Abstract
Let a graph and be a continuous map. Denote by , , and the topological entropy, the set of recurrent points, and the set of special -limit points of , respectively. In this paper, we show that if and only if .
1. Introduction
Let be a metric space. For any , denote by , , and the interior, the boundary, and the closure of in , respectively. For any and any , write . Let be the set of all positive integers and .
Denote by the set of all continuous maps from to . For any , let be the identity map of and the composition map of and . A point is called a periodic point of with period if and for . The orbit of under is the set . Write , called the -limit set of under . In fact, if and only if there exists a sequence of positive integers such that . is called a recurrent point of if . is called a special -limit point of if there exist a sequence of positive integers and a sequence of points such that for any and . Denote by , , and the sets of periodic points, recurrent points, and special -limit points of , respectively. From the definitions it is easy to see that and . Let denote the topological entropy of , for the definition see [1, Chapter VIII].
A metric space is called an arc (resp., an open arc, a circle ) if it is homeomorphic to the interval (resp., the open interval , the unit circle ). Let be an arc and a homeomorphism. The points and are called the endpoints of , and we write . A compact connected metric space is called a graph if there are finitely many arcs () in such that and for all . A graph is called a tree if it contains no circle. A continuous map from a graph (resp., a tree, an interval) to itself is called a graph map (resp., a tree map, an interval map).
Let be a given graph. Take a metric on such that, for any and any , the open ball is always connected. For any finite set , let denote the number of elements of . For any , write , which is called the valence of . is called a branching point (resp., an endpoint) of if (resp., ). Denote by and the sets of endpoints and branching points of , respectively. Take a finite subset of containing such that, for any connected component of , the closure is an arc. Such a subset is called the set of vertexes of , and the closure of every connected component of is called an edge. For any edge of and any , we denote by (or simply if there is no confusion) the smallest connected closed subset of containing , which is called a closed interval of . So, a closed interval is always a subset of an edge. Write and . Let be a graph and closed intervals, and . We write if there exists a closed subinterval such that .
In the study of dynamical systems, recurrent points, topological entropy, and special -limit points play an important role. For interval maps, Hero [2] obtained the following result.
Theorem A (see [2, Corollary]). Let be a compact interval and . Then the following are equivalent: (1)some point that is not recurrent is a special -limit point;(2)some periodic point has period that is not a power of two.
It is known [1, Chapter VIII, Proposition 34] that if and only if some periodic point of has period that is not a power of two for interval map .
In [3], Llibre and Misiurewicz studied the topological entropy of a graph map and obtained the following theorem.
Theorem B (see [3, Theorems 1 and 2]). Let be a graph and . Then if and only if there exist and closed intervals with and such that and .
Recently, there has been a lot of work on the dynamics of graph maps (see [4–13]). In this paper, we will study the topological entropy and special -limit points of graph maps. Our main result is the following theorem.
Theorem 1.1. Let be a graph and . Then if and only if .
2. Proof of Theorem 1.1
In this section, we will prove Theorem 1.1. To do this, we need the following lemmas.
Lemma 2.1 (see [11, Theorem 1]). Let be a graph and . If , then there exist a sequence of positive integers and a sequence of points with such that for any and .
Remark 2.2. The main idea of the proof of Theorem 1 in [11] is similar to the one of Main Theorem in [2].
Lemma 2.3. Let be a graph and . Then .
Proof. Let . Then there exist a sequence of points and a sequence of positive integers such that for every and . Write for . Let be a convergence subsequence of , and let . Then Write Then for any , which implies that and . The proof is completed.
Lemma 2.4 (see [3, Lemma 2.4]). Let be a graph and . Suppose that and are intervals of . If there exist and such that , then or .
Theorem 2.5. Let be a graph and . Then if and only if .
Proof Necessity
If , then take a point . By Lemma 2.3 and , for every , there exists a point such that . Note that are mutually different. Since the numbers of vertexes and edges of are finite, there exists an edge of such that is an infinite set. We can choose integers such that and for every . Take points with such that and for some . Without loss of generality we may assume that and . Since , we can take points and positive integers satisfying the following conditions:(1)the sequence is strictly monotonic with for any and (see Lemma 2.1) and ;(2) for any .
Let and for any . Then and . Noting that , we can assume that , and there exists such that the following conditions hold:
(3) for any ;(4)the sequences and are strictly monotonic, and .
In the following we may consider only the case that is strictly decreasing since the other case that is strictly increasing is similar.
Write for any . Put and for any . Then is a connected set, and
Noting that , we have . Write . Then is a connected set containing and for every .
Since and for any , by Lemma 2.4 it follows that or . There are two cases to consider.
Case 1. There exist such that for every .
Subcase 1.1. There exists such that . Then , and there exist and such that , from which and it follows
Noting and , we have
There exists such that or , which implies
On the other hand, . Thus we can obtain and for some . By Theorem B it follows that .
Subcase 1.2. for all , and there exists such that . Then we can take such that . Thus there exist and such that , which implies . Write . Then or since , which implies
On the other hand, . Thus we can obtain and for some . By Theorem B it follows that .
Subcase 1.3.
One has and for all .
If for some , then there exist and such that since and , which implies . Using arguments similar to ones developed in the proof of Subcase 1.2, we can obtain and for some . By Theorem B it follows that . Now we assume for all . Note since .
If for some , then and . By Theorem B it follows that .
If for some , then there exist and such that since and , which implies . Using arguments similar to ones developed in the proof of Subcase 1.2, we can obtain and for some . By Theorem B it follows that .Case 2. There exists such that for all .
Subcase 2.1. There exist such that for every . Then and . Thus there exist and such that , from which it follows that or . Since , , and
there exists such that or , which implies or . In similar fashion, we can show for some . Thus we get and for some . It follows from Theorem B that .
Subcase 2.2. There exists such that for all and there exists such that and . Take such that for . Then there exist , and such that . Using arguments similar to ones developed in the proof of Subcase 2.1, we can obtain and for some . By Theorem B it follows that .
Subcase 2.3. There exists such that and for all .
If there exist such that for , then there exist , , and such that and , which implies and . Using arguments similar to ones developed in the proof of Subcase 2.1, we can obtain and for some . By Theorem B it follows that . Now we assume that there exists such that for all .
If for all , then using arguments similar to ones developed in the above proof, we can obtain .
If for some , then and . By Theorem B it follows .
Sufficiency
If , then it follows from Theorem B that there exist and closed intervals with and such that and . Without loss of generality we may assume that and and with such that and . By [1, Chapter II, Lemma 2] we can choose with such that one of the following statements holds:(i), , for and for .(ii), , for and for .
We may consider only case (i) since case (ii) is similar. We claim that, for any and any , there exist and such that . In fact, we can choose such that and for any . Thus there exists some . That is, we can choose satisfying , which implies . The claim is proven.
By the above claim we can choose a sequence of positive integers and a sequence of points such that for any and . Note that ; then . The proof is completed.
Acknowledgments
Project Supported by NSF of China (10861002) and NSF of Guangxi (2010GXNSFA013106, 2011GXNSFA018135) and SF of Education Department of Guangxi (200911MS212).