Let and f:Xn→Xn be a continuous map. If f is a second descendible map, then P(f) is closed if and only if one of the following hold: 1) ; 2) For any z ε R (f), there exists a yεw (z,f) ∩ P(f) such that every point of the set orb (y,f) is a isolated point of the set w (z,f); 3) For any z ε R(f), the set w (z,f) is finite; 4) For any z ε R(f), the set w' (z,f) is finite. The consult give another condition of f with closed periodic set other than [1].
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 [
In recent years, many authors studied equivalent conditions of closed periodic points set. Gengrong Zhang [
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.
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 [
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 [
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 [
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 [
. By assumption, the set is closed. Hence for any, the set is closed. Let. According to [
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 [
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 [
. Then there exists a integer such that. Thus for any. By (*), the set is closed.
1) 3) By assumption and according to [
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 [
This work was supported by the NSF of China (No. 11161029), NSF of Guangxi (2010GXNSFA013109, 2012GXNSFDA276040, 2013GXNSFBA019020).