﻿Second Descendible Self-Mapping with Closed Periodic Points Set

Applied Mathematics
Vol.4 No.7(2013), Article ID:33806,3 pages DOI:10.4236/am.2013.47133

Second Descendible Self-Mapping with Closed Periodic Points Set

Gengrong Zhang1, Zhanjiang Ji1, Fanping Zeng1,2

1College of Mathematics and Information Science, Guangxi University, Nanning, China

2Department of Mathematics and Computer Science, Liuzhou Teachers College, Liuzhou, China

Email: zgrzaw@gxu.edu.cn

Copyright © 2013 Gengrong Zhang et al. 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.

Received February 21, 2013; revised March 22, 2013; accepted March 31, 2013

Keywords: Periodic Point; Recurrent Point; w-Limit Point; Second Descendible Map

ABSTRACT

Let and be a continuous map. If f is a second descendible map, then is closed if and only if one of the following hold: 1); 2) For any, there exists a such that every point of the set is a isolated point of the set; 3) For any, the set is finite; 4) For any, the set is finite. The consult give another condition of f with closed periodic set other than [1].

1. Introduction

In this paper, let denote, X denote compact metric space, denote all continuous self-maps on X. The concepts of periodic point, w-limit point of z and the orbit of z are showed by [2]. Denote by the sets of periodic points of f, denote by the w-limit points of z, and denote by the orbit of z. A point is said to be recurrent point if for any neighborhood of x, there exists a positive integer m such that. Let denote the set of recurrent points.

In recent years, many authors studied equivalent conditions of closed periodic points set. Gengrong Zhang [3], Xiong Jincheng [4] and Wang Lidong [5] studied respectively anti-triangular map of X2, continuous self-map of the closed interval and continuous self-map of the circle. They showed equivalent conditions of closed periodic points set (see more detail for [3-5]). Du Ruijin [1] given five equivalent conditions of closed periodic points set if f is a second descendible map of Xn. 1); 2); 3); 4); 5).

In this paper, we will continue to study new equivalent conditions about that the set is closed. The following theorem are given.

Main Theorem Let be a continuous map. If f is a second descendible map, then the following properties are equivalent:

1) The set is closed; 2); 3) For any, there exists a such that every point of the set is a isolated point of the set; 4) For any, the set is finite; 5) For any, the set is finite.

2. Definition and Lemma

Definition 1 For any, let, define:, then pi is said to be canonical projection.

Definition 2 Let, the map f is said to be second descendible if for any, there exists such that . In this case Fi is a descendible group of f.

Lemma 1 [6] Let. Then the following properties are equivalent:

1) is a descendible group of f;

2).

Lemma 2 Let. If f is a second descendible map and is a descendible group of f, then any, we have

.

Proof. Suppose. There exists a positive integer sequence such that. By Lemma 1, we can get

. Hence for any, we have. Thus

. This complete the proof.

Lemma 3 Let. Then if and only if.

Proof. Suppose. For any positive integer k, there exists a positive integer sequence such that

. Hence. Assume

. Then there exists a positive integer sequence such that. By definition,. Hence we complete the proof.

Lemma 4 [5] Let. Then 1) For any, the set is periodic orbit if and only if the set is finite.

2) Let. If y is a isolated point of the set, then we have.

Lemma 5 Let and . If all points of the set are isolated points of the set, then we have.

Proof. Suppose. Then there exists a positive integer l and a sequence such that and. Hence for any, we have. By assumption, for any, the point of is a isolated point of the set. Thus for any

, there exists a neighborhood of such that.

Using the equation of, we have.

By 2) of Lemma 4, we can get that . Hence we have that.

Lemma 6 Let and the set is infinite. Then any, we can get that .

Proof. Assume on the contrary that there exists such that. Thus . Hence the point is a periodic point. Therefore the set is finite, which is impossible. Thus the lemma is proved.

Lemma 7 [5] Let and for any, the set is finite. Then we have.

Lemma 8 Let. If f is a second descendible map and is a descendible group of f, and the set is closed. Then any , we have the set is periodic orbit.

Proof. According to [6], we can get that

. By assumption, the set is closed. Hence for any, the set is closed. Let. According to [4], the set is closed if and only if for any, the set is periodic orbit. Hence for any and any, the set is periodic orbit. Using 1) of Lemma 4, for any and any, the set is finite. The set is finite since. Therefore we have the set is periodic orbit.

3. The Proof of Main Theorem

Main Theorem Let be a continuous map. If f is a second descendible map, then the following properties are equivalent:

1) The set is closed;

2);

3) For any, there exists a such that every point of the set is a isolated point of the set;

4) For any, the set is finite;

5) For any, the set is finite.

Proof. 1) 2) First we will show that the set is closed if and only if for any, (*).

According to [6], we can get that.

Hence the set is closed if and only if for any, the set is closed. Let. It is obvious that the set is closed if and only if. Thus we complete the proof of (*).

Assume. Then there exists a integer such that for any

. Hence. Therefore 1) implies 2).

2) 1) Suppose. For any,. Let. According to [6], we can get that. Hence

. Then there exists a integer such that. Thus for any. By (*), the set is closed.

1) 3) By assumption and according to [1],. For any, let. Thus. By assumption and Lemma 8, the set is periodic orbit. Using 1) of Lemma 4, the set is finite. Hence the set is empty. Thus 1) implies 3).

3) 4) By assumption, for any, there exists a such that every point of the set is a isolated point of the set. By Lemma 5,. Hence the set is finite.

4) 5) It is obvious that 4) implies 5).

5) 1) For any, we have.

Case 1: Suppose that the set is finite. Using 1) of Lemma 4, the set is periodic orbit. So. Thus.

Case 2: Assume that the set is infinite. Then exists a sequence such that the sequence converges to and by Lemma 6, all points of the set are different. Hence. By assumption that the set is finite and Lemma 7, we have that. Thus.

According to [1], the set is closed. Thus we complete the proof of the theorem.

4. Acknowledgements

This work was supported by the NSF of China (No. 11161029), NSF of Guangxi (2010GXNSFA013109, 2012GXNSFDA276040, 2013GXNSFBA019020).

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