A trichotomy for a class of equivalence relations

0.16 MB·English

by  Longyun Ding

A TRICHOTOMY FOR A CLASS OF
EQUIVALENCE RELATIONS
0 LONGYUNDING
1
0
2 Abstract. LetXn, n∈Nbeasequenceofnon-emptysets,ψn :Xn2 →
n R+. We consider the relation E((Xn,ψn)n∈N) on Qn∈NXn by (x,y)∈
a E((Xn,ψn)n∈N)⇔Pn∈Nψn(x(n),y(n))<+∞. IfE((Xn,ψn)n∈N) isa
J Borel equivalence relation, we show a trichotomy that either RN/ℓ1 ≤B
6 E, E1 ≤B E, or E ≤B E0.
We also prove that, for a rather general case, E((Xn,ψn)n∈N) is an
O] equivalence relation iff it is an ℓp-likeequivalence relation.
L
.
h
t 1. Introduction
a
m
AtopologicalspaceiscalledaPolishspaceifitisseparableandcompletely
[ metrizable. Let X,Y be Polish spaces and E,F equivalence relations on

1 X,Y respectively. ABorel reductionofE toF isaBorelfunctionθ : X Y
→
v such that (x,y) E iff (θ(x),θ(y)) F, for all x,y X. We say that E is
4 ∈ ∈ ∈
Borel reducible to F, denoted E F, if there is a Borel reduction of E to
3 ≤B
8 F. If E B F and F B E, we say that E and F are Borel bireducible and
≤ ≤
0 denoteE F. Werefer to[4]and[8]for backgroundonBorel reducibility.
B
. ∼
1 There are several famous dichotomy theorems on Borel reducibility. The
0 first one is the Silver’s dichotomy theorem [13].
0
1 Theorem 1.1 (Silver). Let E be a Π1 equivalence relation. Then E has
1
: either at most countably many or perfectly many equivalence classes, i.e.
v
i E id(N) or id(R) E.
X ≤B ≤B
r There are three dichotomy theorems concerning E0. Before introducing
a these theorems, we recall definitions of equivalence relations E ,E ,Eω.
0 1 0
N
(a) For x,y 2 , (x,y) E m n m(x(n) = y(n)).
0
(b) For x,y ∈ 2N×N, (x,y∈) E⇔ ∃ ∀m ≥n m k(x(n,k) = y(n,k)).
1
(c) For x,y ∈ 2N×N, (x,y) ∈ Eω⇔ ∃k∀m≥n ∀m(x(n,k) =y(n,k)).
∈ ∈ 0 ⇔ ∀ ∃ ∀ ≥
Theorem 1.2. Let E be a Borel equivalence relation. Then
(a) (Harrington-Kechris-Louveau [5]) either E id(R) or E E;
B 0 B
≤ ≤
(b) (Kechris-Louveau [9]) if E E , then E E or E E ;
B 1 B 0 B 1
≤ ≤ ∼
(c) (Hjorth-Kechris [7]) if E Eω, then E E or E Eω.
≤B 0 ≤B 0 ∼B 0
Date: January 6, 2010.
2000 Mathematics Subject Classification. Primary 03E15, 54E35, 46A45.
Research partially supported by the National Natural Science Foundation of China
(Grant No. 10701044).
1
2 LONGYUNDING
AnotherclassofinterestingBorelequivalencerelationscomefromclassical
Banach sequence spaces. Let p 1. For x,y RN, (x,y) RN/ℓ
p
≥ ∈ ∈ ⇔
x y ℓ . It was shown by G. Hjorth [6] that every Borel equivalence
p
rel−ation∈E RN/ℓ is either essentially countable or satisfies E RN/ℓ .
B 1 B 1
Kanovei as≤ked whether the position of RN/ℓ in the -structure∼is similar
1 B
≤
with E and Eω (see [8], Question 5.7.5).
1 0
Question 1.3 (Kanovei). Does every Borel equivalence relation E RN/ℓ
1
satisfy either E E or E RN/ℓ ? ≤
B 0 B 1
≤ ∼
Two kinds of ℓ -like equivalence relations were introduced by T. M´atrai
p
[11] and the author [1]. (1) Let f : [0,1] R+. For x,y [0,1]N, (x,y)
E f(x(n) y(n)) < + . (2) L→et (X ,d ), n ∈N be a sequenc∈e
f ⇔ n∈N | − | ∞ n n ∈
of metrPic spaces, p ≥ 1. For (x,y) ∈ n∈NXn, (x,y) ∈E((Xn,dn)n∈N;p)⇔
n∈Ndn(x(n),y(n))p < +∞. Q
PIn this paper, we introduce a notion surpassing both (1) and (2). Let
X , n N be a sequence of non-empty sets, ψ : X2 R+. For x,y
n n n
∈ → ∈
n∈NXn, (x,y) ∈ E((Xn,ψn)n∈N)⇔ n∈Nψn(x(n),y(n)) < +∞. Though
wQedidnotfindanaturalnecessaryandPsufficientconditionthatE((Xn,ψn)n∈N)
be an equivalence relation, we establish the following trichotomy.
Theorem 1.4. If E =E((Xn,ψn)n∈N) is a Borel equivalence relation, then
either RN/ℓ E, E E, or E E .
1 B 1 B B 0
≤ ≤ ≤
From this trichotomy, we can see that Kanovei’s problem is valid within
equivalence relations of the form E((Xn,ψn)n∈N).
It was shown by R. Dougherty and G. Hjorth [2] that, for p,q 1,
RN/ℓ RN/ℓ iff p q. We complete a different picture for 0 <p ≥1 by
p B q
showin≤g that RN/ℓ ≤ RN/ℓ . ≤
p B 1
∼
Via a process of metrization, we prove that, for a rather general case,
equivalence relations E((Xn,ψn)n∈N) coincide with ℓp-like equivalence rela-
tions E((Xn,dn)n∈N;p).
2. A trichotomy for sum-like equivalence relations
We denote the set of all non-negative real numbers by R+.
Definition 2.1. LetX , n Nbeasequenceofnon-emptysets, ψ : X2
n n n
R+. We define a relation E(∈(Xn,ψn)n∈N) on n∈NXn by →
Q
(x,y) E((Xn,ψn)n∈N) ψn(x(n),y(n)) < +
∈ ⇐⇒ ∞
nX∈N
for x,y X . If (X ,ψ ) = (X,ψ) for every n N, we write
∈ n∈N n n n ∈
E(X,ψ) = EQ((Xn,ψn)n∈N) for the sake of brevity.
Itis hardto findanaturalnecessary and sufficientcondition todetermine
when E((Xn,ψn)n∈N) is an equivalence relation. We need the following
definition:
A TRICHOTOMY FOR A CLASS OF EQUIVALENCE RELATIONS 3
Definition 2.2. If E((Xn,ψn)n∈N) is an equivalence relation, we call it a
sum-like equivalence relation. Furthermore, if X , n N is a sequence of
n
∈
Polish spaces and every ψ is Borel function, it is called a sum-like Borel
n
equivalence relation.
The following easy lemma is very useful for the study of sum-like equiva-
lence relations.
Lemma 2.3. Let E = E((Xn,ψn)n∈N) be a sum-like equivalence relation.
For x,y X , we have
∈ n∈N n
Q
(x,y) E ψ (x(n),y(n)) < + .
n
∈ ⇐⇒ ∞
X
x(n)6=y(n)
Proof. Since E is an equivalence relation, (x,x) E. It follows that
∈
ψ (x(n),y(n)) ψ (x(n),x(n)) < + .
n n
≤ ∞
x(nX)=y(n) nX∈N
Then the lemma follows. (cid:3)
Definition 2.4. Let(X ,d ), n Nbeasequenceof pseudo-metricspaces.
n n
∈
Forp 1,anℓp-likeequivalencerelationE((Xn,dn)n∈N;p)isE((Xn,ψn)n∈N)
≥
where ψ (u,v) = d (u,v)p for u,v X . If (X ,d ) = (X,d) for every
n n n n n
n N, we write E(X,d;p) = E((Xn,∈dn)n∈N;p) for the sake of brevity.
∈
Proposition 2.5. Let E((Xn,dn)n∈N;p) be an ℓp-like equivalence relation.
Then there exists metric d′n on Xn for each n such that E((Xn,d′n)n∈N;p) =
E((Xn,dn)n∈N;p).
Proof. We can see that following d′ ’s meet the requirements.
n
0, u= v,
d′n(u,v) =  2d−(nu,,v), du6=(uv,,vd)n>(u2,−vn),≤ 2−n,
n n
for u,v X .  (cid:3)
n
∈
Proposition 2.6. Let E , i N be asequence of Borel equivalence relations.
i
∈
Then for every p 1 there is an ℓ -like Borel equivalence relation E =
p
E((Xn,dn)n∈N;p) s≥uch that Ei B E for n N.
≤ ∈
Proof. For i N, let E be a Borel equivalence relation on a Polish space
i
Y . We define∈χ :Y2 R+ by
i i i →
0, (u,v) E ,
χi(u,v) = (cid:26) 1, (u,v) ∈/ Eii.
∈
Now fixabijection , :N2 N. For every i,j N, if n= i,j , wedenote
h· ·i → ∈ h i
Xn = Yi and dn = χi. It is easy to verify that E = E((Xn,dn)n∈N;p) is an
ℓ -like Borel equivalence relation. For i N, define θ :Y X by
p ∈ i i → n∈N n
Q
u, k = i,
θ (u)( k,j ) =
i h i (cid:26) ak, k = i,
6
4 LONGYUNDING
where a Y is independent of u. Clearly θ is a Borel reduction of E to
k k i i
∈
E. (cid:3)
Now suppose that E = E((Xn,ψn)n∈N) is a sum-like Borel equivalence
relation. Its position in the -structure of Borel equivalence relations is
B
≤
determined by the following condition:
(ℓ1) c > 0 x,y X such that m n > m(ψ (x(n),y(n)) < c)
∀ ∃ ∈ n∈N n ∃ ∀ n
and Q
ψ (x(n),y(n)) =+ .
n
∞
nX∈N
Lemma 2.7. If (ℓ1) holds, then RN/ℓ E.
1 B
≤
Proof. Firstly, we show a well known fact: RN/ℓ [0,1]N/ℓ . Fix a
1 B 1
bijection , ′ :N Z N. For z RN, we define θ′(≤z) [0,1]N as
h· ·i × → ∈ ∈
0, z(m) < k,
θ′(z)( m,k ′)=  z(m) k, k z(m) < k+1,
h i  1, − k+≤1 z(m).
≤
Then θ′ witness that RN/ℓ  [0,1]N/ℓ .
1 B 1
≤ N
Secondly, we construct a reduction of [0,1] /ℓ to E.
1
For l N, by condition (ℓ1), there exist x ,y X such that
∈ l l ∈ n∈N n
m n> m(ψ (x (n),y (n)) < 2−l)and ψ (x (n),Qy (n)) = + . Then
∃ ∀ n l l n∈N n l l ∞
wecanselecttwosequencesofnaturalnuPmbers(il)l∈N,(jl)l∈N satisfyingthat
(i) i < j < i for l N;
l l l+1
∈
(ii) ψ (x (n),y (n)) < 2−l for i n j ;
n l l l l
≤ ≤
(iii) 1 ψ (x (n),y (n))< 1+2−l.
≤ il≤n≤jl n l l
Fix an elPement a X for every n N. For z [0,1]N, we define
n n
∈ ∈ ∈
ϑ(z) X by
∈ n∈N n
Q
x (n), i n j , ψ (x (m),y (m)) z(l);
l l ≤ ≤ l il≤m≤n m l l ≤
ϑ(z)(n) =  yl(n), il ≤ n≤ jl,Pil≤m≤nψm(xl(m),yl(m)) > z(l);
an, otherwise. P
Note that, for z,w [0,1]N and l N, we have
∈ ∈
z(l) w(l) 2−l < ψ (ϑ(z)(n),ϑ(w)(n)) < z(l) w(l) +2−l.
n
| − |− | − |
X
il≤n≤jl
ϑ(z)(n)6=ϑ(w)(n)
Therefore, by Lemma 2.3,
(ϑ(z),ϑ(w)) E ψ (ϑ(z)(n),ϑ(w)(n)) < +
∈ ⇐⇒ ϑ(z)(n)6=ϑ(w)(n) n ∞
P z(l) w(l) < +
⇐⇒ l∈N| − | ∞
zP w ℓ1.
⇐⇒ − ∈
It follows that [0,1]N/ℓ E. (cid:3)
1 B
≤
A TRICHOTOMY FOR A CLASS OF EQUIVALENCE RELATIONS 5
If (ℓ1) fails, then there exists c > 0 such that
m n> m(ψ (x(n),y(n)) < c) ψ (x(n),y(n)) <+
n n
∃ ∀ ⇒ ∞
nX∈N
for x,y X . Now we denote
∈ n∈N n
Q F = (u,v) X2 : ψ (u,v) < c .
n n n
{ ∈ }
Lemma 2.8. If (ℓ1) fails, let c > 0 and F , n N be defined as above.
n
∈
Then
(1) for x,y X ,
∈ n∈N n
Q
(x,y) E m n> m((x(n),y(n)) F );
n
∈ ⇐⇒ ∃ ∀ ∈
(2) there exists N such that, for n > N , F is a Borel equivalence
0 0 n
relation on X .
n
Proof. Since (ℓ1) fails, clause (1) is trivial.
(2) Firstly, assume for contradiction that, there exist a strictly increasing
cse.quSeenleccetoafnnaxtural numbXers (wnikt)hk∈xN(nan)d=uku∈ Xfonrk ksuchNth.atTψhnekn(uwk,euhk)av≥e
∈ n∈N n k k ∈
n∈Nψn(x(n),x(n)Q) = +∞. This is impossible, since E is an equiva-
lPence relation. So there exists N1 such that, for n > N1,u Xn, we have
∈
ψ (u,u) < c, i.e. (u,u) F .
n n
∈
Secondly, assume for contradiction that, there exist a strictly increasing
sequenceofnaturalnumbers(nk)k∈N anduk,vk ∈ Xnk suchthatψnk(uk,vk) <
c,ψ (v ,u ) c. Select x,y X such that x(n ) = u ,y(n ) = v
for nkk kN kand≥x(n) = y(n) f∈orQotnh∈eNr nn. We have ψn(kx(n),yk(n)) <k c fokr
∈
n > N . Since (ℓ1) fails, ψ (x(n),y(n)) < + , i.e. (x,y) E.
1 n∈N n ∞ ∈
Clearly n∈Nψn(y(n),x(n))P= +∞, so (y,x) ∈/ E. A contradiction! Hence
there exPists N2 such that, for n > N2,u,v Xn, we have ψn(u,v) < c
∈ ⇒
ψ (v,u) <c, i.e. (u,v) F (v,u) F .
n n n
∈ ⇒ ∈
With a similar argument, we can prove that there exists N such that,
3
for n > N ,u,v,r X , we have (u,v),(v,r) F (u,r) F .
3 n n n
∈ ∈ ⇒ ∈
In summary, for n > N = max N ,N ,N , F is an equivalence rela-
0 1 2 3 n
{ }
tion. Since ψ is Borel, so is F . (cid:3)
n n
Recall that E (N) is an equivalence relation on NN similar to E on 2N.
0 0
For x,y NN, (x,y) E (N) m n > m(x(n) = y(n)). It is well known
0
that E ∈ E (N) (se∈e Propos⇔iti∃on 6∀.1.2 of [4]).
0 B 0
∼
Lemma 2.9. For any equivalence relation E on X , if there is a
n∈N n
sequence Fn Xn2, n N such that (1) and (2) ofQLemma 2.8 hold, then
⊆ ∈
either E E, E E , or E is trivial, i.e. all elements in X are
1 ≤B ∼B 0 n∈N n
equivalent. Q
Proof. For n > N , since F is a Borel equivalence relation on X , from
0 n n
the Silver dichotomy theorem [13], either there are at most countably many
F -equivalence classes or there are perfectly many F -equivalence classes.
n n
6 LONGYUNDING
Case 1. There exists a strictly increasing sequence of natural numbers
(nk)k∈N such that there are perfectly many Fnk-equivalence classes. Then
there is a continuous embedding h : 2N X for every k N such that
(h (z),h (w)) F iff z = w. Defikne θ :→2N×Nnk X b∈y
k k ∈ nk → n∈N n
Q
h (x(k, )), n= n ,
θ(x)(n)= k · k
(cid:26) an, otherwise,
where a X is independent of x. By Lemma 2.8.(1), it is straightforward
n n
∈
to check that θ is a reduction of E to E.
1
Case 2. Thereexists N suchthat, forn > N,F has onlyoneequivalence
n
class. From Lemma 2.8.(1), we see that E is trivial.
Case 3. If case 1 fails, then there exists N′ such that, for n > N′, F has
n
at mostcountably many equivalence classes. So E E (N). If case 2 fails,
B 0
≤
then there exists a strictly increasing sequence of natural numbers (nk)k∈N
such that F has more than one equivalence class. Thus E E. Since
E E (Nn)k, we have E E . 0 ≤B (cid:3)
0 B 0 B 0
∼ ∼
Now we have already completed the proof of the following trichotomy.
Theorem 2.10. Let E = E((Xn,ψn)n∈N) be a sum-like Borel equivalence
relation. Then either RN/ℓ E, E E, or E E .
1 B 1 B B 0
≤ ≤ ≤
Corollary 2.11. Let E = E((Xn,ψn)n∈N) be a sum-like Borel equivalence
relation. If E RN/ℓ , then either E E or E RN/ℓ .
B 1 B 0 B 1
≤ ≤ ∼
Proof. It is well known that E RN/ℓ (see [9] Theorem 4.2), so if E
1 B 1 B
RN/ℓ , then E E. Hence th6≤e corollary follows. ≤(cid:3)
1 1 B
6≤
It was shown by R. Dougherty and G. Hjorth [2] that, for p,q 1,
≥
RN/ℓ RN/ℓ p q.
p B q
≤ ⇐⇒ ≤
The following corollary shows that, for p 1, the situation is different.
≤
Corollary 2.12. For 0 < p 1, we have RN/ℓ RN/ℓ .
p B 1
≤ ∼
Proof. Note that RN/ℓ = E(R,ψ) where ψ(u,v) = u v p for u,v R. It
p
is easy to see that (ℓ1) holds for E(R,ψ), so RN/ℓ | −RN|/ℓ . ∈
1 B p
For theother direction, weclaim thatRN/ℓ R≤N/ℓ for 0 < p < q 1.
p B q
We sketch the proof for Theorem 1.1 of [2]≤, that RN/ℓ RN/ℓ≤for
p B q
≤
1 p < q, and check that it is also valid for 0< p < q 1.
≤It will suffice to prove for the case 0 < q < p < q ≤1. We denote ρ = p
2 ≤ q
and r = 4−ρ. Then 1 < ρ < 1 and 1 < r < 1. In [2] p. 1838, the authors
2 4 2
constructed a continuous function K¯ : R R2, and proved that there are
r
→
m′,M′ > 0 such that
m′ s t ρ K¯ (s) K¯ (t) M′ s t ρ,
r r 2
| − | ≤ k − k ≤ | − |
for s,t [i 1,i+1],i Z. And K¯ (s) K¯ (t) 1 if s,t are not in the
r r 2
same in∈terv−al [i 1,i+∈1],i Z. k − k ≥
− ∈
A TRICHOTOMY FOR A CLASS OF EQUIVALENCE RELATIONS 7
Define a mapping θ :RN RN such that, for x RN and k N,
→ ∈ ∈
K¯ (x(k)) = (θ(x)(2k),θ(x)(2k +1)).
r
1
For w =(s,t) R2, denote w q = (s q + t q)q. Note that
∈ k k | | | |
1 1 1
w 2 w ∞ w q 2q w ∞ 2q w 2.
√2k k ≤ k k ≤ k k ≤ k k ≤ k k
For x,y RN, we have
∈
θ(x) θ(y) ℓ K¯ (x(k)) K¯ (y(k)) q < +
− ∈ q ⇐⇐⇒⇒ Pkk∈∈NNkkK¯rr(x(k))−−K¯rr(y(k))kkq2q < +∞∞
⇐⇒ Pk∈N|x(k)−y(k)|p < +∞
xP y ℓp.
⇐⇒ − ∈
Thus, θ is a reduction of RN/ℓ to RN/ℓ . (cid:3)
p q
3. Metrization
Inthissection,weshowthatasum-likeequivalencerelationE((Xn,ψn)n∈N)
coincideswithanℓ -likeequivalencerelation ifthefollowingconditionshold.
p
(m1) Denote X = X . There is a unique function ψ : X2 R+
n∈N n →
such that ψn = ψ ↾ Xn2Sand ψ(u,v) = 1 if no Xn contains both u,v.
(m2) For any u,v,r X, if u,v,r X , then there exists m > n such
n
∈ ∈
that u,v,r X .
m
∈
Letψn = min ψn,1 . WecanseethatE((Xn,ψn)n∈N)= E((Xn,ψn)n∈N).
{ }
Therefore, we may assume ψ 1 if needed.
n
≤
Lemma 3.1. Let (Xn,ψn)n∈N satisfy (m1) and (m2). If ψn 1, then
≤
E((Xn,ψn)n∈N) is an equivalence relation iff the following conditions hold:
(i) ψ(u,u) = 0 for u X;
∈
(ii) there is a C 1 such that for u,v,r X,
≥ ∈
ψ(v,u) Cψ(u,v); ψ(u,r) C(ψ(u,v)+ψ(v,r)).
≤ ≤
Proof. If conditions (i) and (ii) hold, it is trivial that E((Xn,ψn)n∈N) is an
equivalence relation. We only need to prove the other direction.
Now assume that E((Xn,ψn)n∈N) is an equivalence relation.
Firstly, for any u X, by (m2), there is an infinite set I N such that
∈ ⊆
u X for n I. Let x X with x(n) = u for n I. Since
∈ n ∈ ∈ n∈N n ∈
(x,x) E((Xn,ψn)n∈N), we haQve ψ(u,u) = 0.
∈
Secondly, foru,v X, ifψ(u,v) = 0,weclaim thatψ(v,u) = 0. By(m2),
there is an infinite s∈et I N such that u,v X for n I. Therefore, for
n
⊆ ∈ ∈
any x,y X , if x(n) = u,y(n) = v for n I and x(n) = y(n) for
∈ n∈N n ∈
n / I, we hQave (x,y) E((Xn,ψn)n∈N). Hence (y,x) E((Xn,ψn)n∈N). It
∈ ∈ ∈
follows that ψ(v,u) = ψ(y(n),x(n)) < + . Thus ψ(v,u) = 0.
Now assumPenf∈oIr contradictPionn∈Ithat, for every k N∞there are uk,vk X
∈ ∈
such that ψ(v ,u ) > 2kψ(u ,v ) > 0. Since ψ 1, 0 < ψ(u ,v ) < 2−k.
k k k k n k k
≤
By (m2), there are infinitely many n such that u ,v X .
k k n
Select a finite set I N for every k satisfying that∈
k
⊆
8 LONGYUNDING
(i) u ,v X for n I ;
k k n k
∈ ∈
(ii) 2−k I ψ(u ,v ) 2−(k−1);
k k k
≤ | | ≤
(iii) if k < k , then maxI < minI .
1 2 k1 k2
Now we define x,y X by
∈ n∈N n
x(n)Q= u ,y(n) = v , n I ,k N,
k k k
∈ ∈
(cid:26) x(n) = y(n)= an, otherwise,
where a X is independent of x and y. Then we have
n n
∈
ψ(x(n),y(n)) = I ψ(u ,v ) 2−(k−1) < + ,
k k k
| | ≤ ∞
nX∈N kX∈N kX∈N
so (x,y) E((Xn,ψn)n∈N). On the other hand, we have
∈
ψ(y(n),x(n)) = I ψ(v ,u ) > 2k I ψ(u ,v ) 1= + ,
k k k k k k
| | | | ≥ ∞
nX∈N kX∈N kX∈N kX∈N
so (y,x) / E((Xn,ψn)n∈N). A Contradiction! We complete the proof of
∈
that there is C 1 such that ψ(u,v) C ψ(v,u) for u,v X.
1 1
≥ ≤ ∈
With a similar argument, we can prove that there is C 1 such that
2
≥
ψ(u,r) C (ψ(u,v)+ψ(v,r)) for u,v,r X. (cid:3)
2
≤ ∈
Lemma 3.2. Let E((Xn,ψn)n∈N),E((Xn,ϕn)n∈N) be two sum-like equiva-
lence relations, both satisfying (m1) and (m2). If ϕ 1, then
n
≤
E((Xn,ψn)n∈N) E((Xn,ϕn)n∈N)
⊆
A 1 u,v X(ϕ(u,v) Aψ(u,v)).
⇐⇒ ∃ ≥ ∀ ∈ ≤
Proof. “ ” is trivial. “ ” follows similarly as the proof of Lemma 3.1. (cid:3)
⇐ ⇒
Corollary 3.3. Let E((Xn,ψn)n∈N),E((Xn,ϕn)n∈N) be two sum-like equiv-
alence relations, both satisfying (m1) and (m2). If ψ ,ϕ 1, then
n n
≤
E((Xn,ψn)n∈N) = E((Xn,ϕn)n∈N)
A 1 u,v X(A−1ψ(u,v) ϕ(u,v) Aψ(u,v)).
⇐⇒ ∃ ≥ ∀ ∈ ≤ ≤
Before introducing the Metrization lemma, we recall several basic notions
onrelations. LetX beanon-emptyset,wedenote∆(X) = (u,u) :u X .
{ ∈ }
AsubsetU X2 iscalledsymmetricif(u,v) U (v,u) U foru,v X.
⊆ ∈ ⇒ ∈ ∈
For U,V X2 we define U V X2 by
⊆ ◦ ⊆
(u,r) U V v((u,v) U,(v,r) V) ( u,v,r X).
∈ ◦ ⇐⇒ ∃ ∈ ∈ ∀ ∈
Lemma 3.4 (Metrization lemma [10], p. 185). Let U , n N be a sequence
n
∈
of subsets of X2 such that
(i) U = X2;
0
(ii) each U is symmetric and ∆(X) U ;
n n
⊆
(iii) U U U U for each n.
n+1 n+1 n+1 n
◦ ◦ ⊆
Then there is a pseudo-metric d on X satisfying that
U (u,v) : d(u,v) < 2−n U ( n 1).
n n−1
⊆ { } ⊆ ∀ ≥
A TRICHOTOMY FOR A CLASS OF EQUIVALENCE RELATIONS 9
Theorem3.5. Let(Xn,ψn)n∈N satisfy(m1)and(m2). ThenE((Xn,ψn)n∈N)
is an equivalence relation iff it is an ℓ -like equivalence relation.
p
Proof. AssumethatE((Xn,ψn)n∈N)isanequivalencerelation. Withoutloss
of generality, we may assume that ψ 1. From Lemma 3.1, there is C 1
n
≤ ≥
such that, for u,v,r X,
∈
ψ(v,u) Cψ(u,v) and ψ(u,r) C(ψ(u,v)+ψ(v,r)).
≤ ≤
Therefore, if ψ(u,v),ψ(v,r),ψ(r,s) < ε, then
ψ(u,s) C(ψ(u,v)+C(ψ(v,r)+ψ(r,s))) < (2C2+C)ε.
≤
Now denote B = 2C2+C. We define U =X2 and
0
U = (u,v) :ψ(u,v) < B−n,ψ(v,u) < B−n (n 1).
n
{ } ≥
Itfollows that, foreach n, U issymmetricandU U U U . By
n n+1 n+1 n+1 n
◦ ◦ ⊆
Lemma 3.1.(i), ∆(X) U . Then the metrization lemma gives a pseudo-
n
⊆
metric d on X such that U (u,v) :d(u,v) < 2−n U .
n n−1
⊆ { }⊆
It is easy to check that, d(u,v) = 0 iff ψ(u,v) = 0. If ψ(u,v) B−2, then
≥
(u,v) / U , d(u,v) 2−3.
2
∈ ≥
Denote p = log B 1. If 0 < ψ(u,v) < B−2, assume that B−(n+1)
2 ≥ ≤
ψ(u,v) < B−n for some n 2. Then ψ(v,u) < CB−n < B−(n−1). It follows
≥
that (u,v) U ,(u,v) / U , 2−(n+2) d(u,v) < 2−(n−1). So
n−1 n+1
∈ ∈ ≤
B−2d(u,v)p < B−2(2−(n−1))p ψ(u,v) < B2(2−(n+2))p B2d(u,v)p.
≤ ≤
Therefore, we have E((Xn,ψn)n∈N) = E((Xn,d ↾ Xn);p). (cid:3)
Corollary 3.6. Let (Xn,ψn)n∈N satisfy that, for n < m, Xn Xm and
⊆
ψn = ψm ↾ Xn2. Then E((Xn,ψn)n∈N) is an equivalence relation iff it is an
ℓ -like equivalence relation.
p
In particular, E(X,ψ) is an equivalence relation iff there are pseudo-
metric d on X and p 1 such that E(X,ψ) = E(X,d;p).
≥
4. Further remarks
Letusconsideraspecialcaseofsum-likeequivalencerelation. Letψ (u,v) =
f
f(u v ) where f : R+ R+. Then conditions (i) and (ii) for E(R,ψ ) in
f
| − | →
Lemma 3.1 read as (see also [11], Proposition 2).
(i) f(0) = 0;
(ii) there is a C 1 such that for s,t R+,
≥ ∈
f(s+t) C(f(s)+f(t)), f(s) C(f(s+t)+f(t)).
≤ ≤
Denote = x RN : f(x(n)) < + . Then E(R,ψ ) is an
equivalenceNrfelati{on ∈iff isPans∈uNbgr|oup o|f (RN,∞+)}. Furthermore,fanother
f
interesting problem is tNo determine when is a linear subspace of RN.
f
N
This problem was studied by S. Mazur and W. Orlicz (see [12], 1.7). It was
also considered in [12] that when ’s are Banach spaces.
f
N
10 LONGYUNDING
Theorem 4.1 (Mazur-Orlicz). Let f : R+ R+ satisfy that, as n ,
→ → ∞
f(t ) 0 iff t 0. The necessary and sufficient condition for to be a
n n f
→ → N
linear space is that
(a) there exist constants C > 0,ε > 0 such that f(s+t) C(f(s)+f(t))
≤
for s,t < ε;
(b) for every ρ > 0 there are constants D > 0,δ > 0 such that f(s)
≤
Df(t) for t < δ,s < ρt.
Note that for t min ε,δ > 0, n N, we have f(t ) 0. Thus there
n n
≥ { } ∈ 6→
is c > 0 such that f(t) c for t min ε,δ . If we assume that f 1, then
≥ ≥ { } ≤
conditions (a) and (b) in this theorem turn to
(a)’ there exists a constant C′ > 0 such that f(2s) C′f(s) for s R+;
≤ ∈
(b)’ there exists a constant D′ > 0 such that f(s) D′f(t) for s< t.
≤
Foralmostallknownexamplesofsum-likeequivalencerelationsE(R,ψ ),
f
’s are linear spaces. In the end, we present an example in which is
f f
N N
not linear as follows.
Example 4.2. Letg : R+ R+ beanincreasingfunctionwithg(0) = 0,such
→
that g′(t) is decreasing with lim g′(t) = + . For example, g(x) = √x
t→0
∞
is such a function. Let (an)n∈N be a strictly decreasing sequence of positive
numbers with lim a = 0.
n→∞ n
Denote k = g(an). Then k < k . We consider equations y = k x
n an n n+1 n
and y g(a ) = k (x a ). Their solution is (b ,k b ) where
n+1 n+1 n+1 n n n
− − −
b = 2kn+1an+1. We can see that a < b < a .
n kn+kn+1 n+1 n n
Now define f : R+ R+ by
→
0, t = 0,
f(t)=  −knktn,+1(t−an+1)+g(an+1), abnn+1 t≤<ta<n,bn,
≤
g(a ), a t.
0 0
≤

It is easy to checkthat, for s,t R+,
∈
f(s+t) f(s)+f(t), f(s) f(s+t)+f(t).
≤ ≤
Note that f(an+1) = 1 1+ kn+1 . Since lim g(t) = lim g′(t) = + ,
f(bn) 2(cid:16) kn (cid:17) t→0 t t→0 ∞
we can find a sequence (an)n∈N such that knkn+1 → +∞, f(fa(nbn+)1) → +∞ as
n . Then condition (b) in Theorem 4.1 fails.
→ ∞
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