Abstract and Applied Analysis

Volume 2012, Article ID 790758, 14 pages

http://dx.doi.org/10.1155/2012/790758

## The Structure of Disjoint Groups of Continuous Functions

Department of Mathematics, College of Sciences, Shiraz University, Shiraz 71457-44776, Iran

Received 14 February 2012; Accepted 24 April 2012

Academic Editor: Sung Guen Kim

Copyright © 2012 Hojjat Farzadfard and B. Khani Robati. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let *I* 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.

#### 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 such that for all reals and and each 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 paper is an open interval in the real line; the topology of it is considered to be the subspace topology it inherits from . We say that a subset of is a Cantor-like set fitted in if is a nowhere dense set, , and has no lower or upper bound in (here is the set of all limit points of in ). We denote by , , , and to the set of all left sided, right sided, one sided, and two sided limit points of , respectively. Denote by the group of all increasing bijections from onto itself, with composition of functions as its binary operation. There exists a natural ordering relation on the set of all self mappings as follows: given two functions and in define provided and for all . Define if or . With this, is a partial ordering set. A subset of is said to be disjoint if the graphs of any two distinct elements of it do not intersect. If is a disjoint group of continuous functions in , then is a subgroup of and furthermore, is an Archimedean ordered group (see [8, Proposition 3]). Hence according to Hölder’s Theorem there exists an additive subgroup of such that is isomorphic to .

The following two theorems, which have been proved in [5], give an introductory characterization of disjoint subgroups of .

Theorem 1.1. *Let be a disjoint subgroup of . Then *(a)*for all , in one has where (we denote this set by );*(b)*for all , ;*(c)*.*(d)*the set is either an empty set or a Cantor-like set fitted in or .*

Theorem 1.2. *Let be a disjoint subgroup of . Then is a cyclic group if and only if .*

By virtue of these theorems one divides disjoint subgroups of in three classes: The first class consists of the cyclic subgroups; the second one consists of the subgroups of for which is a Cantor-like set fitted in . one names such a group a *spoiled disjoint group*. Finally the third class consists of the subgroups of for which . Such a group is called a *dense disjoint group*. One calls a subgroup of *complete* if for all .

One uses as a tool classes of functions that occur as continuous solutions of simultaneous systems of Abel equations where is a nonempty subset of the group generated by which is noncyclic and disjoint and is a given map. In view of Theorem 6 of [5] either is a homeomorphism or a Cantor function which lives on in the following sense.

*Definition 1.3. *Given a Cantor-like set fitted in we say that is a Cantor function which lives on if (i) is monotone;(ii); (iii) is strictly monotone on .

Such a function is constant on the components of , and is continuous.

Given a function , we say that a function is in the *realm* of if there exists a real number such that for all . Clearly this is unique and we call it the *index of ** with respect to * and denote it by . Denote by the set of all functions that are in the realm of . We say that two functions and are *associate* if for some real constants and . It is easy to see that the associativity is an equivalence relation on the set of all functions from into . Moreover, every two associate functions have the same realms. In particular if and are associate and is a Cantor function which lives on (or a homeomorphism), then is also a Cantor function which lives on (or a homeomorphism). By these notation the following is an immediate consequence of Theorem 10 in [5].

Theorem 1.4. *Let be a disjoint subgroup of . Then for some continuous . If is spoiled, then is a Cantor function which lives on . If is dense, then is a homeomorphism.*

Theorem 12 of [5] can be restate as follows.

Theorem 1.5. *Let be a noncyclic disjoint subgroup of . If and are two continuous functions such that and , then and are associate. In particular .*

In Section 3 we deal with the properties of the elements of when is a Cantor function.

#### 2. Complete and Dense Disjoint Groups

Theorems 2.1 and 2.5 below describe the structure of complete disjoint subgroups of . While Theorem 2.3 describes dense disjoint groups.

Theorem 2.1. *If is a homeomorphism, then is a complete disjoint subgroup of . Moreover, if is increasing, is an isomorphism of onto .**Conversely, to every complete disjoint subgroup of there corresponds a homeomorphism such that .*

*Proof. *For the first part of the theorem (whose proof is straightforward) see [8]. So we prove the converse part. Let be a complete disjoint subgroup of . By Theorem 1.4 one has for some homeomorphism . To show that let . Pick a point in . Then because . It follows that for some . But since , one has . By the first part of the theorem is disjoint; so that . Hence . And we are done.

Our next goal is to present a description of dense disjoint subgroups of . But first a proposition.

Proposition 2.2. *Let and be two disjoint subgroups of and . If is noncyclic, then .*

*Proof. *The proof we present is extracted from the proof of Theorem 3 of the Zdun’s paper [9] with a little modification.

Let . Then . So that . On the other hand by noticing that is Abelian, for every
Since is a homeomorphism,
Fix an in . The by the preceding relation one has for all . Thus . So . Therefore, and the proof is complete.

This proposition yields that if is a noncyclic subgroup of the disjoint group , then either both and are spoiled or they are both dense.

Theorem 2.3. *Let be a noncyclic disjoint subgroup of . Then is a dense group if and only if is embeddable in a complete disjoint subgroup of . Moreover, this complete group is unique.*

*Proof. *If is dense, then by Theorem 1.4, for some homeomorphism . By Theorem 2.1, is complete. Conversely, suppose that for some complete disjoint subgroup of . By Proposition 2.2 one has . Hence, is dense.

The uniqueness part of the theorem follows from Theorems 1.5 and 2.1.

The following proposition will be used now and later.

Proposition 2.4. *Let be a spoiled disjoint subgroup of , and . Then , and . Moreover, is countable.*

*Proof. *The first part of the proposition is in fact Proposition 5 of [5]. To prove the countability of pick a point in . Then . This implies that is countable since is so. On the other hand, and have the same cardinality because is disjoint. Therefore, is countable.

Theorem 2.5. *Let be a disjoint subgroup of . Then is complete if and only if is isomorphic to .*

*Proof. *If is complete, then by Theorem 2.1 one has for some homeomorphism . Since associate functions have the same realms, one can assume that is increasing. Therefore is an isomorphism of onto .

Conversely, suppose that is isomorphic to . Then is uncountable. Since every spoiled group is countable, must be dense. Then by Theorem 2.3 there exists a complete disjoint subgroup of containing . Since by the first part of the theorem is isomorphic to the additive group , is isomorphic to an additive subgroup of via the same isomorphism. But cannot be isomorphic to a strict subgroup of itself. Therefore, and is complete.

#### 3. The Realm of Cantor Functions

Through this section is a Cantor-like set fitted in and is an increasing Cantor function which lives on .

Proposition 3.1. * is a monoid and is a monoid homomorphism from onto .**If is an invertible function in , then and .**If and are in and for some , then .*

*Proof. *It is clear that , the identity functions on , is in and . Let and be in . Then for all
This means that and . That is, is a monoid homomorphism.

To show that is surjective let . Define by
Note that for each , the set is either a component of (which is a closed interval) or a singleton where . So the definition of is meaningful. Since for every , one has . Therefore, and . This proves (a).

For every one has
thus
This means that and .

Let and be in and for some . Then
since is increasing. Therefore, .

For each put provided and let be the component of containing provided . Set , . The following relation makes into a linearly ordered set: given two elements and of define whenever for all , . The topology of is understood to be the order topology. An elementary analysis shows that is a countable dense linearly ordered structure with no minimum or maximum; in other words is of order type (see [10, Chapter 8, Exercises 17 and 18]). In particular is dense in . Since is also of order type , there exists an order preserving bijection of onto . The set is dense in and is dense in ; therefore extends uniquely to an order preserving bijection of onto . This shows that is a linear continuum.

By this notation one can restate the properties of a Cantor function as follows: If is an increasing Cantor function which lives on , then(i) is constant on every ; (ii)if and are in , then if and only if .

We define the canonical map by . Since is an increasing surjection and both and are linear continuums, is continuous.

The following illustrates general properties of the elements of in the language of the preceding notation.

Proposition 3.2. *Let . Then*(a)*if and are in and , then ;*(b)*for every one has ;*(c)*for each there exists a such that .*

*Proof. *Put .(a)From we conclude that . Then
Therefore, .(b)Let . Then . Thus . This implies that .(c)Part (c) is equivalent to saying that for every there exists an such that . To prove this equivalent statement suppose that . Put . Since there exists an such that . thus . So . Now by part (b)

A particular subset of 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. *Let . The following are equivalent. *(a)*.*(b)*.*(c)*.*(d)* maps each component of into another one; moreover for every there exists such that .*

This proposition suggests some notation:

The equivalency of parts (a) and (c) of Proposition 3.3 yields the following.

Corollary 3.4. *With and as above one has
*

*Proof. *Let . Then for some . By Proposition 3.3 we have
Conversely, suppose that is a real number such that . By Proposition 3.1(a) there exists an such that . Proposition 3.3 and the definition of imply that . Therefore .

This completes the proof.

In general if is an Abelian group and a nonempty subset of , we denote
By this notation Corollary 3.4 is restated as . It turns out that is a subgroup of ; we name it the *subgroup laterally generated by *. In fact we have the following.

Proposition 3.5. *Let be an Abelian group and be a nonempty subset of . Then *(a)* is a subgroup of .*(b)* is a subgroup of if and only if .*(c)* (where ).*

*Proof. *Clearly . If and are in , then
Thus, . Moreover, from one concludes . Hence . This proves (a).

If , then by (a), is a subgroup of . Conversely, suppose that is a subgroup of . To show that let . Since one has . So that .To show that let . Since is a group, . Hence . This proves (b).

First we show that . Let . Then . We want to show that . To do this let . If was in , would be in . Since , . Hence would be in which is a contradiction. So we must have . This shows that .

To show that let . Since , one has by applying the preceding discussion on in place of . Multiplying both sides of this relation by it follows that . We have therefore shown that . Or. This shows that .

Next we show that . Plugging in the relation in place of one gets

*Proof of Proposition 3.3. *We proceed as follows: (a)(d)(c)(b)(a).

Put .

(a)(d): suppose that is a component of . Pick in . In this case . By Proposition 3.2, we have . We must show that is a component of . To do this it suffices to show . By (a) one has
so .

Now let . By Proposition 3.2 there exists such that . We seek to show that . Pick in . Then and . So . Hence, . Or . Since , we have and . Therefore . This shows that .

(d)(c): to show that , let . Assume to get a contradiction that . In this case . By (d) one has for some . On the other hand, . Part (a) of Proposition 3.2 implies . But this is impossible because and .

Now we show that . Let . By Proposition 3.2 there exists such that . So that . Hence by (d), cannot be a component of . That is, for some . Clearly . Thus .

(c)(b): To show that let . Then . So

We now show that . Let . Then for some . Since for some . Thus
This proves (b).

(b)(a): noticing that , this follows from part (c) of Proposition 3.5.

Lemma 3.6. *Let and be two Cantor-like sets fitted in . If there exists a Cantor function which lives on both and , then .*

*Proof. *Let . Then lies in the interior of a component of . Since lives on , it is constant on . On the other hand, lives also on . So is included in a component of . Hence . This shows that . By the symmetry we conclude that .

Proposition 3.7. *(a) If is a continuous function in , then the image under of every component of is another one. Moreover, is surjective.**(b) If is a noncyclic disjoint subgroup of such that , then and . Moreover, establishes an isomorphism of onto .*

*Proof. *(a) First we show that is surjective. By Proposition 3.3 we have . Since and , we have and . On the other hand, since is continuous, is connected. Therefore, and is surjective.

Now let a component of . According to Proposition 3.3, for some component of . Since is surjective and since by Proposition 3.2(a) the function does not map any point of into , one concludes .

(b) By Theorems 1.4 and 1.5 the function lives on . On the other hand by the hypothesis lives on . So by Lemma 3.6, . Proposition 2.4 now gives .

Finally, we turn to the map . By Proposition 3.1(a), is a homomorphism of onto . Let and be in and . Pick a point in . Noticing that and are in we have
Therefore . This completes the proof.

#### 4. Spoiled Disjoint Groups

In this section we use Cantor functions to determine the structure of spoiled disjoint (iteration) subgroups of .

Proposition 4.1. *Let and be two spoiled disjoint subgroups of . If there exists a Cantor function such that and are both contained in the realm of and , then and for some which is identity on .*

*Proof. *Since and , we conclude by Proposition 3.7(b) that lives on both and . By Lemma 3.6 we have . Put and suppose that and are as in Section 3. On define provided for some (note that in view of Proposition 3.7 relation *~* is well defined). Clearly *~* is an equivalence relation. Form a set by choosing one element in each equivalence class modulo *~*.

Let and be the restrictions of the map to and , respectively. Proposition 3.7(b) now implies that the maps and are monotone isomorphisms. For each there exists a unique such that ; in fact . Note that the map is therefore an isomorphism of onto . It is easy to see that for each , and belong to the same element of . In particular the functions and agree on . Define the map as follows:

the definition of on : For each define ;

the definition of on : Let . Then there exists a unique element such that . We denote by the unique function in that maps onto . Now for every define
This implies that for all . It follows that maps in a strictly increasing manner onto itself.

We claim that satisfies the conditions of the theorem. First we show that is strictly increasing. Let and be in and . If , then since is strictly increasing on , one has . If , then noticing that and , we get .

Since maps each element of onto itself, it follows that is surjective.

We now show that . It suffices to show that for all , . Or equivalently for all , . To do this let and . First suppose that . Then . So

Next suppose that . Put and . There exists a unique such that . Moreover, for some . We have
Therefore for all , as asserted.

We continue with introducing the piecewise linear group generated by a Cantor function. Let be an increasing Cantor function which lives on a Cantor-like set fitted in . Let . To this we correspond the piecewise linear function as follows. By the definition of there exists an such that . If , put . If let be the line which maps the closed interval increasingly onto the closed interval . We claim that and .

First we must show that the definition of does not depend on the choice of with . To see this let be another element of such that . In this case for every one has hence . In particular and agree on since being elements of , for every we have . Thus is well defined.

Clearly for every . Hence So that and .

That is surjective is clear by Proposition 3.2(c).

To show that is strictly increasing, let and be in and . If , then by the definition of . If , then by Proposition 3.2(a). So that .

Finally, . Therefore as asserted.

Now put . We claim that is a disjoint group and call it the *piecewise linear group generated by *.

Theorem 4.2. *With as above, is a disjoint subgroup of which is isomorphic to as an ordered group. Moreover, if is noncyclic, it is a spoiled disjoint group such that and in particular is a maximal element in the set of all disjoint subgroups of .*

*Proof. *First we show that is a group. Clearly . Let and be in . Proposition 3.1 gives
In other words and are two piecewise linear functions in whose indices with respect to are the same. Since such a function is unique,
Therefore is a subgroup of .

To show that is disjoint, let and be in and for some . Then
hence . So that . It follows that is disjoint.

Suppose that is not cyclic. By Proposition 3.7(b) we get . The same Proposition implies that is an isomorphism of onto .

To show the maximality of when it is noncyclic, suppose to get a contradiction that there exists a disjoint subgroup of which strictly contains . So is spoiled by Proposition 2.2. Thus for some increasing Cantor function . Accordingly is contained in both and . Since is noncyclic, by Theorem 1.5. Thus and . Hence is a strict subset of because . But on the other hand by Proposition 3.7, we have . This gives a contradiction and confirms the maximality of .

##### 4.1. The Structure of Spoiled Disjoint Subgroups of

Let be a Cantor-like set fitted in and be the set of all components of . Let be a countable dense subset of and an order preserving bijection of onto (such a exists for both structures and are of order type ). Since is dense in , extends to an order preserving bijection . Put . Then is an increasing Cantor function which lives on and . Define Note that by Corollary 3.4 we have so is isomorphic to via .

If is an additive subgroup of , then ; therefore is isomorphic to . Combining this fact with Theorem 4.2 the following existence theorem is concluded.

Theorem 4.3. *For every Cantor-like set fitted in and every countable dense subgroup of there exists a maximal spoiled disjoint subgroup of such that is (as an ordered group) isomorphic to and .*

Now one is ready to determine the structure of spoiled disjoint subgroups of .

Theorem 4.4. *Let be a Cantor-like set fitted in and be the set of all components of . The general form of all spoiled disjoint subgroups of such that is given by where is a homeomorphism of onto itself such that for every and is a noncyclic subgroup of for some countable dense subset of for which is noncyclic and some order preserving bijection of onto .*

*Proof. *

Let be a spoiled disjoint subgroup of such that . Then by Theorem 1.4, for some Cantor function which lives on . Put and define by . This is an order preserving bijection (see Section 3). Put . Then is a subgroup of such that . In particular is noncyclic. By Proposition 4.1 there exists a which is identity on and such that .

Suppose that , , , , and are as in the statement of the theorem and assume . Since is dense in and is dense in , the map extends uniquely to an order preserving bijection . Put . Then is an increasing Cantor function which lives on and . Since both and are in and is a monoid, one has .

To show that is disjoint suppose that and are in and for some . Then
thus . The functions and are in and we have
By the definition of for every only one element of has index with respect to . Hence . Canceling and we get . Therefore, is disjoint.

Since is noncyclic and , Proposition 3.7(b) implies .

This completes the proof.

##### 4.2. The Structure of Spoiled Disjoint Iteration Subgroups of

Let be a Cantor-like set fitted in , be a countable dense subset of , and be an additive function such that . Let be an order preserving bijection and be its extension. Put . Then is an increasing Cantor function which lives on and . So that . This implies that for all , . This allows us to set for every . Put The group is a spoiled disjoint iteration subgroup of such that : because for all reals and

If is in addition a divisible additive subgroup of , then there exists an additive function such that . We have . So is isomorphic, as an ordered group, to . This yields the following existence theorem.

Theorem 4.5. *For every Cantor-like set fitted in and every countable divisible subgroup of there exists a spoiled disjoint iteration subgroup of such that is (as an ordered group) isomorphic to and .*

The next theorem determines the structure of spoiled disjoint iteration subgroups of . 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. *Let be a Cantor-like set fitted in and be the set of all components of . The general form of all spoiled disjoint iteration subgroups of such that is given by the relation where is a homeomorphism of onto itself such that for all and for some countable dense subset of , an order preserving bijection and an additive function such that .*

*Proof. *

Let be a spoiled disjoint iteration subgroup of such that . By Theorem 4.4, where is a homeomorphism in such that , and is a noncyclic subgroup of for some countable dense subset of such that is noncyclic and some order preserving bijection . Then
For every define . It is straightforward to see that is an iteration group. Put . Then the map is in isomorphism of onto an additive subgroup of . Define by . Then . Moreover, is an additive function such that is an additive subgroup of . So .

Suppose that where and , , , and are as in the statement of the theorem. Then by Theorem 4.4, is a spoiled disjoint subgroup of and . For every put . Then for all reals and , . Furthermore, .

This completes the proof.

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