Continuous Maps of the Interval Whose Periodic Points Form a Closed Set

We show that for a continuous map of a closed interval to itself, if the set of periodic points is closed, then every recurrent point is periodic. If, furthermore, the set of least periods of the periodic points is finite, then every nonwandering point is periodic. This answers a question of L. Block [Proc. Amer. Math. Soc. 67 (1977), 357-360]. Introduction. In [1] and [2], L. Block studied continuous maps of a closed interval to itself with finitely many periodic points. He showed [1, Theorem B] that if, in addition, every periodic point is fixed, then every nonwandering point is periodic, and [2, Theorem A] that the same conclusion holds if the set of nonwandering points is finite. In [1] he raised the question of whether this conclusion follows just from the assumption that the set of periodic points is finite. The main result of this paper is the following, which answers Block's question in the affirmative. Theorem 4. For a continuous map of a closed interval to itself, if the set of least periods of the periodic points is finite, then every nonwandering point is periodic. In the course of deriving this result, we show (Theorem 3) that Block's result of [1] is true without the assumption that there are only finitely many periodic points. We also show (Theorem 1) that if the set of periodic points is closed, then every recurrent point is periodic, and (Corollary to Theorem 2) that if the set of least periods of the periodic points is finite, then every orbit is asymptotic to a periodic orbit. Notation and terminology. Throughout this paper, /: [a, b] -* [a, b] will be continuous. A point x E [a, b] is fixed provided that fix) = x; x is periodic provided that there exists « > 1 such that f"(x) = x (any such n is called a period of x); x is recurrent provided that for every neighborhood U of x, there exists « > 1 such that f(x) E U; x is nonwandering provided that for every neighborhood U of x, there exists « > 1 such that f"(U) n U ¥= 0. Let F(f), P(f), R(f) and NW(f) denote the sets of fixed, periodic, recurrent and nonwandering points, respectively. Each of Received by the editors February 12, 1979 and, in revised form, May 26, 1979. AMS (MOS) subject classifications (1970). Primary 54H20. © 1980 American Mathematical Society 0002-9939/80/0000-O225/$02.7S 127 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 128 E. M. COVEN AND G. A. HEDLUND these sets is invariant, i.e., / maps each set to itself. Furthermore, 0 ¥= F(f) C P(f) Ç R(f) ç NW(f), F(f) and NW(f) are closed, and for each n > 1, P(f) = P(fn) and [3] R(f) = R(f). Let NP(f), NR(f) and W(f) denote the sets of nonperiodic, nonrecurrent and wandering (i.e., not nonwandering) points, respectively. Let C C [a, b]. We say that/ is completely positive on C provided th&tf;(x) > x for all x G C and all n > 1 ; / is completely negative on C provided that f(x) < x for all x G C and all n > 1. Let I,JC [a, b\. We write I a. Results. If P(f) is closed, then the components of NP(f) are of the form (/?, a), [a, q) or (/?, ¿?] where/?, q G P(/). Note that P(f) is closed if P(/) = F(f), or more generally, if the set of least periods of the periodic points is finite. Lemma 1. IfC is a component of NP(f), then for each n > 1, either f(x) > x for allx E C orf(x) < x for all x E C. In particular, if a E C, then f is completely positive on C and if b E C, then f is completely negative on C. Proof. It suffices to consider the case where C is an interval. Let n > 1 and suppose that there exist x, y G C such that/"(x) > x and f(y) a for all n > 1. Hence / is completely positive on C. Similarly, if b E C, then / is completely negative on C. □ Lemma 2. If C = (/?, a) is a component of NP(f) and p and q are fixed, then f is either completely positive on C or completely negative on C. Proof. By Lemma 1, either fix) > x for all x E C or fix) < x for all x G C. We assume the former. Let n > 1. Since C is a component of NP(f) = NP(f"), it follows from Lemma 1 that either/"(x) > x for all x E C or/"(x) < x for all x G C. But/'(/?) = /? for all i > 1, so if x G C is sufficiently close to /?, then x x for all x E C. Since n was arbitrary,/is completely positive on C. Similarly, if fix) < x for all x G C, then/is completely negative on C. □ An immediate consequence of Lemmas 1 and 2 is the following. Lemma 3. If P(f) = F(f) and C is a component of NP(f), then f is either completely positive on C or completely negative on C. □ Theorem 1. Iff: [a, b]—>[a, b] is continuous and P(f) is closed, then R(f) =