Second Descendible Self-Mapping with Closed Periodic Points Set ()
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).