# The Structure of Disjoint Groups of Continuous Functions

Authors: Hojjat Farzadfard; B. Khani Robati
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

LetI be an open interval. We describe the general structure of groups of continuous self functions on I which are disjoint, that is, the graphs of any two distinct elements of them do not intersect. Initially the class of all disjoint groups of continuous functions is divided in three subclasses: cyclic groups, groups the limit points of their orbits are Cantor-like sets, and finally those the limit points of their orbits are the whole interval I. We will show that (1) each group of the second type is conjugate, via a specific homeomorphism, to a piecewise linear group of the same type; (2) each group of the third type is a subgroup of a continuous disjoint iteration group. We conclude the Zdun's result on the structure of disjoint iteration groups of continuous functions as special case of our results.


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## Body

## 1. Introduction

The problem of characterizing disjoint groups of continuous functions appears in connection with the issues, such as, describing the solution of the simultaneous systems of Abel’s functional equations (mainly in [1–3]) and systems of differential equations with several deviations (see [1, 4]). In a recent paper on the simultaneous systems of Abel’s equations ([5]) this problem has been paid a new attention. Zdun in [6] has investigated the structure of disjoint iteration groups of homeomorphism on an open interval, that is, disjoint groups of the form G={ft:t∈ℝ} such that fs∘ft=fs+t for all reals s and t and each ft is a homeomorphism of an open interval onto itself. This problem was studies earlier by Domsta ([7, Section 4.2]) but not so general. In the present paper we characterize not only disjoint iteration groups but all the disjoint groups of continuous functions on an open interval in a more general setting.Throughout this paperI is an open interval in the real line; the topology of it is considered to be the subspace topology it inherits from ℝ. We say that a subset L of I is a Cantor-like set fitted in I if L is a nowhere dense set, L′=L, and L has no lower or upper bound in I (here L′ is the set of all limit points of L in I). We denote by L-*, L+*, L*, and L** to the set of all left sided, right sided, one sided, and two sided limit points of L, respectively. Denote by Bi(I) the group of all increasing bijections from I onto itself, with composition of functions as its binary operation. There exists a natural ordering relation on the set 𝔉(I) of all self mappings as follows: given two functions f and g in 𝔉(I) define f<g provided f≠g and f(x)≤g(x) for all x∈I. Define f≤g if f<g or f=g. With this, (𝔉(I),<) is a partial ordering set. A subset of 𝔉(I) is said to be disjoint if the graphs of any two distinct elements of it do not intersect. If G is a disjoint group of continuous functions in 𝔉(I), then G is a subgroup of Bi(I) and furthermore, (G,∘,<) is an Archimedean ordered group (see [8, Proposition 3]). Hence according to Hölder’s Theorem there exists an additive subgroup A of ℝ such that (G,∘,<) is isomorphic to (A,+,<).The following two theorems, which have been proved in [5], give an introductory characterization of disjoint subgroups of Bi(I).Theorem 1.1.
LetG be a disjoint subgroup of  Bi(I). Then (a)
for allx, y in I one has G(x)′=G(y)′ where G(x)≔{f(x):f∈G} (we denote this set by L(G));(b)
for allf∈G, f(L(G))=L(G);(c)
L(G)′=L(G).(d)
the setL(G) is either an empty set or a Cantor-like set fitted in I or L(G)=I.Theorem 1.2.
LetG be a disjoint subgroup of  Bi(I). Then  G is a cyclic group if and only if  L(G)=∅.By virtue of these theorems one divides disjoint subgroups ofBi(I) in three classes: The first class consists of the cyclic subgroups; the second one consists of the subgroups G of Bi(I) for which L(G) is a Cantor-like set fitted in I. one names such a group a spoiled disjoint group. Finally the third class consists of the subgroups G of Bi(I) for which L(G)=I. Such a group is called a dense disjoint group. One calls a subgroup G of Bi(I)  complete if G(x)=I for all x∈I.One uses as a tool classes of functionsϕ:I→ℝ that occur as continuous solutions of simultaneous systems of Abel equations
(1.1)ϕ(f(x))=ϕ(x)+λ(f),f∈S,x∈I,
where 𝒮 is a nonempty subset of Bi(I) the group generated by which is noncyclic and disjoint and λ:𝒮→ℝ is a given map. In view of Theorem 6 of [5] either ϕ is a homeomorphism or a Cantor function which lives on L(G) in the following sense.Definition 1.3.
Given a Cantor-like setL fitted in I we say that ϕ:I→ℝ is a Cantor function which lives on L if (i)
ϕ is monotone;(ii)
ϕ(L)=ℝ;(iii)
ϕ is strictly monotone on L**.Such a functionϕ is constant on the components of I-L**, ϕ(L**)⋂ϕ(I-L**)=∅ and ϕ is continuous.Given a functionϕ:I→ℝ, we say that a function f:I→I is in the realm of ϕ if there exists a real number α such that ϕ(f(x))=ϕ(x)+α for all x∈I. Clearly this α is unique and we call it the index offwith respect toϕ and denote it by indϕ(f). Denote by Realm(ϕ) the set of all functions f:I→I that are in the realm of ϕ. We say that two functions ϕ:I→ℝ and ψ:I→ℝ are associate if ψ=rϕ+s for some real constants r≠0 and s. It is easy to see that the associativity is an equivalence relation on the set of all functions from I into ℝ. Moreover, every two associate functions have the same realms. In particular if ϕ and ψ are associate and ϕ is a Cantor function which lives on L (or a homeomorphism), then ψ is also a Cantor function which lives on L (or a homeomorphism). By these notation the following is an immediate consequence of Theorem  10 in [5].Theorem 1.4.
LetG be a disjoint subgroup of Bi(I). Then G⊆Realm(ϕ) for some continuous ϕ:I→ℝ. If G is spoiled, then ϕ is a Cantor function which lives on L(G). If G is dense, then ϕ is a homeomorphism.Theorem 12 of [5] can be restate as follows.Theorem 1.5.
LetG be a noncyclic disjoint subgroup of  Bi(I). If  ϕ1:I→ℝ and  ϕ2:I→ℝ are two continuous functions such that  G⊆Realm(ϕ1) and  G⊆Realm(ϕ2), then  ϕ1 and  ϕ2 are associate. In particular  Realm(ϕ1)=Realm(ϕ2).In Section3 we deal with the properties of the elements of Realm(ϕ) when ϕ is a Cantor function.


## 2.  Complete and Dense Disjoint Groups

Theorems2.1 and 2.5 below describe the structure of complete disjoint subgroups of Bi(I). While Theorem 2.3 describes dense disjoint groups.Theorem 2.1.
Ifϕ:I→ℝ is a homeomorphism, then Realm(ϕ) is a complete disjoint subgroup of Bi(I). Moreover, if ϕ is increasing, indϕ is an isomorphism of (Realm(ϕ),∘,<) onto (ℝ,+,<).
Conversely, to every complete disjoint subgroupG of Bi(I) there corresponds a homeomorphism ϕ:I→ℝ such that G=Realm(ϕ).Proof.
For the first part of the theorem (whose proof is straightforward) see [8]. So we prove the converse part. Let G be a complete disjoint subgroup of Bi(I). By Theorem 1.4 one has G⊆Realm(ϕ) for some homeomorphism ϕ:I→ℝ. To show that Realm(ϕ)⊆G let f∈Realm(ϕ). Pick a point a in I. Then f(a)∈G(a) because G(a)=I. It follows that f(a)=g(a) for some g∈G. But since G⊆Realm(ϕ), one has g∈Realm(ϕ). By the first part of the theorem Realm(ϕ) is disjoint; so that f=g. Hence f∈G. And we are done.Our next goal is to present a description of dense disjoint subgroups ofBi(I). But first a proposition.Proposition 2.2.
LetG and H be two disjoint subgroups of Bi(I) and H⊆G. If H is noncyclic, then L(H)=L(G).Proof.
The proof we present is extracted from the proof of Theorem 3 of the Zdun’s paper [9] with a little modification.
Letx∈I. Then H(x)⊆G(x). So that L(H)=H(x)′⊆G(x)′=L(G). On the other hand by noticing that G is Abelian, for every f∈G(2.1)f(H(x))={f(g(x)):g∈H}={g(f(x)):g∈H}=H(f(x)).
Since f is a homeomorphism,
(2.2)f(L(H))=f(H(x)′)=(f(H(x)))′=(H(f(x)))′=L(H).
Fix an a in L(H). The by the preceding relation one has f(a)∈L(H) for all f∈G. Thus G(a)⊆L(H). So L(G)=G(a)′⊆L(H). Therefore, L(H)=L(G) and the proof is complete.This proposition yields that ifH is a noncyclic subgroup of the disjoint group G⊆Bi(I), then either both G and H are spoiled or they are both dense.Theorem 2.3.
LetG be a noncyclic disjoint subgroup of Bi(I). Then G is a dense group if and only if G is embeddable in a complete disjoint subgroup of Bi(I). Moreover, this complete group is unique.Proof.
IfG is dense, then by Theorem 1.4, G⊆Realm(ϕ) for some homeomorphism ϕ:I→ℝ. By Theorem 2.1, Realm(ϕ) is complete. Conversely, suppose that G⊆P for some complete disjoint subgroup P of Bi(I). By Proposition 2.2 one has L(G)=L(P)=I. Hence, G is dense.
The uniqueness part of the theorem follows from Theorems1.5 and 2.1.The following proposition will be used now and later.Proposition 2.4.
LetG be a spoiled disjoint subgroup of  Bi(I),  L≔L(G) and  f∈G. Then  f(L-*)=L-*,  f(L+*)=L+* and  f(L**)=L**. Moreover,  G is countable.Proof.
The first part of the proposition is in fact Proposition 5 of [5]. To prove the countability of G pick a point a in L-*. Then G(a)⊆L-*. This implies that G(a) is countable since L-* is so. On the other hand, G and G(a) have the same cardinality because G is disjoint. Therefore, G is countable.Theorem 2.5.
LetG be a disjoint subgroup of  Bi(I). Then  G is complete if and only if  (G,∘,<) is isomorphic to  (ℝ,+,<).Proof.
IfG is complete, then by Theorem 2.1 one has G=Realm(ϕ) for some homeomorphism ϕ:I→ℝ. Since associate functions have the same realms, one can assume that ϕ is increasing. Therefore indϕ is an isomorphism of (G,∘,<) onto (ℝ,+,<).
Conversely, suppose that(G,∘,<) is isomorphic to (ℝ,+,<). Then G is uncountable. Since every spoiled group is countable, G must be dense. Then by Theorem 2.3 there exists a complete disjoint subgroup P of Bi(I) containing G. Since by the first part of the theorem P is isomorphic to the additive group ℝ, G is isomorphic to an additive subgroup of ℝ via the same isomorphism. But ℝ cannot be isomorphic to a strict subgroup of itself. Therefore, G=P and G is complete.


## 3.  The Realm of Cantor Functions

Through this sectionL is a Cantor-like set fitted in I and ϕ:I→ℝ is an increasing Cantor function which lives on L.Proposition 3.1.
(Realm(ϕ),∘) is a monoid and indϕ is a monoid homomorphism from (Realm(ϕ),∘) onto (ℝ,+).
Iff is an invertible function in Realm(ϕ), then f-1∈Realm(ϕ) and indϕ(f-1)=-indϕ(f).
Iff and g are in Realm(ϕ) and f(a)≤g(a) for some a∈I, then indϕ(f)≤indϕ(g).Proof.
It is clear thati, the identity functions on I, is in Realm(ϕ) and indϕ(i)=0. Let f and g be in Realm(ϕ). Then for all x∈I(3.1)ϕ(f∘g(x))=ϕ(f(g(x)))=ϕ(g(x))+indϕ(f)=ϕ(x)+indϕ(g)+indϕ(f).
This means that f∘g∈Realm(ϕ) and indϕ(f∘g)=indϕ(f)+indϕ(g). That is, indϕ:Realm(ϕ)→ℝ is a monoid homomorphism.
To show thatindϕ:Realm(ϕ)→ℝ is surjective let α∈ℝ. Define f:I→I by
(3.2)f(x)≔min⁡ϕ-1({ϕ(x)+α}).
Note that for each t∈ℝ, the set ϕ-1({t}) is either a component of I-L** (which is a closed interval) or a singleton {x} where x∈L**. So the definition of f is meaningful. Since f(x)∈ϕ-1({ϕ(x)+α}) for every x∈ℝ, one has ϕ(f(x))=ϕ(x)+α. Therefore, f∈Realm(ϕ) and indϕ(f)=α. This proves (a).
For everyx∈I one has
(3.3)ϕ(x)=ϕ(f(f-1(x)))=ϕ(f-1(x))+indϕ(f),
thus
(3.4)ϕ(f-1(x))=ϕ(x)-indϕ(f).
This means that f-1∈Realm(ϕ) and indϕ(f-1)=-indϕ(f).
Letf and g be in Realm(ϕ) and f(a)≤g(a) for some a∈I. Then
(3.5)ϕ(a)+indϕ(f)=ϕ(f(a))≤ϕ(g(a))=ϕ(a)+indϕ(g),
since ϕ is increasing. Therefore, indϕ(f)≤indϕ(g).For eachx∈I put [x]≔{x} provided x∈L** and let [x] be the component of I-L** containing x provided x∈I-L**. Set Î≔{[x]:x∈I}, 𝒞≔{[x]:x∈I-L**}. The following relation makes Î into a linearly ordered set: given two elements J and K of Î define J<K whenever x<y for all x∈J, y∈K. The topology of Î is understood to be the order topology. An elementary analysis shows that (𝒞,<) is a countable dense linearly ordered structure with no minimum or maximum; in other words (𝒞,<) is of order type η (see [10, Chapter 8, Exercises 17 and 18]). In particular 𝒞 is dense in Î. Since ℚ is also of order type η, there exists an order preserving bijection θ of (𝒞,<) onto (ℚ,<). The set 𝒞 is dense in Î and ℚ is dense in ℝ; therefore θ extends uniquely to an order preserving bijection θ¯ of Î onto ℝ. This shows that Î is a linear continuum.By this notation one can restate the properties of a Cantor function as follows: Ifψ:I→ℝ is an increasing Cantor function which lives on L, then(i)
ψ is constant on every J∈Î;(ii)
ifx and y are in I, then [x]<[y] if and only if ψ(x)<ψ(y).We define the canonical mapπ:I→Î by π(x)=[x]. Since π is an increasing surjection and both I and Î are linear continuums, π is continuous.The following illustrates general properties of the elements ofRealm(ϕ) in the language of the preceding notation.Proposition 3.2.
Letf∈Realm(ϕ). Then(a)
ifx and y are in I and [x]<[y], then [f(x)]<[f(y)];(b)
for everyx∈I one has f([x])⊆[f(x)];(c)
for eachK∈Î there exists a J∈Î such that f(J)⊆K.Proof.
Putα≔indϕ(f).(a)
From[x]<[y] we conclude that ϕ(x)<ϕ(y). Then
(3.6)ϕ(f(x))=ϕ(x)+α<ϕ(y)+α=ϕ(f(y)).
Therefore, [f(x)]<[f(y)].(b)
Lety∈[x]. Then ϕ(y)=ϕ(x). Thus ϕ(f(y))=ϕ(y)+α=ϕ(x)+α=ϕ(f(x)). This implies that f(y)∈[f(x)].(c)
Part (c) is equivalent to saying that for everyy∈I there exists an x∈I such that f([x])⊆[y]. To prove this equivalent statement suppose that y∈I. Put t=ϕ(y)-α. Since ϕ(I)=ℝ there exists an x∈I such that ϕ(x)=t=ϕ(y)-α. thus ϕ(f(x))=ϕ(x)+α=ϕ(y). So [f(x)]=[y]. Now by part (b)(3.7)f([x])⊆[f(x)]=[y].A particular subset ofRealm(ϕ) is of special importance for us since it is useful for constructing spoiled disjoint groups. To indicate it we need the following.Proposition 3.3.
Letf∈Realm(ϕ). The following are equivalent. (a)
ϕ(L*)+indϕ(f)=ϕ(L*).(b)
ϕ(L**)+indϕ(f)=ϕ(L**).(c)
f(L**)=L**.(d)
f maps each component of I-L** into another one; moreover for every K∈𝒞 there exists J∈𝒞 such that f(J)⊆K.This proposition suggests some notation:(3.8)ρ(ϕ)≔{f∈Realm(ϕ):f(L**)=L**},Ind(ϕ)≔{indϕ(f):f∈ρ(ϕ)}=indϕ(ρ(ϕ)).The equivalency of parts (a) and (c) of Proposition3.3 yields the following.Corollary 3.4.
Withϕ and  L as above one has
(3.9)Ind(ϕ)={α∈R:α+ϕ(L*)=ϕ(L*)}.Proof.
Letα∈Ind(ϕ). Then α=indϕ(f) for some f∈ρ(ϕ). By Proposition 3.3 we have
(3.10)α+ϕ(L*)=indϕ(f)+ϕ(L*)=ϕ(L*).
Conversely, suppose that α is a real number such that α+ϕ(L*)=ϕ(L*). By Proposition 3.1(a) there exists an f∈Realm(ϕ) such that α=indϕ(f). Proposition 3.3 and the definition of ρ(ϕ) imply that f∈ρ(ϕ). Therefore α∈Ind(ϕ).
This completes the proof.In general ifG is an Abelian group and E a nonempty subset of G, we denote(3.11)LasG(E)≔{x∈G:xE=E}.
By this notation Corollary 3.4 is restated as Ind(ϕ)=Lasℝ(ϕ(L*)). It turns out that LasG(E) is a subgroup of G; we name it the subgroup laterally generated byE. In fact we have the following.Proposition 3.5.
LetG be an Abelian group and  E be a nonempty subset of  G. Then (a)
LasG(E) is a subgroup of G.(b)
E is a subgroup of G if and only if LasG(E)=E.(c)
LasG(E)=LasG(Ec) (where Ec=G-E).Proof.
Clearlye∈LasG(E). If x and y are in LasG(E), then
(3.12)(xy)E=x(yE)=xE=E.
Thus, xy∈LasG(E). Moreover, from xE=E one concludes E=x-1E. Hence x-1∈LasG(E). This proves (a).
IfLasG(E)=E, then by (a), E is a subgroup of G. Conversely, suppose that E is a subgroup of G. To show that LasG(E)⊆E let x∈LasG(E). Since e∈E one has xe∈xE. So that x∈E.To show that E⊆LasG(E) let x∈E. Since E is a group, xE=E. Hence x∈LasG(E). This proves (b).
First we show thatLasG(E)⊆LasG(Ec). Let a∈LasG(E). Then aE=E. We want to show that aEc⊆Ec. To do this let x∈Ec. If ax was in E, a-1(ax) would be in a-1E. Since a-1∈LasG(E), a-1E=E. Hence x would be in E which is a contradiction. So we must have ax∈Ec. This shows that aEc⊆Ec.
To show thatEc⊆aEc let a∈LasG(E). Since a-1∈LasG(E), one has a-1Ec⊆Ec by applying the preceding discussion on a-1 in place of a. Multiplying both sides of this relation by a it follows that Ec⊆aEc. We have therefore shown that aEc=Ec. Ora∈LasG(Ec). This shows that LasG(E)⊆LasG(Ec).
Next we show thatLasG(Ec)⊆LasG(E). Plugging Ec in the relation LasG(E)⊆LasG(Ec) in place of E one gets
(3.13)LasG(Ec)⊆LasG((Ec)c)=LasG(E).Proof of Proposition3.3.
We proceed as follows: (a)⇒(d)⇒(c)⇒(b)⇒(a).
Putα≔indϕ(f).
(a)⇒(d): suppose that J is a component of I-L**. Pick x in J. In this case J=[x]. By Proposition 3.2, we have f([x])⊆[f(x)]. We must show that [f(x)] is a component of I-L**. To do this it suffices to show f(x)∈I-L**. By (a) one has
(3.14)ϕ(f(x))=ϕ(x)+α∈ϕ(L*)+α=ϕ(L*),
so f(x)∈I-L**.
Now letK∈𝒞. By Proposition 3.2 there exists J∈Î such that f(J)⊆K. We seek to show that J∈𝒞. Pick x in J. Then J=[x] and K=[f(x)]. So f(x)∈I-L**. Hence, ϕ(x)+α=ϕ(f(x))∈ϕ(L*). Or ϕ(x)∈ϕ(L*)-α. Since -α∈Lasℝ(ϕ(L*)), we have ϕ(L*)-α=ϕ(L*) and ϕ(x)∈ϕ(L*). Therefore x∈I-L**. This shows that J∈𝒞.
(d)⇒(c): to show that f(L**)⊆L**, let x∈L**. Assume to get a contradiction that f(x)∈I-L**. In this case [f(x)]∈𝒞. By (d) one has f(J)⊆[f(x)] for some J∈𝒞. On the other hand, f([x])⊆[f(x)]. Part (a) of Proposition 3.2 implies J=[x]={x}. But this is impossible because J⊆I-L** and x∈L**.
Now we show thatL**⊆f(L**). Let y∈L**. By Proposition 3.2 there exists J∈Î such that f(J)⊆[y]. So that f(J)⊆L**. Hence by (d), J cannot be a component of I-L**. That is, J=[x]={x} for some x∈L**. Clearly y=f(x). Thus y∈f(L**).
(c)⇒(b): To show that ϕ(L**)+α⊆ϕ(L**) let x∈L**. Then f(x)∈L**. So
(3.15)ϕ(x)+α=ϕ(f(x))∈ϕ(L**).
We now show thatϕ(L**)⊆ϕ(L**)+α. Let y∈ϕ(L**). Then y=ϕ(x) for some x∈L**. Since f(L**)=L**,x=f(a) for some a∈L**. Thus
(3.16)y=ϕ(x)=ϕ(f(a))=ϕ(a)+α∈ϕ(L**)+α.
This proves (b).
(b)⇒(a): noticing that ϕ(L*)=ℝ-ϕ(L**), this follows from part (c) of Proposition 3.5.Lemma 3.6.
LetL and M be two Cantor-like sets fitted in I. If there exists a Cantor function ψ:I→ℝ which lives on both L and M, then L=M.Proof.
Letx∈I-L. Then x lies in the interior of a component J of I-L**. Since ψ lives on L, it is constant on J. On the other hand, ψ lives also on M. So J is included in a component of I-M**. Hence x∈I-M. This shows that M⊆L. By the symmetry we conclude that L=M.Proposition 3.7.
(a) Iff is a continuous function in ρ(ϕ), then the image under f of every component of I-L** is another one. Moreover, f is surjective.
(b) IfG is a noncyclic disjoint subgroup of Bi(I) such that G⊆Realm(ϕ), then G⊆ρ(ϕ) and L(G)=L. Moreover, indϕ establishes an isomorphism of (G,∘,<) onto (indϕ(G),+,<).Proof.
(a) First we show thatf is surjective. By Proposition 3.3 we have L**=f(L**)⊆f(I). Since supL**=supI and infL**=infI, we have supf(I)=supI and inff(I)=infI. On the other hand, since f is continuous, f(I) is connected. Therefore, f(I)=I and f is surjective.
Now letJ a component of I-L**. According to Proposition 3.3, f(J)⊆K for some component K of I-L**. Since f is surjective and since by Proposition 3.2(a) the function f does not map any point of I-J into K, one concludes f(J)=K.
(b) By Theorems1.4 and 1.5 the function ϕ lives on L(G). On the other hand by the hypothesis ϕ lives on L. So by Lemma 3.6, L(G)=L. Proposition 2.4 now gives G⊆ρ(ϕ).
Finally, we turn to the mapindϕ. By Proposition 3.1(a), indϕ is a homomorphism of (G,∘) onto (indϕ(G),+). Let f and g be in G and f<g. Pick a point a in L**. Noticing that f(a) and g(a) are in L** we have
(3.17)ϕ(a)+indϕ(f)=ϕ(f(a))<ϕ(g(a))=ϕ(a)+indϕ(g).
Therefore indϕ(f)<indϕ(g). This completes the proof.


## 4.  Spoiled Disjoint Groups

In this section we use Cantor functions to determine the structure of spoiled disjoint (iteration) subgroups ofBi(I).Proposition 4.1.
LetG and  H be two spoiled disjoint subgroups of  Bi(I). If there exists a Cantor function  ϕ:I→ℝ such that  G and  H are both contained in the realm of  ϕ and  indϕ(G)=indϕ(H), then  L(G)=L(H) and  H=γ∘G∘γ-1 for some  γ∈Bi(I) which is identity on  L(G).Proof.
SinceG⊆Realm(ϕ) and H⊆Realm(ϕ), we conclude by Proposition 3.7(b) that ϕ lives on both L(G) and L(H). By Lemma 3.6 we have L(G)=L(H). Put L≔L(G) and suppose that Î and 𝒞 are as in Section 3. On 𝒞 define J~K provided f(J)=K for some f∈G (note that in view of Proposition 3.7 relation ~ is well defined). Clearly ~ is an equivalence relation. Form a set 𝒜 by choosing one element in each equivalence class modulo ~.
Letλ1 and λ2 be the restrictions of the map indϕ to G and H, respectively. Proposition 3.7(b) now implies that the maps λ1 and λ2 are monotone isomorphisms. For each f∈G there exists a unique f0∈H such that indϕ(f)=indϕ(f0); in fact f0=λ2-1∘λ1(f). Note that the map f↦f0 is therefore an isomorphism of G onto H. It is easy to see that for each x∈I, f(x) and f0(x) belong to the same element of Î. In particular the functions f and f0 agree on L. Define the map γ as follows:
the definition ofγ on L**: For each x∈L** define γ(x)=x;
the definition ofγ on I-L**: Let K∈𝒞. Then there exists a unique element J∈𝒜 such that J~K. We denote by fJ,K the unique function in G that maps J onto K. Now for every x∈J define
(4.1)γ(fJ,K(x))=(fJ,K)0(x).
This implies that γ(y)=(fJ,K)0∘(fJ,K)-1(y) for all y∈K. It follows that γ maps K in a strictly increasing manner onto itself.
We claim thatγ satisfies the conditions of the theorem. First we show that γ is strictly increasing. Let x and y be in I and x<y. If [x]=[y], then since γ is strictly increasing on [x], one has γ(x)<γ(y). If [x]<[y], then noticing that γ(x)∈[x] and γ(y)∈[y], we get γ(x)<γ(y).
Sinceγ maps each element of Î onto itself, it follows that γ is surjective.
We now show thatγ∘G∘γ-1=H. It suffices to show that γ∘f∘γ-1(y)=f0(y) for all f∈G, y∈I. Or equivalently γ∘f(y)=f0∘γ(y) for all f∈G, y∈I. To do this let f∈G and y∈I. First suppose that y∈L**. Then f(y)∈L**. So
(4.2)γ∘f(y)=γ(f(y))=f(y)=f0(y)=f0(γ(y))=f0∘γ(y).
Next suppose thaty∈I-L**. Put K=[y] and M=f(K). There exists a unique J∈𝒜 such that J~K. Moreover, y=fJ,K(x) for some x∈J. We have
(4.3)γ∘f(y)=γ∘f∘fJ,K(x)=γ∘fJ,M(x)=(fJ,M)0(x)=(f∘fJ,K)0(x)=f0∘(fJ,K)0(x)=f0∘γ(fJ,K(x))=f0∘γ(y).
Therefore γ∘f(y)=f0∘γ(y) for all f∈G, y∈I as asserted.We continue with introducing the piecewise linear group generated by a Cantor function. Letϕ:I→ℝ be an increasing Cantor function which lives on a Cantor-like set L fitted in I. Let α∈Ind(ϕ). To this α we correspond the piecewise linear function δϕ,α as follows. By the definition of Ind(ϕ) there exists an f∈ρ(ϕ) such that indϕ(f)=α. If x∈L**, put δϕ,α(x)=f(x). If x∈I-L** let δϕ,α be the line which maps the closed interval [x] increasingly onto the closed interval [f(x)]. We claim that δϕ,α∈ρ(ϕ)⋂Bi(I) and indϕ(δϕ,α)=α.First we must show that the definition ofδϕ,α does not depend on the choice of f with indϕ(f)=α. To see this let g be another element of ρ(ϕ) such that indϕ(g)=α. In this case for every x∈I one has
(4.4)ϕ(f(x))=ϕ(x)+α=ϕ(g(x)),
hence [f(x)]=[g(x)]. In particular f and g agree on L** since being elements of ρ(ϕ), for every x∈L** we have {f(x)}=[f(x)]=[g(x)]={g(x)}. Thus δϕ,α is well defined.Clearly[δϕ,α(x)]=[f(x)] for every x∈I. Hence
(4.5)ϕ(δϕ,α(x))=ϕ(f(x))=ϕ(x)+α.
So that δϕ,α∈Realm(ϕ) and indϕ(δϕ,α)=α.Thatδϕ,α is surjective is clear by Proposition 3.2(c).To show thatδϕ,α is strictly increasing, let x and y be in I and x<y. If [x]=[y], then δϕ,α(x)<δϕ,α(y) by the definition of δϕ,α. If [x]<[y], then [δϕ,α(x)]<[δϕ,α(y)] by Proposition 3.2(a). So that δϕ,α(x)<δϕ,α(y).Finally,δϕ,α(L**)=f(L**)=L**. Therefore δϕ,α∈ρ(ϕ)⋂Bi(I) as asserted.Now putΔ(ϕ)≔{δϕ,α:α∈Ind(ϕ)}. We claim that Δ(ϕ) is a disjoint group and call it the piecewise linear group generated byϕ.Theorem 4.2.
Withϕ as above,  Δ(ϕ) is a disjoint subgroup of  Bi(I) which is isomorphic to  Ind(ϕ) as an ordered group. Moreover, if  Δ(ϕ) is noncyclic, it is a spoiled disjoint group such that  L(Δ(ϕ))=L and in particular  Δ(ϕ) is a maximal element in the set of all disjoint subgroups of  Bi(I).Proof.
First we show thatΔ(ϕ) is a group. Clearly δϕ,0=i∈Δ(ϕ). Let α and β be in Ind(ϕ). Proposition 3.1 gives
(4.6)indϕ(δϕ,α∘δϕ,β-1)=α-β=indϕ(δϕ,α-β).
In other words δϕ,α∘δϕ,β-1 and δϕ,α-β are two piecewise linear functions in Realm(ϕ) whose indices with respect to ϕ are the same. Since such a function is unique,
(4.7)δϕ,α∘δϕ,β-1=δϕ,α-β∈Δ(ϕ).
Therefore Δ(ϕ) is a subgroup of Bi(I).
To show thatΔ(ϕ) is disjoint, let α and β be in Ind(ϕ) and δϕ,α(a)=δϕ,β(a) for some a∈I. Then
(4.8)ϕ(a)+α=ϕ(δϕ,α(a))=ϕ(δϕ,β(a))=ϕ(a)+β,
hence α=β. So that δϕ,α=δϕ,β. It follows that Δ(ϕ) is disjoint.
Suppose thatΔ(ϕ) is not cyclic. By Proposition 3.7(b) we get L(Δ(ϕ))=L. The same Proposition implies that indϕ is an isomorphism of (Δ(ϕ),∘,<) onto (Ind(ϕ),+,<).
To show the maximality ofΔ(ϕ) when it is noncyclic, suppose to get a contradiction that there exists a disjoint subgroup G of Bi(I) which strictly contains Δ(ϕ). So G is spoiled by Proposition 2.2. Thus G⊆Realm(ψ) for some increasing Cantor function ψ:I→ℝ. Accordingly Δ(ϕ) is contained in both Realm(ϕ) and Realm(ψ). Since Δ(ϕ) is noncyclic, Realm(ϕ)=Realm(ψ) by Theorem 1.5. Thus Realm(ϕ)=Realm(ψ) and indϕ=indψ. Hence Ind(ϕ) is a strict subset of indϕ(G) because Ind(ϕ)=indϕ(Δ(ϕ)). But on the other hand by Proposition 3.7, we have indϕ(G)⊆Ind(ϕ). This gives a contradiction and confirms the maximality of Δ(ϕ).


### 4.1.  The Structure of Spoiled Disjoint Subgroups ofBi(I)

LetL be a Cantor-like set fitted in I and 𝒞 be the set of all components of I-L**. Let E be a countable dense subset of ℝ and θ an order preserving bijection of 𝒞 onto E (such a θ exists for both structures (𝒞,<) and (E,<) are of order type η). Since 𝒞 is dense in Î={[x]:x∈I}, θ extends to an order preserving bijection θ¯:Î→ℝ. Put ϕ=θ¯∘π. Then ϕ is an increasing Cantor function which lives on L and ϕ(L*)=E. Define
(4.9)Δ(L,E,θ)≔Δ(ϕ).
Note that by Corollary 3.4 we have
(4.10)indϕ(Δ(ϕ))=Ind(ϕ)=LasR(ϕ(L*))=LasR(E),
so (Δ(ϕ),∘,<) is isomorphic to (Lasℝ(E),+<) via indϕ.IfE is an additive subgroup of ℝ, then Lasℝ(E)=E; therefore (Δ(ϕ),∘,<) is isomorphic to (E,+,<). Combining this fact with Theorem 4.2 the following existence theorem is concluded.Theorem 4.3.
For every Cantor-like setL fitted in  I and every countable dense subgroup  A of  (ℝ,+) there exists a maximal spoiled disjoint subgroup  G of  Bi(I) such that  G is (as an ordered group) isomorphic to  A and  L(G)=L.Now one is ready to determine the structure of spoiled disjoint subgroups ofBi(I).Theorem 4.4.
LetL be a Cantor-like set fitted in  I and  𝒞 be the set of all components of  I-L**. The general form of all spoiled disjoint subgroups of  Bi(I) such that  L(G)=L is given by  G=γ∘Δ∘γ-1 where  γ is a homeomorphism of  I onto itself such that  γ(x)=x for every  x∈L and  Δ is a noncyclic subgroup of  Δ(L,E,θ) for some countable dense subset  E of  ℝ for which  Lasℝ(E) is noncyclic and some order preserving bijection  θ of  𝒞 onto  E.Proof.
Part1
LetG be a spoiled disjoint subgroup of Bi(I) such that L(G)=L. Then by Theorem 1.4, G⊆Realm(ϕ) for some Cantor function ϕ:I→ℝ which lives on L. Put E≔ϕ(L*) and define θ:𝒞→E by θ([x])=ϕ(x). This θ is an order preserving bijection (see Section 3). Put Δ≔{δϕ,α:α∈indϕ(G)}. Then Δ is a subgroup of Δ(L,E,θ) such that indϕ(G)=indϕ(Δ). In particular Δ is noncyclic. By Proposition 4.1 there exists a γ∈Bi(I) which is identity on L and such that G=γ∘Δ∘γ-1.Part2
Suppose thatγ, Δ, L, E, and θ are as in the statement of the theorem and assume G=γ∘Δ∘γ-1. Since 𝒞 is dense in Î and E is dense in ℝ, the map θ extends uniquely to an order preserving bijection θ¯:Î→ℝ. Put ϕ=θ¯∘π. Then ϕ is an increasing Cantor function which lives on L and Δ(L,E,θ)=Δ(ϕ)⊆Realm(ϕ). Since both γ and γ-1 are in Realm(ϕ) and Realm(ϕ) is a monoid, one has G⊆Realm(ϕ).
To show thatG is disjoint suppose that f and g are in G and f(a)=g(a) for some a∈I. Then(4.11)ϕ(a)+indϕ(f)=ϕ(f(a))=ϕ(g(a))=ϕ(a)+indϕ(g),
thus indϕ(f)=indϕ(g). The functions γ-1∘f∘γ and γ-1∘g∘γ are in Δ and we have
(4.12)indϕ(γ-1∘f∘γ)=-indϕ(γ)+indϕ(f)+indϕ(γ)=indϕ(f)=indϕ(g)=-indϕ(γ)+indϕ(g)+indϕ(γ)=indϕ(γ-1∘g∘γ).
By the definition of Δ(ϕ) for every α∈Ind(ϕ) only one element of Δ(ϕ) has index α with respect to ϕ. Hence γ-1∘f∘γ=γ-1∘g∘γ. Canceling γ and γ-1 we get f=g. Therefore, G is disjoint.
SinceG is noncyclic and G⊆Realm(ϕ), Proposition 3.7(b) implies L(G)=L.
This completes the proof.


### 4.2.  The Structure of Spoiled Disjoint Iteration Subgroups ofBi(I)

LetL be a Cantor-like set fitted in I, E be a countable dense subset of ℝ, and c:ℝ→ℝ be an additive function such that E+Im(c)=E. Let θ:𝒞→E be an order preserving bijection and θ¯:Î→ℝ be its extension. Put ϕ=θ¯∘π. Then ϕ is an increasing Cantor function which lives on L(G) and ϕ(L*)=E. So that Im(c)⊆Lasℝ(E)=Ind(ϕ). This implies that for all t∈ℝ, c(t)∈Ind(ϕ). This allows us to set δt:=δϕ,c(t) for every t∈ℝ. Put
(4.13)Δ(L,E,θ,c)≔{δt:t∈R}.
The group Δ≔Δ(L,E,θ,c) is a spoiled disjoint iteration subgroup of Bi(I) such that L(Δ)=L: because for all reals s and t(4.14)δs∘δt=δϕ,c(s)∘δϕ,c(t)=δϕ,c(s)+c(t)=δs+t.IfE is in addition a divisible additive subgroup of ℝ, then there exists an additive function c:ℝ→ℝ such that Im(c)=E. We have E=Im(c)=Lasℝ(E)=Ind(ϕ). So Δ(L,E,θ,c) is isomorphic, as an ordered group, to E. This yields the following existence theorem.Theorem 4.5.
For every Cantor-like setL fitted in  I and every countable divisible subgroup  A of  (ℝ,+) there exists a spoiled disjoint iteration subgroup  G={ft:t∈ℝ} of  Bi(I) such that  G is (as an ordered group) isomorphic to  A and  L(G)=L.The next theorem determines the structure of spoiled disjoint iteration subgroups ofBi(I). It was already stated and proved by Zdun in [6, Theorem 4]. Here is given a new proof as well as a little modification in the statement.Theorem 4.6.
LetL be a Cantor-like set fitted in  I and  𝒞 be the set of all components of  I-L**. The general form of all spoiled disjoint iteration subgroups  G={ft:t∈ℝ} of  Bi(I) such that  L(G)=L is given by the relation  ft=γ∘δt∘γ-1(t∈ℝ) where  γ is a homeomorphism of  I onto itself such that  γ(x)=x for all  x∈L and  {δt:t∈ℝ}=Δ(L,E,θ,c) for some countable dense subset  E of  ℝ, an order preserving bijection  θ:𝒞→E and an additive function  c:ℝ→ℝ such that  E+Im(c)=E.Proof.
Part1
LetG={ft:t∈ℝ} be a spoiled disjoint iteration subgroup of Bi(I) such that L(G)=L. By Theorem 4.4, G=γ∘Δ∘γ-1 where γ is a homeomorphism in 𝔉(I) such that γ|L=i|L, and Δ is a noncyclic subgroup of Δ(L,E,θ) for some countable dense subset E of ℝ such that Lasℝ(E) is noncyclic and some order preserving bijection θ:𝒞→E. Then
(4.15)Δ=γ-1∘G∘γ={γ-1∘ft∘γ:t∈R}.
For every t∈ℝ define δt≔γ-1∘ft∘γ. It is straightforward to see that Δ={δt:t∈ℝ} is an iteration group. Put ϕ≔θ¯∘π. Then the map indϕ is in isomorphism of (Δ,∘,<) onto an additive subgroup of Lasℝ(E). Define c:ℝ→ℝ by c(t)≔indϕ(δt). Then δt=δϕ,c(t). Moreover, c is an additive function such that Im(c) is an additive subgroup of Lasℝ(E). So E+Im(c)=E.Part2
Suppose thatG=γ∘Δ∘γ-1 where Δ=Δ(L,E,θ,c) and γ, E, θ, and c are as in the statement of the theorem. Then by Theorem 4.4, G is a spoiled disjoint subgroup of G and L(G)=L. For every t∈ℝ put ft≔γ∘δt∘γ-1. Then for all reals s and t, fs∘ft=fs+t. Furthermore, G={ft:t∈ℝ}.
This completes the proof.


## 4.1.  The Structure of Spoiled Disjoint Subgroups ofBi(I)

LetL be a Cantor-like set fitted in I and 𝒞 be the set of all components of I-L**. Let E be a countable dense subset of ℝ and θ an order preserving bijection of 𝒞 onto E (such a θ exists for both structures (𝒞,<) and (E,<) are of order type η). Since 𝒞 is dense in Î={[x]:x∈I}, θ extends to an order preserving bijection θ¯:Î→ℝ. Put ϕ=θ¯∘π. Then ϕ is an increasing Cantor function which lives on L and ϕ(L*)=E. Define
(4.9)Δ(L,E,θ)≔Δ(ϕ).
Note that by Corollary 3.4 we have
(4.10)indϕ(Δ(ϕ))=Ind(ϕ)=LasR(ϕ(L*))=LasR(E),
so (Δ(ϕ),∘,<) is isomorphic to (Lasℝ(E),+<) via indϕ.IfE is an additive subgroup of ℝ, then Lasℝ(E)=E; therefore (Δ(ϕ),∘,<) is isomorphic to (E,+,<). Combining this fact with Theorem 4.2 the following existence theorem is concluded.Theorem 4.3.
For every Cantor-like setL fitted in  I and every countable dense subgroup  A of  (ℝ,+) there exists a maximal spoiled disjoint subgroup  G of  Bi(I) such that  G is (as an ordered group) isomorphic to  A and  L(G)=L.Now one is ready to determine the structure of spoiled disjoint subgroups ofBi(I).Theorem 4.4.
LetL be a Cantor-like set fitted in  I and  𝒞 be the set of all components of  I-L**. The general form of all spoiled disjoint subgroups of  Bi(I) such that  L(G)=L is given by  G=γ∘Δ∘γ-1 where  γ is a homeomorphism of  I onto itself such that  γ(x)=x for every  x∈L and  Δ is a noncyclic subgroup of  Δ(L,E,θ) for some countable dense subset  E of  ℝ for which  Lasℝ(E) is noncyclic and some order preserving bijection  θ of  𝒞 onto  E.Proof.
Part1
LetG be a spoiled disjoint subgroup of Bi(I) such that L(G)=L. Then by Theorem 1.4, G⊆Realm(ϕ) for some Cantor function ϕ:I→ℝ which lives on L. Put E≔ϕ(L*) and define θ:𝒞→E by θ([x])=ϕ(x). This θ is an order preserving bijection (see Section 3). Put Δ≔{δϕ,α:α∈indϕ(G)}. Then Δ is a subgroup of Δ(L,E,θ) such that indϕ(G)=indϕ(Δ). In particular Δ is noncyclic. By Proposition 4.1 there exists a γ∈Bi(I) which is identity on L and such that G=γ∘Δ∘γ-1.Part2
Suppose thatγ, Δ, L, E, and θ are as in the statement of the theorem and assume G=γ∘Δ∘γ-1. Since 𝒞 is dense in Î and E is dense in ℝ, the map θ extends uniquely to an order preserving bijection θ¯:Î→ℝ. Put ϕ=θ¯∘π. Then ϕ is an increasing Cantor function which lives on L and Δ(L,E,θ)=Δ(ϕ)⊆Realm(ϕ). Since both γ and γ-1 are in Realm(ϕ) and Realm(ϕ) is a monoid, one has G⊆Realm(ϕ).
To show thatG is disjoint suppose that f and g are in G and f(a)=g(a) for some a∈I. Then(4.11)ϕ(a)+indϕ(f)=ϕ(f(a))=ϕ(g(a))=ϕ(a)+indϕ(g),
thus indϕ(f)=indϕ(g). The functions γ-1∘f∘γ and γ-1∘g∘γ are in Δ and we have
(4.12)indϕ(γ-1∘f∘γ)=-indϕ(γ)+indϕ(f)+indϕ(γ)=indϕ(f)=indϕ(g)=-indϕ(γ)+indϕ(g)+indϕ(γ)=indϕ(γ-1∘g∘γ).
By the definition of Δ(ϕ) for every α∈Ind(ϕ) only one element of Δ(ϕ) has index α with respect to ϕ. Hence γ-1∘f∘γ=γ-1∘g∘γ. Canceling γ and γ-1 we get f=g. Therefore, G is disjoint.
SinceG is noncyclic and G⊆Realm(ϕ), Proposition 3.7(b) implies L(G)=L.
This completes the proof.


## 4.2.  The Structure of Spoiled Disjoint Iteration Subgroups ofBi(I)

LetL be a Cantor-like set fitted in I, E be a countable dense subset of ℝ, and c:ℝ→ℝ be an additive function such that E+Im(c)=E. Let θ:𝒞→E be an order preserving bijection and θ¯:Î→ℝ be its extension. Put ϕ=θ¯∘π. Then ϕ is an increasing Cantor function which lives on L(G) and ϕ(L*)=E. So that Im(c)⊆Lasℝ(E)=Ind(ϕ). This implies that for all t∈ℝ, c(t)∈Ind(ϕ). This allows us to set δt:=δϕ,c(t) for every t∈ℝ. Put
(4.13)Δ(L,E,θ,c)≔{δt:t∈R}.
The group Δ≔Δ(L,E,θ,c) is a spoiled disjoint iteration subgroup of Bi(I) such that L(Δ)=L: because for all reals s and t(4.14)δs∘δt=δϕ,c(s)∘δϕ,c(t)=δϕ,c(s)+c(t)=δs+t.IfE is in addition a divisible additive subgroup of ℝ, then there exists an additive function c:ℝ→ℝ such that Im(c)=E. We have E=Im(c)=Lasℝ(E)=Ind(ϕ). So Δ(L,E,θ,c) is isomorphic, as an ordered group, to E. This yields the following existence theorem.Theorem 4.5.
For every Cantor-like setL fitted in  I and every countable divisible subgroup  A of  (ℝ,+) there exists a spoiled disjoint iteration subgroup  G={ft:t∈ℝ} of  Bi(I) such that  G is (as an ordered group) isomorphic to  A and  L(G)=L.The next theorem determines the structure of spoiled disjoint iteration subgroups ofBi(I). It was already stated and proved by Zdun in [6, Theorem 4]. Here is given a new proof as well as a little modification in the statement.Theorem 4.6.
LetL be a Cantor-like set fitted in  I and  𝒞 be the set of all components of  I-L**. The general form of all spoiled disjoint iteration subgroups  G={ft:t∈ℝ} of  Bi(I) such that  L(G)=L is given by the relation  ft=γ∘δt∘γ-1(t∈ℝ) where  γ is a homeomorphism of  I onto itself such that  γ(x)=x for all  x∈L and  {δt:t∈ℝ}=Δ(L,E,θ,c) for some countable dense subset  E of  ℝ, an order preserving bijection  θ:𝒞→E and an additive function  c:ℝ→ℝ such that  E+Im(c)=E.Proof.
Part1
LetG={ft:t∈ℝ} be a spoiled disjoint iteration subgroup of Bi(I) such that L(G)=L. By Theorem 4.4, G=γ∘Δ∘γ-1 where γ is a homeomorphism in 𝔉(I) such that γ|L=i|L, and Δ is a noncyclic subgroup of Δ(L,E,θ) for some countable dense subset E of ℝ such that Lasℝ(E) is noncyclic and some order preserving bijection θ:𝒞→E. Then
(4.15)Δ=γ-1∘G∘γ={γ-1∘ft∘γ:t∈R}.
For every t∈ℝ define δt≔γ-1∘ft∘γ. It is straightforward to see that Δ={δt:t∈ℝ} is an iteration group. Put ϕ≔θ¯∘π. Then the map indϕ is in isomorphism of (Δ,∘,<) onto an additive subgroup of Lasℝ(E). Define c:ℝ→ℝ by c(t)≔indϕ(δt). Then δt=δϕ,c(t). Moreover, c is an additive function such that Im(c) is an additive subgroup of Lasℝ(E). So E+Im(c)=E.Part2
Suppose thatG=γ∘Δ∘γ-1 where Δ=Δ(L,E,θ,c) and γ, E, θ, and c are as in the statement of the theorem. Then by Theorem 4.4, G is a spoiled disjoint subgroup of G and L(G)=L. For every t∈ℝ put ft≔γ∘δt∘γ-1. Then for all reals s and t, fs∘ft=fs+t. Furthermore, G={ft:t∈ℝ}.
This completes the proof.



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*Source: 790758-2012-06-10.xml*