Journal of Chaos

Volume 2013 (2013), Article ID 181238, 4 pages

http://dx.doi.org/10.1155/2013/181238

## Unstable Manifolds of Continuous Self-Mappings

^{1}School of Mathematics Sciences, University of Electronic Science and Technology of China, Chengdu 611731, China^{2}Department of Mathematics, Sichuan University of Science and Engineering, Zigong 643000, China

Received 5 March 2013; Revised 16 May 2013; Accepted 23 May 2013

Academic Editor: Tao Xiang

Copyright © 2013 Tianxiu Lu and Peiyong Zhu. 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

Unstable manifolds of continuous self-mappings on completely densely ordered linear ordered topological spaces (CDOLOTS) are discussed. Let be a continuous self-map. First, the interval with endpoints of two adjacent fixed points is contained in the unilateral unstable manifold of one of the endpoints. Then, by using the above conclusion, we prove that periodic points of not belong to the unstable manifold of their iteration points of (for some ), unless the iteration points are themselves.

#### 1. Introduction

The complexity of a dynamical system is a central topic of research since the introduction of the term of chaos in 1975 by Li and Yorke [1], known as Li-Yorke chaos today. As concepts relate to chaotic, generalized periodic points are intensively discussed. Throughout this paper, the sets of fixed points, periodic points, -limit points, and nonwandering points of a system are denoted by , , , and , respectively.

Considering a continuous self-map on a compact interval , Block [2] proved the following results.(1-1) If has finitely many periodic points, then the period of each periodic point is a power of 2.(1-2) If is finite, then .

Based on Block’s results, Xiong [3] obtained that(1-3) if and only if is closed,(1-4) if is closed, then for every point in , .

In 2002, Ding and Nadler [4] showed that the invariant set of an -contractive map on a compact metric space is the same as the set of periodic points of . Furthermore, the set of periodic points of is finite and, only assuming that is locally compact, there is at most one periodic point in each component of . Forti et al. [5] gave an example of a triangular map of the unite square, , possessing periodic orbits of all periods and such that no infinite -limit set of contains a periodic point. Moreover, Forti show that there is a triangular map of type monotone on the fibres such that any recurrent point of is uniformly recurrent. And restricted to the set of its recurrent point is chaotic in the sense of Li and Yorke. Mai and Shao [6] obtained a structure theorem of graph maps without periodic points, which states that any graph map without periodic points must be topologically conjugate to one of the described class. Recently, Abbas and Rhoades [7] proved that some fixed point theorems in cone metric spaces gave the fact that in a cone with only a partial ordering, the continuous maps have no nontrivial periodic points.

However, the research of generalized periodic points on topological space is very few. The current paper studies continuous self-maps on CDOLOTS. To characterize the properties of continuous self-maps, it is necessary to study unstable manifolds on CDOLOTS. The conclusions in this paper is the generalization of the ones on real line (see [2, 8]). Furthermore, this paper gave a counterexample to examine that a condition of Theorem 8 is indispensable.

#### 2. Preliminaries

Throughout this paper, the order relationship of linear ordered set is denoted by “”. And denotes “ or ”. is the system of all open neighborhoods of a point . Definitions and notations of intervals in are similar to the real line. Infinite intervals are denoted by , , and . If is empty, will be called for immediate predecessor of and will be called for immediate successor of . If there exists an element such that for all , will be called for the largest element of and denoted by . If there exists an element such that for all , will be called for smallest element of and denoted by . If is a topological space with the subbase , then is called a linear ordered topological space. We say that is densely ordered if whenever with , there is an element such that . We say that is complete if every nonempty set with upper bound has a least upper bound in . A space is called a completely densely ordered linear ordered topological space (CDOLOTS), if it is a linear ordered topological space and its order relation is complete and densely ordered. In Munkres [9], there are some examples of CDOLOTS which are different from real line. A separation of is a pair , of disjoint nonempty open subsets of whose union is . The space is said to be connected if there does not exist a separation of . According to [9], if be a CDOLOTS, then is connected, and so are integers and rays in . A connected set is an interval if it includes more than one point. The definitions of other basic concepts (e.g., continuous self-mapping, periodic point, and periodic orbit) are as usual (see [9–11]).

Let be a continuous map. Then the image of a closed interval in is a closed interval. Let . The unstable manifold is defined as follows. Denote if for any neighborhood of , for some positive integer . If is a fixed point of , unilateral unstable manifolds and are defined as follows. Denote if for every interval with left endpoint , for some positive integer . Denote if for every interval with right endpoint , for some positive integer . By [8], the unstable manifold (or unilateral unstable manifold) of a fixed point on is connected.

The following property follows easily from the definitions.

Lemma 1. *Let be a CDOLOTS and a continuous map. If is a fixed point of , then .*

#### 3. Inclusion Relationship

Lemma 2. *Let be a general topological space and a totally ordered set. and are continuous maps. Then is an open set in .*

*Proof. *Given any , it is clear that and . Now we consider two cases.*Case 1*. If is an immediate predecessor of , by continuity of , it follows that . For any , we have that and , then , . Noting the fact that is immediate predecessor of , one has . This implies that .*Case 2*. If is not an immediate predecessor of , then there exist some such that . Clearly, and . Thus it is not difficult to check that and .

Summing up Cases 1 and 2, according to the arbitrariness of , it follows that is an open set in .

In what follows, is a CDOLOTS and a continuous map.

Lemma 3. *Let with . If , then one of the following two cases holds:*(i)* for any ;*(ii)* for any .*

*Proof. *Suppose that there exist , in such that and . Define a continuous map by for any . Let us take
Clearly, and . Combining this with Lemma 2, it is not difficult to check that and constitute a separation of , which contradicts the connectivity of .

Theorem 4. *If and are two adjacent fixed points of with , then
*

*Proof. *Since and are two adjacent fixed points of with , applying Lemma 3, it follows that one of the following cases holds:(i) for any ;(ii) for any .

Without loss of generality, we may assume that case (i) holds.

Now we assert that . That is, given any fixed , for any , there exists a such that . Take . To prove this, we consider three cases.*Case 1* . Noting that holds for any and that , one has that
(1-1) If , then .(1-2) If , we have that for some .

In fact, suppose that holds for any , and then

For any , as , we have . Combining this with the fact that holds for any , it follows that
Take . Clearly, . Meanwhile, it is easy to see that
*Case 2* . Observing that and , we know that .*Case 3* . Then .

According to Lemma 1 and Theorem 4, Corollary 5 follows immediately.

Corollary 5. *If and are two adjacent fixed points of with , then
*

#### 4. Separability

Lemma 6. *Let be a nonempty closed interval such that or . Then .*

*Proof. *Without loss of generality, we may assume that .

When , this holds trivially. It remains to consider the case that .

Since , then there exist some such that
(1)If or , the conclusion is clear.(2)If and , then and . is an open set with endpoints and . Then, there exists an element in such that . Otherwise, it contradicts that is a connect set.

Lemma 7. *Let is a fixed point of and . Then if and if .*

*Proof. *Suppose that (the case is similar).*Case 1*. If is the smallest element of , then for arbitrary in . Since , one has for some integer . Thus, .*Case 2*. If is not the smallest element of , let is a fixed point adjacent to with (or let arbitrary in which satisfied if there are no fixed points which is ahead of ). Since , then.

(i) for every in or (ii) for every in .

And because is a fixed point of , one has for every positive integer .

We prove by contradiction.

Suppose , then . There exists a positive integer such that . Then , a contradiction.

Thus .

Theorem 8. *If has finitely many periodic points. is a periodic orbit of with period . And are distinct elements on the periodic orbit. Then .*

*Proof. *According to the connectivity of the unstable manifold at a fixed point, it is obvious that is an interval. By resorting, one can assume that . Use reduction to absurdity.

Suppose there exist such that .

For arbitrary .

If , then
If , then for some . Since is a continuous map, there exists a neighborhood of such that for every neighborhood of . Because , there exists an integer such that . Then
One has

Since and , one has . Then
Combining with (9) and (12), clearly, for arbitrary , contains at least one element of . Thus is an interval which contains and . Similarly, one has
Next we will show that .

In fact, by Lemma 7, if , then and . One has for some integer and some in .

Put

For arbitrary in , there exists an element in such that (otherwise it contradicts the definition of ). Then contains an interval of the form . Since , for the above interval , there exists a positive integer such that . And because , then
By , one has . Then
So has a periodic point in by Lemma 6. Since is an arbitrary point with , then has infinitely many periodic points, a contradiction.

Therefore, by (13), .

Similar to the above discussion, it follows that
In particular, . But is an interval containing and some elements of , then . By the same argument as the preceding paragraphs, it follows that has infinitely many periodic points, a contradiction.

We thus conclude that .

*Remark 9. *Since is the most natural example of CDOLOTS, Lemma 4 in [8], Lemma 3 in [8], Lemma 6 in [2], and Theorem 8 in [2] are corollaries of Theorem 4, Lemma 6, Lemma 7, and Theorem 8 in this paper, respectively.

*Remark 10. *Practically, the results of Theorems 4 and 8 are ture for continuous maps on compact intervals. This paper makes it clear that topological structure of spaces has no impact on these results, if the spaces are completely and densely ordered.

*Remark 11. *The condition “ has finitely many periodic points” in Theorem 8 cannot be removed. We will give an example to show it.

*Example 12. *Triangular tent map by for and for .

As one knows, for every positive integer , has periodic points with period . So has infinitely many periodic points.

According the image of for every positive integer , is a periodic function with period . was divided into subintervals which have the same length . That is, , , , , . The image of on every subinterval is the same. The image on the first subinterval was observed, with increasing from 0 to 1 with slope on and decreasing from 1 to 0 with slope on . So, it is easy to check that there exists a positive integer such that for every open integer (there ).

Now, we study the 2-periodic orbit of . For every , let (there ). By the previous paragraph, there exists a positive integer such that . Then there exists an even such that . That is, .

#### Acknowledgments

This project was supported by the Fundamental Research Funds for the Central Universities, the Scientific Research Fund of Sichuan Provincial Education Department (no. 12ZA098), and Artificial Intelligence of Key Laboratory of Sichuan Province (no. 2012RYY04).

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