0,j (sj+1 )FN U 1,j (tj+1 ).) Then apply 2.5 to (T, U 0,j , M ),
A ⊆ U 0,j (sj+1 ) and (T, U 1,j , M ), B ⊆ U 1,j (tj+1 ) to obtain good refinements
(T, U 0,j+1 , M ), (T, U 1,j+1 , M ) resp., with U 0,j+1 (sj+1 ) = A, U 1,j+1 (tj+1 ) = B.
Consider finally the case when K > 1, i.e., n < L. We will define by induction
on 1 ≤ j ≤ K good tree structures (T, U 0,j , M ), (T, U 1,j , M ) such that:
A) U 0,0 = U 1,0 = U ;
B) (T, U i,j+1 , M ) refines (T, U i,j , M ) for i = 0, 1, j < K;
C) if s, t ∈ Cj0 , j 0 ≤ j, then U 0,j (s)FN U 1,j (t);
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230
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
D) if s, t ∈ Cj , then (U 0,j (s) × U 1,j (t)) ∩ Fn+1 = ∅;
E) if j 0 > j, then for s ∈ Cj0 , we have U 0,j (s) = U 1,j (s).
We then put Us0 = U 0,K (s), Us1 = U 1,K (s). This clearly works.
We are given U 0,0 , U 1,0 by A). Assume now U i,j has been defined for i = 0, 1.
We will define U i,j+1
Let C = Cj+1 . Then by E) U 0,j |C = U 1,j |C. Since Fn+1 is
S . i,j
meager in FN on {U (s) : s ∈ C}, we can apply the previous case (i.e., K = 1)
to define U 0,j+1 (s), U 1,j+1 (s) for s ∈ C, which are good refinements of U 0,j |C,
U 1,j |C resp., and satisfy C), D) for s, t ∈ C.
We can now use the same argument as in 2.5 to define U 0,j+1 (s), U 1,j+1 (s) for
s 6∈ C. For such an s there is a unique shortest path to some point in C of length
l(s, C). We define U i,j+1 (x) inductively on l(s, C): If s0 is the next vertex in the
shortest path from s to C, we can assume by induction that U i,j+1 (s0 ) has been
defined and we let
U i,j+1 (s) = U i,j (s) ∩ [U i,j+1 (s0 )]FM (k)
k
if s — s0 .
Clearly B) is satisfied, and so is D), for j + 1.
To prove C), we note that it is clear if s, t ∈ Cj+1 by construction. So assume
s, t ∈ Cj0 , j 0 ≤ j. Since (T, U i,j+1 , M ) are good and N ≥ M (n), we have, by
transitivity, U i,j+1 (s)FN U i,j+1 (t). So it is enough to show that
U 0,j+1 (s)FN U 1,j+1 (s), ∀s ∈ V.
So we prove, by induction on l(s, C), that U 0,j+1 (s)FN U 1,j+1 (s). We let l(s, C) = 0,
if s ∈ C. This is clear then for l(s, C) = 0, by construction. Else let s0 be as before,
k
so by the induction hypothesis, U 0,j+1 (s0 )FN U 1,j+1 (s0 ). If k ≤ n (where s — s0 ),
then
U 0,j+1 (s)FM(k) U 0,j+1 (s0 )FN U 1,j+1 (s0 )FM(k) U 0,j+1 (s0 ),
so we are done as M (k) ≤ M (n) ≤ N , by transitivity. Else k > n. Then let
x ∈ U 0,j+1 (s). Since U 0,j (s)FN U 1,j (s) (by C), E) for j), let y ∈ U 1,j (s) be such
that xFN y. Let also x0 ∈ U 0,j+1 (s0 ) with x0 FM(k) x and y 0 ∈ U 1,j+1 (s0 ) with x0 FN y 0 .
Then yFM(k) y 0 , as N ≤ M (k), so y ∈ U 1,j+1 (s) by definition. So, reversing also
the roles of U 0,j+1 (s) and U 1,j+1 (s), we have U 0,j+1 (s)FN U 1,j+1 (s).
Finally, we prove E), i.e.,
(∗)
s ∈ Cj0 , j 0 > j + 1 ⇒ U 0,j+1 (s) = U 1,j+1 (s).
k
This is again by induction on l(s, C). Let s0 be as before, s — s0 , and assume (∗)
holds for s0 . If s0 ∈ Cj0 , j 0 > j + 1, then we are clearly done, since
U 0,j+1 (s) = U 0,j (s) ∩ [U 0,j+1 (s0 )]FM (k)
= U 1,j (s) ∩ [U 1,j+1 (s0 )]FM (k)
(by the induction hypothesis for s0 and E) for j)
= U 1,j+1 (s).
Otherwise, s ∈ Cj0 , for j 0 ≤ j + 1, so that in particular k > n, and thus M (k) ≥ N .
Then, by C),
U 0,j+1 (s0 )FN U 1,j+1 (s0 ),
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
231
so
U 0,j+1 (s0 )FM(k) U 1,j+1 (s0 ),
i.e.,
[U 0,j+1 (s0 )]FM (k) = [U 1,j+1 (s0 )]FM (k) ,
and we are done as above.
We are now ready to proceed
construction of Us for s ∈ 2p+1 satisfying
S to the
p0
R(1)–(4), assuming Us , for s ∈ p0 ≤p 2 , are given satisfying R(1)–(4) for all p0 < p.
Lemma 2.9. Let A be a finite set, ∼0 ⊆ ∼1 ⊆ · · · ⊆ ∼k a sequence of equivalence
relations on A, with ∼0 = equality and ∼k = A × A. Then there is a labelled tree T
with set of vertices A and labels in the set {0, . . . , k} such that
a ∼j b ⇔ a, b are j-connected.
Proof. By induction on k. For k = 0, this is obvious. Assume it true for k = p.
Let k = p + 1. Pick a point ai ∈ Ci , where {Ci }qi=1 are the ∼p -equivalence classes.
For each Ci there is, by the induction hypothesis, a labelled tree Ti with set of
vertices Ci and labels {0, . . . , p} satisfying the above for ∼j |Ci , 0 ≤ j ≤ p. Define
T , with set of vertices A, by adding to the edges of the Ti ’s the edges (ai , ai+1 ) for
i = 1, . . . , q − 1 with label p + 1 = k. This clearly works.
Apply this lemma now to A = 2p and ∼0 , ∼1 , . . . , ∼L(p) . Call the resulting
labelled tree T . Consider the tree structure (T, U, M ), where U is as given by the
induction hypothesis and M (n) = N (n) for n ≤ L(p), which again is given by the
induction hypothesis. Note that M is monotone (in fact strictly increasing).
Note now that condition R(4) for p0 = p − 1 implies that (T, U, M ) is good:
0
Indeed, let p0 = hn0 , k 0 i. Let s, t ∈ 2p be such that s = sbi, t = tbi0 , s, t ∈ 2p . Let
j ≤ L(p) = L(p0 + 1) be such that s ∼j t. There are two cases: (A) k 0 = 0, so that
L(p0 ) = n0 , L(p0 + 1) = L(p) = n0 + 1; (B) k 0 > 0, so that n0 < L(p0 ) = L(p0 + 1). In
case (A), if j = L(p) = n0 +1 then Us FN (j) Ut by R(4) (a), (b) for p0 , transitivity, and
the monotonicity of N . If j < L(p), i.e., j ≤ n0 , then i = i0 , so since also j ≤ L(p0 ),
we have by R(4) (a) for p0 that Us FN (j) Ut . In case (B), if j = L(p) we are done
as before. If j < L(p) = L(p0 ), then either i = i0 , and since s ∼j t as well, we are
done, by R(4) (a) for p0 ; or else i 6= i0 in which case j > n0 , so N (j) ≥ N (n0 + 1).
Again s ∼j t, so Usbi FN (j) Utbi by R(4) (a) for p0 and Utbi0 FN (n0 +1) Utbi by R(4)
(b) for p0 , thus Usbi FN (j) Utbi0 again.
We now have two cases for p.
Case (α) : p = hn, 0i, so that L(p) = n and L(p + 1) = n + 1.
In this case we have to define also N (n + 1). For that we apply 2.7: We can
find a refinement (T, US∗ , M ) of (T, U, M ) and N (n + 1) > N (n) such that Fn+1 is
meager in FN (n+1) on s∈2p Us∗ . (Note that if Fn+1 is meager in Fm on A, it is also
meager in Fm0 on A for m0 ≥ m.) By applying also 2.5, repeatedly, we can assume
τ
that d(Us∗ ) ≤ 2−p and Us∗ ⊆ Us .
Case (β) : p = hn, ki for k > 0, so that n < L(p) = L(p + 1).
In this case, we do not have to define a new value of N . Also,
S by the induction hypothesis, R(3) implies that Fn+1 is meager inFN (n+1) on s∈2p Us . So let
τ
(T, U ∗ , M ) be a good refinement of (T, U, M ) so that d(Us∗ ) ≤ 2−p and Us∗ ⊆ Us
(by 2.5 again).
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232
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
So in either case we have a good tree structure (T, U ∗ , M ) refining (T, U, M )
S
τ
with Fn+1 meager in FN (n+1) on s∈2p Us∗ and d(Us∗ ) ≤ 2−p , Ux∗ ⊆ Us . Let N =
N (n + 1). Thus M (n) ≤ N ≤ M (n0 ) for any n0 > n, n0 ≤ L(p) (if such exists). By
2.8 then, there are good refinements (T, U 0 , M ), (T, U 1 , M ) of (T, U ∗ , M ) satisfying
Us0 FN Us1 , ∀s ∈ 2p and (Us0 × Ut1 ) ∩ Fn+1 = ∅, for any two vertices s, t of T which
are n-connected, i.e., by 2.9, s ∼n t. We put now
Usbi = Usi .
Clearly R(1), (2) are satisfied (note that Usb0 ∩ Usb1 = ∅ follows from R(2)).
Moreover, R(3) holds by the choice of N (n + 1) and the induction hypothesis.
Next, R(4) (a) is true since (T, Us0 , M ), (T, Us1 , M ) are good and M is monotone.
Finally, R(4) (b) holds because of (i) of 2.8.
Thus the construction is complete, and so is the proof of 2.2
Let us note also the following corollary of the main result, which points out
another interesting property of E1 .
Theorem 2.10. Let E be a Borel equivalence relation. Then
E1 ≤ E ⇔ E1 vi E.
Proof. Let f : (2N )N → X be such that xE1 y ⇔ f (x)Ef (y). Let X ∗ = f [(2N )N ] and
X ∗∗ = [X ∗ ]E . We claim first that X ∗∗ is Borel. Indeed, let
R(x, y) ⇔ f (y)Ex.
Then R is Borel with Kσ sections, so since x ∈ X ∗∗ ⇔ ∃yR(x, y), X ∗∗ is Borel.
Moreover, there is a Borel function ϕ : X ∗∗ → (2N )N such that R(x, ϕ(x)), i.e.,f ◦
ϕ(x)Ex. Then ϕ is a reduction of E|X ∗∗ into E1 , so from 2.1 it follows that
E1 v E|X ∗∗ ; thus E1 v E. So we can assume above that f is 1-1. We can also
suppose that X = 2N .
Now define g : X ∗∗ → (2N )N by g(x) = (x, ϕ(x)1 , ϕ(x)2 , . . . ). Then g is 1-1 and
g(x)E1 ϕ(x), so f ◦ g(x)Ex. Also g ◦ f (y)E1 y. Now apply Schroeder-Bernstein to
f, g to show that E1 ∼
= E1 |X ∗∗ , thus E1 vi E.
3. Stable equivalence and isomorphism
As an application of the result in Section 2 we can also classify hypersmooth
Borel equivalence relations, at least under further mild regularity assumptions,
with respect to stable equivalence, where we call two Borel equivalence relations
E, F stably equivalent if
E × I2N ∼
= F × I2N .
Let us say that a Borel equivalence relation E on X is strongly
smooth if it
S
admits a Borel selector, and strongly hypersmooth if E = n Fn , with F0 ⊆
F1 ⊆ · · · and Fn strongly smooth.
There are easy examples of smooth Borel E which are not strongly smooth (see,
e.g., [K3, 18.D]), and from 3.8 below these are not even strongly hypersmooth.
However most natural examples of smooth E are actually strongly smooth. Also
every smooth E with Kσ equivalence classes is strongly smooth.
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
233
We have
Theorem 3.1. Let E be a non-smooth strongly hypersmooth Borel equivalence relation. Then E is stably equivalent to exactly one of E0 × I2N , E1 .
The same conclusion holds if E is hypersmooth with Kσ equivalence classes, in
the sense that E is a Borel equivalence relation on a Borel set B in some Polish
space Y and all E-equivalence classes are Kσ in Y .
Proof. We will need the following two lemmas:
Lemma 3.2. Let E be a Borel equivalence relation with Kσ classes, i.e., E is
a Borel equivalence relation on a Borel set B in a Polish space Y and each Eequivalence class is Kσ in Y . Let X be a standard Borel space and F a Borel
equivalence relation on X. If E × I2N v F , then E × I2N vi F .
Proof. Let the Borel function f embed E × I2N into F and define X ∗ , X ∗∗ as in
the proof of 2.10. As in that proof, X ∗∗ is Borel and there is a Borel function ϕ :
X ∗∗ → B × 2N with f ◦ ϕ(x)F x. Assuming, without loss of generality, that X = 2N ,
define g : X ∗∗ → B × 2N as follows: If ϕ(x) = (b, z), then g(x) = (b, x). Then
ϕ(x)E × I2N g(x), so f ◦ g(x)F x. Also g is 1-1. Now apply Schroeder-Bernstein.
Lemma 3.3. Let E be Borel and strongly hypersmooth. Then E × I2N vi E1 .
e = E × I2N , so that E
e is also strongly hypersmooth. We can of
Proof. Put E
S
e
e =
course assume that the space of E is 2N . Let E
n Fn , F0 ⊆ F1 ⊆ · · · , with
F0 = ∆(2N ), Fn strongly smooth with Borel selector fn . Consider the canonical
e into E1 , and define X ∗ , X ∗∗ for this f as in the
embedding f (x) = (fn (x)) of E
proof of 2.10. We claim that X ∗∗ is Borel and there is Borel ϕ : X ∗∗ → 2N with
f ◦ ϕ((xm ))E1 (xm ). The proof then can be completed as in 3.2.
To see that X ∗∗ is Borel, we verify that
(xm ) ∈ X ∗∗ ⇔ ∃n∀m ≥ n(xm = fm (xn )).
⇒: Pick x so that (xm )E1 f (x) and n so that ∀m ≥ n(xm = fm (x)). Then
xn = fn (x) and xn Fn x, so xn Fm x for m ≥ n and fm (xn ) = fm (x) for m ≥ n; thus
xm = fm (xn ), ∀m ≥ n.
⇐: Fix n with xm = fm (xn ) for m ≥ n. Then f (xn )E1 (xm ) and (xm ) ∈ X ∗∗ .
Finally, if (xm ) ∈ X ∗∗ , let
ϕ((xm )) = xn ,
where n is least with ∀m ≥ n(xm = fm (xn )). Then f ◦ ϕ((m ))E1 (xm ).
We now complete the proof of 3.1. Since E is not smooth, E0 v E, so E0 × I2N v
E × I2N . Since E0 × I2N has Kσ equivalence classes and E0 × I2N × I2N ∼
= E0 × I2N ,
we have by 3.2 that E0 × I2N vi E × I2N .
Since E is hypersmooth, E ≤ E0 or else E1 v E. In the first case we have that
E × I2N v E0 × I2N . (Indeed, if g reduces E to E0 , the function
f (x, α) = (g(x), hx, αi),
N
N
where h i : X × 2 → 2 is a Borel injection and X the space of E, embeds E × I2N
into E0 × I2N .) If now E has Kσ equivalence classes, E × I2N vi E0 × I2N by 3.2, so
E ×I2N ∼
= E0 ×I2N . On the other hand, if E is strongly hypersmooth, E ×I2N vi E1
by 3.3, so, replacing E × I2N by an isomorphic copy, we can assume that it has Kσ
equivalence classes, so, by 3.2 again, E × I2N vi E0 × I2N and we have shown that
E × I2N ∼
= E0 × I2N .
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234
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
In the case when E1 v E, we have by 2.10 that E1 vi E × I2N . By 1.3 and 3.2 or
3.3, depending on whether E has Kσ equivalence classes or is strongly hypersmooth,
we also have E × I2N vi E1 , so E1 ∼
= E × IR .
Let us call a Borel equivalence relation E uniformly continuous if E ∼
= E ×I2N .
In view of 3.1, the only uniformly continuous, non-smooth, strongly hypersmooth
Borel equivalence relations are E0 × I2N and E1 . This should be compared with
analogous measure theoretic results of Vershik and Vinokurov-Ganikhodzhaev (see
[V], [VF], [VG].)
The following criterion can be useful in verifying whether a given E is uniformly
continuous.
Proposition 3.4. Let E be a Borel equivalence relation. Then the following are
equivalent:
(i) E is uniformly continuous.
(ii) There is a smooth Borel F ⊆ E which has uniformly continuum-size equivalence classes, i.e., there is a Borel function f : X × 2N → X such that xF y →
f (x, α) = f (y, α)F x and α 6= β ⇒ f (x, α) 6= f (x, β) (where X is the space of E).
Proof. (i)⇒(ii): It is enough to show that E × I2N satisfies (ii). Indeed, let F ⊆
E × I2N be given by
(x, α)F (y, β) ⇔ x = y.
Clearly F is smooth. Put also
f ((x, α), r) = (x, r).
(ii)⇒(i): Let g(x) = (x, 0), so that g embeds E into E × I2N (where 0 = 000 · · · ).
Let f be given now by (ii). Fix a Borel injection h i : X × 2N → 2N and define
h : X × 2N → 2N by
h(x, α) = f (x, hx, αi).
Then h is 1-1 and, since h(x, α)Ex,
g ◦ h(x, α)E × I2N (x, α),
while
h ◦ g(x)Ex,
so by applying Schroeder-Bernstein to g, h we have that E ∼
= E × I2N .
As an application, we see that
E1 (= E0 (2N )) ∼
= Et (2N ).
This is because Et (2N ) is hypersmooth, uniformly continuous (as it contains F ,
where xF y ⇔ ∀n ≥ 1(xn = yn )) and E1 v Et (2N ) (via the map x ∈ (2N )N 7→
(hn, xn i) ∈ (2N )N , where n = 0n b1∞ , h i : (2N )2 → 2N is a Borel injection); thus
0
Et (2N ) 6≤ E0 . In fact, more generally, if E is Borel hypersmooth with E1 ⊆ E ⊆ EC
,
0
N N
where EC on (2 ) is defined by
0
(xn )EC
(yn ) ⇔ {xn : n ∈ N}∆{yn : n ∈ N} is finite,
then E ∼
= E1 . This is because any such E is not reducible to a countable Borel
equivalence relation, by 1.5, and is uniformly continuous, since it contains F as
above.
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
235
As another application, let U : X → X be Borel and uniformly 2ℵ0 -to-1, i.e.,
assume there is a function V : U [X] × 2N → X with analytic graph such that
U (V (y, α)) = y for all α ∈ 2N , and α 6= β ⇒ V (y, α) 6= V (y, β). Then E0 (U ), Et (V )
are either smooth or else Borel isomorphic to one of E1 , E0 × I2N . This is because
F ⊆ E0 (U ) ⊆ Et (U ), where xF y ⇔ U (x) = U (y), and F together with f (x, α) =
V (U (x), α) satisfies 3.4, (ii). (Of course, all these cases can occur, as we can see
by taking U to be the restriction of the shift on (2N )N to various Borel invariant
subsets.)
In [K2] it is shown that for any Borel equivalence relation of the form ER (i.e.,
induced by a Borel flow) none of whose equivalence classes is a singleton, we have
that either ER is smooth or else ER ∼
= E0 ×I2N . By the results in [JKL] it follows also
that if G is compactly generated of polynomial growth and EG has all equivalence
classes uncountable, then again EG is smooth or else EG ∼
= E0 × I2N .
Finally, we can apply also the preceding methods to classify EH , and therefore
the coset spaces G/H, for subgroups H of Polish groups G, which can be written
as unions of increasing sequences of closed subgroups. The result is as follows:
S
Theorem 3.5. Let G be a Polish group, H = n Hn , where (Hn ) is a increasing
sequence of closed subgroups. Denote by EH the equivalence relation
xEH y ⇔ xH = yH.
Then exactly one of the following holds:
(i) H is closed and EH is smooth.
(ii) H is not closed but, for sufficiently large n, Hn+1 /Hn is countable. Then if
H is uncountable, EH ∼
= E0 × I2N , while if H is countable EH ≈ E0 .
(iii) For infinitely many n, Hn+1 /Hn is uncountable and EH ∼
= E1 .
Proof. Assume that H is not closed, so EH is not smooth. Then E0 v EH .
Consider first the case where for sufficiently large n, Hn+1 /Hn is countable. If all
Hn are countable, so is H and thus EH is countable. Since EH is hypersmooth, by
[DJK], EH is then hyperfinite and EH ≈ E0 . So assume some Hn is uncountable.
Renumber so that H0 is uncountable and Hn+1 /Hn is countable for all n, i.e.,
Hn /H0 is countable for all n and so H/H0 is countable. Then EH0 is strongly
smooth and EH /EH0 is countable. It follows that EH ≤ E0 . Let X0 be a Borel
transversal for EH0 . Define Fn on X0 by
S
Fn = EHn X0
and F = n Fn , so F = EH X0 . Since F is countable Borel and hypersmooth,
it is hyperfinite, so F ≤ E0 . But EH ≤ F by the map: h(x) = unique element of
X0 ∩ [x]EH0 .
By 3.1 then, EH × IR ∼
= E0 × IR . But we claim that EH is uniformly continuous,
so EH ∼
= E0 × IR . For that we use 3.4. Let F = EH0 . Since H0 is uncountable,
let ϕ : R → H0 be a Borel bijection. Let X0 be a Borel transversal for EH0 and let
h(x) be defined as above. Put f (x, α) = h(x) · ϕ(α).
Now consider the case when for infinitely many n, Hn+1 /Hn is uncountable. By
renumbering, we can assume that H0 is uncountable and Hn+1 /Hn is uncountable
for each n. We will show then that E1 v EH . As before EH is strongly hypersmooth
and uniformly continuous, so EH ∼
= E1 .
Since Hn+1 /Hn is uncountable, we claim that Hn is meager in Hn+1 . Indeed,
otherwise Hn = Hn · Hn−1 would contain an open nbhd of the identity in Hn+1 , so
Hn would be open in Hn+1 and Hn+1 /Hn would be countable.
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236
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
From this we can derive the following.
Lemma. For each n, EHn is meager in EHn+1 (equipped with the relativized product
topology from G2 ; EHn+1 is closed in G2 ).
Proof. Put Fn = EHn+1 . Let U, V be open in G with (U × V ) ∩ Fn+1 6= ∅. We will
find open U 0 ⊆ U , V 0 ⊆ V such that (U 0 × V 0 ) ∩ Fn+1 6= ∅, but (U 0 × V 0 ) ∩ Fn = ∅.
Consider L = {g ∈ G : ∃x ∈ U (xg ∈ V )}. This is open in G and L ∩ Hn+1 6= ∅.
So find open L0 ⊆ L with L0 ∩ Hn+1 6= ∅, but L0 ∩ Hn = ∅. Fix g0 ∈ Hn+1 , g0 ∈ L0
and x0 ∈ U with x0 g0 = y0 ∈ V . Let
T = {(x, y) : x ∈ U & y ∈ V & x−1 y ∈ L0 }.
This is an open nbhd of (x0 , y0 ), so we can find U 0 ⊆ U , V 0 ⊆ V with (x0 , y0 ) ∈
U 0 × V 0 , U 0 × V 0 ⊆ T . Since (x0 , y0 ) ∈ Fn+1 , we have (U 0 × V 0 ) ∩ Fn+1 6= ∅. If
(U 0 × V 0 ) ∩ Fn 6= ∅, let x0 ∈ U 0 , y 0 ∈ V 0 , g 0 ∈ Hn be such that x0 g 0 = y 0 . Then
(x0 )−1 y 0 = g 0 ∈ L0 ∩ Hn , a contradiction. So (U 0 × V 0 ) ∩ Fn = ∅.
We can now repeat the argument of Case II in the proof of 2.1. Instead of
the Gandy-Harrington topology we work with the Polish topology of G. The two
relevant points are:
(i) EHn is meager in EHm for any n < m.
(ii) The EHn -saturation of an open set in G is also open, being a union of
translates of this open set.
Thus in the construction of CaseII we can take Us to be open sets in G.
If we are wiling to allow a wider class of isomorphisms than Borel, we actually
have a simpler formulation of the preceding results. Recall that a function is Cmeasurable if it is measurable with respect to the smallest σ-algebra containing
the Borel sets and closed under the Souslin operation A. We also call sets in this
class C-measurable.
Theorem 3.6. Let E be a nonsmooth, hypersmooth Borel equivalence relation. If
every E-equivalence class is uncountable, then E is isomorphic by a C-measurable
isomorphism to exactly one of E1 or E0 × I2N .
(Here, to say that an isomorphism f between standard Borel spaces is C-measurable, means that both f and f −1 are C-measurable.)
Proof. By 3.1 it is enough to show that if E is as in the hypothesis of 3.6, then E
is isomorphic to E × I2N by a C-measurable isomorphism.
Let X, Y be standard Borel spaces. A bijection f : A → B, where A ⊆ X, B ⊆ Y
and f [A] = B, will be called C-measurable if f, f −1 , A, B are C-measurable. Note
that by the usual Schroeder-Bernstein argument if f : X → A ⊆ Y , g : Y → B ⊆ X
are C-measurable bijections, then there is a C-measurable bijection h : X → Y .
Consider now E (on X) and E ×I2N (on X ×2N ). Clearly the map f (x) = (x, 0) is
a C-measurable bijection of X with X ×{0}. If we can find a C-measurable bijection
g : X × 2N A ⊆ X such that g(x, α)Ex, then by applying Schroeder-Bernstein
to f, g we obtain
S a C-measurable isomorphism of E with E × I2N .
Let E = n Fn , F0 ⊆ F1 ⊆ · · · , where F0 is equality and Fn is smooth. Let
gn : X → X be a C-measurable selector for Fn . Define a new C-measurable selector
fn of Fn by letting f0 = identity, fn+1 = fn ◦ gn+1 . If Tn = {x : fn (x) = x} is the
corresponding transversal for fn , then Tn is C-measurable and Tn+1 ⊆ Tn .
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
237
Lemma 3.7. Let An = {x : x ∈ Tn & [x]Fn is uncountable}. Then there is a
C-measurable bijection Rn : An × 2N → Bn ⊆ X such that
(i) Rn (x, α)Fn x,
(ii) Bn ∩ Bm = ∅, if n 6= m.
Granting this lemma, we complete the proof as follows: For eachSx ∈ X, let
n(x) = least n such that [x]Fn is uncountable. This exists as [x]E = n [x]Fn and
[x]E is uncountable. Put
g(x, α) = Rn(x) (fn(x) (x), hx, αi)
(where h i : X × 2N → 2N is a Borel bijection.)
First we check that g is 1-1: If g(x, α) = g(x0 , α0 ), then as the ranges of Rn are
disjoint, n(x) = n(x0 ) (= n.) So Rn (fn (x), hx, αi) = Rn (fn (x0 ), hx0 , α0 i) = y. Then
hx, αi = hx0 , α0 i.
Clearly G is C-measurable—note that x 7→ n(x) is C-measurable. Let B =
range(g). We check that B is C-measurable: For α ∈ 2N , let α = h(α)0 , (α)1 i.
Then if π0 , π1 are the two projections of X × 2N ,
y ∈ range(g) ⇔ ∃n[y ∈ range(Rn ) &
n((π1 (Rn−1 (y)))0 ) = n &
π0 (Rn−1 (y)) = fn ((π1 (Rn−1 (y)))0 )].
Also for y ∈ range(g), A ⊆ X × R, A a Borel set,
g −1 (y) ∈ A ⇔ ∃n[y ∈ range(Rn ) &
n((π1 (Rn−1 (y)))0 ) = n &
π0 (Rn−1 (y)) = fn ((π1 (Rn−1 (y)))0 ) &
((π1 (Rn−1 (y)))0 , (π1 (Rn−1 (y)))1 ) ∈ A].
So g −1 is also C-measurable.
Finally, g(x, α)Fn(x) fn(x) (x)Fn(x) x, so g(x, α)Ex.
Proof of Lemma 3.7. Clearly R0 = ∅. By a standard result of descriptive set
theory, we can find R1∗ satisfying the conditions of the lemma for F1 . Put R1 (x, α) =
R1∗ (x, h1, αi), where h i is a Borel bijection of N × 2N with 2N . This works as well.
Now consider A2 and split it in two parts: A02 = {x ∈ A2 : ∃y[y ∈ [x]F2 & [y]F1
is uncountable}, A002 = A2 \A02 . So A02 is Σ11 and A002 is in Π11 & Σ11 , so certainly
both are C-measurable. Let g2 be a C-measurable function with domain A02 , such
that g2 (x) is a y witnessing that x ∈ A02 . Let f1 (y) = z ∈ T1 . Thus [z]F1 is
uncountable. Put R2∗∗ (x, α) = R1∗ (z, h2, αi). Then R2∗∗ is a C-measurable bijection
with domain A02 . (Note that given z, x can be determined as x = f2 (z).) Let R2∗∗∗
be a C-measurable injection with domain A002 satisfying (i) of the lemma for x ∈ A002 ,
n = 2. Put R2∗ = R2∗∗ ∪ R2∗∗∗ . Clearly R2∗ satisfies all the conditions of the lemma
for F2 . Put
R2 (x, α) = R2∗ (x, h1, αi).
Next, split A3 into two parts: A03 = {x ∈ A3 : ∃y[y ∈ [x]F2 & [y]F2 is uncountable]}, A003 = A3 \A03 . For A003 , define R3∗∗∗ as before. For A03 , let g3 be C-measurable
choosing a witness y = g3 (x) to the fact that x ∈ A03 . Let f2 (y) = z ∈ T2 . Thus
z ∈ A2 . If z ∈ A002 , let R3∗∗ (x, α) = R2∗∗∗ (z, h2, αi). If z ∈ A02 , let f1 (g2 (z)) = w ∈ T1
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238
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
and let R3∗∗ (x, α) = R1∗ (w, h3, αi). Finally, put R3∗ = R3∗∗ ∪ R3∗∗∗ and R3 (x, α) =
R3∗ (x, h1, αi).
Proceed this way by induction on n.
We conclude this section with an additional result that further clarifies the role
of strong hypersmoothness.
Theorem 3.8. If the Borel equivalence relation E is smooth and strongly hypersmooth, then E is strongly smooth.
Remark. Note that this also implies that the assumption that E is strongly hypersmooth is essential in 3.3, because by 3.8 (and the remarks preceding 3.1) there is a
smooth but not strongly hypersmooth E. Then E × I2N vi E1 fails, since otherwise
E × I2N would be strongly hypersmooth, as E1 is, and so would be E.
Proof (of 3.8). By relativization, it is enough to assume that E is a ∆11 equivalence
relation on N , S
(En ) is a ∆11 -sequence of Π01 equivalence relations on N with E0 ⊆
E1 ⊆ · · · and n En = E, f : N → N is ∆11 such that xEy ⇔ f (x) = f (y), and
(fn ) is a ∆11 -sequence of ∆11 functions fn : N → N so that fn is a selector for En .
Let f (N ) = A, which is a Σ11 subset of N . To show that E is strongly smooth, it
is enough to show that for each z ∈ A there is x ∈ ∆11 (z)Swith f (x) = z.
Let C = f −1 [{z}]. If y ∈ C, then [y]E = C, so C = n [y]En . Apply the Baire
Category Theorem in the Gandy-Harrington topology relativized to z (i.e., the
topology τ z whose basic nbhds are the Σ11 (z) subsets of N ), noticing that C ∈ τ z
and [y]En are closed in N , so τ z -closed. We can then find S ⊆ [y]En , for some n,
with S 6= ∅, S ∈ Σ11 (z). Now fn (x0 ) = fn (y), ∀x0 ∈ S, so x = fn (y) ∈ ∆11 (z) ∩ C
and we are done.
4. Embedding E1
It is a delicate question to decide, for a given Borel equivalence relation E,
whether E1 ≤ E. The only obstruction we know is given in 4.1 below.
Let E be a Borel equivalence relation on X. We call E idealistic if there is a
map C ∈ X/E 7→ Ic , assigning to each E-equivalence class C a σ-ideal IC of subsets
of C, with C 6∈ IC , such that IC satisfies the ccc (countable chain condition), i.e.,
any collection of pairwise disjoint subsets of C which are not in IC is countable,
and moreover the map C 7→ IC is Borel in the following sense:
For each Borel A ⊆ X 2 the set AI defined by
x ∈ AI ⇔ {y ∈ [x]E : A(x, y)} ∈ I[x]E
is Borel.
Examples of idealistic E include those induced by Borel actions of Polish groups
and the measured ones, i.e., those for which there is a Borel assignment x 7→ µx of
probability measures such that µx ([x]E ) = 1 and xEy ⇒ µx ∼ µy (see for example
[K1]).
Theorem 4.1. Let E be an idealistic Borel equivalence relation. Then E1 6≤ E.
Proof. If E1 ≤ E, then by 2.10, E1 vi E, so E1 must be idealistic too. But E1 is
the union of a sequence of smooth Borel equivalence relations, so by Theorem 1.5
of [K1], E1 ≤ F for a countable Borel equivalence relation F , contradicting 1.4.
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
239
If G is a Polish group acting in a Borel way on a standard Borel space X and
EG is the corresponding equivalence relation, then E1 6≤ EG by 4.1 provided EG is
Borel. However by a modification of the proof of 4.1 we do not need to impose this
restriction.
Theorem 4.2. Let EG be the equivalence relation induced by a Borel action of a
Polish group. Then E1 6≤ EG .
Proof. Assume E1 ≤ EG via the Borel function f : (2N )N → X, where EG lives on
X. Let f ((2N )N ) = Y and Z = [Y ]EG , so that Y, Z are Σ11 .
Let g : Y → (2N )N be a C-measurable inverse for f and define the equivalence
relation F on Y by
S
yF y 0 ⇔ g(y)E1 g(y 0 ) (⇔ yEG y 0 ).
Then F = n Fn , where F0 ⊆ F1 ⊆ · · · are equivalence relations on Y which are
C-measurable smooth, i.e., for each n there is a C-measurable function sn : Y → 2N
with yFn y 0 ⇔ sn (y) = sn (y 0 ).
Now for each D ∈ X/EG let ID be the canonical σ-ideal on D (see, e.g., [K1])
given by
A ∈ ID ⇔ {g ∈ G : g · x ∈ A} is meager in G,
where (g, x) 7→ g · x is the action and x ∈ D. (This definition is independent of x.)
Clearly ID is ccc. Fix a C-measurable function h : Z → Y such that h(z)EG z. For
each C ∈ Y /F , define the following σ-ideal JC on C :
A ∈ JC ⇔ h−1 (A) ∈ I[C]EG .
It is clearly ccc, and C 6∈ JC . Also C 7→ JC satisfies the following:
For each C-measurable C ⊆ Y 2 , the set CJ defined by
y ∈ CJ ⇔ {y 0 ∈ [y]F : C(y, y 0 )} ∈ J[y]F
is ∆12 .
We can now repeat the argument for the proof of 1.5, (ii)⇒(i) in [K1], to show
that there is a Σ12 set A ⊆ Y which meets every F -equivalence class in a countable
nonempty set. Let F 0 = F |A. It follows that there is a ∆12 function H : (2N )N → X
such that xE1 y ⇔ H(x)F 0 H(y). Then we can repeat the proof of 1.5 (with H
replacing f there) to reach a contradiction. The only additional fact that is needed
in the present case is that H is Baire measurable, i.e., we need to know that Σ12 sets
have the Baire property. However, since the result we want to prove (i.e., 4.2) is
equivalent (in ZFC) to a Π13 sentence, it is enough, by standard metamathematical
results, to prove it assuming additionally MA +¬ CH, which implies that all Σ12
sets have the property of Baire, and we are done.
We can use these results to discuss various classes of examples. Let us consider
first equivalence relations generated by filters on N; see [L]. For E a Borel equivalence relation on X and F a Borel filter on N, denote by E F the following Borel
equivalence relation on X N :
(xn )E F (yn ) ⇔ {n : xn Eyn } ∈ F.
If E = ∆2 is the equality relation on 2 = {0, 1}, we write 2F instead of ∆F
2 .
0
If E has uncountably many equivalence classes, then ∆2N ≤ E, so E1 = ∆N
2N ≤
N0
E , where N0 = the Fréchet filter = {A ⊆ N : A is cofinite}. Given two filters
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240
ALEXANDER S. KECHRIS AND ALAIN LOUVEAU
F, G, let F ≤ G iff there is ϕ : N → N with F = ϕG = {A ⊆ N : ϕ−1 [A] ∈ G}. It is
well-known that if F is a free Borel filter, then N0 ≤ F, and it is easy to see that
F ≤ G ⇒ EF ≤ EG .
So E N0 ≤ E F for any free Borel F, and thus for any Borel E with uncountably
many equivalence classes we have E1 ≤ E F .
The situation with 2F is quite different. Clearly E1 6≤ 2N0 = E0 . We denote by
N also the filter {N}. Then E1 ∼
= 2N×N0 , so E1 ≤ 2F for any F with N × N0 ≤ F.
Since N ≤ N0 , it follows that N × N0 ≤ N0 × N0 = N1 and so E1 ≤ 2N and thus
E1 ≤ 2Nξ , where Nξ are the iterated Fréchet filters; see [L].
On the other hand, every ideal on N is also a subgroup of the compact group
(ZN
2 , +), where + is coordinatewise addition. If F is a Borel filter and I its dual
F
ideal, then I is a Borel subgroup of ZN
is exactly the Borel equivalence
2 and 2
N
relation given by the cosets of I in Z2 , thus is generated by a Borel I-action.
Recall now that a standard Borel group G (i.e., a group which is a standard Borel
space for which multiplication and inverse are Borel) is called Polishable if there
is a (necessarily unique) Polish topology on G with the same Borel structure, under
which G becomes a topological group. Thus if I is Polishable, E1 6≤ 2F by 4.1.
This has an interesting application concerning ideals:
If I is a Borel ideal, F its dual filter and N × N0 ≤ F, then I is not Polishable.
On the other hand, there are interesting Polishable ideals I (for which therefore
E1 6≤ 2F ). For instance, let k0 = 0 < k1 < k2 < · · · be such that kn+1 = kn + n
and let In = [kn , kn+1 ), so that card(In ) = n. Define, for p ∈ [1, ∞), the ideal Ip
by
A ∈ Ip ⇔ (card(A ∩ In )/n)n≥1 ∈ lp .
Another class of E for which E1 6≤ E comes from model theory. If E is the
isomorphism relation on a Borel (invariant under isomorphism) class of countable
structures, then by 4.2, E1 6≤ E. In particular, this applies to the examples discussed in [FS]. For each Borel equivalence relation E on X, we denote, as in [L],
by E + the following equivalence relation on X N :
(xn )E + (yn ) ⇔ ∀n∃m(xn Eym ) & ∀n∃m(xm Eyn ).
Then ∆+
, ∆++
, . . . (to the transfinite) can be Borel reduced to the equivalence
2N
2N
relation of isomorphism of trees discussed in [FS], so E1 embeds in none of them
either.
Another important Borel equivalence relation is measure equivalence ∼. (Here
X is the space of probability measures on an uncountable standard Borel space,
and µ ∼ ν ⇔ µ ν & ν µ.) By the Spectral Theorem, ∼ ≤ E, where E is the
equivalence relation of unitary isomorphism of normal operators on Hilbert space.
By 4.2, E1 6≤ E, so E1 6≤ ∼. We can reduce to ∼ various other Borel equivalence
relations and thus use this to give alternative proofs that E1 P
cannot be reduced to
them. For example, consider EC = ∆+
≤
∼
(use
(x
)
→
7
2−n δxn , with δx =
n
N
2
the Dirac measure on x). Now if F is a countable Borel equivalence relation, then
N
F ≤ EC (send any x to an enumeration of [x]F in a Borel way) and easily EC
≤ EC .
N
It follows that E1 6≤ EC and E1 6≤ F , for any countable F .
Finally, note that in [K1] an example is given of a Kσ subgroup H of TN which
is Polishable but EH is not comparable in ≤ with E1 . Another example would be
Ip . So 3.5 does not extend to Fσ subgroups.
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HYPERSMOOTH BOREL EQUIVALENCE RELATIONS
241
We conclude with the following problems:
Problem. If E is a Borel equivalence relation, is it true that either E1 ≤ E or E
is idealistic?
Problem. Let F be a Borel filter on N and I its dual ideal. Is it true that either
E1 ≤ 2F or I is Polishable? (This has been recently solved affirmatively by Solecki.)
5. Global effects
Although the preceding results are “local”, being concerned with Borel equivalence relations which are ≤ E1 , they have a surprising “global” consequence about
the structure of the class of all Borel equivalence relations.
Given a pair (E, E ∗ ) of Borel equivalence relations with E < E ∗ , we say that
(E, E ∗ ) satisfies the dichotomy property if for any Borel equivalence relation F
we have F ≤ E or E ∗ ≤ F .
Thus Silver’s Theorem (see [S]) asserts that (∆N , ∆2N ) satisfies the dichotomy
property, and the Glimm-Effros type dichotomy proved in [HKL] implies that
(∆2N , E0 ) satisfies the dichotomy property. Notice also that trivially (∆n , ∆n+1 )
(n = 1, 2, . . . ) satisfy the dichotomy property. If (E, E ∗ ) satisfies the dichotomy
property, then E, E ∗ are nodes in ≤, i.e., every Borel equivalence relation is comparable in ≤ to each one of them.
We have, now,
Theorem 5.1. Let E be a Borel equivalence relation which is a node for ≤, i.e.,
for any Borel equivalence relation F we have E ≤ F or F ≤ E. Then E ≈∗ ∆n
(n = 1, 2, . . . ), ∆N , ∆2N or E0 .
In particular, the only pairs (E, E ∗ ) satisfying the dichotomy property are
(∆n , ∆n+1 ) (n = 1, 2, . . . ), (∆N , ∆2N ), (∆2N , E0 ).
Proof. Call a class U of Borel equivalence relations unbounded (in ≤) if there
is no Borel equivalence relation E such that ∀F ∈ U (F ≤ E). We will use the
following result.
Theorem 5.2 (Harrington, unpublished). There is a nonempty class U of Borel
equivalence relations which is unbounded, and every F ∈ U is induced by a Borel
action of a Polish group (so in particular is idealistic).
Let E be a Borel equivalence relation which is such that E ≤ F or F ≤ E for
all Borel equivalence relations F . By applying 5.2, we obtain that E ≤ F for some
F ∈ U . We also have E1 ≤ E or E ≤ E1 . If E1 ≤ E then E1 ≤ F , which violates
4.1, so E ≤ E1 . Then by 2.1, and the remarks following it, either E ≈∗ E1 , which
is impossible, or E ≈∗ E0 or E is smooth. In the last case, clearly (by Silver’s
Theorem) E ≈∗ ∆2N or E ≤ EN , and if E ≤ EN , either E ≈∗ En , n = 1, 2, . . . , or
E ≈∗ EN .
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Department of Mathematics, A. P. Sloan Laboratory of Mathematics and Statistics,
California Institute of Technology, Pasadena, California 91125
E-mail address: [email protected]
Equipe d’Analyse, Université Paris VI, 4, Place Jussieu, 75230 Paris Cedex 05, France
E-mail address: [email protected]
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