Inverse Limits: From Continua to Chaos

Developments in Mathematics VOLUME 25 Series Editors: Krishnaswami Alladi, University of Florida Hershel M. Farkas, Heb...

Author: Tom Ingram | William S. Mahavier

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Developments in Mathematics VOLUME 25 Series Editors: Krishnaswami Alladi, University of Florida Hershel M. Farkas, Hebrew University of Jerusalem Robert Guralnick, University of Southern California

For further volumes: http://www.springer.com/series/5834

W.T. Ingram • William S. Mahavier

Inverse Limits From Continua to Chaos

W.T. Ingram Professor Emeritus Mathematics and Statistics Missouri University of Science and Technology Rolla, Missouri USA [email protected]

William S. Mahavier Professor Emeritus Emory University Atlanta, Georgia USA

ISSN 1389-2177 e-ISBN 978-1-4614-1797-2 ISBN 978-1-4614-1796-5 DOI 10.1007/978-1-4614-1797-2 Springer New York Dordrecht Heidelberg London Library of Congress Control Number: 2011940293 Mathematics Subject Classification (2010): 54F15, 37B45, 54H20, 37E05 © Springer Science+Business Media, LLC 2012 All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)

To Barbara and Jean without whose support this book would not have been possible.

Preface

Inverse limits have played a crucial role in the development of the theory of continua in the past 50 years or so. Particularly useful is their inherent ability to produce complicated spaces from simple ones. Nowhere is this feature more evident than in a couple of papers, one by G. W. Henderson and one by Howard Cook. Henderson shows that the pseudo-arc (an hereditarily indecomposable chainable continuum) is the inverse limit on the interval [0, 1] with a single bonding map. Cook employs inverse limits to produce a continuum having only one nonconstant self-map, the identity. In dynamics the inverse limit construction allows the study of a dynamical system consisting of a topological space and a map of that space into itself to be turned into the study of a (likely more complicated) space and one of its self-homeomorphisms. We believe one of the best ways to learn about inverse limits is by means of a thorough study of inverse limits on [0, 1]. For this reason, we begin this monograph with a detailed look at inverse limits on [0, 1] in our ﬁrst chapter. There we develop the basic properties of inverse limits employing only theorems from topology normally found in a senior-level undergraduate or a beginning-level graduate course in topology. However, for the convenience of the reader, most of the background is developed in an appendix. The origins of the study of inverse limits date back to the 1920s and 1930s. In the 1950s and 1960s the ﬁeld exploded with a torrent of results. In 1954 C. E. Capel showed that monotone maps on arcs (respectively, simple closed curves) produce arcs (respectively, simple closed curves). An extremely important paper by R. D. Anderson and Gustav Choquet appeared in 1959 showing just how useful inverse limits can be in describing complicated examples. They constructed an example of a planar tree-like continuum no two of whose nondegenerate subcontinua are homeomorphic. Employing the techniques of Anderson and Choquet, J. J. Andrews produced a chainable continuum with the same property. In 1967 Mahavier showed that the Andrews example is not homeomorphic to an inverse limit on [0, 1] using a single vii

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Preface

bonding map. The example of Cook mentioned earlier also appeared in 1967. In the early 1970s Ingram used inverse limits to describe an example of a nonchainable tree-like continuum such that all of its nondegenerate proper subcontinua are arcs (and consequently the continuum does not contain a triod). Late in the decade of the 1970s Bellamy used inverse limits in producing his example of a tree-like continuum without the ﬁxed point property. R. F. Williams introduced many dynamicists to the value of inverse limits in his 1967 paper on nonwandering sets. In 1990 Marcy Barge and Joe Martin promoted interest in inverse limits in dynamics using inverse limits to construct global attractors in the plane. Earlier they had demonstrated connections between dynamics and continuum theory showing, for example, that an inverse limit on [0, 1] with a single bonding map contains an indecomposable continuum if the bonding map has a periodic point whose period is not a power of 2. Studies of inverse limits on intervals with a single bonding map chosen from the logistic family (or the tent family) have led to interesting results. Major unsolved problems remain in determining the nature of inverse limits with bonding maps from these simple families of interval maps, ˇ although quite recently Barge, Bruin, and Stimac have solved one of these by showing that two tent maps on [0, 1] that have diﬀerent maximum values in the interval [1/2, 1] produce nonhomeomorphic inverse limits. A fundamental example in dynamics is the Smale horseshoe. The continuum most naturally associated with the horseshoe is a chainable continuum ﬁrst described by Janiszewski who was reacting to the ﬁrst example of an indecomposable continuum that Brouwer had constructed a few years earlier. Subsequently, Knaster gave us a beautiful geometric description of Janiszewski’s example that we depict in Section 3 of Chapter 1. This example remains at the heart of the connection between dynamical systems and continuum theory. Recent work of Judy Kennedy, David Stockman, and Jim Yorke as well as work of Alfredo Medio and Brian Raines has connected inverse limits to the theories of backward dynamics in economics. Some recent research in inverse limits has been, in part, devoted to studying inverse limits on compact Hausdorﬀ spaces with upper semi-continuous bonding functions. Inverse limits of systems over directed sets other than the set of positive integers have proved useful at times, as well. Chapter 2 includes a development of the theory of inverse limits in this very general setting where the factor spaces are compact Hausdorﬀ spaces, the bonding functions are upper semi-continuous, and the underlying directed set may be any directed set. Anyone interested in dynamical systems or continuum theory or in exploring relationships between the two disciplines could beneﬁt from a careful study of the ﬁrst chapter. The reader who is already familiar with inverse limits may choose to skim through or skip the ﬁrst chapter and move directly to the second. The third chapter is devoted to a few topics in continuum

Preface

ix

theory not addressed in the ﬁrst two chapters, and Chapter 4 is devoted to a fundamental approximation theorem of Morton Brown that has seen much use in inverse limits over the years. Spring Branch, TX San Leon, TX

W. T. Ingram William S. Mahavier September, 2010

Acknowledgments

The authors appreciate the assistance of Michel Smith and Scott Varagona of Auburn University and Thelma West of the University of Louisiana at Lafayette who read large portions of this monograph. Without their diligent help, there would have been more errors in this book than there are. Of course, we are responsible for all errors which, but for their help, would have been more numerous.

Note: Sadly, during the late stages of the preparation of the manuscript for this volume, on October 8, 2010, William S. (Bill) Mahavier died. This work was truly a joint eﬀort of both authors. We both agreed that what we were producing was better than either of us would have achieved alone. Although Bill did not live to see the publication of this book, those who knew him will see the results of his handiwork throughout. I count it a privilege to have participated with him in writing it. W. T. (Tom) Ingram

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Contents

1

Inverse Limits on Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Basic properties of inverse limits on the interval [0, 1] . . . . . . . 1.2 Examples and remainders of topological rays . . . . . . . . . . . . . . . 1.3 Inverse limits on [0, 1] with only one bonding map . . . . . . . . . . 1.4 Period three implies indecomposability . . . . . . . . . . . . . . . . . . . . 1.5 Inverse limits with only one map of an interval . . . . . . . . . . . . . 1.6 Inverse limits on intervals with sequences of maps . . . . . . . . . . 1.7 Inverse limits with unimodal bonding maps . . . . . . . . . . . . . . . . 1.8 Logistic maps and their inverse limits . . . . . . . . . . . . . . . . . . . . . 1.9 The piecewise linear family of unimodal maps fa b . . . . . . . . . . 1.10 The tent family . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 Other families of mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11.1 The family F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11.2 The family G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11.3 Markov maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11.4 Permutation maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.12 Characterization of inverse limits on [0, 1] as chainable continua . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.13 An inverse limit homeomorphic to a sin(1/x)-curve . . . . . . . .

1 1 6 13 19 22 26 27 31 48 52 55 55 56 57 58 59 67

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 2

Inverse Limits in a General Setting . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Deﬁnitions and a basic theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Graphs of upper semi-continuous functions . . . . . . . . . . . . . . . . 2.4 Consistent systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Compact inverse limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Connected inverse limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Systems in which all of the bonding functions are mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75 75 76 78 79 80 83 85 xiii

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2.6.2 Systems in which the directed set is totally ordered . . . Examples in the special case that each factor space is [0, 1] . . Mapping theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Upper semi-continuous functions that are unions of functions Inverse limit systems with mappings . . . . . . . . . . . . . . . . . . . . . . 2.10.1 A basis for the topology . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10.2 Closed subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10.3 Closed subsets of a system with upper semi-continuous bonding functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10.4 Intersections of closed subsets of the inverse limit . . . . . 2.10.5 The subsequence theorem . . . . . . . . . . . . . . . . . . . . . . . . . 2.10.6 Other induced homeomorphisms . . . . . . . . . . . . . . . . . . . . 2.10.7 Inverse limits as sequential limiting sets . . . . . . . . . . . . . 2.10.8 Inverse limits as intersections of closed sets . . . . . . . . . . 2.11 Inverse limits with metric factor spaces . . . . . . . . . . . . . . . . . . . . 2.12 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 2.8 2.9 2.10

86 90 104 111 115 115 115 117 117 119 120 121 122 122 124

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 3

Inverse Limits in Continuum Theory . . . . . . . . . . . . . . . . . . . . . 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Indecomposability, atriodicity, and unicoherence . . . . . . . . . . . . 3.2.1 Indecomposability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Triods and atriodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Unicoherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4 Irreducibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Monotone bonding maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Irreducibility and indecomposability . . . . . . . . . . . . . . . . . . . . . . 3.5 Closed subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 The full projection property . . . . . . . . . . . . . . . . . . . . . . . 3.6 Indecomposability of inverse limits with upper semicontinuous bonding functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Continua that cannot be obtained with one bonding function 3.8 Additional topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8.1 Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8.2 Property of Kelley . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8.3 Fixed point property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8.4 Hyperspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

131 131 131 132 133 134 135 137 138 142 142 146 147 148 149 150 151 152

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 4

Brown’s Approximation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Brown’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 An application of Brown’s theorem . . . . . . . . . . . . . . . . . . . . . . .

155 155 155 161

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References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 5

Appendix: An Introduction to the Hilbert Cube . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 A brief introduction to topology . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 The Hilbert cube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 The metric topology for Q . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 The product topology for Q . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 The metric topology and the product topology for Q are identical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 A countable basis for the topology of Q . . . . . . . . . . . . . . . . . . . 5.5 Q is compact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Q is connected . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Consequences of compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.10 Boundary bumping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

167 167 167 169 170 170 171 171 172 175 177 178 179 180

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

Chapter 1

Inverse Limits on Intervals

Abstract This chapter provides an introduction to inverse limits for anyone who has completed a basic course in topology. We begin with some of the fundamental properties of inverse limits on the interval [0, 1] and we include numerous instructive examples. This introduction culminates with a brief study of inverse limits on [0, 1] with a single bonding map. Much of the remainder of the chapter is devoted to inverse limits as they relate to dynamical systems. We begin this with a look at period 3 showing that an inverse limit on [0, 1] with a single bonding map having a periodic point of period 3 contains an indecomposable continuum. We investigate inverse limits with unimodal bonding maps, logistic bonding maps, tent maps as bonding maps, and certain other families of bonding maps. For the continuum theorists reading this chapter we prove that inverse limits on [0, 1] are characterized by chainability. However, this proof provides insight into the geometric realization of inverse limits as continua and, as such, can increase one’s understanding of the variety and complexity of the objects produced by the inverse limit construction. We close the chapter with a proof that the inverse limit of a certain unimodal map is the familiar sin(1/x)-curve.

1.1 Basic properties of inverse limits on the interval [0, 1]

Our emphasis in this chapter is on simple examples and fundamental properties of inverse limits on intervals. We begin with inverse limits on [0, 1]. For the sake of completeness we include most topological properties we need in an appendix but most of them can be found in any introductory text on general topology. It is our hope that many readers will ﬁnd that this chapter contains all they need to get started doing research on inverse limits on [0, 1] and the

W.T. Ingram and W.S. Mahavier, Inverse Limits: From Continua to Chaos, Developments in Mathematics 25, DOI 10.1007/978-1-4614-1797-2_1, © Springer Science+Business Media, LLC 2012

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1 Inverse Limits on Intervals

remainder of the monograph can be used as a reference to other inverse limit topics. We use I to denote [0, 1] and N to denote the set of positive integers. Our inverse limits are subsets of the inﬁnite product Q = I ∞ with the product topology. We frequently use the well-known fact (Theorem 239) [12, Theorem 4.2.2, p. 259] that this topology is the same as the metric topology on Q given by the following formula. If each of x = (x1 , x2 , x3 , . . .) and y = (y1 , y2 , y3 , . . .) is in Q then d(x, y) =

|xi − yi | i>0

2i

.

By a mapping, or a map for short, we mean a continuous function and a map from I onto I is one with both domain and range I whereas by a map from I into I we mean one with domain I and range a subset of I. A map from I onto I is also called a surjection. A map from I into I is said to be 1-1 or an injection if and only if it is true that if f (x) = f (y), then x = y. If f is a map and X is a subset of the domain of f , then f (X) is the set of all points f (p) for all points p ∈ X. If f is a mapping and p is a point such that f (p) = p then p is called a ﬁxed point for f . For the remainder of this section we assume that we have a sequence f1 , f2 , f3 , . . . of maps from I into I and we adopt the notation that boldface Roman characters denote a sequence whereas Roman characters denote the terms of the sequence. Using this convention we have f = f1 , f2 , f3 , . . .. The inverse limit of the sequence f is denoted by lim f ←− and is the set of all sequences x ∈ Q such that for each positive integer n, fn (xn+1 ) = xn . The maps fi are call bonding maps. In the study of inverse limits it is conventional to use πn to denote the function, called the projection, from lim f into I such that if x = (x1 , x2 , x3 , . . .) ∈ lim f then πn (x) = xn . ←− ←− Thus πn (x) is the nth coordinate of x. As a simple example illustrating these projections, we provide the following. Example 1 Let f1 be the map from I into I such that if x ∈ I, f1 (x) = 0, and if i > 1, fi (x) = x. In this example π1 (lim f ) = {0}, and πi (lim f ) = I for i > 1. Thus ←− ←− = lim f . But π2−1 (x) = (0, x, x, x, . . .) is degenerate for each x in [0, 1] ←− so π2 is 1-1. By Theorem 4 below, π2 is continuous and by Theorem 6 below, lim f is compact so by Theorem 259, π2 is a homeomorphism and lim f is ←− ←− homeomorphic to I. π1−1 (0)

If n and m are positive integers and n < m, then fn m denotes the composite map fn ◦ fn+1 ◦ · · · ◦ fm−1 from I into I. Using this notation fn = fn n+1

1.1 Basic properties of inverse limits on the interval [0, 1]

3

and, for convenience, fn n denotes the identity on I (i.e., fn n (x) = x for each x ∈ I). The following is an immediate consequence of our deﬁnitions. Theorem 2 If x ∈ lim f , n and m are positive integers, and n < m then ←− πn (x) = fn m (πm (x)). The study of inverse limits is simpliﬁed by observing that the topology of the inverse limit results from a very simple set of basis elements as described in the following deﬁnition and theorem. The statement that R is a region means that there exist a positive integer n and an open subset O of I such that R = πn−1 (O). Theorem 3 The set of all regions is a basis for the topology of lim f as a ←− subspace of Q. Proof. Let x ∈ lim f and let O be a basic open set in Q containing x. There ←− exist an integer n and a ﬁnite collection O1 , O2 , O3 , . . . , On of open subsets of I such that O = O1 × O2 × O3 × · · · × On × Q. By Theorem 2, fi n (xn ) = xi for each i ≤ n. Because each fi n is continuous, it follows that there is an open subset Ui of I containing xn such that fi n (Ui ) ⊆ Oi for 1 ≤ i ≤ n. If U = i>0 Ui then πn−1 (U ) is a region containing x that is a subset of O. Theorem 4 If n ∈ N, πn is continuous. Proof. Let x ∈ lim f . Then πn (x) = xn and xn ∈ I. If O is an open set ←− in I containing xn , then πn−1 (O) is a region and thus open in lim f and ←− πn (πn−1 (O)) ⊆ O. We next turn to showing that our inverse limits are nonempty and compact. We frequently use the following approximations to our inverse limits. If n ∈ N, let Gn denote the set of all points in Q such that if i ≤ n, fi (xi+1 ) = xi . Clearly lim f = i>0 Gi . In the next theorem and frequently thereafter we ←− use the fact that Q is compact, Theorem 246 (see also [12, Theorem 3.2.4, p. 138]) being the product of compact spaces. Theorem 5 If n is a positive integer, Gn is a nonempty compact set. Proof. We ﬁrst prove that Gn is a closed subset of Q. To see this let x be a point of Q − Gn and let i ≤ n be a positive integer such that xi = fi (xi+1 ). There are mutually exclusive open sets U and V in I containing xi and fi (xi+1 ), respectively, and an open set W containing xi+1 such that fi (W ) ⊆ V . The set of all points y of Q such that yi ∈ U and yi+1 ∈ W is an open set in Q containing x but no point of Gn so x is not a limit point of Gn . It follows that Gn is compact by Theorem 255 being a closed subset of the compact set Q. That Gn = ∅ follows from the fact that if x ∈ I, (f1 n+1 (x), f2 n+1 (x), f3 n+1 (x), . . . , fn n+1 (x), x, x, x, . . . ) ∈ Gn .

4

1 Inverse Limits on Intervals

Theorem 6 lim f is a nonempty compact set. ←− Proof. Because i>0 Gi = lim f and, for each i ∈ N, Gi is a nonempty com←− pact set and Gi+1 ⊆ Gi , by Theorem 255, lim f is nonempty and compact. ←− The preceding proof is short and standard but it uses the Tychonoﬀ theorem that the product of compact spaces is compact and that in turn depends on the axiom of choice. We provide a proof that does not use the axiom of choice. This proof is longer and somewhat tedious but it is instructive in studying inverse limits. We do not provide this proof to avoid the use of the axiom of choice but instead to illustrate a method of using sequences that is common in the study of inverse limits. Proof. It follows from Theorem 5 without the use of the axiom of choice that each Gn is closed so lim f is closed by Theorem 256. We prove that each ←− inﬁnite subset of lim f has a limit point. This implies by Theorem 245 that ←− lim f is compact. We assume our maps are surjective so we are assured that ←− lim f is nonempty. Let M = x1 , x2 , x3 , . . . be a sequence of distinct points ←− in lim f and for each n ∈ N, let Mn = πn (M ). We ﬁrst assume that for each ←− n ∈ N, Mn is ﬁnite. Because M1 is ﬁnite there is a point p1 ∈ M1 such that K1 = π1−1 (p1 ) ∩ M is inﬁnite. Now M2 is ﬁnite, and π2 (K1 ) ⊆ M2 , so there is a point p2 ∈ M2 such that K2 = π2−1 (p2 ) ∩ K1 is inﬁnite. Now p2 is the second term of some member of K1 and that member of K1 has p1 as its ﬁrst term so f1 (p2 ) = p1 . This process can be continued to determine a sequence p1 , p2 , p3 , . . . of points such that for each n ∈ N, πn−1 (pn ) ∩ M is inﬁnite, and pn = fn (pn+1 ). Now, p = (p1 , p2 , p3 , . . .) is a limit point of M because if O is open in I and contains pn , then the region πn−1 (O) contains inﬁnitely many points of M . Next let m be the least positive integer such that Mm is inﬁnite. Let pm be a limit point of Mm and let sm be a sequence of distinct points of Mm converging to pm . There is a sequence sm+1 of points of Mm+1 that converges to a point pm+1 such that fm (sm+1 ) is a subsequence of sm . Because fm is continuous, fm (pm+1 ) = pm . Similarly there is a sequence sm+2 of points of Mm+2 that converges to a point pm+2 such that fm+1 (sm+2 ) is a subsequence of sm+1 and fm m+2 (sm+2 ) is a subsequence of sm . This process may be continued to deﬁne for each positive integer n a sequence sm+n of points of Mm+n that converges to a point pm+n and such that for 0 ≤ i < n, fm+i m+n (sm+n ) is a subsequence of sm+i . The point (f1 m (pm ), f2 m (pm ), . . . , fm−1 m (pm ), pm , pm+1 , pm+2 , . . .) is a limit point of lim f . ←− Now that we know our inverse limits are compact we turn to the problem of showing that they are connected. Just as we did in the case of compactness

1.1 Basic properties of inverse limits on the interval [0, 1]

5

we ﬁrst give a short proof and then a longer one that illustrates a technique that is useful in the study of inverse limits. Theorem 7 If n is a positive integer, Gn is connected. Proof. Let n ∈ N. If x ∈ Q let h(x) = (f1 n+1 (xn+1 ), f2 n+1 (xn+1 ), . . . , fn n+1 (xn+1 ), xn+1 , xn+2 , . . . ). The fact that h is continuous follows from the continuity of the functions fj n+1 for j ≤ n + 1. Now Q is connected by Theorem 253 so Gn = h(Q) is connected being the image of a connected set by Theorem 257. Theorem 8 lim f is connected. ←− Proof. lim f = i>0 Gn so lim f is connected by Theorem 269 inasmuch as ←− ←− it is the intersection of a nested sequence of compact connected sets. Thus we have that our inverse limits are continua, a continuum being a compact connected set. As indicated earlier we next provide a diﬀerent proof for Theorem 8. We assume our maps are surjective. Proof. Assume lim f is the union of two compact sets H and K. For each ←− n ∈ N, let Mn = πn (H) ∩ πn (K). Mn = ∅ because I is connected and our bonding maps are surjective. Let An = m>n fn m (Mm ). An = ∅ as it is the intersection of a nested sequence of nonempty compact sets. Let x1 ∈ A1 . For each n > 1 there is a point pn in Mn such that f1 n (pn ) = x1 . Let of {yi }i>1 converges to yn = f2 n (pn ). By Theorem 243 some subsequence a point x2 and f1 (x2 ) = x1 . Now for each n > 1, i≥n {yi } is a subset of f2 n (Mn ) so x2 ∈ A2 . Similarly there is a point x3 ∈ A3 such that f2 (x3 ) = x2 . We may continue to establish the existence of a point (x1 , x2 , x3 , . . .) ∈ lim f ←− that is a point or a limit point of each of H and K and thus in both H and K. This contradiction implies that lim f is connected. ←− By a separating point of a connected point set M we mean a point p of M such that M − {p} is not connected. Some authors use the term cut point of M for what we are calling a separating point. By an arc we mean a continuum having only two nonseparating points. We call the nonseparating points of an arc its endpoints. In the appendix we show that every continuum has at least two nonseparating points (Theorem 262) and that any continuum homeomorphic to I is an arc (Theorem 263). Many authors deﬁne an arc as a continuum homeomorphic to I. In this book we do not prove that the two deﬁnitions of the term are equivalent. For a proof the reader is referred to [37, Theorem 6.17, p. 96] or [12, Section 6.38, p. 375].

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1 Inverse Limits on Intervals

Theorem 9 If for each i, fi is a homeomorphism of I onto I then lim f is ←− an arc. Proof. For each number x1 ∈ I, there is one and only one point x ∈ lim f such that π1 (x) = x1 beause each fi is a homeomorphism and fi (I) = I. Thus π1 is 1-1 and π1 is continuous by Theorem 4 from the compact set lim f onto I so, by Theorem 259, π1 is a homeomorphism. ←− Theorem 10 If X is a closed subset of lim f and for each i, πi (X) = I, ←− then X = lim f . ←− Proof. Let x ∈ lim f and let R be a region containing x. There exist an n ∈ N ←− and an open subset O of I such that πn (x) ∈ O and R = πn−1 (O). Because πn (X) = I, if t ∈ O and t = xn there is a point y ∈ X such that πn (y) = t so y ∈ R and y = x. Thus x is a limit point of X and in X inasmuch as X is closed. In the preceding theorem the assumption that X is closed is necessary as can be seen by considering Example 11 in the next section. If X = lim f − ←− {(1, 1, 1, . . .)} then for each i, πi (X) = I but X is a proper subset of lim f . ←−

1.2 Examples and remainders of topological rays We next provide some simple examples. We frequently consider the inverse limit of a sequence where there is a function f such that, for each positive integer n, fn = f . In this case, we use the notation lim f to mean the inverse ←− limit of the constant sequence f, f, f, . . . and we frequently refer to lim f as ←− an inverse limit with only one bonding map. Example 11 If f (x) = 2x if 0 ≤ x ≤ 1/2 and f (x) = 1 if 1/2 < x ≤ 1, then lim f is an arc (see Figure 1.1). ←− Proof. Assume x ∈ lim f and x = (0, 0, 0, . . .) and x = (1, 1, 1, . . .). We ←− prove that x is a separating point of lim f . There is an n ∈ N such that ←− xn ∈ {0, 1}. x is the only point p of M such that πn (p) = xn because if i < n, xi = fi n (xn ) and, if i > n, xi = xn /2i−n . Thus we have that I − {xn } is the union of the two sets [0, xn ) and (xn , 1] each open in I and that lim f − {x} is ←− the union of the two open sets πn−1 ([0, xn )) and πn−1 ((xn , 1]) and is therefore not connected. Our next example, like the previous one, has an arc as the inverse limit but the proof is slightly more diﬃcult.

1.2 Examples and remainders of topological rays (1/2,1)

7 (1,1)

(0,0)

Fig. 1.1 A map that is not a homeomorphism whose inverse limit is an arc

Example 12 If a is a number, 1/2 < a < 1, f (x) = 2x if 0 ≤ x ≤ 1/2, and f (x) = 2(a − 1)(x − 1/2) + 1 if 1/2 ≤ x ≤ 1, then lim f is an arc (see Figure ←− 1.2). Proof. Note that f has exactly two ﬁxed points, 0 and p > 1/2. As in the previous example we prove that M = lim f has at most two nonseparat←− ing points. We begin by assuming that x ∈ lim f and x = (0, 0, 0, . . .) or ←− (p, p, p, . . .). Observe that if there is an n such that xn < a then πn−1 (xn ) = x. Furthermore xn = 0 so M − {x} is the union of the two mutually exclusive open sets πn−1 ([0, xn )) and πn−1 ((xn , 1]) so x is a separating point of M . Next we prove that if x ∈ M and x = (p, p, p . . .) then xn < a for some n so x = (0, 0, 0, . . .) or is a separating point of M . To this end we assume that x ∈ M and xn ≥ a for each n. Because a ≤ x2 ≤ 1 and f is decreasing on [a, 1], f (a) ≥ x1 ≥ a so x1 ∈ [a, f (a)]. Similarly x3 ∈ [a, 1] and f (a) ∈ (a, 1) so f is decreasing on [a, f (a)]. It follows that f (a) ≥ x2 ≥ a and x1 ∈ [f 2 (a), f (a)]. One may proceed by induction to show that, for each n, x1 is between f n (a) and f n+1 (a). Inasmuch as f (x) < 0 for x > a the sequence {f i (a)}i>0 converges to the ﬁxed point p, so we have that x1 = p. But x2 = p because x2 > a and f (x2 ) = p. It follows by induction that xn = p for each n and x = (p, p, p, . . .).

8

1 Inverse Limits on Intervals (1/2,1)

(1,a)

(0,0)

Fig. 1.2 A nonmonotone map whose inverse limit is an arc

In our discussion of the next example we describe some subcontinua of our inverse limit. Here f = f1 , f2 , f3 , . . . and we assume our maps are surjective. One way to determine subcontinua of M = lim f is to start with ←− a subinterval of I, say [a1 , b1 ]. There is a subinterval [a2 , b2 ] such that f1 ([a2 , b2 ]) = [a1 , b1 ]. Continuing inductively for each n select a subinterval [an+1 , bn+1 ] of I such that fn ([an+1 , bn+1 ]) = [an , bn ]. Now for each point x1 ∈ [a1 , b1 ] our construction ensures that there is a point x ∈ M such that for each n, xn ∈ [an , bn ]. Let K be the set of all such points. Thus we have that K = M ∩ ([a1 , b1 ] × [a2 , b2 ] × [a3 , b3 ] × · · · ). At this point we could extend our deﬁnition of an inverse limit to include sequences such as f1 | [a2 , b2 ], f2 | [a3 , b3 ], f3 | [a4 , b4 ], . . . as described above. Instead we show that it is possible, for each n, to create a map gn from I onto I that is a copy of fn | [an+1 , bn+1 ] and such that lim g is homeomorphic to K. This is ←− the content of our next theorem. Theorem 13 Let M = lim f and let J be a sequence of subintervals of I ←− such that for each i, fi (Ji+1 ) = Ji = [ai , bi ]. For x ∈ [0, 1] let hi (x) = ai + x(bi − ai ), let K be the set of all points x of M such that, for each i, xi ∈ Ji , and let yi = h−1 i (xi ). Finally, for each i and each x ∈ I, deﬁne gi (x) = h−1 i (fi (hi+1 (x))). Then gi maps I onto I, gi (yi+1 ) = yi and K is homeomorphic to lim g. ←−

1.2 Examples and remainders of topological rays

9

Proof. hi is a homeomorphism of I onto Ji , therefore yi = h−1 i (xi ) is unique for each xi ∈ Ji . Moreover f (Ji+1 ) = Ji so gi maps I onto −1 −1 I. gi (yi+1 ) = h−1 i (fi (hi+1 (yi+1 ))) = hi (fi (xi+1 )) = hi (xi ) = yi . If y ∈ lim g, deﬁne ϕ(y) = (h1 (y1 ), h2 (y2 ), h3 (y3 ) . . .) = x ∈ K. Now ϕ ←− is 1-1 because each hi is 1-1 and ϕ(lim g) = K because if x ∈ K, then ←− −1 −1 ϕ(h−1 1 (x1 ), h2 (x2 )h3 (x3 ) . . .) = x. To see that ϕ is continuous at y, let U be a region in lim f containing ϕ(y) = x. Thus there exist an i and an open ←− set O in I containing xi such that U is the set of all points z of lim f such ←− that zi ∈ O. Because hi is continuous we have that h−1 i (O) is open in I. Moreover, if w ∈ lim g and wi ∈ h−1 i (O) and ϕ(w) = z then zi ∈ O. ←− A very important special case of Theorem 13 which we frequently use is the case where we have only one bonding map f and there is an interval [a, b] ⊆ I such that f ([a, b]) = [a, b]. As an immediate consequence of Theorem 13 we have the following. Corollary 14 If f maps I into I, [a, b] ⊆ I and f ([a, b]) = [a, b], then lim f ∩ ←− [a, b]∞ is a continuum. Proof. One needs only to note that the set of all points x from Theorem 13 such that for each i, xi ∈ [ai , bi ] is lim f ∩ ([a1 , b1 ] × [a2 , b2 ] × · · · ). Because ←− [ai , bi ] = [a, b] for each i this is lim f ∩ [a, b]∞ . ←− It is immediate from the deﬁnition of lim f that if n ∈ N and x ∈ I, then ←− if x ∈ lim f and πn (x) = x, the ﬁrst n coordinates of x are uniquely given ←− by xi = fi n (x) for i ≤ n. Because of this we can ignore the ﬁrst few terms of our inverse limit and observe that if, for each i > 0, we deﬁne gi = fn+i , then lim g is homeomorphic to lim f . This fact is a special case of the following ←− ←− theorem. Theorem 15 If n1 , n2 , n3 , . . . is an increasing sequence of positive integers, and for each i ∈ N gi = fni ni+1 , then K = lim g is homeomorphic to M = ←− lim f . ←− Proof. Let ϕ be the function from M into K such that if x ∈ M , ϕ(x) = (xn1 , xn2 , xn3 , . . .). To see that ϕ is 1-1 we ﬁrst note that if ϕ(x) = ϕ(y) then for each i, yni = xni . Now if k is an integer that is not ni for any i then j = 1 and k < nj or there is an integer j > 1 such that k is between nj−1 and nj . In either case we have xk = fk nj (xnj ) = fk nj (ynj ) = yk . To see that ϕ is continuous at x, let k ∈ N, let O be open in I and contain xnk , and let R be the region containing ϕ(x) consisting of all points of K whose kth coordinate is in O. Now the set of all points T in M whose nk th coordinate is in O is a region in M and if y ∈ T , then ϕ(y) ∈ R. Let y = xn1 , xn2 , . . . be in K. If i < n1 , let xi = fi n1 , and if nj 0, let xi = fi nj+1 . It follows that ϕ(x) = y so ϕ(M ) = K. Also M is compact so by Theorem 259, ϕ is a homeomorphism.

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1 Inverse Limits on Intervals

By a topological ray we mean the homeomorphic image of the nonnegative real numbers and the image of 0 under the homeomorphism is called the endpoint of the ray. We are now ready to consider our next example. The inverse limit of this example is homeomorphic to the closure of the graph of h where h(x) = sin(1/x) for 0 < x ≤ 2/π, but we prove a weaker theorem inasmuch as we feel our theorem illustrates some basic properties of inverse limits whereas these properties might be harder to see if we were trying to tediously construct a homeomorphism from our example onto the closure of the graph of h. We hope that a reader who understands our proof could construct such a homeomorphism. It is worth noting that Nadler [36] has shown that there are uncountably many nonhomeomorphic continua each of which is the closure of a topological ray R such that R − R is an arc. Here we use the notation that if M is a point set, M is the closure of M . At times the closure of M may be denoted Cl(M ).

(1/2,1)

(1,1/2)

(0,0)

Fig. 1.3 A simple example whose inverse limit is homeomorphic to the closure of the graph of sin(1/x) for x ∈ (0, π]

1.2 Examples and remainders of topological rays

11

Example 16 Let f (x) = 2x for 0 ≤ x ≤ 1/2 and let f (x) = 3/2 − x for 1/2 ≤ x ≤ 1. lim f is homeomorphic to the closure of a topological ray R and ←− R − R is an arc (see Figure 1.3). Proof. Let M = lim f . First note that for each i, f ([1/2, 1]) = [1/2, 1] and ←− let A be the set of all points x ∈ M such that for each i, xi ∈ [1/2, 1]. For each i the graph of fi restricted to [1/2, 1] is a straight line interval from the point (1/2, 1) to (1, 1/2). The corresponding map gi from I onto I given in Theorem 13 is gi (x) = 1 − x and is a homeomorphism so, by Theorem 9, A is an arc. We show it is the limiting arc of a topological ray. Observe that for each i > 0, f [0, 1/2i] = [0, 1/2i−1 ]. We ﬁrst consider the set A1 of all points x ∈ M such that for each i > 0, xi ∈ [0, 1/2i−1]. For each i, f | [0, 1/2i ] is a homeomorphism and each of the maps hi from Theorem 13 is the identity on I. So A1 is an arc. Moreover, its endpoints are (0, 0, 0, . . .) and (1, 1/2, 1/4, . . .). Next consider the set A2 of all points x ∈ M such that xi ∈ [1/2i−1 , 1/2i−2 ] for i > 1. Each of the maps hi for i > 1 from Theorem 13 is the identity so with the aid of Theorem 15 it follows that A2 is an arc. Moreover its endpoints are (1, 1/2, 1/4, . . .) and (1/2, 1, 1/2, 1/4, . . .). Now if x < 1/2, then f −1 (x) = x/2 so if x ∈ M and xi < 1/2 then there is only one choice for xi+1 , but if 1/2 ≤ x < 1 there are two choices, one in [1/2, 1] and one in [0, 1/2]. So if we start with x1 ∈ [1/2, 1], if n ∈ N, we may select each of x2 , x3 , . . . , xn in [1/2, 1] and then select xn+1 in [1/4, 1/2]. In this case xn+i = xn /2i . That the set An of all such points is an arc can be seen again with the use of Theorems 13 and 15. Now each of the arcs A1 , A2 , A3 , . . . is of length at least 1/4, the lengths actually approaching 1/2 as n → ∞. Also, if i and j are two positive integers, then Ai and Aj do not intersect unless |j − i| = 1. Furthermore, Ai has one endpoint in common with Ai+1and, if i > 1, the other endpoint in common with Ai−1 . It follows that R = i>0 Ai is a topological ray. It remains to show that each point of A is a limit point of R. Let x ∈ A and ε > 0. Let n ∈ N such that 1/2n < ε. From the deﬁnition of An , there is a point y ∈ An such that πi (y) = πi (x) for i ≤ n so d(x, y) ≤ 1/2n < ε so that x is a limit point of R. Our next theorem is from an excellent master’s thesis by Ralph Bennett at the University of Tennessee in 1962 [5]. It was motivated by the preceding example. Theorem 17 (Bennett) [5] Suppose g is a sequence of functions from I onto I and f is the sequence of maps from I onto I such that if n ∈ N 1. fn (0) = 0; fn (1/4) = 1 2. fn is linear on [0, 1/4] and [1/4/1/2] 3. For x ∈ [1/2, 1], fn (x) = gn (2x − 1)/2 + 1/2. Then lim f is the union of a topological ray R and a continuum K such that ←− K is homeomorphic to lim g and R − R = K. ←−

12

1 Inverse Limits on Intervals

The proof of this theorem is much the same as that of the previous example, the main diﬀerence being that the arc that is the closure of the ray in the example is replaced by lim g. Instead of proving this theorem we establish ←− a modiﬁcation of Bennett’s theorem due to Tom Ingram. First we want to establish a slight generalization of Theorem 9. A map is monotone if and only if it is nondecreasing or nonincreasing. Theorem 18 If, for each i, gi is a monotone map from I onto I then lim g ←− is an arc. Proof. For each positive integer i, gi (0) is 0 or 1. If there is an integer n such that gi (0) = 0 for i > n, deﬁne fi = gn+i for each i > 0. Otherwise let n1 , n2 , n3 , . . . be the increasing sequence of all positive integers such that gni (0) = 1 for each i > 0, and deﬁne fi = gni ni+1 for each i > 0. In either case we have that each fi is nondecreasing for each i and it follows from Theorem 15 that lim f is homeomorphic to lim g. Now, for each i fi (0) = 0 ←− ←− and fi (1) = 1. Let x be a point of lim f diﬀerent from (0, 0, 0, . . .) and ←− (1, 1, 1, . . .). Let H be the union of all the open sets πi−1 [0, xi ) for all i for which xi = 0. Similarly let K be the union of all open sets πi−1 (xi , 1] for all i for which xi = 1. If y ∈ lim f and y = x, then there is an i such that ←− πi (y) = πi (x) so if xi ∈ {0, 1} then yi ∈ [0, xi ) or yi ∈ (xi , 1] and y ∈ H ∪ K. If xi = 0, y ∈ K and if xi = 1, y ∈ H. Thus we have that lim f − {x} is ←− the union of the two open sets H and K. We claim H and K are mutually exclusive. Suppose that y ∈ H ∩K. Then there exist two integers i and j such that yj ∈ (xj , 1] and yi ∈ [0, xi ). If i < j, because xj < yj , from Theorem 2 and the fact that fi j is nondecreasing, we have fi j (xj ) = xi ≤ fi j (yj ) = yi , but yi < xi . This contradiction implies that y is a separating point of lim f if ←− i < j. A similar contradiction occurs if j < i. Thus lim f and therefore lim g ←− ←− is an arc inasmuch as it has at most two nonseparating points. If R is a topological ray and K is a continuum such that R − R = K then K is said to be a remainder of R or frequently just a remainder for short. If n ∈ N, and f is a map from I into I then f n denotes the n-fold composite of f with itself. That is, f 1 = f , and for n > 1, f n = f ◦ f n−1 . For convenience f 0 represents the identity map on I. Theorem 19 (Ingram) [17] Suppose f maps I onto Iand c ∈ (0, 1), 1. f ([c, 1]) ⊆ [c, 1] 2. f | [0, c] is monotone 3. There is a positive integer j such that f j ([0, c]) = I. Then lim f is the closure of a topological ray R having remainder a continuum ←− K. Moreover K is the set of all points x ∈ lim f such that πi (x) ∈ [c, 1] for ←− each i ∈ N.

1.3 Inverse limits on [0, 1] with only one bonding map

13

Proof. Let M = i≥0 f i ([c, 1]). Because M is the intersection of a nested sequence of points or closed intervals M is a point or a closed interval. Now f (M ) = M and if x ∈ lim f then πi (x) ∈ [c, 1] for each i ≥ 0 if and only if ←− πi (x) ∈ M for each i ≥ 0. It follows from Corollary 14 that K is a continuum. Let Rn be the set of all points x ∈ lim f such that πi (x) ∈ [0, c] if i ≥ n. ←− Using Theorem 13 one sees that R1 is homeomorphic to lim g where for each ←− i, gi is monotone from I onto I and thus by Theorem 18 is an arc. If n > 1 then again using Theorem 13 we have that Rn is homeomorphic to lim g ←− where, for each i, gi maps I onto I and if i ≥ n, gi is monotone. By Theorem 15 we can ignore the ﬁrst n − 1 terms of g so that lim g is homeomorphic to ←− the inverse limit of the sequence gn , gn+1 , gn+2 , . . . of monotone maps. But this inverse limit is an arc by Theorem 18, so Rn is an arc. Moreover Rn is a subset of Rn+1 so R = i>0 Ri is a topological ray. It remains to show that K is the remainder of R. First note that if x ∈ lim f − K then there is a ←− positive integer n such that πn (x) < c so x ∈ Rn . Thus M = R ∪ K and R ∩ K = ∅ so we must show that each point of K is a limit point of R. By hypothesis there is a positive integer j such that f j ([0, c]) = I. If g = f j , then from Theorem 15 there is a homeomorphism ϕ from lim f onto lim g. ←− ←− Moreover if ϕ(K) = K , then K is the set of all points of x ∈ lim g such that ← − πi (x) ∈ [c, 1] for each i ∈ N. We show that each point of K is a limit point of R = ϕ(R). To this end let x be a point of K . Let O be a region containing x. There is a positive integer m and an open set U in I such that O = π −1 (U ) and πm (x) ∈ U . Because g([0, c]) = I, there is a point ym+1 ∈ [0, c] such that g(ym+1 ) = xm . There is a sequence of points ym+1 , ym+2 , ym+3 , . . . such that for each i ≥ 1, g(ym+i+1 ) = ym+i . The point z such that πi (z) = πi (x) if i ≤ m and πi (z) = yi if i > m is a point of R in O so x is a limit point of R .

1.3 Inverse limits on [0, 1] with only one bonding map Much of the work with inverse limits on I has been done with a single bonding map. With only one bonding map f , then we can deﬁne a shift map ϕ from lim f into lim f by ϕ(x) = (f (x1 ), f (x2 ), f (x3 ), . . .) = (f (x1 ), x1 , x2 , . . .). In ←− ←− the case where the shift map is a homeomorphism, we call it a shift homeomorphism. This map plays a signiﬁcant role in the study of inverse limits with only one bonding map. One reason for its importance is given in the following theorem. Theorem 20 If M = lim f is an inverse limit with only one bonding map ←− f , then the shift map ϕ(x) = (f (x1 ), x1 , x2 , x3 , . . .) is a homeomorphism of M onto M . Moreover if f (I) = I, then ϕ is the identity if and only if f is the identity.

14

1 Inverse Limits on Intervals

Proof. Clearly ϕ(x) ∈ M and it is easy to see that ϕ is continuous. It is 1-1 because if ϕ(x) = ϕ(y), then xi = yi for each i. Thus by Theorem 259, ϕ is a homeomorphism. Also if x ∈ M , then ϕ(x2 , x3 , x4 , . . .) = (x1 , x2 , x3 . . .) so ϕ maps M onto M . If f is the identity on I and x1 ∈ I, then f −1 (x1 ) = x1 so x2 = x1 and continuing we see that ϕ(x) = (x1 , x1 , x1 , . . .) so ϕ is the identity. If ϕ is the identity map then we have ϕ(x) = (f (x1 ), x1 , x2 , . . .) = x = (x1 , x2 , x3 , . . .) so f is the identity. Often, a continuum homeomorphic to an inverse limit on I is called a chainable continuum. Not all chainable continua can be represented as an inverse limit with only one bonding map and this is our next observation. Theorem 21 (Mahavier) [29] There is a chainable continuum X such that if f maps I onto I, then M = lim f is not homeomorphic to X. ←− Proof. J. J. Andrews [1] has given an example of a chainable continuum X such that no two of its nondegenerate subcontinua are homeomorphic. We show that if M = lim f is an inverse limit with only one bonding map then ←− M must contain two nondegenerate homeomorphic subcontinua and thus M could not be homeomorphic to X. If f is the identity map then M is an arc and clearly contains distinct subarcs. If f is not the identity, then by Theorem 20 the shift map ϕ is not the identity so there is a point x such that ϕ(x) = y = x. Let O be an open set containing y whose closure does not contain x. Let U be an open set containing x whose closure does not intersect O and such that ϕ(U ) ⊆ O. Let H be the component of U ∩ M that contains x. By Theorem 276, H contains a point of the boundary of U so H is nondegenerate and h(H) is homeomorphic to H. But h(H) = H because x ∈ H but x ∈ h(H). For a constructive example of a chainable continuum that is not an inverse limit on I with only one bonding map see Marsh [31]. Next we turn our attention to questions of decomposability and indecomposability of our inverse limits. A continuum M is said to be decomposable if and only if it is the union of two proper subcontinua, and it is said to be indecomposable otherwise. If M is degenerate, that is, it consists of only one point, then M is indecomposable. It is not trivial to describe a nondegenerate indecomposable continuum. Our next example is just such a continuum. Our map is called a tent map. We do not intend to prove this in this chapter but the continuum given in Example 22 is homeomorphic to a continuum that is well known to topologists and is arguably the simplest example of an indecomposable continuum. We include below a geometric description of this continuum in the plane. The example may be found in [26]. The Smale horseshoe map [11, Figure 2.3,

1.3 Inverse limits on [0, 1] with only one bonding map

15

p. 179] is a homeomorphism h from a topological disk D in a Euclidean plane into itself. Ifthat map is described carefully enough and then iterated, the intersection i>0 hi (D) will be homeomorphic to the continuum described below and to Example 22.

(1/2,1)

(0,0)

(1,0)

Fig. 1.4 A simple example whose inverse limit is an indecomposable continuum

Example 22 Let f be the map from I onto I such that if x ∈ [0, 1/2], f (x) = 2x and if x ∈ [1/2, 1] then f (x) = 2(1 − x). Then lim f is an indecomposable ←− continuum (see Figure 1.4 for the map and Figure 1.5 for a depiction of the inverse limit). Proof. Suppose that lim f is the union of two proper subcontinua, H and K. ←− By Theorem 10 there must be some positive integer i such that πi (H) = I, and thus πk (H) = I if k ≥ i. There also must be some integer j such that if k ≥ j, then πj (K) = I. Assume that i < j so that neither πj (K) nor πj (H) is I. Now 0 is in πj+1 (H) or πj+1 (K), say πj+1 (H). Then 1 is not in πj+1 (H) inasmuch as πj+1 (H) is connected and not I, and 1/2 ∈ πj+1 (H) because this would imply that πj (H) = I. But this implies that both 1/2 and 1 are in πj+1 (K) so 0 and 1 must be in πj (K) and thus I = πj (K). This is a contradiction.

16

1 Inverse Limits on Intervals

Kuratowski [27] provides a picture of this continuum much like the depiction shown in Figure 1.5, and in [26] he also provides a picture and a geometric description that he attributes to Knaster. The ﬁrst example of an indecomposable continuum was given by Brouwer [7]. An accessible description of Brouwer’s example is found in [24]. An inverse limit description of Brouwer’s example is given in [22]. Kuratowski attributes the origin of Example 22 to Janiszewski in his dissertation and remarks that Janiszewski’s work had close links to that of Brouwer. Because of Knaster’s beautiful geometric description of Janiszewski’s example, some authors refer to this example as a Knaster continuum and others refer to it as the buckethandle. We call any continuum homeomorphic to this continuum a BJK horseshoe. Let C denote the “middle-third” Cantor set in the interval [0, 1] on the x-axis. Let C0 be the set of all semicircles lying in the upper half-plane (on or above the x-axis) with center the point (1/2, 0) and with endpoints lying on C. Let C1 denote a similar collection of semicircles lying in the lower halfplane with center the midpoint of the interval [2/3, 1] and with endpoints in C ∩ [2/3, 1]. Continuing, we let C2 be the set of all semicircles with center the midpoint of [2/32 , 3/32 ] and with endpoints in [2/32 , 3/32 ] ∩ C. For each i let Ci be the set of all semicircles with center the midpoint of [2/3i , 3/3i ] and with endpoints in [2/3i , 3/3i ] ∩ C. The union of all these semicircles is our indecomposable continuum.

Fig. 1.5 A partial picture of the continuum from Example 22

Example 22 leads us to the following theorem. Theorem 23 If f maps I into I and there are three numbers a < b < c in I such that f (a) = f (c) = 0 and f (b) = 1 or f (a) = f (c) = 1 and f (b) = 0, then lim f is indecomposable. ←−

1.3 Inverse limits on [0, 1] with only one bonding map

17

Proof. The proof here is almost identical to showing Example 22 is indecomposable. We assume that f (a) = f (c) = 0 and f (b) = 1. The other case is similar. Assume that lim f is the union of two proper subcontinua, H and ←− K. Because f (I) = I, it follows from Theorem 10 that there is an integer i such that neither πi (H) nor πi (K) is I. Assume without loss of generality that a ∈ πi+1 (H). If b ∈ πi+1 (H) then πi (H) = I by Theorem 2 contrary to the choice of i. But πi+1 (H) is connected so if c ∈ πi+1 (H), then b ∈ πi+1 (H) and this is a contradiction. So both b and c must be in πi+1 (K) which implies that πi (K) = I and this is again a contradiction. Next we have a generalization of the preceding theorem. Theorem 24 (Bennett)[5] If f maps I onto I and for each positive number ε there are numbers a < b < c such that each of f (a) and f (c) is within ε of 0 and f (b) is within ε of 1, or each of f (a) and f (c) is within ε of 1 and f (b) is within ε of 0, then lim f is indecomposable. ←− Proof. For each positive integer n let an , bn , and cn be numbers such that (a) f (an ) < 1/n, f (cn ) < 1/n, and 1 − f (bn ) < 1/n or (b) 1 − f (an ) < 1/n, 1 − f (cn) < 1/n, and f (bn ) < 1/n. Assume that inﬁnitely many of the sets of numbers satisfy condition (a). There is an increasing sequence n1 , n2 , n3 , . . . of positive integers such that each of the sequences {ani }i>0 , {bni }i>0 , and {cni }i>0 converges to the points a, b, and c, respectively. We have that a ≤ b ≤ c but because f (a) = f (c) = 0 and f (b) = 1, then a = b and c = b so a < b < c and, by Theorem 23, lim f is indecomposable. ←− Various special cases have been considered where there was only one bonding map. Some examples follow. Theorem 25 (Mahavier) [30] If M is an inverse limit with only one piecewise monotone bonding map, then every subcontinuum of M contains an arc. Some rather complicated continua can be obtained as an inverse limit with only one bonding map as can be seen by the following examples. Example 26 (Bennett)[5] There is a map f of I onto I whose inverse limit is a continuum M containing only two topologically diﬀerent nondegenerate continua and that is the closure of a topological ray R such that R − R is homeomorphic to M (see Figure 1.6). In this example f is the union of an inﬁnite sequence of straight line intervals together with the point (1, 1). The restriction of f to [0, 1/2] is the union of two intervals, one with endpoints (0, 0) and (1/4, 1) and the other with endpoints (1/4, 1) and (1/2, 1/2). The restriction of f to [1/2, 3/4] is the

18

1 Inverse Limits on Intervals

(1/4,1)

(1,1)

(5/8,1)

(3/4,3/4)

(1/2,1/2)

(0,0)

Fig. 1.6 A simple example whose inverse limit contains only two topologically diﬀerent nondegeneaate subcontinua

union of two similar intervals from (1/2, 1/2) to (5/8, 1) and from (5/8, 1) to (3/4, 3/4). This process can be continued adding pairs of intervals to deﬁne f on [3/4, 7/8], [7/8, 15/16], . . .. Finally deﬁne f (1) = 1. That lim f has the ←− required properties is a direct application of Theorems 13 and 17. A continuum M is said to be hereditarily indecomposable if every subcontinuum of M is indecomposable. A pseudo-arc is a chainable hereditarily indecomposable continuum. For an example of such a continuum see Bing [6]. As far as we know, the ﬁrst example of such a continuum was given by B. Knaster [25]. We see from the following examples that the pseudo-arc can be obtained as an inverse limit on I with only one bonding map. Theorem 27 (Henderson [13]) There is a map from I onto I whose inverse limit is a pseudo-arc. The statement that the map f from I into I is transitive means that if U and V are open sets then there is a positive integer n such that f n (U ) ∩ V = ∅. Theorem 28 (Minc and Transue [34]) There is a transitive map from I onto I whose inverse limit is a pseudo-arc.

1.4 Period three implies indecomposability

19

1.4 Period three implies indecomposability

In this section and subsequently we frequently consider subsets M of an inverse limit with a single bonding map f consisting of all points x ∈ lim f ←− such that πi (x) ∈ J for some interval J ⊆ I such that f (J) = J. Thus we have M = lim f ∩ J ∞ . It is an immediate consequence of Theorem 13 that ←− M is a continuum and we have used this fact earlier. There are several other consequences of Theorem 13 that we use and we state some of these now. Theorem 29 If f maps I onto I, J = [u, v] is a subinterval of I such that f (J) = J, and M = lim f ∩ J ∞ , then ←− 1. If there are three numbers a < b < c in J such that f (a) = f (c) = u and f (b) = v or f (a) = f (c) = v and f (b) = u, then M is indecomposable. 2. If for each positive number ε there are numbers a < b < c in J such that each of f (a) and f (c) is within ε of u and |f (b) − v| < ε or each of f (a) and f (c) is within ε of v and |f (b) − u| < ε, then M is indecomposable. 3. If k is a positive integer and N = lim f k ∩ J ∞ then M and N are homeo←− morphic. Proof of 1. This theorem is an analogue of Theorem 23. In our special case, for each i the map hi from Theorem 13 is h where h(x) = u + x(v − u) for each x in I. Assume f (a) = f (c) = u and f (b) = v. Now h is a homeomorphism so there are unique points a , b , and c such that h(a ) = a, h(b ) = b, and h(c ) = c. It is easy to show that for the map g = h−1 ◦ f ◦ h from Theorem 13 we have that g(a ) = g(c ) = 0 and g(b ) = 1 so lim g is indecomposable ←− by Theorem 23. But M is homeomorphic to lim g so M is indecomposable. ←− The other case is similar. It is worth noting that for most of these theorems one could either appeal to Theorem 13 as we just did or prove the theorem directly by simply mimicking the proof of the analogous theorem for an inverse limit on I. We do this in the next proof. Proof of 2. We only brieﬂy sketch this proof inasmuch as it is exactly like that of Theorem 24. Assume that there are sequences {ai }i>0 , {bi }i>0 , and {ci }i>0 such that, for each i, ai < bi < c1 , |f (ai ) − u| < 1/i, |f (ci ) − u| < 1/i, and |f (bi ) − v| < 1/i. Next we ﬁnd an increasing sequence of positive integers {ni }i>0 such that {ani }i>0 converges to a, {bni }i>0 converges to b, and {cni }i>0 converges to c. Now f (a) = f (c) = u, f (b) = v, and a < b < c so M is indecomposable by Theorem 29.1. The other case where we have |f (ai ) − v| < 1/i, |f (ci ) − v| < 1/i, and |f (bi ) − u| < 1/i is similar. Proof of 3. This theorem is analogous to a special case of Theorem 15 and our proof is a simpliﬁcation of our proof for that theorem. If x ∈ M , we deﬁne

20

1 Inverse Limits on Intervals

ϕ(x) = (x1 , xk+1 , x2k+1 , x3k+1 , . . .). Now ϕ(x) ∈ N because f k (xnk+1 ) = x(n−1)k+1 for n > 0 by Theorem 2. To see that ϕ is an injection assume that ϕ(x) = ϕ(y) and note that if n is a positive integer, then there is a positive integer m with (m−1)k+1 ≤ n < (m+1)k+1. Because ϕ(x) = ϕ(y), we have x(m+1)k+1 = y(m+1)k+1 so xn = fn mk+1 (xnmk+1 ) = fn mk+1 (ymk+1 ) = yn by Theorem 2. To see that ϕ is continuous at x, let R be a region containing ϕ(x). By deﬁnition there exist a nonnegative integer i and an open subset −1 (O). U is a O of I containing xik+1 such that R = πi−1 (O). Let U = πik+1 region containing x and ϕ(U ) ⊆ R. We have that ϕ(M ) = N and because M is compact, ϕ is a homeomorphism by Theorem 259. Next we examine how properties from dynamical systems on a function f inﬂuence lim f . We begin with a look at periodicity. Li and Yorke piqued ←− interest in interval maps with periodic points of period three with their article in the American Mathematical Monthly, “Period three implies chaos” [28]. We begin with a theorem that might be labeled “period three implies indecomposable subcontinua.” Suppose X is a topological space, f is a mapping from X into X, p is a point of X, and n is a positive integer. We say that p is periodic of period n for f provided f n (p) = p and if 1 ≤ j < n then f j (p) = p. In [11] under these conditions p is called periodic of prime period n. Theorem 30 If f is a mapping from I into I having a periodic point of period three then lim f contains an indecomposable continuum. ←− In the proof of this theorem we introduce the notation cl(M ) to mean the closure of the set M . Proof. Suppose f has a periodic point p of period 3. Let x be the smallest of p, f (p), and f 2 (p) so that we have two cases to consider depending on which of f (x) and f 2 (x) is smaller. We consider the case where x < f (x) < f 2 (x). The other case can be gotten from our case by considering f 2 because x is a periodic point of period 3 for f 2 and x < f 2 (x) < f 4 (x) = f (x). Let J = [x, f (x)], let J = [f (x), f 2 (x)], let K = cl(∪i>0 f i (J)), and let K = cl(∪i>0 f i (J )). We show that K = K . Observe that J ⊆ [x, f 2 (x)] ⊆ f (J ) because f (J ) contains x and f 2 (x). Thus for each i > 0 f i (J) ⊆ f i+1 (J ) so K ⊆ K . But similarly J ⊆ f (J) so K ⊆ K and K = K . It follows from the deﬁnitions of J and J that if ε > 0, there is a positive integer n such i m i that if m ≥ n then d(t, ∪m i=1 (f (J))) < ε and d(t, ∪i=1 (f (J ))) < ε for each t ∈ K. f 2 (J) contains both x and f 2 (x), thus it contains [x, f 2 (x)] so J ⊆ f 2 (J). Thus f (J) ⊆ f 3 (J). Also J ⊆ f 3 (J) so we have that J ∪ f (J) ⊆ f 3 (J). Whence f (J) ∪ f 2 (J) ⊆ f 4 (J). This gives that f 2 (J) ∪ f 3 (J) ⊆ f 5 (J) and because f 3 (J) contains f (J) we have ∪3i=1 f i (J) ⊆ f 5 (J). It follows by induction that if k is a positive integer ∪ki=1 f i (J) ⊆ f k+2 (J). Now f (J ) contains

1.4 Period three implies indecomposability

21

J because it contains x and f 2 (x). Thus f (J ) ⊆ f 2 (J ) so J ∪ f (J ) ⊆ f (J ) ⊆ f 2 (J ). Similarly J ∪ f (J ) ∪ f 2 (J ) ⊆ f (J ) ∪ f 2 (J ) ⊆ f 3 (J ). By induction we conclude that for each positive integer k, ∪ki=1 f i (J ) ⊆ f k+1 (J ). Let ε > 0. As observed above, there is a positive integer n such that if m ≥ n i m i then d(t, ∪m i=1 (f (J))) < ε and d(t, ∪i=1 (f (J ))) < ε for each t ∈ K. By k choosing k = n + 2 we have d(t, f (J)) < ε and d(t, f k (J )) < ε for each t ∈ K. Let K = [c, d] and let ε > 0. There are points p and q in J and r and s in J such that |f k (p) − c| < ε, |f k (q) − d| < ε, |f k (r) − c| < ε, and |f k (s) − d| < ε. By Theorem 13 the set M of all points x such that πi (x) ∈ K for each i is a continuum. If p < q, then we have f k (p) and f k (r) are within ε of c, f k (q) is within ε of d, and p < q < r. Otherwise we have q < p < s with f k (q) and f k (s) within ε of d and f k (p) within ε of c. Now, by Theorem 29.2, N = lim f k ∩ J ∞ is indecomposable and, by Theorem 29.3, N is homeomorphic ←− to lim f ∩ J ∞ . ←− Sarkovskii’s theorem is an extremely useful theorem about periodicity in mappings of the reals. It has been extended, perhaps most notably for continuum theorists, to hereditarily decomposable chainable continua [33] (a continuum is hereditarily decomposable provided each of its nondegenerate subcontinua is decomposable). A proof of Sarkovskii’s theorem for the reals is readily accessible in [11, Theorem 10.2, p. 62]. Consider the following ordering of the positive integers, 3 5 7 · · · 2 · 3 2 · 5 2 · 7 · · · 22 · 3 22 · 5 · · · 23 · 3 23 · 5 · · · · · · 23 22 2 1. In the following R denotes the set of real numbers. Theorem 31 (Sarkovskii) Suppose f is a mapping from R into R and f has a periodic point of period k. If k j then f has a periodic point of period j. We make use of Sarkovskii’s theorem and Theorem 30 in the following theorem. This theorem ﬁrst appeared in [4] and independently in [16]. Theorem 32 If f is a mapping from I into I having a periodic point whose period is not a power of 2 then lim f contains an indecomposable continuum. ←− Proof. Suppose f has a periodic point whose period is 2k (2n+1) where k ≥ 0 k and n ≥ 1. Then f 2 has a periodic point of period 2n + 1, so by Sarkovskii’s k theorem f 2 has a periodic point of period 6. Thus the composite function k k+1 (f 2 )2 has a periodic point of period 3. By Theorem 30, lim f 2 contains an ←−

22

1 Inverse Limits on Intervals

indecomposable subcontinuum, but, by Theorem 15, lim f is homeomorphic ←− k+1 to lim f 2 . ←− There is no hope for a converse to Theorem 32 inasmuch as George W. Henderson has shown that the pseudo-arc, an hereditarily indecomposable chainable continuum, is the inverse limit on [0, 1] using a single bonding map that has no periodic points except ﬁxed points [13]. By way of contrast to Theorem 32 we remark that there is a mapping f from an interval into itself having the properties that if n ≥ 0 then f has a periodic point whose period is 2n , if x is a periodic point for f then there is a nonnegative integer n such that the period of x is 2n , and every nondegenerate subcontinuum of lim f is decomposable. See Example 76. ←−

1.5 Inverse limits with only one map of an interval

In this section we extend our inverse limits to consider inverse limits with a single bonding map f from an interval [a, b] into [a, b] where [a, b] is not necessarily I or a subset of I. The deﬁnition of lim f is unchanged and the ←− only diﬀerence is that if x ∈ lim f then xi ∈ [a, b]. The distance function ←− is unchanged. This extension of our deﬁnition to inverse limits with a single bonding map on an interval [a, b] is another step in the process begun in Theorem 13 and broadened in Section 1.4. Proofs of the existence, compactness, and connectedness of these inverse limits are virtually the same as the proofs of the corresponding theorems for inverse limits on I. Those proofs relied for the most part on the compactness and connectedness of Q. Because H:Q→ → [a, b]∞ given by H(x) = y where yi = a + (b − a)xi for each positive integer i is a surjective homeomorphism, we have [a, b]∞ is compact and connected. Here, and for the remainder of this monograph, when f is a map from a set H into a set K, we use the notation f : H → K. Furthermore we use the notation f : H → → K for a map from H into K such that f (H) = K, that is, when f is a surjection. Theorems 30 and 32 from the previous section hold in the case where the bonding maps are maps of intervals using essentially the same proofs. Further extensions are made in a later section. In later chapters, for example, extensions are made to include inverse limits of a sequence of maps between metric spaces and inverse limits with upper semi-continuous set-valued bonding functions. The following theorem provides a bridge between inverse limits with a single bonding map on I and these inverse limits with a single bonding map

1.5 Inverse limits with only one map of an interval

23

on [a, b]. If f : [a, b] → [a, b] and g : [c, d] → [c, d] are maps of intervals, then f and g are said to be topologically conjugate if there is a homeomorphism h from [c, d] onto [a, b] such that if x ∈ [c, d] then h(g(x)) = f (h(x)). Theorem 33 If f : [a, b] → [a, b] and g : [c, d] → [c, d] are maps of intervals that are topologically conjugate, then lim f is homeomorphic to lim g. ←− ←− Proof. Let h be a homeomorphism such that h ◦ g = f ◦ h and let ϕ be the function deﬁned on lim g given by ϕ(x) = y where yi = h(xi ) for each ←− positive integer i. Because h is a homeomorphism, it is clear that ϕ is 1-1. Suppose x ∈ lim g and V = V1 × V2 × · · · × Vn × [a, b]∞ is a basic open set ←− containing ϕ(x). For each integer i, 1 ≤ i ≤ n, there is set Ui open in [c, d] that contains xi and h(Ui ) ⊆ Vi . If U = U1 × U2 × · · · × Un × [c, d]∞ then U is an open set containing x and ϕ(U ) ⊆ V , so ϕ is continuous at x. Inasmuch as ϕ is 1-1 and continuous and lim g is compact, ϕ is a homeomorphism. ←− Showing that ϕ(lim g) = lim f completes the proof. Let x be a point of ←− ←− lim g and suppose y = ϕ(x). Then, f (yi+1 ) = f (h(xi+1 )) = h(g(xi+1 )) = ←− h(xi ) = yi . So, ϕ(lim g) ⊆ lim f . Suppose y ∈ lim f and let x be the point ←− ←− ←− of [c, d]∞ such that xi = h−1 (yi ) for each i. Then, g(xi+1 ) = g(h−1 (yi+1 )) = h−1 (f (h(h−1 (yi+1 )))) = h−1 (f (yi+1 )) = h−1 (yi ) = xi . Thus, x ∈ lim g ←− and h(x) = y so we have ϕ is surjective. Often it is diﬃcult if not impossible to obtain a conjugacy between two mappings even if the mappings produce homeomorphic inverse limits. There are other ways of obtaining a homeomorphism between inverse limits. Although this topic is explored more thoroughly in later chapters, we present one theorem along this line. We use the theorem that follows it later in this chapter. Theorem 34 Suppose f : [a, b] → → [a, b] is a mapping from the interval [a, b] onto [a, b] and g is a mapping from the interval [c, d] onto [c, d]. If ϕ1 , ϕ2 , ϕ3 , . . . is a sequence of homeomorphisms of [a, b] onto [c, d] such that, → lim g given for each positive integer i, ϕi ◦ f = g ◦ ϕi+1 then ϕ : lim f → ←− ←− by ϕ(x) = (ϕ1 (x1 ), ϕ2 (x2 ), ϕ3 (x3 ), . . . ) is a homeomorphism of lim f onto ←− lim g. ←− Proof. If x ∈ lim f , then g(ϕ(xi+1 )) = ϕi (f (xi+1 )) = ϕi (xi ) so ϕ(x) ∈ ←− lim g. To see that ϕ is surjective, let y be a point of lim g. For each positive ←− ←− −1 −1 integer i, let xi = ϕ−1 i (yi ). Then, f (xi+1 ) = f (ϕi+1 (yi+1 )) = ϕi (g(yi+1 )) = ϕ−1 f . To see that ϕ is 1-1, suppose x = y. There is i (yi ) = xi . Thus x ∈ lim ←− a positive integer j such that xj = tj so ϕj (xj ) = ϕj (tj ) and consequently ϕ(x) = ϕ(y). Finally, to see that ϕ is continuous, let V be a basic open set containing ϕ(x). There are a positive integer i and an open set O containing ϕ(xi ) such that V = πi−1 (O). There is an open set R containing xi such that ϕi (R) ⊆ O. Then, πi−1 (R) is an open set containing x and ϕ(R) ⊆ V . Thus,

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1 Inverse Limits on Intervals

ϕ is a 1-1 mapping from lim f onto the continuum lim g, so by Theorem 259, ←− ←− ϕ is a homeomorphism. Recall that a BJK horseshoe is any continuum homeomorphic to the continuum of Example 22, that is, homeomorphic to lim f where f (x) = 2x for ←− 0 ≤ x ≤ 1/2 and f (x) = 2 − 2x for 1/2 ≤ x ≤ 1. Theorem 35 Suppose g : [a, b] → → [a, b] is a mapping from the interval [a, b] onto itself such that g(a) = a, c is a point between a and b, and h1 : [a, c] → → → [a, b] are homeomorphisms such that h1 (a) = h2 (b) = a [a, b] and h2 : [c, b] → and hj (c) = b for j = 1, 2 and if x ∈ [a, c] then g(x) = h1 (x) whereas if x ∈ [c, b] then g(x) = h2 (x). Then, lim g is a BJK horseshoe. ←− Proof. Let f : [0, 1] → → [0, 1] be given by f (x) = 2x for 0 ≤ x ≤ 1/2 and f (x) = 2 − 2x for 1/2 ≤ x ≤ 1. Then lim f is the BJK horseshoe. Let ϕ1 : ←− [0, 1] → → [a, b] be a linear homeomorphism such that ϕ1 (0) = a. Inductively, −1 let ϕn+1 (x) = h−1 1 (ϕn (f (x))) for 0 ≤ x ≤ 1/2 and ϕn+1 (x) = h2 (ϕn (f (x))) for 1/2 ≤ x ≤ 1. Note that ϕn is a homeomorphism of [0, 1] onto [a, b] for each positive integer n and ϕn ◦ f = g ◦ ϕn+1 . By Theorem 34, ϕ is a homeomorphism.

(1/2,1)

(0,0)

(c,b)

(1,0)

(a,a)

(b,a)

Fig. 1.7 The BJK map (on the left) and a typical map producing the BJK horseshoe as in Theorem 35

We now state some analogues of theorems for inverse limits on I. Our ﬁrst theorem is an analogue of Theorem 23. Theorem 36 If f : [a, b] → [a, b] and there are three numbers x < y < z in [a, b] such that f (x) = f (z) = a and f (y) = b or f (x) = f (z) = b and f (y) = a, then lim f is indecomposable. ←−

1.5 Inverse limits with only one map of an interval

25

Proof. Let h : I → → [a, b] be the linear homeomorphism given by h(x) = a + (b − a)x and let g = h−1 f h. Then g : I → → I and hg = f h so f and g are topologically conjugate. Note that h−1 (x), h−1 (y), and h−1 (z) are three points of I such that h−1 (x) < h−1 (y) < h−1 (z) where g(h−1 (x)) = g(h−1 (z)) = 0 and g(h−1 (y)) = 1 or g(h−1 (x)) = g(h−1 (z)) = 1 and g(h−1 (y)) = 0. By Theorem 23, lim g is indecomposable. It is easy to see that the image of an ←− indecomposable continuum under a homeomorphism is indecomposable. By Theorem 33, lim f and lim g are homeomorphic, so lim f is indecomposable. ←− ←− ←− Our next theorem is a very useful special case of Theorem 15 in the setting of inverse limits on [a, b] with a single bonding map. Theorem 37 If f : [a, b] → [a, b] and n is a positive integer, then lim f n is ←− homeomorphic to lim f . ←− Proof. Let h : I → → [a, b] be the linear homeomorphism given by h(x) = a + (b − a)x and let g = h−1 f h. Then f and g are topologically conjugate. By Theorem 33, lim f and lim g are homeomorphic. Note that g n = h−1 f n h so ←− ←− f n and g n are topologically conjugate and so lim f n and lim g n are homeo←− ←− morphic. It is a consequence of Theorem 15 that lim g and lim g n are home←− ←− omorphic and, consequently, it follows that lim f and lim f n are homeomor←− ←− phic. Lemma 38 If f : [a, b] → [a, b] is a mapping, [c, d] ⊆ [a, b] is a subinterval such that f ([c, d]) ⊆ [c, d], and k = f | [c, d] then lim k = {x ∈ lim f | xi ∈ ←− ←− [c, d] for each positive integer i}. Proof. Let K = {x ∈ lim f | xi ∈ [c, d] for each positive integer i}. If p ∈ ←− lim k then p ∈ lim f and pi ∈ [c, d] for each positive integer i so lim k ⊆ K. ←− ←− ←− On the other hand, if pi ∈ [c, d] for each i and p ∈ lim f then k(pi+1 ) = pi ←− for each i and we have K ⊆ lim k. ←− The next theorem is an analogue of Theorem 19. Its proof makes use of the same conjugacy employed in the previous two theorems and ultimately relies on the observation that under a homeomorphism the image of a topological ray is a topological ray and the image of the remainder of a ray is the remainder of a ray. The proof of the second part of its conclusion makes use of Lemma 38. We leave the details to the reader. Theorem 39 Suppose f : [a, b] → → [a, b] is a mapping and c is a number, a < c < b, such that 1. f ([c, b]) ⊆ [c, b] 2. f | [a, c] is monotone 3. There is a positive integer j such that f j ([a, c]) = [a, b].

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Then, lim f is the closure of a ray having remainder a continuum K. ←− Moreover, K = lim k where k = f | [c, b]. ←− We have seen several theorems that give suﬃcient conditions for an inverse limit to be an indecomposable continuum. We end this section with a couple of theorems that furnish a suﬃcient condition for an inverse limit with a single bonding map to be decomposable. As a matter of convenience in the next two theorems, we allow [c, d] to denote a degenerate set in the case where c = d. Theorem 40 Suppose f : [a, b] → → [a, b] is a mapping and each of c and d is a point of (a, b) with c ≤ d. If f ([a, d]) = [a, d], f ([c, b]) = [c, b], and f ([c, d]) ⊆ [c, d], then lim f is decomposable. ←− Proof. Let M = lim f . Suppose x ∈ M and x ∈ / [c, d]∞ . Then there is a ←− positive integer n such that xn ∈ / [c, d]. If xn ∈ [a, c) then xj ∈ [a, d] for j ≥ n so x ∈ [a, d]∞ . In a similar manner, if xn ∈ (d, b] then x ∈ [c, b]∞ . It follows that lim f is the union of two of its subcontinua, H = lim f ∩ [a, d]∞ and ←− ←− K = lim f ∩ [c, b]∞ observing that H and K are continua by Corollary 14. ←− Note that because f is surjective it has ﬁxed points p1 ∈ [a, c) and p2 ∈ (d, b]. Then x = (p1 , p1 , p1 , . . . ) is in H but not in K whereas y = (p2 , p2 , p2 , . . . ) is in K but not in H. Thus, H and K are proper subcontinua of M whose union is M so M is decomposable. Theorem 41 Suppose f : [a, b] → → [a, b] is a mapping and each of c and d is a point of (a, b) with c ≤ d. If f ([a, d]) = [c, b], f ([c, b]) = [a, d], and f ([c, d]) ⊆ [c, d], then lim f is decomposable. ←− Proof. f 2 satisﬁes the hypothesis of Theorem 40, therefore this is an immediate consequence of Theorem 37.

1.6 Inverse limits on intervals with sequences of maps

In order to make this book as accessible as possible to as wide an audience as possible we began our discussion of inverse limits by restricting our attention to the interval [0, 1]. To analyze the subsets of inverse limits on [0, 1] it quickly became necessary to discuss inverse limits where the bonding maps are restrictions of the original bonding maps to subintervals of [0, 1]. See Theorem 13. We further extended the deﬁnition of inverse limit to inverse limits on an interval that is not necessarily [0, 1] with a single bonding map in the last section. In this brief section, we discuss completing this process with a

1.7 Inverse limits with unimodal bonding maps

27

look at inverse limits on intervals with sequences of maps. Details of this are so similar to things we have already done that we only sketch the process here. We believe interested readers can supply their own details if they wish. In Chapter 2, we provide a detailed investigation of inverse limits in very general settings that subsumes this particular discussion. Suppose [a1 , b1 ], [a2 , b2 ], [a3 , b3 ], . . . is a sequence of intervals and, for each positive integer n, fn : [an+1 , bn+1 ] → [an , bn ] is a mapping. By deﬁning lim f ←− to be the set of all sequences x = (x1 , x2 , x3 , . . . ) such that xi = fi (xi+1 ) for each positive integer i the deﬁnition of the inverse limit remains unchanged. Topologically, the inverse limit is a subset of the compact product space P = [a1 , b1 ] × [a2 , b2 ] × [a3 , b3 ] × · · · so by using the relative topology and making an appropriate minor adjustment to the deﬁnition of region, proofs of Theorems 2 through 8 carry over more or less unchanged. Some of those proofs relied on the compactness of Q but P is compact being a product of compact spaces. It is also true that P is homeomorphic to Q under a homeomorphism deﬁned coordinatewise using homeomorphisms of [an , bn ] onto [0, 1]. For the reader relying on the metric for Q to get a metric for inverse limits on [0, 1], there is somewhat of a problem inasmuch as for some x and y in P, i>0 |xi − yi |/2i may not converge. One indirect means of dealing with this diﬃculty would be to employ a homeomorphism between P and Q and measure the distance between points of P by the distance between their images under the homeomorphism. Another solution in the case where all of the intervals [an , bn ] are nondegenerate, would be to use d(x, y) = i i>0 |xi − yi |/2 |ai − bi | as a metric for the inverse limit. Theorems 9, 10, and 15 also hold in this setting using more or less the same proofs and we use the corresponding more general results as needed without additional proof.

1.7 Inverse limits with unimodal bonding maps

In this section we present some results on inverse limits on intervals with unimodal bonding maps. Unimodal maps have been studied extensively in the literature on dynamical systems. Here we concentrate on characterizing indecomposability of inverse limits on intervals with unimodal bonding maps. Recall that a mapping is monotone provided the inverse image of each point is connected. A map f : [a, b] → [a, b] is called unimodal provided f is

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not monotone and there is a point c, a < c < b, such that f | [a, c] and f | [c, b] are monotone. In the next few theorems, if f : [a, b] → → [a, b] is a unimodal map from [a, b] onto itself such that f (b) = a and f (c) = b, much use is made of the ﬁrst ﬁxed point for f 2 in [c, b]. Often, but not always, this is the ﬁxed point for f in [c, b]. Figure 1.8 depicts a unimodal map f and a point q of [c, b] such that f 2 (q) = q but q is not a ﬁxed point for f .

(c,b) (q,f(q) (p,p) (f(q),q)

(a,f(a))

(b,a)

Fig. 1.8 A unimodal map with a period 2 point

Lemma 42 Suppose f : [a, b] → [a, b] is unimodal, f (c) = b, and f (a) = a or f (b) = a. Let q be the ﬁrst ﬁxed point for f 2 in [c, b]. If f 2 (b) < q then there is a positive integer k such that f 2k+1 (c) < c. Proof. If f 2 (b) < c then f 3 (c) < c because f 3 (c) = f 2 (f (c)) = f 2 (b). Suppose f 2 (b) ≥ c. Because f is nonincreasing on [c, b], f (b) ≤ f (q). If f (b) ≥ c, we have f 2 (b) ≥ f 2 (q) = q contrary to hypothesis, so f 2 (c) = f (b) < c. The graph of f 2 is below the identity on [c, q) and c ≤ f 2 (b) < q, thus we have f 2 (f 2 (b)) < f 2 (b). It follows that if k is an integer such that c ≤ f 2k (b) then f 2 (f 2k (b)) < f 2k (b). If f 2k (b) ≥ c for every k, the sequence f 2 (b), f 4 (b), f 6 (b), . . . converges to a point r, c ≤ r < q. But f 2 (r) = r which contradicts the hypothesis that q is the ﬁrst ﬁxed point for f 2 in [c, b]. Thus, there is a k such that f 2k (b) < c so f 2k+1 (c) < c.

1.7 Inverse limits with unimodal bonding maps

29

Our next theorem gives a characterization of indecomposability of the inverse limit for unimodal mappings such that f (c) = b and f (b) = a. Theorem 43 Suppose f : [a, b] → → [a, b] is a unimodal mapping, f (c) = b, f (b) = a, and q is the ﬁrst ﬁxed point for f 2 in [c, b]. Then, lim f is ←− indecomposable if and only if f (a) < q. Proof. Suppose f (a) < q. Then, f ([a, c]) contains [q, b] and therefore [p, b] where p is the ﬁxed point for f in [c, b]. For each positive integer n, p is in f n ([a, c]). Because f 2 (c) = a, both a and p belong to f 2 ([a, c]). Because [a, c] ⊆ [a, p], [a, c] ⊆ f 2 ([a, c]). It follows that [a, c] ⊆ f 2i ([a, c]) for i ∈ N. Because f (a) = f 2 (b) < q, by Lemma 42, there is a positive integer k such that f 2k+1 (c) < c. Because f 2 (c) = a, f 2k−1 (a) < c. Thus, [c, p] ⊂ f 2k−1 ([a, c]), so b ∈ f 2k ([a, c]) and we have [a, b] = f 2k ([a, c]). Now f n ([c, b]) = [a, b] for every n so we have that f 2k ([a, c]) = [a, b] and f 2k ([c, b]) = [a, b]. This implies that there are points x < y < z such that f (x) = f (z) = a and f (y) = b, or f (x) = f (z) = b and f (y) = a so that lim f 2k is indecomposable by Theorem ←− 36. But this implies that lim f is indecomposable by Theorem 37. ←− On the other hand, if f (a) ≥ q, because f 2 (q) = q and q > c, we have f ([a, f (q)]) = [q, b] and f ([q, b]) = [a, f (q)]. Because f ([q, f (q)]) = [q, f (q)], it follows from Theorem 41 that lim f is decomposable. ←− If f : [a, b] → → [a, b] is a unimodal map with f (c) = b and f (b) = a and h : [a, b] → → [a, b] is the order-reversing homeomorphism given by h(x) = a+ b − x then g is conjugate to f if g(x) = h(f (h−1 (x))) = a+ b − f (a+ b − x). Because g(a) = b and g(a + b − c) = a, we obtain the following theorem from Theorems 43 and 33. Theorem 44 Suppose f : [a, b] → → [a, b] is a unimodal mapping, f (c) = a, f (a) = b, and q is the last ﬁxed point for f 2 in [a, c]. Then, lim f is ←− indecomposable if and only if f (b) > q. Our next theorem relates the characterizing condition f (a) < q from Theorem 43 to the existence of odd periodicity in f. If f (c) = b and f (b) = a, this condition is equivalent to f 2 (b) < q. We use the condition f 2 (b) < q in order to handle the case where f (a) = a in addition to the case where f (b) = a Theorem 45 Suppose f : [a, b] → → [a, b] is a unimodal map such that f (c) = b, f (a) = a (or f (b) = a), and q is the ﬁrst ﬁxed point for f 2 in [c, b]. Then, f has a periodic point of odd period greater than 1 if and only if f 2 (b) < q. Proof. Suppose f 2 (b) < q. By Lemma 42 there is a positive integer k such that f 2k+1 (c) < c. Because f is not monotone, a < c < b. If f has a ﬁxed point in [a, c], let r denote its last ﬁxed point in [a, c]. Let y1 = c. For each n > 1 there

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is a point yn such that r < yn < yn−1 ≤ c and f (yn ) = yn−1 , so f n (yn ) = b for n = 1, 2, 3, . . .. In particular, f 2k+1 (y2k+1 ) = b and f 2k+1 (c) < c so f 2k+1 has a ﬁxed point s between y2k+1 and c. Then, s is periodic for f of odd period greater than 1 because s > r. (The period of the periodic point may not be 2k + 1 but is a factor of 2k + 1.) Assume that f has no ﬁxed point in [a, c]. Now f 2k+1 [a, b] = [a, b] so f (a) = a or f 2k+1 (a) > a. Because f 2k+1 (c) < c, if f 2k+1 (a) > a then 2k+1 has a ﬁxed point between a and c so f has a periodic point of odd f period. If f 2k+1 (a) = a then a is a periodic point of f of odd period. In each of these cases because f has no ﬁxed point in [a, c], the periodic point has period greater than 1. 2k+1

Conversely, suppose f 2 (b) ≥ q. There are two cases to consider: (1) f (b) = a and (2) f (a) = a. In case (1), we have f (a) ≥ q, so f ([a, f (q)]) = [q, b] and f ([q, b]) = [a, f (q)]. Now if p is the ﬁxed point of f in [c, b] then q ≤ p and f is nonincreasing on [c.b] so f (q) ≥ f (p) ≥ q. If f (q) = q then f ([a, q)) = (q, b] and f ((q, b]) = [a, q) so if x = q, then, for odd values of n, f n (x) = x because x ∈ [a, q) and f n (x) ∈ (q, b], or x ∈ (q, b] and f n (x) ∈ [a, q). It follows that q is the only periodic point of f of odd period. A similar case holds if q < f (q). Inasmuch as f ([q, f (q)]) ⊆ [q, f (q)], if x is periodic and not in [q, f (q)] then no point of the orbit of x can be in [q, f (q)] so the orbit of x alternates between [a, q) and (f (q), b] and the period of x is even. Here by the orbit of a point x under a map f we mean the sequence x, f (x), f 2 (x), . . .. We now have that if f has a periodic point of odd period it is in [q, f (q)]. But f is nonincreasing on [q, f (q)] so the only periodic points for f in [q, f (q)] are the ﬁxed point for f in that interval and, possibly, points of period two. In case (2), let r denote the last ﬁxed point for f in [a, c]. If a = r, because a is a ﬁxed point it has period 1. If a < r, let x ∈ [a, r]. If x is not a ﬁxed point and x < f (x) then because f is nondecreasing on [a, r], for some n, f n (x) is a ﬁxed point greater than x or the orbit of x under f is an increasing sequence that converges to the ﬁrst ﬁxed point of f that is greater than x. In either case x is not periodic. If x > f (x) a similar process shows that x is not periodic. Assume that f (b) < c and that r < x < f (b). Because r < x < c, [x, c] contains no ﬁxed point of f so x < f (x). If fact, if n ≥ 0 is an integer and f n (x) < c then f n (x) < f n+1 (x). It cannot be that each term of the orbit of x under f is less than c for this would imply it converges to a ﬁxed point greater than r. So there is an integer k such that f k (x) ≥ c > f (b) and we have that x < f (x) < f 2 (x) · · · < f k (x). Because f (b) < c < q ≤ f 2 (b), we have that f 2 (b) > f (b), so f ([f (b), b]) = [f (b), b] and if t ≥ f (b), then f (t) ≥ f (b). So if m > k, f m (x) = x because f m (x) ≥ f (b). Now if x ∈ [f (b), b] we have that g = f | [f (b), b] is a unimodal map from [f (b), b] onto [f (b), b], g(b) = f (b), f (b) < c < b, and g(c) = b so the proof from case (1) applies to show that f has no periodic points of odd except ﬁxed points.

1.8 Logistic maps and their inverse limits

31

Finally we consider the case where f (b) ≥ c. Let p be the ﬁxed point of f in [c, b]. We have f ([c, p]) = [p, b] and f ([p, b]) ⊆ [c, p] so if x = p and x is periodic the orbit of x alternates between [c, p) and (p, b] so its period is not odd. In fact, if x = p, x ∈ [c, d], and x is periodic then the period is 2. Again by conjugacy, we have the following corollary. Corollary 46 Suppose f : [a, b] → → [a, b] is a unimodal map such that f (c) = a, f (a) = b (or f (b) = b), and q is the last ﬁxed point for f 2 in [a, c]. Then, f has a periodic point of odd period greater than 1 if and only if f 2 (a) > q. If f : [a, b] → → [a, b] is unimodal and f (a) = a then the core of lim f ←− is lim g where g = f | [f (b), b] whereas if f (b) = b the core is lim g where ←− ←− g = f | [a, f (a)]. Often we refer to the map g as the core of the map f . In the second part of the following theorem, it could be helpful to note that the core of the map is unimodal of the type discussed in the ﬁrst part of the theorem. Theorem 47 If f : [a, b] → → [a, b] is a unimodal map such that f (c) = b and f (b) = a (or f (c) = a and f (a) = b), then lim f is indecomposable if and ←− only if f has a periodic point of odd period greater than 1. If f : [a, b] → → [a, b] is a unimodal map such that f (a) = a (or f (b) = b) then the core of lim f ←− is indecomposable if and only if f has a periodic point of odd period greater than 1.

1.8 Some relationships between the dynamics of logistic maps and their inverse limits

The logistic family of mappings has a rich history in one-dimensional dynamics. This parameterized family has been studied in various forms in the literature. In [8] it is the family of mappings fμ given by fμ (x) = 1 − μx2 on [−1, 1] for 0 < μ ≤ 2 whereas in [11] it takes the form fμ (x) = μx(1 − x) on [0, 1] with the parameter μ satisfying 0 ≤ μ ≤ 4. Here we consider the logistic family to be the parameterized family of mappings fλ : [0, 1] → [0, 1] given by fλ (x) = 4λx(1 − x) where the parameter λ satisﬁes 0 ≤ λ ≤ 1. The study of this family from a dynamical systems point of view is normally concerned with such properties as periodicity, attracting periodic points, and bifurcations among others. In this section we consider inverse limits on [0, 1] with a single bonding map chosen from the logistic family. In order to understand some of these inverse limits we need information about the maps in the family that derives from varying the parameter λ. Although these maps are simple

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quadratics, determining essential properties such as periodic points and the orbit of the critical point that allow us to determine the nature of the inverse limit requires information that quickly becomes computationally diﬃcult to obtain explicitly. Much of our treatment is self-contained. Exceptions to this include a couple of early references to Devaney’s book [11]. Near the end of the proof of Theorem 65 there is a crucial reference to a theorem of Hubbard via some of Milnor’s writing [32] for which we were unable to obtain an elementary proof. One speciﬁc property of the logistic family that facilitates some of our computations is the property of having a negative Schwarzian derivative (deﬁned below); see [8] and [11]. We employ two salient features of the Schwarzian derivative: (1) if f has a negative Schwarzian derivative then so does f n for each positive integer n and (2) if f is positive over an interval where f has a negative Schwarzian derivative then f cannot decrease and then later increase over that interval. Suppose f is a function and x is a point at which f exists. The Schwarzian derivative of f at x is denoted Sf (x) and is given by f (x) 3 f (x) 2 − . Sf (x) = f (x) 2 f (x) A mapping f : [a, b] → [a, b] that is three times diﬀerentiable on [a, b] is said to have a negative Schwarzian derivative on [a, b] provided Sf (x) < 0 for each x ∈ [a, b]. It customary to allow Sf (x) to have value −∞. It should be observed that if f has negative Schwarzian derivative, f, f , and f are all continuous. In the proof of Theorem 65 we make use of the continuity of f . Throughout this section we say that a map f deﬁned on an interval [p, q] is strictly unimodal on [a, b] ⊆ [p, q] provided f ([a, b]) ⊆ [a, b] and f is unimodal on [a, b] with critical point c, f (a) = f (b) = a, f (x) > 0 for a ≤ x < c, f (x) < 0 for c < x ≤ b, and Sf (x) < 0 for x ∈ [a, b]. It should be observed that if f is strictly unimodal on [a, b] such that f (c) > c and f −1 (c) = {d1 , c1 }, then (f 2 ) (x) = 0 only for x ∈ {d1 , c1 , c}. We begin with several preliminary results. Theorem 48 [11, Prop. 11.3] If f : [a, b] → [a, b] and g : [a, b] → [a, b] are mappings and x is a point of [a, b] such that Sf (x) < 0 and Sg(x) < 0, then S(f ◦ g)(x) < 0. For the logistic family, S(fλ ) is negative because fλ (x) = 0 for each x ∈ [0, 1]. Thus, we have the following corollary. Corollary 49 For the logistic family, Sfλn (x) < 0 for each x ∈ [0, 1] and each positive integer n.

1.8 Logistic maps and their inverse limits

33

Theorem 50 [11, Lemma 11.5] If Sf (x) < 0 over an interval J and f (x) > 0 for x ∈ J then f does not have a local minimum in the interior of J. Lemma 51 Suppose f : [a, b] → R is a mapping such that Sf (x) < 0 for a ≤ x ≤ b. If f (x) > 0 for x ∈ [a, b] and f (a) ≥ 1 and f (b) ≥ 1, then f (x) > 1 for a < x < b. Proof. Sf (x) < 0 for each x ∈ [a, b], thus f is not constant on any subinterval of [a, b]. If follows from Theorem 50 that if a < t 1. Similarly, if a < t 1. If f (t) ≤ 1 for some t, x < t < y, then f has a local minimum in the interior of [a, b] contrary to Theorem 50. Lemma 52 Suppose f : [a, b] → [a, b] is unimodal with critical point c, f (a) = f (b) = a, and f (c) > c. Suppose f is strictly decreasing on [c, b] and let c1 = (f | [c, b])−1 (c). 1. If f 2 (c) > c then f 2 (c1 ) < c1 . 2. If f 2 (c) = c then f 2 (c1 ) = c1 . 3. If f 2 (c) < c then f 2 (c1 ) > c1 . Proof. Let p be the ﬁxed point for f in [c, b]. Because c p. Inasmuch as c < f (c) = f 2 (c1 ) we see that both c1 and f 2 (c1 ) belong to [c, b]. Because c = f (c1 ) and f 2 (c) = f 2 (f (c1 )) = f (f 2 (c1 )), and f | [c, b] is a strictly decreasing homeomorphism, 1. If f 2 (c) > c, c1 > f 2 (c1 ). 2. If f 2 (c) = c, c1 = f 2 (c1 ). 3. If f 2 (c) < c, c1 < f 2 (c1 ).

Lemma 53 Suppose f : [a, b] → [a, b] is strictly unimodal with critical point c, f (c) > c, and p is the ﬁxed point for f in [c, b]. If f 2 (c) ≤ c, f (p) < −1. Proof. f is strictly unimodal, therefore Theorem 48 yields that f 2 has a negative Schwarzian derivative. If f 2 (c) = c, then f 2 (f (c)) = f (c), so c, p, and f (c) are ﬁxed points for f 2 . By the mean value theorem there exist points r and s, c < r f 2 (c) = c. Thus, if c < x < f (c), both f (x) and f (f (x)) are negative so (f 2 ) (x) > 0. Because (f 2 ) > 0 on [r, s], (f 2 ) (p) > 1 by Lemma 51. But, (f 2 ) (p) = (f (p))2 , so f (p) < −1. If f 2 (c) < c and c1 = (f | [c, b])−1 (c), f 2 (c1 ) > c1 by Lemma 52. The slope of the line joining (c, f 2 (c)) and (p, p) is greater than 1, so by the mean value theorem there is a point r, c < r < p, such that (f 2 ) (r) > 1. Similarly, the slope of the line joining (p, p) and (c1 , f 2 (c1 )) is greater than 1 thus there is a point s, p < s < c1 , such that (f 2 ) (s) > 1. Again, from Lemma 51, (f 2 ) (p) > 1 and thus f (p) < −1.

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Lemma 54 Suppose f : [a, b] → [a, b] is strictly unimodal with critical point c, f (c) > c, and p is the ﬁxed point for f in [c, b]. If −1 ≤ f (p) < 0 then f 3 (c) > p. Proof. By Lemma 53, f 2 (c) > c. Because f decreases on [c, b] and c < p, f (c) > p and, therefore, f 2 (c) < p. Because c < f 2 (c) < p, f 3 (c) > p. Lemma 55 Suppose f : [a, b] → [a, b] is strictly unimodal with critical point c, f (c) > c, and p is the ﬁxed point for f in [c, b]. If f (p) ≥ −1 and c1 = (f | [c, b])−1 (c) then f 2 (c1 ) < c1 . Proof. Note that inasmuch as f is strictly unimodal (f 2 ) (x) > 0 for c < x < c1 and (f 2 ) (x) < 0 for c1 < x < b. Because f (p) ≥ −1, f 2 (c) > c by Lemma 53. By Lemma 52, f 2 (c1 ) < c1 . Theorem 56 Suppose f : [a, b] → [a, b] is strictly unimodal with critical point c, f (c) > c, and p is the ﬁxed point for f in [c, b]. If f (p) ≥ −1 then f does not have a period 2 point in [c, p]. Proof. Suppose q is a point such that c < q p is also a ﬁxed point for f 2 . Let c1 = (f | [c, b])−1 (c). By Lemma 55, f 2 (c1 ) < c1 . Because c < q, f (q) < f (c) = f 2 (c1 ) < c1 . Thus, p < f (q) < c1 . Noting that (f 2 ) (x) > 0 for c < x < c1 and q, p, and f (q) are ﬁxed points for f 2 , by the mean value theorem, there are points r and s, q < r 0 for r ≤ x ≤ s, we may use Lemma 51 to conclude that (f 2 ) (p) > 1. Because (f 2 ) (p) = (f (p))2 , we obtain f (p) < −1. This is a contradiction. Theorem 57 Suppose f : [a, b] → [a, b] is strictly unimodal with critical point c, f (c) > c, and p is the ﬁxed point for f in [c, b]. If f (p) ≥ −1 then the sequence c, f (c), f 2 (c), . . . converges to p. Proof. f 2 (c) > c and, by Theorem 56, f 2 has no ﬁxed point between c and p, therefore f 2 (x) > x for c ≤ x < p. The sequence f 2 (c), f 4 (c), f 6 (c), . . . is increasing and bounded above by p so it converges to a ﬁxed point for f 2 . Because f 2 has no ﬁxed point between c and p, we see that this sequence converges to p. Because f (p) = p, the sequence f 3 (c), f 5 (c), f 7 (c), . . . also converges to p and the conclusion follows. Lemma 58 If f : [a, b] → [a, b] is strictly unimodal and f (a) > 1, then f has a ﬁxed point p > a. Proof. f (a) > 1, thus there is a number x > a such that f (x) > x. Also, because f (b) = a < b, there is a ﬁxed point for f between x and b.

1.8 Logistic maps and their inverse limits

35

We now turn to some preliminary theorems on inverse limits that we use in obtaining some inverse limit results for the logistic family. Our ﬁrst result is a theorem (Theorem 60) on inverse limits on intervals that is a special case of a theorem from a later chapter that holds in a much more general setting. Its proof depends on the following. Lemma 59 If f : [a, b] → [a, b] is a mapping and [c, d] = n>0 f n ([a, b]), then f ([c, d]) = [c, d]. Proof. Because f n+1 ([a, b]) ⊆ f n ([a, b]) for each positive integer n, f ([c, d]) ⊆ [c, d]. Let y ∈ [c, d]. For each i, there is a point zi ∈ [a, b] such that f i (zi ) = y. For i ≥ 2, let xi = f i−1 (zi ) and note that f (xi ) = y. Some subsequence of x2 , x3 , x4 , . . . converges to a point x of [a, b]. Because f (xi ) = y for each i, f (x) = y. Let n be a positive integer and let U be an open interval containing x. There exists a positive integer j ≥ n + 1 such that xj ∈ U . Then, xj ∈ f j−1 ([a, b]) ⊆ f n ([a, b]). It follows that x ∈ f n ([a, b]) = f n ([a, b]). Because x ∈ f n ([a, b]) for each n, x ∈ [c, d]. An inverse limit on an interval with a single bonding map that is not surjective is an inverse limit on a (possibly degenerate) subinterval with a single surjective bonding map as may be seen in the following theorem. Theorem 60 If f : [a, b] → [a, b] is a mapping, [c, d] = n>0 f n ([a, b]), and g = f | [c, d], then lim f = lim g. ←− ←− Proof. If x ∈ lim f and each of i and n is a positive integer, xi ∈ f n ([a, b]) ←− because xi = f n (xi+n ). Thus, for each i, xi ∈ [c, d] and we have lim f ⊆ lim g. ←− ←− Because lim g is clearly a subset of lim f , the conclusion follows. ←− ←− Theorem 61 Suppose f : [a, b] → [a, b] is strictly unimodal, f (a) > 1, and p > a is the last ﬁxed point for f in [a, b]. If f (p) ≥ −1 then lim f is an arc. ←− Proof. There are two cases: (1) p≤ c and (2) p > c. In case p ≤ c, because c, f (c), f 2 (c), . . . converges to p, i>0 f i ([a, b]) = [a, p]. Thus, g = f | [a, p] is a homeomorphism, so lim g is an arc. By Theorem 60, lim f = lim g. ←− ←− ←− Suppose p > c. By Lemma 53, f 2 (c) > c so f ([c, f (c)]) ⊆ [c, f (c)]. By Theorem 57, the sequence c, f (c), f 2 (c), . . . converges to p. It follows that i i>0 f ([c, f (c)]) = {p}. By Theorem 39, the inverse limit is the closure of a ray with remainder a point, so lim f is an arc. ←− For 1/4 < λ ≤ 3/4, fλ is strictly unimodal, fλ (0) > 1, and pλ = 1 − 1/(4λ) > 0 is the positive ﬁxed point for fλ , and fλ (pλ ) ≥ −1. Thus, as an application of Theorem 61 and Corollary 49, we have the following example.

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Example 62 For 1/4 < λ ≤ 3/4, lim fλ is an arc. ←− Example 62 could have been dealt with directly, but we wish to obtain information about inverse limits in the logistic family for λ > 3/4. The machinery we developed leading to Example 62 plays a key role in this endeavor. Lemma 63 is useful in the proof of Theorem 65. Lemma 63 Suppose f : [a, b] → [a, b] is strictly unimodal with critical point c and f has a ﬁxed point p such that f (p) ≤ −1. Then f (c) > c, c < p, and there are points b1 = (f | [c, b])−1 ((f | [a, c])−1 (p)) and c1 = (f | [c, b])−1 (c) such that p < c1 < b1 0 for p < x < c1 , (f 2 ) (c1 ) = 0, and (f 2 ) (x) < 0 for c1 < x < b. Moreover, if f (c) ≤ b1 then f 2 is strictly unimodal on [p, b1 ] with critical point c1 . Proof. Note that c < p because f (p) < 0. Inasmuch as f is strictly unimodal, p < f (c) so c < f (c). For b1 = (f | [c, b])−1 ((f | [a, c])−1 (p)) observe that p < b1 < b. For c1 = (f | [c, b])−1 (c) note that p < c1 < b1 . If p ≤ x < c1 , (f 2 ) (x) = f (f (x))f (x) so (f 2 ) (x) > 0 because x > c and f (x) > c. On the other hand, if c1 < x ≤ b, (f 2 ) (x) < 0 because x > c and f (x) < c. Finally, (f 2 ) (c1 ) = 0 because f (c1 ) = c and f (c) = 0. Suppose f (c) ≤ b1 . Then f 2 (c1 ) ≤ b1 . Note that f 2 (p) = p and f 2 (b1 ) = p. In the previous paragraph we showed that (f 2 ) (x) < 0 for p ≤ x < c1 and (f 2 ) (x) < 0 for c1 < x ≤ b. It follows that f 2 ([p, b1 ]) ⊆ [p, b1 ]. Because S(f ) < 0, S(f 2 ) < 0. Thus, f 2 is strictly unimodal on [p, b1 ] with critical point c1 . With Devaney [11, p. 79], by a one-parameter family of functions gλ we mean a continuous function G deﬁned on a product of intervals such that (1) G(x, λ) = gλ (x), (2) the partial derivative of G in its second variable is continuous, and (3) gλ is a C ∞ function for each value of λ. The logistic family G(x, λ) = 4λx(1 − x) over [0, 1] × [0, 1] is a prime example of such a one-parameter family of functions. The following statement is Theorem 12.5 from [11, p. 85]. Theorem 64 Let gλ be a one-parameter family of functions and suppose that gλ0 (x0 ) = x0 and gλ 0 (x0 ) = 1. Then there are intervals J containing x0 in its interior and N containing λ0 in its interior and a smooth function p : N → J such that p(λ0 ) = x0 and gλ (p(λ)) = p(λ). Moreover, gλ has no other ﬁxed points in J. We now turn speciﬁcally to the logistic family, fλ (x) = 4λx(1 − x) with 0 ≤ x ≤ 1 and 0 ≤ λ ≤ 1. Let λ0 = 1/4, λ1 = 3/4, μ0 = 1, and p0 (λ) = 1 − 1/(4λ) for λ ≥ λ0 . Let a0 (λ) = 0, b0 (λ) = 1, and c0 (λ) = 1/2 for each λ. For convenience of notation let p−1 (λ) = 0, c−1 (λ) = λ, and b−1 (λ) = 2 for each λ.

1.8 Logistic maps and their inverse limits

37

Theorem 65 For the logistic family there exist an increasing sequence λ and a decreasing sequence μ of parameter values such that for each nonnegative integer n: 1. λn+1 < μn 2. There exist functions an , bn , and cn such that each is continuous on n (λn , μn ] and, for λn < λ ≤ μn , fλ2 is strictly unimodal on [an (λ), bn (λ)] ⊆ n [0, 1] with critical point cn (λ) such that fλ2 (cn (λ)) = λ ≤ bn (λ) < bn−1 (λ) 3. There exists a function pn , continuous on (λn , μn ], so that, for λn < λ ≤ n μn , pn (λ) is the last ﬁxed point for fλ2 on [0, 1], and an (λ) < pn (λ) < bn (λ) with an (λ) = pn−1 (λ) n n 4. For λ = μn , fλ2 (λ) = pn−1 (λ) whereas fλ2 (λ) > pn−1 (λ) for λn < λ < μn n n 5. (fλ2n+1 ) (pn (λn+1 )) = −1 whereas (fλ2 ) (pn (λ)) < −1 for λn+1 < λ ≤ μn n and (fλ2 ) (pn (λ)) > −1 for λn < λ < λn+1 . Proof. We proceed inductively. Recall that we adopted the following notation. p−1 (λ) = 0, c−1 (λ) = λ, c0 (λ) = 1/2, a0 (λ) = 0, b−1 (λ) = 2, and b0 (λ) = 1 for each λ. Furthermore, we let λ0 = 1/4, λ1 = 3/4, and μ0 = 1. Note that λ1 < μ0 . Being constant functions, a0 , b0 , and c0 are continuous at every real number. For λ0 < λ ≤ μ0 , fλ is strictly unimodal on [a0 (λ), b0 (λ)] (see Corollary 49) with critical point c0 (λ) = 1/2 such that fλ (c0 (λ)) = λ ≤ b0 (λ) < b−1 (λ). For p0 (λ) = 1 − 1/(4λ), p0 is continuous on (λ0 , μ0 ] and fλ (p0 (λ)) = p0 (λ) for λ0 < λ ≤ μ0 . Moreover, p0 (λ) is the last ﬁxed point for fλ on [0, 1] for λ0 < λ ≤ μ0 , a0 (λ) < p0 (λ) < b0 (λ), and a0 (λ) = p−1 (λ) inasmuch as both are 0. Also, for λ = μ0 , fλ (λ) = 0 = p−1 (λ) and fλ (λ) > 0 = p−1 (λ) for λ0 < λ < μ0 . Finally, p0 (λ1 ) = 2/3 and fλ 1 (2/3) = −1 whereas fλ (p0 (λ)) < −1 for λ > λ1 and fλ (p0 (λ)) > −1 for λ < λ1 so the statement holds for n = 0. Suppose k is a nonnegative integer and λ0 < λ1 < · · · < λk < λk+1 and μ0 > μ1 > μ2 > · · · > μk have been deﬁned such that for 0 ≤ i ≤ k, 1. λi+1 < μi 2. There exist functions ai , bi , and ci such that each is continuous on (λi , μi ], i and, for λi < λ ≤ μi , fλ2 is strictly unimodal on [ai (λ), bi (λ)] ⊆ [0, 1] with i critical point ci (λ) such that fλ2 (ci (λ)) = λ ≤ bi (λ) < bi−1 (λ) 3. There is a continuous function pi on (λi , μi ] such that, if λi < λ ≤ μi , i pi (λ) is the last ﬁxed point for fλ2 in [0, 1], and ai (λ) < pi (λ) < bi (λ) with ai (λ) = pi−1 (λ) i i 4. For λ = μi , fλ2 (λ) = pi−1 (λ) whereas fλ2 (λ) > pi−1 (λ) for λi < λ < μi i i 5. (fλ2i+1 ) (pi (λi+1 )) = −1 whereas (fλ2 ) (pi (λ)) < −1 for λi+1 < λ ≤ μi and i

(fλ2 ) (pi (λ)) > −1 for λi < λ < λi+1 . k+1

We begin with the deﬁnition of μk+1 . Observe that, for λ = μk , fλ2 (λ) < k pk (λ). To see this, recall that, for λ = μk , fλ2 (λ) = pk−1 (λ), so that

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k

fλ2 (λ) = fλ2 (pk−1 (λ)) = pk−1 (λ) = ak (λ) < pk (λ). For λ = λk+1 , inask+1 k k k much as fλ2 (λ) = (fλ2 )2 (fλ2 (ck (λ)) = (fλ2 )3 (ck (λ)) and, by Lemma 54 k (the graph of fλ2 is above the diagonal at ck (λk+1 ) because its derivative k k+1 at pk (λk+1 ) is −1), (fλ2 )3 (ck (λ)) > pk (λ), we have fλ2 (λ) > pk (λ). It now follows that between λk+1 and μk there is a parameter value λ such k+1 that fλ2 (λ) = pk (λ). Choose μk+1 to be the parameter value such that k+1 k+1 fλ2 (λ) = pk (λ) for λ = μk+1 and we have that fλ2 (λ) > pk (λ) for λk+1 < λ < μk+1 . Note that λk+1 < μk+1 < μk . Let ak+1 (λ) = pk (λ) for λk+1 < λ ≤ μk+1 . Note that ak (λ) < ak+1 (λ) k and ak+1 is continuous on (λk+1 , μk+1 ]. Because fλ2 is strictly unimodal on k [ak (λ), bk (λ)] and (fλ2 ) (pk (λ)) < −1 on (λk+1 , μk ], then for λk+1 < λ ≤ μk , k k let bk+1 (λ) = (fλ2 | [ck (λ), bk (λ)])−1 ((fλ2 | [ak (λ), ck (λ)])−1 (pk (λ))). So bk+1 is continuous on (λk+1 , μk+1 ] because it is continuous on (λk+1 , μk ]. Morek+1 over, fλ2 (bk+1 (λ)) = ak+1 (λ) and bk+1 (λ) < bk (λ). We now show that λ ≤ bk+1 (λ) for λk+1 < λ ≤ μk+1 . For if not and bk+1 (λ) < λ ≤ k k k+1 bk (λ), then fλ2 (bk+1 (λ)) > fλ2 (λ) and so fλ2 (λ) < pk (λ) contrary to the choice of μk+1 . Furthermore, for λk+1 < λ ≤ μk , let ck+1 (λ) = k (fλ2 | [ck (λ), bk (λ)])−1 (ck (λ)) so ck+1 is continuous on (λk+1 , μk+1 ] because it is continuous on (λk+1 , μk ]. Also, note that by the choice of ck+1 (λ), k+1 k k fλ2 (ck+1 (λ)) = fλ2 (ck (λ)) = λ. For λk+1 < λ ≤ μk+1 , (fλ2 ) (pk (λ)) < −1 k k+1 and fλ2 (ck (λ)) = λ ≤ bk+1 (λ) so Lemma 63 yields that fλ2 is strictly unimodal on [ak+1 (λ), bk+1 (λ)] with critical point ck+1 (λ). k

We now turn to deﬁning pk+1 . For λ = λk+1 , (fλ2 ) (pk (λ)) = −1 k so ck (λ) < pk (λ) and thus fλ2 (ck (λ)) > ck (λ). By Lemma 55, we have k+1 k+1 fλ2 (ck+1 (λ)) < ck+1 (λ). We now show that, for λ = μk , fλ2 (ck+1 (λ)) > k ck+1 (λ). Recall that λ ≤ bk (λ) for λk < λ ≤ μk . For λ = μk , fλ2 (λ) = k pk−1 (λ) = ak (λ). Moreover, fλ2 (bk (λ)) = ak (λ) for λk < λ ≤ μk . For λ = μk , each of λ and bk (λ) is a point of [ak (λ), bk (λ)] greater than ck (λ), thus it follows that bk (λ) = λ (i.e., bk (μk ) = μk ). ck+1 (μk ) < bk (μk ) and k+1 k+1 μk = fμ2k (ck+1 (μk )), thus we obtain that ck+1 (μk ) < fμ2k (ck+1 (μk )). It now follows that there is a parameter value ν, λk+1 < ν < μk , such that k+1 k+1 fν2 (ck+1 (ν)) = ck+1 (ν); that is, ck+1 (ν) is a ﬁxed point for fν2 . Morek+1 k+1 over, (fν2 ) (ck+1 (ν)) = 0 because ck+1 (ν) is a critical point for fν2 . We may now apply Theorem 64 to obtain intervals J with ck+1 (ν) in its interior and N with ν in its interior and a smooth function q : N → J such k+1 that q(ν) = ck+1 (ν) and fλ2 (q(λ)) = q(λ) for λ ∈ N . The uniqueness of q guaranteed in Theorem 64 allows us to extend the domain of q to contain k+1 (λk+1 , μk ] provided, if λk+1 < λ ≤ μk , (fλ2 ) (r) = 1 when r is a ﬁxed point k+1 k+1 for fλ2 . Suppose λ ∈ (λk+1 , μk ]. If x > ck+1 (λ) then (fλ2 ) (x) < 0 so k+1 fλ2 cannot have derivative 1 at a ﬁxed point that is greater than ck+1 (λ).

1.8 Logistic maps and their inverse limits

39

k+1

Suppose s is a ﬁxed point for fλ2 between pk (λ) and ck+1 (λ). By the mean k+1 value theorem, between the two ﬁxed points s and pk (λ) for fλ2 is a point r k+1 k k+1 2 2 2 such that (fλ ) (r) = 1. Because (fλ ) (pk (λ)) = −1, (fλ ) (pk (λ)) = 1. k+1 k+1 is greater than Thus, (fλ2 ) (s) < 1 for otherwise the derivative of fλ2 or equal to 1 at both pk (λ) and s and 1 at the point r between them contrary to Theorem 51. To see that q may be extended so that its domain contains (λk+1 , μk ], suppose it has been extended so that its domain N has greatest lower bound α > λk+1 . Let α1 , α2 , α3 , . . . be a sequence of elements of N that converges to α. Because q(α1 ), q(α2 ), q(α3 ), . . . is a sequence of points of [0, 1] some subsequence converges to a point x1 in [0, 1]. For convenience we may assume that this subsequence is q(α1 ), q(α2 ), q(α3 ), . . . . For each positive integer i, ak+1 (αi ) < q(αi ) < bk+1 (αi ) and, by the continuity of ak+1 and bk+1 , ak+1 (α1 ), ak+1 (α2 ), . . . converges to ak+1 (α) and bk+1 (α1 ), bk+1 (α2 ), . . . converges to bk+1 (α). So, x1 ∈ [ak+1 (α), bk+1 (α)]. k+1 Inasmuch as fλ2 (x) is continuous in both λ and x and fαi (q(αi )) = q(αi ) k+1 k+1 for each i, it follows that fα2 (x1 ) = x1 . Because (fλ2 ) (x) is continuous k+1 k+1 in both λ and x and (fα2i ) (qi (αi )) < 1, (fα2 ) (x1 ) ≤ 1. ak+1 (α) = pk (α) k+1 and (fα2 ) (pk (α)) > 1, thus ak+1 (α) < x1 . As shown above using Theok+1 k+1 at x1 , we obtain rem 51, (fα2 ) (x1 ) < 1. Applying Theorem 64 to fα2 an interval M containing α in its interior and and interval K containing x1 k+1 in its interior and a function s : M → K such that fλ2 (s(λ)) = s(λ) for k+1 each λ ∈ M and fλ2 has no other ﬁxed point in K. Note that M ∩ N = ∅ k+1 because M ∩ N contains at least one αi . If β ∈ M ∩ N , fβ2 (s(β)) = s(β) k+1

and fβ2 (q(β)) = q(β), so s(β) = q(β). Thus, using s we may extend q so that its domain contains M , a contradiction to the choice of α. Similarly, we may extend the domain of q to contain μk . Thus, we obtain a continuous k+1 function q on (λk+1 , μk ] such that fλ2 (q(λ)) = q(λ) for λk+1 < λ ≤ μk . Let pk+1 = q. We now show that if λk+1 < λ ≤ μk then pk+1 (λ) is the last ﬁxed point k+1 k+1 on [0, 1]. Let λ ∈ (λk+1 , μk ]. Because fλ2 is strictly unimodal on for fλ2 k+1 [ak+1 (λ), bk+1 (λ)], λ = fλ2 (ck+1 (λ)) ≤ bk+1 (λ) and ak+1 (λ) and pk+1 (λ) k+1 are its only ﬁxed points in this interval. The maximum value on [0, 1] for fλ2 k+1 k+1 is λ, so, if x > bk+1 (λ) then fλ2 (x) ≤ λ ≤ bk+1 (λ). Thus, fλ2 (x) < x k+1 and x is not a ﬁxed point for fλ2 . This yields that pk+1 (λ) is the last ﬁxed k+1 point for fλ2 on [0, 1] for λk+1 < λ ≤ μk . Next we obtain λk+2 . Recall that ν (deﬁned in the construction of pk+1 ) k+1 is a value of λ such that (fν2 ) (pk+1 (ν)) = 0. On the other hand, for k+1 λ = μk+1 , we have (fλ2 ) (pk+1 (λ)) < −1. To see this, note that, for k+1 k+1 λ = μk+1 , fλ2 (λ) = pk (λ) = ak+1 (λ) < ck+1 (λ). But, fλ2 (λ) = k+1 k+1 (fλ2 )2 (ck+1 (λ)), so (fλ2 )2 (ck+1 (λ)) < ck+1 (λ). By Lemma 53, it follows

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1 Inverse Limits on Intervals k+1

that (fλ2 ) (pk+1 (λ)) < −1. Thus, between ν and μk+1 there is a parameter k+1 value λ such that (fλ2 ) (pk+1 (λ)) = −1. Denote this parameter value by λk+2 . Finally, it is a consequence of Theorem 1.2 of [32] that, if λk+2 < λ ≤ μk+1 , k+1 < −1 whereas (fλ2 ) (pk+1 (λ)) > −1 if λk+1 < λ < λk+2 .

k+1 (fλ2 ) (pk+1 (λ))

One consequence of Theorem 65 is that the sequence λ converges as does the sequence μ. Denote by λc the limit of the sequence λ. This is often called the Feigenbaum limit. We do not show it here, but the sequence μ also converges to λc ; see [8, p. 46]. The sequences λ and μ mark signiﬁcant changes in the dynamics of the logistic family. These two sequences also mark signiﬁcant changes in inverse limits in the logistic family. However, before delving deeper into inverse limits on [0, 1] with logistic bonding maps, we include some general properties of inverse limits on intervals that are useful in our investigation. Theorem 66 Suppose f : [a, b] → [a, b] is unimodal with critical point c such that c < f (c), ﬁxed point p > c, f (a) = a, f 2 (c) = a and f has no ﬁxed point between a and c. Then lim f is the closure of a ray with remainder C = lim r ←− ←− where r = f | [f 2 (c), f (c)]. In addition, if f 3 (c) ≥ p, then C is the union of two continua intersecting only at (p, p, p, . . . ) and each is homeomorphic to lim ϕ where ϕ = f 2 | [p, f (c)]. ←− Proof. The maximum value of f on [a, b] is f (c), so f (c) ≥ f 2 (c). Now, we show that if x ∈ [f 2 (c), f (c)] then f (x) ∈ [f 2 (c), f (c)]. Suppose x ∈ [f 2 (c), f (c)]. If x ≥ c then f (x) ≥ f 2 (c) because f decreases on [c, b]. If x < c then f 3 (c) ≤ f (x) because f increases on [a, c]. However, f has no ﬁxed point between a and c so the graph of f lies above the identity on (a, c]. It follows that f 2 (c) < f 3 (c). In either case, f (x) ∈ [f 2 (c), f (c)]. To complete the proof of the ﬁrst part of the theorem we make use of Theorem 39. There are two cases: f 2 (c) ≥ c and f 2 (c) < c. If f 2 (c) ≥ c, then f ([c, f (c)]) ⊆ [c, f (c)]. From the fact that f ([a, c]) = [a, f (c)], Theorem 39 applied to f | [a, f (c)] yields that lim f is the closure of a ←− ray with remainder lim g where g = f | [c, f (c)]. Because f is nonincreasing on ←− 2 [c, f (c)], f ([c, f (c)]) = [f (c), f (c)]. It follows from Theorem 60 that lim g = ←− lim r where r = f | [f 2 (c), f (c)]. ←− If f 2 (c) < c, observe that there is an integer n ≥ 3 such that f n (c) ≥ c. If not, the sequence f 2 (c), f 3 (c), f 4 (c), . . . is an increasing sequence every term of which is less than c so the sequence converges to a point q ≤ c. But,

1.8 Logistic maps and their inverse limits

41

f (q) = q contrary to the hypothesis that f has no ﬁxed point between a and c and c < f (c). Then, f n+1 ([a, c]) = [a, f (c)] and, as in the previous case, employing Theorems 39 and 60 we conclude that C is the remainder of a ray. By hypothesis c < f (c) and f decreases on [c, b], so f ([p, f (c)]) = [f 2 (c), p]. If f 3 (c) > p then f ([f 2 (c), p]) ⊆ [p, f (c)] so f 2 ([p, f (c)] ⊆ [p, f (c)] and f 2 ([f 2 (c), p]) ⊆ [f 2 (c), p]. Let K1 = lim ϕ and K2 = lim ψ where ϕ = ←− ←− f 2 | [p, f (c)] and ψ = f 2 | [f 2 (c), p]. Then K1 and K2 are homeomorphic to subcontinua C1 and C2 of C, respectively, such that C = C1 ∪ C2 , C1 ∩ C2 = {(p, p, p, . . . )}, and C1 and C2 are homeomorphic under the shift homeomorphism, fˆ. In the following theorem and subsequent examples we adopt the notation from Theorem 65. Theorem 67 For the logistic family, suppose n ≥ 0 and λn+1 < λ < μn . Let n n n g = fλ2 | [an (λ), bn (λ)] and r = fλ2 | [fλ2 (λ), λ]. Then, lim g is the closure of ←− a ray with remainder C = lim r. If λn+1 < λ ≤ μn+1 then C is the union of ←− n+1 two continua each homeomorphic to lim ϕ where ϕ = fλ2 | [pn (λ), λ] and ←− intersecting only at (pn (λ), pn (λ), pn (λ), . . . ) whereas, if λn+1 < λ ≤ λn+2 , C is an arc. n

Proof. We employ Theorem 66 on fλ2 . To that end, suppose n ≥ 0 and λn+1 < λ < μn and let M = lim g. ←− First observe that conditions (2) and (3) of the conclusion of Theorem 65 n yield that fλ2 is strictly unimodal on [an (λ), bn (λ)] where an (λ) = pn−1 (λ) n and on that interval fλ2 has critical point cn (λ), ﬁxed point pn (λ), and n fλ2 (cn (λ)) = λ. Furthermore, by condition (5) of the conclusion of Theorem n n 65, (fλ2 ) (pn (λ)) < −1 so, by Lemma 63, fλ2 (cn (λ)) > cn (λ). n

Next, we show that fλ2 has no ﬁxed point between an (λ) and cn (λ). We n start by showing that (fλ2 ) (an (λ)) > 1. If n = 0, because a0 (λ) = 0 for all λ and fλ (0) > 1 for λ > 1/4 and λ1 = 3/4, we have fλ (a0 (λ)) > 1 for λ1 < λ < n−1 μ0 = 1. If n > 0, by Theorem 65, (fλ2 ) (pn−1 (λ)) < −1 for λn < λ ≤ μn−1 n n from which it follows that (fλ2 ) (an (λ)) > 1. If fλ2 has a ﬁxed point between n an (λ) and cn (λ), denote the ﬁrst one by s. Because (fλ2 ) (an (λ)) > 1, we n have fλ2 (t) > t for an (λ) < t < s, so it follows from Corollary 49, Lemma n 51, and the mean value theorem that (fλ2 ) (s) < 1. From the inequality n fλ2 (cn (λ)) > cn (λ) and the mean value theorem it follows that there is a n n point y, s < y < cn (λ), such that (fλ2 ) (y) > 1. These values of (fλ2 ) at an (λ), s, and y contradict Lemma 51. n

We now show that (fλ2 )2 (cn (λ)) > an (λ). If n = 0, c0 (λ) = 1/2 so 2 fλ (c0 (λ)) = fλ2 (1/2) = fλ (λ) > 0 if 0 < λ < 1. If n > 0, it follows from

42

1 Inverse Limits on Intervals

condition (4) of the conclusion of Theorem 65 that, for λn < λ < μn , n n fλ2 (λ) > pn−1 (λ); that is, (fλ2 )2 (cn (λ)) > an (λ). n

fλ2 satisﬁes the hypothesis of the ﬁrst part of Theorem 66, therefore we obtain that M is the closure of a ray with remainder C. To determine the nature of the remainder C for λn+1 < λ ≤ μn+1 (the reader should note that here the upper limit on λ is μn+1 , not μn ) we use n the second part of Theorem 66 so we must show that (fλ2 )3 (cn (λ)) ≥ pn (λ). Again, appealing to condition (4) of the conclusion of Theorem 65, for n ≥ n+1 n 0 and λn+1 < λ ≤ μn+1 , fλ2 (λ) ≥ pn (λ). Because fλ2 (cn (λ)) = λ, it n follows that (fλ2 )3 (cn (λ)) ≥ pn (λ). Thus, by the second part of Theorem 66, for λn+1 < λ ≤ μn+1 , the remainder C is the union of two continua each n+1 homeomorphic to lim ϕ where ϕ = fλ2 | [pn (λ), λ]. ←− n+1

Finally, suppose λn+1 < λ ≤ λn+2 . Theorem 65 yields that fλ2 is strictly unimodal on [an+1 (λ), bn+1 (λ)] with critical point cn+1 (λ) and ﬁxed n+1 point pn+1 (λ) such that (fλ2 ) (pn+1 (λ)) ≥ −1. Furthermore, inasmuch n as (fλ2 ) (pn (λ)) < −1 for λn+1 < λ ≤ μn and an+1 (λ) = pn (λ), we n+1 see that (fλ2 ) (an+1 (λ)) > 1. By Theorem 61, A = lim ψ is an arc ←− n+1 | [an+1 (λ), bn+1 (λ)]. But, ψ([an+1 (λ), bn+1 (λ)]) = [pn (λ), λ] where ψ = fλ2 and ψ([pn (λ), λ]) = [pn (λ), λ], so by Theorem 60, A = lim ϕ. Because ←− C is the union of two homeomorphic copies of A intersecting only at (pn (λ), pn (λ), pn (λ), . . . ), C is an arc. Some remarks are in order. When λ = μn , the remainder C in the previous theorem is the union of two BJK horseshoes but we only show this for n = 1 in Example 73 (see Figure 1.11). Also, although we do not prove it here, if μn+1 < λ < μn , the remainder C is indecomposable. Theorem 68 For 1/4 < λ < 1, lim f λ is a decomposable continuum. ←− Proof. Let M = lim f λ . For 1/4 < λ ≤ λ1 = 3/4, M is an arc; see Example ←− 62. For λ1 < λ < 1, by Theorem 67, M is the closure of a ray R with remainder C = lim r where r = fλ | [fλ (λ), λ]. Because x = (0, 0, 0, . . . ) is a ←− point of M that is not in C, x lies in an arc α that is a subset of R, and α does not intersect C. Then, α and M − α are proper subcontinua of M whose union is M . We now turn to some examples of inverse limits using a single logistic bonding map. The key to realizing what these inverse limits are is Theorem 67.

1.8 Logistic maps and their inverse limits

43

Example 69 For the logistic family, if λ1 < λ ≤ λ2 , lim fλ is the closure of ←− a ray with remainder an arc. Although we do not show it here, if λ1 < λ ≤ λ2 , lim fλ is homeomorphic ←− to a sin(1/x)-curve. Example 70 For the logistic family, if λ2 < λ ≤ λ3 , lim fλ is the closure ←− of a ray having remainder the union of two homeomorphic closures of rays intersecting only at (p0 (λ), p0 (λ), p0 (λ), . . . ) and each of these two rays has remainder an arc (see Figure 1.9). Proof. According to Theorem 67 with n = 0, for λ1 < λ < μ0 , lim fλ is the ←− closure of a ray with remainder C where C is the union of two continua each homeomorphic to lim ϕ where ϕ = fλ2 | [p0 (λ), λ]. Because fλ2 ([a1 (λ), b1 (λ)]) = ←− [p0 (λ), λ] and fλ2 ([p0 (λ), λ]) = [p0 (λ), λ], it follows from Theorem 60 that lim ϕ = lim ψ where ψ = fλ2 | [a1 (λ), b1 (λ)]. By Theorem 67 with n = 1, for ←− ←− λ2 < λ ≤ λ3 , lim ψ is the closure of a ray with remainder an arc. ←−

Fig. 1.9 The inverse limit from Example 70

In Example 70, for λ2 < λ ≤ λ3 , the inverse limit is the closure of a ray with remainder the union of two sin(1/x)-curves joined at a common endpoint of their rays. (A point p of a continuum M is an endpoint of M provided if each of H and K is a subcontinuum of M containing p then H ⊆ K or K ⊆ H.) In light of Examples 69 and 70 the reader may correctly conclude that, for λ3 < λ ≤ λ4 , the inverse limit is the closure of a ray with a remainder that is the union of two rays joined at a common endpoint and each of these two rays has remainder the union of two sin(1/x)-curves joined at a common endpoint of their rays (for a total of four sin(1/x)-curves in the inverse limit). In other words, the inverse limit is the closure of a ray with remainder the union of two copies of the inverse limit occurring in Example 70 (see Figure 1.10). We include the following example without proof.

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1 Inverse Limits on Intervals

Fig. 1.10 The inverse limit from Example 71 with n = 3

Example 71 For the logistic family, if n is an integer such that n ≥ 2 and λn < λ ≤ λn+1 , lim fλ is the closure of a ray R having remain←− der the union of two homeomorphic closures of rays intersecting only at (p0 (λ), p0 (λ), p0 (λ), . . . ). This inverse limit contains 2n−1 mutually exclusive arcs each of which is the remainder of a ray that intersects only one other such ray at a single point that is a common endpoint of these two rays. The remainder of R is the union of two copies of lim fλ where λn−1 < λ ≤ λn ←− whose intersection is a single point, the endpoint of the dense ray of each copy of lim fλ . ←− Theorem 72 If n ≥ 0 and λn < λ ≤ λn+1 , then every subcontinuum of lim f λ is decomposable. ←− In a later chapter we show that if f : [a, b] → [a, b] is a unimodal map such that f (a) = f (b) = a and f (c) = b (or f (a) = f (b) = b and f (c) = a), then lim f is homeomorphic to the BJK horseshoe, see Figure 1.5. This was ﬁrst ←− proved by James F. Davis [9]. With that information and Theorem 67 we describe the following examples. Example 73 For the logistic family, if λ = μ1 , lim fλ is the closure of a ←− ray with remainder the union of two BJK horseshoes intersecting only at (p0 (λ), p0 (λ), p0 (λ), . . . ) where this point is a common endpoint of the two BJK horseshoes (see Figure 1.11).

1.8 Logistic maps and their inverse limits

45

Proof. Applying Theorem 67 with n = 0, we conclude that lim fλ is the ←− closure of a ray with remainder C = lim r where r = fλ | [fλ (λ), λ]. Moreover, ←− C is the union of two continua each homeomorphic to lim ϕ where ϕ = ←− fλ2 | [p0 (λ), λ]. The graph of fλ2 is shown in Figure 1.12. Davis’ theorem [9] yields that lim ϕ is a BJK horseshoe. ←− We believe the reader may now correctly conclude that the inverse limit for λ = μ2 is the closure of a ray with remainder the union of two copies of lim fμ1 . We have stated this without proof in Example 74. We include ←− Example 75, also without proof, for information. Example 74 For the logistic family, if λ = μ2 , lim fλ is the closure of a ←− ray with remainder two homeomorphic closures of rays intersecting only at (p0 (λ), p0 (λ), p0 (λ), . . . ). Each of these two rays has remainder the union of two BJK horseshoes intersecting only at a single common endpoint.

Fig. 1.11 Two BJK horseshoes joined at a common endpoint

Example 75 For the logistic family, if n is an integer such that n ≥ 2 and λ = μn , lim fλ is the closure of a ray R with remainder two homeomorphic ←− closures of rays intersecting only at (p0 (λ), p0 (λ), p0 (λ), . . . ). This inverse limit contains 2n−1 mutually exclusive continua each of which is the union of two BJK horseshoes intersecting only at a single common endpoint. Furthermore, each of the two homeomorphic continua constituting the remainder of R is homeomorphic to lim fμn−1 . ←− We invite the reader to contrast lim fλ for λn < λ ≤ λn+1 and lim fμn . ←− ←− Each of the 2n−1 arcs in the former has been replaced by a pair of BJK

46

1 Inverse Limits on Intervals (b,b)

(p,p)

(a,a)

Fig. 1.12 The graph of fλ2 for λ = μ1 where a = fλ (λ), b = λ, and p = p0 (λ)

horseshoes intersecting at a single common endpoint in the latter. Thus on the left side of the Feigenbaum limit λc , as the parameter increases to λc in the inverse limit we obtain more and more complicated continua containing only decomposable nondegenerate subcontinua whose subcontinua consist of arcs or closures of rays or unions of closures of rays of varying degrees of complexity. On the right side of the Feigenbaum limit along the sequence μ we ﬁnd inverse limits containing nondegenerate indecomposable subcontinua. In fact, although we do not show this except for λ > μ1 (see Theorem 77) and λ ∈ {1, μ1 }, for λ > λc , lim fλ contains an indecomposable continuum ←− [3, Theorem 4, p. 166]. It is only natural that we next mention the nature of the inverse limit, lim fλc . ←− Example 76 Let f be the member of the logistic family at the Feigenbaum limit λ = λc . Then, every nondegenerate subcontinuum of M = lim f is de←− composable [15, Theorem 8, p. 56]. Moreover, the inverse limit is the closure of a ray with remainder the union of two homeomorphic copies of M intersecting only at the common endpoint of their respective dense rays [15, Theorem 9, p. 58]. In [15, Theorem 11, p. 60] it is shown that virtually any unimodal mapping g : [a, b] → [a, b] such that g(b) = a and having the property that g has periodic points of periods all the powers of 2 and no others has an inverse limit homeomorphic to the remainder of the ray from Example 76. At some parameter values above λc we can say more about the inverse limit. Here we limit our discussion to the nature of some of the remainders

1.8 Logistic maps and their inverse limits

47

in the inverse limit. In the following proof, we make use of the fact that, for λ > μ1 , fλ2 (λ) < p0 (λ). This is not obvious but showing it is true can be accomplished by showing that fλ2 (λ) is a decreasing function of λ for λ ≥ μ1 (actually it decreases for λ ≥ η where η ≈ 0.8426) and the ﬁxed point function, p0 (λ) = 1 − 1/(4λ), is increasing over (1/4, 1]. Theorem 77 For the logistic family, if μ1 < λ < 1, lim fλ the closure of a ←− ray with remainder an indecomposable continuum. Proof. For λ > μ1 , fλ2 (λ) < p0 (λ) inasmuch as fλ2 (λ) = 4λ2 (1 − λ) is decreasing if λ > μ1 . Because fλ has no period 2 point between 1/2 and p0 (λ), by Theorem 43, using q = p0 (λ), lim g is indecomposable where ←− g = fλ | [fλ (λ), λ]. But, by Theorem 66, lim fλ is the closure of a ray with ←− remainder lim g. ←− When one views a bifurcation diagram generated by the logistic family (see, e.g., [8, p. 26]), one is likely to notice the stable period 3 “window” that occurs for parameter values near λ = 0.957. Our ﬁnal example in this section deals with inverse limits of logistic maps for parameters chosen from that window. Example 78 For the logistic family, if λ is chosen so that fλ has a periodic point x of period 3 such that |(fλ3 ) (x)| < 1 then lim fλ is the closure of a ray ←− with remainder an indecomposable continuum H such that H has only three endpoints and every proper subcontinuum of H is an arc. Example 78 is considered in full detail in [3, Theorem 8, p. 168] as are a number of the examples of this section. Additional material on inverse limits with logistic bonding maps may be found there as well. Also, the members of the logistic family are unimodal, therefore there is additional information on their inverse limits in the section on unimodal maps. The indecomposable continuum H that is the remainder in Example 78 is homeomorphic to the familiar indecomposable continuum described in [37, Example 1.10, pp. 7– 8]; see also [14, Figure 3–21, pp. 141–142]. The indecomposable continuum H has only three endpoints and each of its proper subcontinua is an arc. Finally, before the period 3 “window” of the bifurcation diagram, one may ﬁnd a period 5 window (in fact, there are three of them). For parameter values from those windows the remainder in the inverse limit has ﬁve endpoints and every proper subcontinuum is an arc. Furthermore, for parameters chosen from two diﬀerent period 5 windows, the remainders of the corresponding inverse limits are not homeomorphic. Similar remarks apply to other windows of periodicity.

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1 Inverse Limits on Intervals

1.9 The piecewise linear family of unimodal maps fa b In this section we look at a piecewise linear family of maps parameterized by two parameters, a and b. These maps are deﬁned on [0, 1] and their graphs consist of two straight line intervals: one from (0, b) to (a, 1) and the other from (a, 1) to (1, 0). Speciﬁcally, for 0 < a < 1 and 0 ≤ b ≤ 1, the mappings of the family are deﬁned by (1 − b)x/a + b, 0 ≤ x ≤ a fa b (x) = (x − 1)/(a − 1), a ≤ x ≤ 1. For 0 ≤ b < 1, fa b is a unimodal map. Our goal is to determine in terms of the parameters when the inverse limit using one of these mappings as a single bonding map produces an indecomposable inverse limit, when the inverse limit contains an indecomposable continuum, and speciﬁcally what the inverse limit is when it is hereditarily decomposable. These results are summarized in a picture of parameter space labeled Figure 1.14. Lemma 79 If f : [c, d] → → [c, d] is a piecewise linear unimodal map such that f (c) = c, t is a number, c < t < d, such that f (t) = d and the slope of the graph of f on [t, d] is −1, then lim f is the closure of a ray with an arc as ←− remainder. Proof. Using Theorem 39, we see that lim f is the closure of a ray with ←− remainder lim g where g = f | [t, d]. Because g is a homeomorphism of [t, d] ←− onto itself, the remainder is an arc. It is signiﬁcant in the inverse limit of the mappings fa b if the product of the slopes of the two pieces of the map is −1. This product is −1 when b = a2 − a + 1, and Theorem 80 provides information on the inverse limit at such pairs of parameter values. As we show in Theorem 88, the curve b = a2 − a + 1 in parameter space marks the transition from simple inverse limits to complicated ones. In the following, it may be helpful for the reader to recall that the point 1/(2 − a) is the ﬁxed point for fa b . In the previous theorem, although we did not show it, the inverse limit is actually a sin(1/x)-curve. Assuming this, we have the following theorem. Theorem 80 If b = a2 − a + 1 then lim fa b is the union of two sin(1/x)←− curves intersecting at a common endpoint of their rays.

1.9 The piecewise linear family of unimodal maps fa b

49

Proof. Let p = 1/(2 − a). For b = a2 − a + 1, g1 = f 2 | [p, 1] is a piecewise linear unimodal map and the slope of its second piece is −1. By Lemma 79, H = lim g1 is the closure of a ray with an arc as remainder, in this case ←− a sin(1/x)-curve. Using the shift homeomorphism as we did in the proof of Theorem 66 we see that lim f is the union of two continua homeomorphic to ←− H intersecting only at the common endpoint (p, p, p, . . . ) of the two rays. Theorem 81 If b > a2 − a + 1 then lim fa b is an arc. ←− 2 2 Proof. If b > a − a + 1, g1 = fa b | [1/(2 − a), 1] is a piecewise linear unimodal 2 map. If r denotes the solution to fna b (x) = 1, the slope of g1 on [r, 1] has absolute value less than 1 so n>0 g1 ([r, 1]) is a single point. Using Theorem 39, we see lim g1 is the closure of a ray with remainder a single point; that ←− is, it is an arc. Proceeding as in the proof of Theorem 80 we have lim fa b is ←− an arc. Having dealt with parameters satisfying b ≥ 1 − a + a2 , we now turn our attention to the case that b < 1 − a + a2 . We show that for parameters satisfying this inequality, lim fa b contains an indecomposable continuum and, ←− in fact, is indecomposable if b < 1/(2−a). In order to deal with parameters in this range, our strategy is to employ Theorem 43 and Corollary 44. Theorems 87 and 89 deal with b = 1/(2 − a) and b < 1/(2 − a), respectively. It is somewhat more complicated to deal with parameters in the range 1/(2 − a) < b < 1 − a + a2 . Here we obtain an interval J and a positive integer n n depending on a and b so that lim fa2 b | J is indecomposable. Speciﬁcally, such ←− n an interval J is [0, fa2 b (0)]. n

For 1/(2 − a) < b < 1 − a + a2 , the graph of fa2 b over a small interval containing 0 resembles the mapping g introduced in Lemma 82. (This actually occurs over a slightly larger range of values for the parameters, namely a < b < 1 − a + a2 .) Lemma 82 is used in the proof of Theorem 88. The conditions added to g in Lemma 83 guarantee that the graph of g 2 | [0, g 2 (0)] looks n much like the graph of g. We apply Lemma 83 to fa2 b in Lemma 86, and, ﬁnally, in Theorem 88 demonstrate that lim fa b contains an indecomposable ←− continuum for 1/(2 − a) < b < 1 − a + a2 . Lemma 82 Suppose 0 < m , 1 < m, and 0 < B ≤ 1, and g is a piecewise linear map from [0, B] onto itself given by −mx + B, 0 ≤ x ≤ B/m g(x) = m (x − B/m), B/m ≤ x ≤ B. Then, g 2 (0) < B/(m + 1) if and only if m2 < 1 + m/m . Proof. g 2 (0) = m B(1 − 1/m), thus g 2 (0) < B/(m + 1) if and only if m ((m − 1)/m) < 1/(m+ 1) which occurs if and only if m /m(m2 − 1) < 1. The lemma follows from these observations.

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B/(m+1)

g(B)

B/(m+1) g(B)

0 B/m

B

0

B/m

B

Fig. 1.13 Typical maps g and g 2 from Lemma 83

The reader will note that the assumption that g : [0, B] → → [0, B] in Lemma 82 actually imposes an unstated condition on the slopes m and m , namely that m ≤ m/(m − 1). However, when we apply this lemma in the proof of Theorem 88, we have that the interval on which it is used is invariant with respect to the mapping under consideration for other more restrictive reasons. For the following lemma, it may be helpful to observe that if g is the mapping deﬁned in Lemma 82, its ﬁxed point is B/(m+ 1) and g 2 (0) = g(B). Lemma 83 Suppose 1 < m, 0 < m , 0 < B ≤ 1, and g is the mapping from Lemma 82. If g(B) < B/m then the graph of g 2 consists of three straight line intervals of slopes −mm , m2 , and −mm . Moreover, if 1 < mm and g(B) ≤ B/(m + 1), g 2 ([0, g 2 (0)]) = [0, g 2 (0)]. Proof. The graph of g 2 consists of three straight line intervals because g(B) < B/m. The slope of the ﬁrst piece of the graph is −mm and mm > 1 yields that this piece hits the bottom of [0, B]2 at a point with ﬁrst coordinate t less than g 2 (0) and g 2 (0) = g(B) ≤ B/(m + 1). The ﬁrst ﬁxed point for g 2 following t is B/(m + 1) so g 2 (g(B)) ≤ g 2 (0). Let ϕ be the integer-valued function deﬁned inductively on the set of nonnegative integers by ϕ(0) = 0, ϕ(1) = 1, and ϕ(i + 1) = ϕ(i) + 2ϕ(i − 1) for each positive integer i. Lemma 84 ϕ(i) − 2ϕ(i − 1) = 1 if i is odd and − 1 if i is even. Proof. ϕ(1) − 2ϕ(0) = 1. By deﬁnition ϕ(n + 1) = ϕ(n) + 2ϕ(n − 1) so ϕ(n + 1) − 2ϕ(n) = −ϕ(n) + 2ϕ(n − 1). Lemma 85 Suppose 0 ≤ b ≤ 1 − a + a2 , 0 < a < 1, m1 = (1 − b)/a, and ϕ(n) ϕ(n+1) m2 = 1/(a − 1). Then, for each positive integer n, |m1 m2 | > 1.

1.9 The piecewise linear family of unimodal maps fa b

51 ϕ(1)

ϕ(2)

Proof. We proceed by induction. For n = 1, |m1 m2 | = |m1 m2 | = ((1 − b)/a)(1/(1 − a)). But (1 − b)/(a − a2 ) > 1 if and only if b < ϕ(2) ϕ(3) 1 − a + a2 . For n = 2, |m1 m2 | = |m1 m32 | = |m1 m2 |(1/(1 − a))2 > 2 1/(1 − a) > 1 because 0 < a < 1. Suppose that the statement is true for n ∈ {1, 2, . . . , k} where k ≥ 2. Then, using the deﬁnition of ϕ, ϕ(k+1) ϕ(k+2) ϕ(k) ϕ(k+1) 2ϕ(k−1) 2ϕ(k) ϕ(k) ϕ(k+1) m2 | = |m1 m2 ||m1 m2 |. Both |m1 m2 | |m1 2ϕ(k−1) 2ϕ(k) and |m1 m2 | are greater than 1 by the inductive hypothesis. We now return to the mappings fa b recalling that their graphs consist of two straight line intervals having slopes m1 = (1 − b)/a and m2 = 1/(a − 1). Lemma 86 Suppose 1/(2 − a) ≤ b < 1 − a + a2 and n is a positive integer. n Suppose further that if n ≥ 2 then fa2 b (0) is less than the ﬁrst ﬁxed point for n−1 n n n fa2 b . Then, fa2 b : [0, fa2 b (0)] → → [0, fa2 b (0)] and the slopes of the ﬁrst two n ϕ(n) ϕ(n+1) 2ϕ(n−1) 2ϕ(n) and m1 m2 , respectively. pieces of fa2 b are given by m1 m2 Proof. Again we proceed by induction. For n = 1, it is easy to see that fa2 b ([0, fa2 b (0)]) = [0, fa2 b (0)] and the slopes of the two pieces of fa2 b | [0, fa2 b (0)] ϕ(1) ϕ(2) 2ϕ(0) 2ϕ(1) are m1 m2 and m1 m2 , respectively. Inductively, assume k ≥ 1, k k k k 2 2 2k fa b ([0, fa b (0)]) = [0, fa b (0)], and the slopes of the two pieces of fa2 b | [0, fa2 b (0)] ϕ(k) ϕ(k+1) 2ϕ(k−1) 2ϕ(k) are given by m1 m2 and m1 m2 , respectively. To verify the (k+1) 2 statement for n = k + 1, assume fa b (0) is less than the ﬁrst ﬁxed point for k ϕ(k) ϕ(k+1) | > 1, so we may apply Lemma fa2 b . From Lemma 85, we have |m1 m2 (k+1) (k+1) (k+1) 2 2 83 to conclude that fa b ([0, fa b (0)]) = [0, fa2 b (0)]. Also, by the ﬁrst (k+1) (k+1) part of Lemma 83, the slopes of the two pieces of fa2 b | [0, fa2 b (0)] are ϕ(k)+2ϕ(k−1) ϕ(k+1)+2ϕ(k) 2ϕ(k) 2ϕ(k+1) given by m1 m2 and m1 m2 , respectively. Howϕ(k)+2ϕ(k−1) ϕ(k+1)+2ϕ(k) ever, by the deﬁnition of ϕ we have that m1 m2 = ϕ(k+1) ϕ(k+2) m2 . m1 As we have seen the ﬁrst ﬁxed point for f 2 after the critical point plays a critical role in the nature of the inverse limit for a unimodal map f . For the maps fa b , the ﬁrst ﬁxed point for fa2 b in [b, 1] is simply its ﬁxed point, 1/(2 − a). Theorem 87 If b = 1/(2 − a) then lim fa b is the union of two BJK horse←− shoes intersecting at a common endpoint. Proof. For b = 1/(2−a), g1 = fa2 b | [1/(2−a), 1] is a piecewise linear unimodal surjection that maps both endpoints to 1/(2−a) and an intermediate point to b. By Theorem 35, H = lim g1 is a BJK horseshoe. The shift homeomorphism ←− fˆa b maps H onto K = lim g2 where g2 = fa2 b | [0, 1/(2 − a)]. The union of H ← − and K is lim f 2 which is homeomorphic to lim f . The only point common to ←− ←− H and K is (1/(2 − a), 1/(2 − a), 1/(2 − a), . . . ), an endpoint of each of H and K.

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1 Inverse Limits on Intervals

Theorem 88 If 1/(2 − a) < b < a2 − a + 1 then lim fa b contains an inde←− composable continuum. Proof. Let m1 = (1 − b)/a and m2 = 1/(a − 1) (the slopes of the two pieces of fa b ). We show that there is a positive integer n so that Corollary 44 apn n n plies to fa2 b on [0, fa2 b (0)]. Because b < 1 − a + a2 , |m1 m2 | > 1. If fa2 b (0) n−1 is less than the ﬁrst ﬁxed point for fa2 b each n ≥ 2, then by Lemma 86, n ϕ(n) ϕ(n+1) the slopes of the ﬁrst two pieces of fa2 b are given by mn = m1 m2 2ϕ(n−1) 2ϕ(n) and mn = m1 m2 , respectively (slightly abusing the notation). By (ϕ(n)−2ϕ(n−1)) (ϕ(n+1)−2ϕ(n)) Lemma 82, m2n < 1 + |mn /mn | = 1 + |m1 m2 |. 2 Lemma 84 yields mn < 1 + |m1 /m2 | if n is odd and m2n < 1 + |m2 /m1 | if n is even. But m2n increases without bound as n increases without bound because 2ϕ(n) 2ϕ(n+1) 2(ϕ(n+1)−ϕ(n)) m2 = (m1 m2 )2ϕ(n) m2 , |m1 m2 | > 1, |m2 | > m2n = m1 1, and both ϕ(n) and ϕ(n + 1) − ϕ(n) increase without bound as n increases without bound, a contradiction. Thus, there is a positive integer n such that n n−1 n fa2 b (0) is not less than the ﬁrst ﬁxed point for fa2 b . If f 2 (0) is greater n−1 n n than the ﬁrst ﬁxed point for f 2 , by Corollary 44, lim fa2 b | [0, fa2 b (0)] is ← − n n−1 indecomposable. If f 2 (0) is equal to the ﬁrst ﬁxed point for f 2 , just as n n we saw in the proof of Theorem 87, lim fa2 b | [0, fa2 b (0)] is the union of two ←−n BJK horseshoes. In either case, lim fa2 b , and therefore lim fa b , contains an ←− ←− indecomposable continuum. Theorem 89 If b < 1/(2 − a) then lim fa b is indecomposable. ←− Proof. This is a direct consequence of Theorem 43 inasmuch as f (0) = b and 1/(2 − a) is the ﬁxed point for fa b .

Additional information on the family of maps fa b is available in the literature. See [21] as well as [19] where the members of the family are denoted gb c .

1.10 The tent family

In this section we apply some results from the previous section to the tent family, T . The tent mappings have been studied in dynamics for a long time. The members of the tent family of mappings are piecewise linear unimodal mappings for which the absolute values of the slopes of the two pieces are equal. One representation of the members of this parameterized family of mappings is given by

1.10 The tent family

53

1

b=1-a+a2 b>1-a+a2 1/(2-a)
b

b<1/(2-a)

0

a

1

Fig. 1.14 Parameter space: for b > 1 − a + a2 the inverse limit of fa b is an arc whereas for b = 1 − a + a2 the inverse limit is a pair of sin(1/x)-curves. The inverse limit contains an indecomposable continuum for 1/(2 − a) ≤ b < 1 − a + a2 . If b = 1/(2 − a) the inverse limit is a pair of BJK horseshoes and below that curve the inverse limit is indecomposable

Tλ (x) =

2λx, 0 ≤ x ≤ 1/2, 2λ(1 − x), 1/2 ≤ x ≤ 1,

for 0 ≤ λ ≤ 1. Tλ is unimodal and Tλ (0) = 0, therefore its core is lim g where g = ←− Tλ | [Tλ (λ), λ]. A typical member of the tent family and the core of the map are shown in Figure 1.15. Theorem 90 For the family T 1. If 0 ≤ λ < 1/2, lim Tλ is degenerate ←− 2. If λ = 1/2, lim Tλ is an arc ←− 3. If 1/2 < λ ≤ 1, lim Tλ contains an indecomposable continuum ←−

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1 Inverse Limits on Intervals

(1/2,lambda)

(0,0)

(1,0)

Fig. 1.15 A typical tent map with its core (shown in the inner box)

√ 4. If λ = 1/ 2, the core of lim Tλ is the union of two BJK horseshoes inter←− secting √ only at a common endpoint 5. If 1/ 2 < λ < 1, the core of lim Tλ is indecomposable ←− 6. If λ = 1, the inverse limit is a BJK horseshoe. Proof. For items (1) and (2) the reader needs only to observe that for 0 ≤ λ < 1/2, lim Tλ = {(0, 0, 0, . . . )} and for λ = 1/2 lim Tλ = lim Id | [0, 1/2]. ←− ←− ←− Item (6) is included only for completeness of the statement of the theorem. For 1/2 < λ ≤ 1, the core of lim Tλ is topologically conjugate to fa b for a = ←− 1−1/(2λ) and b = 2−2λ. Eliminating λ yields b = (2a−1)/(a−1). It is easy to check that (2a − 1)/(a − 1) < a2 − a + 1 for 0 < a < 1. By Theorem 88, lim Tλ ←− contains √ an indecomposable continuum. That (2a − 1)/(a − 1) = 1/(2 − a) for yields item (4) by Theorem 87. Finally, (2a− 1)/(a− 1) < 1/(2 − a) λ = 1/ 2 √ for λ > 1/ 2, thus item (5) follows by Theorem 89. In this section we have merely scratched the surface of what is known about inverse limits with a single bonding map chosen from the tent family. Over the past several years, a considerable body of literature has been produced on this subject. A large portion of it has grown out of work on a question that ˇ was recently settled by Barge, Bruin, and Stimac when they proved Theorem 91 below. See [2] for details of its proof as well as some of the literature on the question. Theorem 91 If 1/2 ≤ λ1 ≤ 1, 1/2 ≤ λ2 ≤ 1, and λ1 = λ2 then lim Tλ1 and ←− lim Tλ2 are not homeomorphic. ←−

1.11 Other families of mappings

55

A family of maps of “constant slope” containing a subfamily of maps from [0, 1] onto [0, 1] conjugate to the tent family was studied in [18]. The maps of this family that are conjugate to tent maps are those passing through (0, 0), (1/m, 1), and (1, 2 − m) for 1 ≤ m ≤ 2. Inverse limits with maps from this larger family when the absolute value of the slope is an integer greater than 1 were considered by Watkins [39] and Debski [10].

1.11 Other families of mappings In this section we mention results for some other families of mappings that have attracted attention. We state the results in the theorems but we leave the proofs to the reader as exercises in employing theorems from this chapter.

1.11.1 The family F The family F is the parameterized family of mappings ft given by 2x, 0 ≤ x ≤ 1/2, ft (x) = 2(1 − t)(1 − x) + t, 1/2 ≤ x ≤ 1, where the parameter t satisﬁes 0 ≤ t ≤ 1. A typical member of the family F is depicted in Figure 1.16. For 0 < t < 1/2, lim ft is the closure of a ray with its core as remainder; ←− that is, the remainder is lim ft | [t, 1]. It is not diﬃcult to verify that ft | [t, 1] ←− for 0 < t < 1/2 is conjugate to fa b for b = 1 − 2a with a = t. Theorem 92 For the family F 1. 2. 3. 4.

If If If If

1/2 < t ≤ 1, lim ft is an arc ←− t = 1/2, lim ft is the closure of a ray with an arc as remainder ←− 0 ≤ t < 1/2, lim ft contains an indecomposable continuum ←− t = 0, the inverse limit is a BJK horseshoe.

It should be noted that lim ft never contains two sin(1/x)-curves intersect←− ing at a common endpoint because the curves b = 1 − 2a and b = 1 − a + a2 do not intersect for 0 < a < 1/2 (see Figure 1.18).

56

1 Inverse Limits on Intervals (1/2,1)

(1,t)

(0,0)

Fig. 1.16 A typical map of the family F with its core (shown in the inner box)

1.11.2 The family G The family G is the parameterized family of mappings gt given by 2(1 − t)x + t, 0 ≤ x ≤ 1/2 gt (x) = 2(1 − x), 1/2 ≤ x ≤ 1, where the parameter t satisﬁes 0 ≤ t ≤ 1. A typical member of the family G is shown in Figure 1.17. Each map gt is a map fa b for b = t and a = 1/2. Theorem 93 For the family G 1. If 3/4 < t ≤ 1, lim gt is an arc ←− 2. If t = 3/4, lim gt is the union of two sin(1/x)-curves intersecting at a ←− common endpoint of their rays 3. If 0 ≤ t < 3/4, lim gt contains an indecomposable continuum ←− 4. If t = 2/3, lim gt is the union of two BJK horseshoes intersecting only at ←− a common endpoint 5. If 0 ≤ t < 2/3, lim gt is indecomposable ←− 6. If t = 0, the inverse limit is a BJK horseshoe. Figure 1.18 is an enhancement of Figure 1.14 and depicts a summary of the results on piecewise linear unimodal maps from Sections 1.9 through 1.11.

1.11 Other families of mappings

57 (1/2,1)

(0,t)

(1,0)

Fig. 1.17 A typical map of the family G

1.11.3 Markov maps One large family of maps we have not mentioned explicitly is the family of Markov maps. However, many of the maps we have considered are Markov maps. A map f : [0, 1] → [0, 1] from an interval [0, 1] into [0, 1] is called a Markov map provided there is a ﬁnite subset A of [0, 1] containing 0 and 1 such that f (A) ⊆ A (i.e., A is invariant under f ) and if x and y are elements of A with no element of A between them then the restriction of f to [x, y] is monotone. The set A is called a Markov partition for f . For instance, surjective unimodal maps having their critical point in a periodic orbit are Markov maps as are all of the permutation maps (discussed in the next subsection). The following theorem due to Brian Raines [38, Theorem 3.1] is quite useful in the theory of inverse limits with Markov maps. Theorem 94 (Raines) Suppose each of f and g is a Markov map of [0, 1] with Markov partitions A = {c0 < c1 < · · · < cn } and B = {d0 < d1 < · · · < dn }, respectively, and suppose further that f (ci ) = cj if and only if g(di ) = dj . Then, lim f is homeomorphic to lim g. ←− ←− Raines has extended the notion of Markov map to maps of graphs and generalized certain results for Markov maps of intervals to Markov maps of graphs notably including Theorem 94 [38].

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1 Inverse Limits on Intervals

1 pair of sin(1/x) curves

pair of BJK horseshoes

b

cores of the tent family cores of family F

three end-point indecomposable continuum

family G BJK horseshoe 0

a

1

Fig. 1.18 Parameter space: if (a, b) is above the curve along which the inverse limit is a pair of sin(1/x)-curves, the inverse limit of fa b is an arc. For (a, b) between that curve and the curve where pairs of BJK horseshoes occur, the inverse limit contains an indecomposable continuum. For (a, b) below the curve where pairs of BJK horseshoes occur, the inverse limit is an indecomposable continuum

1.11.4 Permutation maps We conclude this section by mentioning one additional family of mappings whose inverse limits have been studied. These are permutation maps as introduced in [20]. Suppose n is a positive integer and σ is a permutation on {1, 2, . . . , n}; that is, σ is a function from {1, 2, . . . , n} onto itself. Using σ, we construct a map fσ from [0, 1] onto [0, 1] as follows. For each positive integer i, 1 ≤ i ≤ n, let ai = (i − 1)/(n − 1) and fσ (ai ) = aσ(i) ; then extend the function linearly to a map of [0, 1]. The following theorem illustrates the type of theorem known about permutation maps [20, Theorem 15, p. 400]. Theorem 95 Suppose n ∈ N, σ is a permutation on {1, 2, . . . , n} and the orbit of 1 under σ is {1, 2, . . . , n}. If fσk (0) = 1/(n − 1) and n and k are

1.12 Characterization of inverse limits on [0, 1] as chainable continua

59

relatively prime, the lim fσ is an indecomposable continuum having n and ←− only n endpoints and every nondegenerate subcontinuum of it is an arc.

1.12 Characterization of inverse limits on [0, 1] as chainable continua

Next we turn to the problem of identifying the types of continua that can be obtained as inverse limits on I. We show that these comprise the set of all chainable continua. A chain is a ﬁnite sequence C = (C1 , C2 , . . . , Cn ) of open sets such that Ci ∩ Cj = ∅ if and only if |i − j| ≤ 1. The sets C1 , C2 , . . . , Cn are called links of C. If ε > 0, a chain is an ε-chain provided the diameter of each link of C is less than ε. A continuum M is chainable provided for each ε > 0 there is an ε-chain covering M . We show that a nondegenerate continuum M is chainable if and only if there exists a sequence f of piecewise linear mappings of the interval [0, 1] onto itself such that M is homeomorphic to lim f . The reader should note that the links of a chain are deﬁned to be open ←− sets without speciﬁcity as to the space in which they are open. Normally, the links are open in some space containing the continuum they cover; in the case of inverse limits on [0, 1] this space is Q. However, for the most part in this section, when M is a chainable continuum, we assume that M is our space so the links are open subsets of M . Theorem 96 If f is a sequence of mappings such that, for each positive integer i, fi : [0, 1] → [0, 1] then lim f is chainable. ←− Proof. Suppose ε > 0. There is a positive integer n such that i>n 2−i < ε/3. For each i, 1 ≤ i ≤ n, fi n is uniformly continuous. It follows that there is a positive number δ such that, for each i, 1 ≤ i ≤ n, if |s − t| < δ then |fi n (s) − fi n (t)| < ε/3. If x, y ∈ lim f and |xn − yn | < δ then d(x, y) = ←− ∞ n n |xi − yi |/2i = i=1 |xi − yi |/2i + i=n+1 |xi − yi |/2i < i=1 ε/(3 · i>0 2i )+ i>n 2−i < (2ε)/3. Thus, if A ⊆ [0, 1] and the diameter of A is less than δ then the diameter of πn−1 (A) < ε. Let D = (D1 , D2 , . . . , Dm ) be a δ-chain covering [0, 1] and let Ci = πn−1 (Di ), 1 ≤ i ≤ m. Then C = (C1 , C2 , . . . , Cm ) is an ε-chain covering lim f . ←− Theorem 96 is a special case of a more general theorem, Theorem 174 of Chapter 2, that we include once we have extended our deﬁnition to inverse limits on sequences of continua.

60

1 Inverse Limits on Intervals

It is considerably more diﬃcult to show that each nondegenerate chainable continuum is homeomorphic to an inverse limit on intervals with piecewise linear surjective bonding maps. In order to achieve this we use the following lemma. A chain C is called taut provided Ci ∩ Cj = ∅ if and only if |i − j| ≤ 1. Lemma 97 If M is a chainable continuum and ε > 0 there exists a taut ε-chain C covering M such that each link of C is open in M and contains a point of M that is not in the closure of any other link of C. Proof. We may assume that if D = (D1 , D2 , . . . , Dn ) is a chain covering the continuum M , each link of D intersects M (the links of D are not assumed to be open in M ). Moreover, we may assume that each link of D contains a point of M that is not in any other link of D. For example, if D1 ⊆ D2 , we may simply remove D1 from D and renumber. Similarly, if Dn ⊆ Dn−1 , we may remove Dn from D. If 1 < i < n and Di is a subset of the union of all of the other links of D, then M is a subset of the union of two mutually exclusive open sets D1 ∪ D2 ∪ · · · ∪ Di−1 and Di+1 ∪ Di+2 ∪ · · · ∪ Dn and yet intersects both of them. To complete the proof we employ a technique similar to one found in [37, p. 235]. Let D = (D1 , D2 , . . . , Dm ) be an ε/3-chain covering M such that each link of D contains a point of M that belongs to no other link of D and let n = m/2 if m is even whereas n = (m − 1)/2 if m is odd. For 1 ≤ i < n let Ci = (D2i−1 ∪ D2i ) ∩ M and Cn = (Dm−1 ∪ Dm ) ∩ M if m is even whereas Cn = (Dm−2 ∪ Dm−1 ∪ Dm ) ∩ M if m is odd. The chain C is a taut ε-chain covering M each link of which is open in M and contains a point of M not in the closure of any other link of C. In the proof of Theorem 99, we make use of Lebesgue’s covering lemma [23, p. 154]. If x is a point of a metric space (X, d) and ε > 0, let U (x, ε) = {p ∈ X | d(p, x) < ε}. Theorem 98 (Lebesgue’s Covering Lemma) If A is a compact subset of a metric space and G is a collection of open sets covering A, then there is a positive number ε such that if x is in A then U (x, ε) is a subset of some member of G. If G is a collection of sets, G ∗ denotes the union of all of the sets in G. A chain D is said to reﬁne the chain C provided each link of D is a subset of some link of C and D strongly reﬁnes C provided the closure of each link of D is a subset of some link of C. The mesh of a chain is the largest of the diameters of its links. Theorem 99 If M is a chainable continuum, there exists a sequence C1 , C2 , C3 , . . . such that for each positive integer n

1.12 Characterization of inverse limits on [0, 1] as chainable continua

61

1. Cn is a taut chain covering M 2. Each link of Cn is open in M and contains a point of M not in the closure of any other link of Cn 3. The mesh of Cn is less than 1/n 4. Cn+1 strongly reﬁnes Cn 5. M = n>0 Cn∗ . Proof. Use Lemma 97 to obtain a taut 1-chain C1 covering M such that each link of C1 is open in M and contains a point of M not in the closure of any other link of C1 . By using Theorem 98 and Lemma 97 we obtain a taut (1/2)-chain C2 covering M that strongly reﬁnes C1 such that each link of C2 is open in M and contains a point of M not in the closure of any other link of C2 . Continuing inductively we obtain a sequence C1 , C2 , C3 , . . . of taut chains covering M and satisfying conditions (1)-(4). M is a continuum, therefore condition (5) is an immediate consequence of conditions (2) and (3). If M is a chainable continuum, a sequence of chains satisfying the conclusion of Theorem 99 is called a deﬁning sequence for M (even if the links of the chains are open in the space containing M ). The next lemma is rather technical but it is used to construct bonding maps with inverse limit homeomorphic to M . In Lemma 100, we construct a function that depends on two chains we start with and a positive number δ. Each time we use the lemma in the proof of Theorem 102 speciﬁc choices of the chains and δ are made. Lemma 100 Suppose C = {C1 , C2 , . . . , Cn } and D = {D1 , D2 , . . . , Dm } are taut chains such that D strongly reﬁnes C and δ > 0. Suppose further that there are points p of C1 and q of Cn and a positive number σ such that d(Ci , Cj ) > σ if |i − j| > 1, d(p, Ci ) > σ if i = 1, and d(q, Ci ) > σ if i = n. If the mesh of D is less than (nσδ)/2 and p and q are in D∗ , then there is a piecewise linear mapping f : [0, 1] → → [0, 1] from [0, 1] onto [0, 1] that is not constant on any subinterval of [0, 1] and 1. If 1 ≤ j ≤ m, there is a positive integer i, 1 ≤ i ≤ n, such that f ([(j − 1)/m, j/m]) ⊆ [(i − 1)/n, i/n] 2. If 1 ≤ j < m and f ([(j − 1)/m, j/m]) ⊆ [(i − 1)/n, i/n] where 1 ≤ i ≤ n, then Dj ⊆ Ci 3. If 1 ≤ j ≤ m, 1 ≤ i ≤ n, Dj ⊆ Ci , and l is an integer such that Dl ∩ Dj = ∅ then f ([(l − 1)/m, l/m]) ⊆ [(i − 2)/n, (i + 1)/n] 4. If 1 ≤ j ≤ m and s and t are points of [(j −1)/m, j/m] then |f (s)−f (t)| < δ. Proof. This proof is somewhat long, therefore we oﬀer an outline. The proof begins with a partition of the chain D into subchains each of which lies in a single link of C (see Figure 1.19 where there are 17 subchains in the partition). Then the function f is deﬁned based on the behavior of the chosen partition

62

1 Inverse Limits on Intervals C1

C2

C3

C4 1

2

3 4

9

C6

C5

17

16

q

p

Fig. 1.19 A schematic representation of the chain D and its partition reﬁning the chain C 1

5/6

4/6

3/6

2/6

1/6

0

0

m1/m

m 9 /m

m 16 /m

1

Fig. 1.20 A mapping based on the chains depicted in Figure 1.19. Here n = 6 and k = 17

within the chain C (see Figure 1.20 for a picture of a mapping f based on the chains in Figure 1.19). The deﬁnition of f is given in several parts. First, the domain interval [0, 1] is subdivided into intervals each of length 1/m so we have a subinterval for each link of D. Similarly, the range interval [0, 1] is subdivided into intervals each of length 1/n, one for each link of the chain C. Our goal in deﬁning f is to mimic the embeddings of the links of D in the links of C with the manner in which f maps subintervals of the domain

1.12 Characterization of inverse limits on [0, 1] as chainable continua

63

into subintervals of the range. Initially the function is only deﬁned at certain points of the subdivision that are determined by the numbers of links of the elements of the partition of D. The values of the function at these points are determined both by the way D reﬁnes C and the direction the chain D is “moving” in C. Next, the function is extended to the entire interval and shown to be surjective. Some care is taken when D turns in C in order to control the slopes of the linear pieces of the function as these slopes relate to the number δ. The number δ is included in this lemma for technical reasons that become apparent only when this lemma is used in the proof of Theorem 102. Partition of D. Partition D into mutually exclusive subchains B1 , B2 , . . . , Bk such that if 1 ≤ j ≤ k there is an integer ij , 1 ≤ ij ≤ n, such that Bj∗ ⊆ Cij and |ij+1 − ij | = 1. For each j, 1 ≤ j ≤ k, let hj denote the number of links in Bj and mj = h1 + · · · + hj . Note that the mesh of D is less than (nσδ)/2, thus more than 2/(nδ) links are required for a subchain of D to cross a link of C or to run from p (respectively, q) to a link of C not containing p (respectively, q). Thus, if Bj∗ intersects Cij −1 and Cij +1 or p ∈ Bj∗ or q ∈ Bj∗ then hj > 2/(nδ). Deﬁnition of the function f at mj /m for 1 ≤ j < k. We now turn to the deﬁnition of f . For 1 ≤ j < k, let ij /n if ij < ij+1 f (mj /m) = (ij − 1)/n if ij+1 < ij . We now extend f to [0, 1]. This is achieved by considering several cases: extension to [mj−1 /m, mj /m] for 1 < j < k when f (mj−1 /m) = f (mj /m), extension to [mj−1 /m, mj /m] for 1 < j < k and f (mj−1 /m) = f (mj /m), extension to [0, m1 /m], and extension to [mk−1 /m, 1]. Extension of f to [mj−1 /m, mj /m] for 1 < j < k, part (1). Suppose 1 < j < k and f (mj−1 /m) = f (mj /m). Extend f linearly on [mj−1 /m, mj /m]. In this case we have ij−1 < ij < ij+1 or ij−1 > ij > ij+1 but, whichever occurs, it follows that f ([mj−1 /m, mj /m]) = [(ij −1)/n, ij /n]. Moreover, if mj−1 < l ≤ mj then f ([(l − 1)/m, l/m]) ⊆ [(ij − 1)/n, ij /n] and Dl ⊆ Cij . Furthermore, because hj > 2/(nδ) the absolute value of the slope of f on the interval [mj−1 /m, mj /m] is less than mδ (in fact the absolute value of the slope is actually less than (mδ)/2), so if s and t are in [mj−1 /m, mj /m] and |s − t| ≤ 1/m then |f (s) − f (t)| < δ/2 < δ. Extension of f to [mj−1 /m, mj /m] for 1 < j < k, part (2). Suppose 1 < j < k and f (mj−1 /m) = f (mj /m). Let rj denote the midpoint of the interval [mj−1 /m, mj /m]. In order that f (mj−1 /m) = f (mj /m), either ij is greater than both ij−1 and ij+1 or ij is less than both ij−1 and ij+1 .

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Suppose ij > ij−1 and ij > ij+1 . Then, f (mj /m) = (ij − 1)/n = ij−1 /n = f (mj−1 /m). In the case where hj > 2/(nδ), let f (rj ) = ij /n. If hj ≤ 2/(nδ), let f (rj ) = (ij − 1)/n + hj δ/2 (so f (rj ) ≤ ij /n). Suppose ij < ij−1 and ij < ij+1 . Then, f (mj /m) = ij /n = (ij−1 − 1)/n = f (mj−1 /m). In the case where hj > 2/(nδ), let f (rj ) = (ij − 1)/n. If hj ≤ 2/(nδ), let f (rj ) = ij /n − hj δ/2 (so f (rj ) ≥ (ij − 1)/n). Let f be linear on [mj−1 , rj ] and [rj , mj ]. Note that f (rj ) is deﬁned so that the absolute value of the slope of each of the two pieces of f on the interval [mj−1 /m, mj /m] is less than mδ. So, if s and t are in [mj−1 /m, mj /m] and |s − t| ≤ 1/m then |f (s) − f (t)| < δ. Moreover, if mj−1 < l ≤ mj then f ([(l − 1)/m, l/m]) ⊆ [(ij − 1)/n, ij /n] and Dl ⊆ Cij . Extension of f to [0, m1 /m]. Suppose i1 < i2 . If h1 > 2/(nδ), let f (0) = (i1 − 1)/n. If h1 ≤ 2/(nδ), let f (0) = i1 /n − h1 δ/2 (so f (0) ≥ (i1 − 1)/n). Suppose i1 > i2 . If h1 > 2/(nδ), let f (0) = i1 /n. If h1 ≤ 2/(nδ), let f (0) = (i1 − 1)/n + h1 δ/2 (so f (0) ≤ i1 /n). Let f be linear on [0, m1 /m]. Note that f (0) is deﬁned so that the absolute value of the slope of f on the interval [0, m1 /m] is less than mδ/2. So, if s and t are in [0, m1 /m] and |s − t| ≤ 1/m then |f (s) − f (t)| < δ/2 < δ. Moreover, if 1 ≤ l ≤ m1 then f ([(l − 1)/m, l/m]) ⊆ [(i1 − 1)/n, i1 /n] and Dl ⊆ Ci1 . Extension of f to [mk−1 /m, 1]. Suppose ik−1 < ik . If hk > 2/(nδ), let f (1) = (ik−1 + 1)/n = ik /n. If hk ≤ 2/(nδ), let f (1) = ik−1 /n + hk δ/2 (so f (1) ≤ ik /n). Suppose ik−1 > ik so ik−1 − 1 = ik . If hk > 2/(nδ), let f (1) = (ik − 1)/n. If hk ≤ 2/(nδ), let f (1) = ik /n − hk δ/2 (so f (1) ≥ (ik − 1)/n). Let f be linear on [mk−1 /m, 1]. Note that f (1) is deﬁned so that the absolute value of the slope of f on the interval [mk−1 /m, 1] is less than mδ/2. So, if s and t are in [mk−1 /m, 1] and |s − t| ≤ 1/m then |f (s) − f (t)| < δ/2 < δ. Moreover, if mk−1 < l ≤ mk then f ([(l − 1)/m, l/m]) ⊆ [(ik − 1)/n, ik /n] and Dl ⊆ Cik . This completes the deﬁnition of f . The construction ensures that f is a piecewise linear mapping that is constant on no subinterval of [0, 1]. As f was being deﬁned we made sure that condition (1) was satisﬁed and checked that conditions (2) and (4) were satisﬁed. All that remains to be veriﬁed is that f is surjective and condition (3) is satisﬁed. The function f is surjective. The points p and q are used to ensure that there are points a and b so that f (a) = 0 and f (b) = 1. The point p is in Bj∗ for some j and, because C1 is the only link of C containing p, ij = 1. Because d(p, Ci ) > σ for i = 1, hj > 2/(nδ). If j = 1, then i1 < i2 and f (0) = (i1 − 1)/n = 0. If j = k, then ik = 1 and ik−1 = 2 so f (mk−1 /m) = 1/n and f (1) = (ik − 1)/n = 0. If 1 < j < k, then ij−1 = ij+1 = 2 so f (mj−1 /m) = f (mj /m) = 1/n and f (rj ) = 0 where rj is the midpoint of the interval [mj−1 /m, mj /m]. That f achieves a value of 1 is handled similarly employing the point q in place of p in the argument.

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Veriﬁcation of condition (3). Finally, to see that condition (3) holds, suppose 1 ≤ j ≤ m, 1 ≤ i ≤ n, Dj ⊆ Ci and l is an integer such that Dl ∩ Dj = ∅. There exists an integer g, 1 ≤ g ≤ k, such that Dl belongs to Bg . There is an integer ig , 1 ≤ ig ≤ n, such that (Bg )∗ ⊆ Cig . Because Dl ⊆ Cig we have Ci ∩ Cig = ∅ and thus |i − ig | ≤ 1. By the deﬁnition of f , f ([(l − 1)/m, l/m]) ⊆ [(ig − 1)/n, ig /n], but [(ig − 1)/n, ig /n] ⊆ [(i − 2)/n, (i + 1)/n]. In the proof of Theorem 102 we make use of the following theorem found in [35, Theorem 114, p. 47]. Theorem 101 If Q is a property and, for each positive integer n, Hn is a ﬁnite set and, for each n, each element of Hn+1 has property Q with respect to some element of Hn then there is a sequence J1 , J2 , J3 , . . . such that, for each n, Jn ∈ Hn and Jn+1 has property Q with respect to Jn . A sequence δ of positive numbers is a Lebesgue sequence for an inverse limit sequence f provided for each i and j with i < j, if x, y ∈ Xj and dj (x, y) < δj then di (fi j (x), fi j (y)) < 2−j . Theorem 102 If M is a nondegenerate chainable continuum, there exists an inverse limit sequence f of piecewise linear maps of [0, 1] onto itself such that M is homeomorphic to lim f and no map in the sequence f is constant ←− on any subinterval of [0, 1]. Proof. The proof is broken into several parts: the construction of the inverse limit sequence, the deﬁnition of the homeomorphism h, and the proof that h is continuous, one-to-one, and surjective. Construction of the inverse limit sequence. Let C1 , C2 , C3 , . . . be a deﬁning sequence of chains for M obtained from Theorem 99 with Cn = n ) for each n. (C1n , C2n , . . . , Cm n Let p1 be a point of M ∩ C11 such that p1 does not belong to Ci1 for i = 1 1 and q1 be a point of M ∩ Cm such that q1 does not belong to Ci1 for i = m1 . 1 There is a positive number σ1 such that d(p1 , Ci1 ) > σ1 for 1 σ1 for 1 ≤ i < m1 , and d(Ci1 , Cj1 ) > σ1 for |i − j| > 1. Let δ1 = 1/2. There exists an integer n2 > 1 such that the mesh of Cn2 is less than (m1 σ1 δ1 )/2. For convenience of notation we assume n2 = 2. Using δ1 = δ, σ1 = σ, C1 = C, C2 = D, p1 = p, and q1 = q in Lemma 100 we obtain a piecewise linear mapping f1 : [0, 1] → → [0, 1] satisfying the conclusion of that lemma. f1 is uniformly continuous, thus for ε = 2−2 there is a positive number δ2 such that if |s − t| < δ2 then |f1 (s) − f1 (t)| < 2−2 . Choose p2 ∈ M ∩ C12 and 2 q2 ∈ M ∩Cm such that p2 ∈ / Ci2 for i > 1 and q2 ∈ / Ci2 for i < m2 . There exists 2

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a positive number σ2 such that d(p2 , Ci2 ) > σ2 for i > 1, d(q2 , Ci2 ) > σ2 for i < m2 and d(Ci2 , Cj2 ) > σ2 for |i − j| > 1. There is a positive integer n3 > 2 such that the mesh of Cn3 is less than (m2 σ2 δ2 )/2. Again, for convenience of notation we assume n3 = 3. Using δ2 , σ2 , C2 , C3 , p2 , and q2 appropriately in → [0, 1] satisfying Lemma 100 we obtain a piecewise linear mapping f2 : [0, 1] → the conclusion of that lemma. Inasmuch as f2 3 = f2 , and f1 3 = f1 ◦ f2 are uniformly continuous, for ε = 2−3 , there is a positive number δ3 such that if |s − t| < δ3 then |fi j (s) − fi j (t)| < 2−3 for j = 3 and 1 ≤ i < j. Continuing inductively, we obtain a sequence f of piecewise linear mappings of [0, 1] onto itself and a Lebesgue sequence δ for f such that for each positive integer n 1. If 1 ≤ j ≤ mn+1 there is a positive integer i, 1 ≤ i ≤ mn , such that fn ([(j − 1)/mn+1 , j/mn+1 ]) ⊆ [(i − 1)/mn , i/mn ] 2. If 1 ≤ j ≤ mn+1 , 1 ≤ i ≤ mn , and fn ([(j − 1)/mn+1 , j/mn+1 ]) ⊆ [(i − 1)/mn , i/mn ] then Cjn+1 ⊆ Cin 3. If 1 ≤ j ≤ mn+1 , 1 ≤ i ≤ mn , Cjn+1 ⊆ Cin , and l is an integer such that Cln+1 ∩ Cjn+1 = ∅, then fn ([(l − 1)/mn+1 , l/mn+1 ]) ⊆ [(i − 2)/mn , (i + 1)/mn ] 4. If 1 ≤ j ≤ mn+1 and s, t ∈ [(j−1)/mn+1, j/mn+1 ] then |fn (s)−fn (t)| < δn 5. fn is not constant on any subinterval of [0, 1]. Deﬁnition of h. Suppose x ∈ M . There exists a sequence Ci11 , Ci22 ,

⊆ Ci33 , . . . of links closing down on x, that is, such that Cinn ∈ Cn , Cin+1 n+1 Cinn , and {x} = n>0 Cinn . From condition (3) it follows that fn ([(in+1 − 2)/mn+1 , (in+1 + 1)/mn+1 ]) ⊆ [(in − 2)/mn , (in + 1)/mn ]. Therefore, we have −1 that πn+1 ([(in+1 − 2)/mn+1 , (in+1 + 1)/mn+1 ]) ⊆ πn−1 ([(in − 2)/mn , (in + 1)/mn ]). From condition (4) and the fact that δ is a Lebesgue sequence for f it follows that for 1 ≤ l ≤ mn the diameter of πn−1 ([(l − 1)/mn , l/mn]) n does not exceed 6/2 z, y ∈ πn−1 ([(l − 1)/mn , l/mn]). . To see this,i suppose −(n−1) (1/2 + 1/4 + · · · + 1/2n−2 ) + Then, d(z, y) = i>0 |zi − yi |/2 < 2 −(n−1) n n 2 (1 + 1/2 + 1/4 + . . . ) < 2/2 + 4/2 . Thus, only one point belongs to −1 π n>0 n ([(in − 2)/mn , (in + 1)/mn ]). Let this point be h(x). Note that the deﬁnition of h(x) is independent of the choice of the sequence of links Ci11 , Ci22 , Ci33 , . . . closing down on x. Indeed, suppose Ci11 , Ci22 , Ci33 , . . . and Ck11 , Ck22 , Ck33 , . . . are sequences of links closing down on x and suppose n is a positive integer such that πn−1 ([(kn − 2)/mn , (kn + 1)/mn ]) ∩ πn−1 ([(in − 2)/mn , (in +1)/mn ]) = ∅. Again, from condition (3) it follows that fn ([(in+1 − 2)/mn+1 , (in+1 + 1)/mn+1 ]) ⊆ [(in − 2)/mn , (in + 1)/mn ] and fn ([(kn+1 − 2)/mn+1 , (kn+1 + 1)/mn+1 ]) ⊆ [(kn − 2)/mn , (kn + 1)/mn ]. Because x ∈ Cin+1 ∩Ckn+1 , |in+1 −kn+1 | ≤ 1 so [(kn+1 −1)/mn+1 , kn+1 /mn+1 ] and [(in+1 − n+1 n+1 1)/mn+1 , in+1 /mn+1 ] have a point in common. Thus, there is a point s of

1.13 An inverse limit homeomorphic to a sin(1/x)-curve

67

[(in − 2)/mn , (in + 1)/mn ] ∩ [(kn − 2)/mn , (kn + 1)/mn ]. The maps in the sequence f are surjective, therefore there is a point p of lim f such that pn = ←− s. This contradicts the assumption that πn−1 ([(kn − 2)/mn , (kn + 1)/mn ]) ∩ πn−1 ([(in − 2)/mn , (in + 1)/mn ]) = ∅. The function h is one-to-one. Suppose x and y are points of M and x = y. Let Ci11 , Ci22 , Ci33 , . . . and Cj11 , Cj22 , Cj33 , . . . be sequences such that ⊆ C n and Cjn+1 ⊆ Cjnn for each n, and Cinn , Cjnn ∈ Cn for each n, Cin+1 n+1 n+1 n in n {x} = n>0 Cin and {y} = n>0 Cjn . Because x = y, there exists a positive integer n such that |in − jn | ≥ 4. Then [(in − 2)/mn , (in + 1)/mn ] ∩ [(jn − 2)/mn , (jn + 1)/mn ] = ∅. It follows that h(x) = h(y). The function h is surjective. Suppose y ∈ lim f . We now make ←− use of Theorem 101. For each positive integer n let Hn = {J | J = [(l − 1)/mn , l/mn ] for some l, 1 ≤ l ≤ mn , and yn ∈ J}. For each n, Hn contains no more than two elements. If K ∈ Hn+1 , by condition (1), there is an element J of Hn such that f (K) ⊆ J. It follows from Theorem 101 that there exists a sequence l1 , l2 , l3 , . . . such that yi ∈ [(li − 1)/mi , li /mi ] and fi ([(li+1 − 1)/mi+1 , li+1 /mi+1 ]) ⊆ [(li − 1)/mi , li /mi ] for each i. By condition ⊆ Clii for each i. There is a point x of M such that {x} = n>0 Clnn . (2), Cli+1 i+1 Then, h(x) = y. The function h is continuous. Suppose x ∈ M . Let Ci11 , Ci22 , Ci33 , . . . ⊆ Cinn for each n be a sequence of links each containing x such that Cin+1 n+1 −1 and {h(x)} = n>0 πn ([(in − 2)/mn , (in + 1)/mn ]). Let V be an open set containing h(x). There is a positive integer k such that πk−1 ([(ik −3)/mk , (ik + 2)/mk ]) ⊆ V . Suppose y ∈ Cikk . There is a sequence Cl11 , Cl22 , Cl33 , . . . of links each containing y such that Cln+1 ⊆ Clnn for each n and {h(y)} = n+1 −1 −1 n>0 πn ([(ln −2)/mn , (ln +1)/mn ]). Because |lk −ik | ≤ 1, h(y) ∈ πk ([(ik − 3)/mk , (ik + 2)/mk ]) and so h(y) ∈ V . M and lim f are compact metric spaces and h is one-to-one, continuous, ←− and surjective, therefore h is a homeomorphism from M onto lim f by The←− orem 259.

1.13 An inverse limit homeomorphic to a sin(1/x)-curve

We close this chapter with a proof that one can obtain a sin(1/x)-curve as an inverse limit on [0, 1] with a single bonding map. Let S denote the closure

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of the graph of y = sin(1/x) on (0, 1]. In this section we show that the inverse limit on [0, 1] of the function g whose graph consists of two linear pieces passing through (0, 0), (1/2, 1), and (1, 1/2) is homeomorphic to S. In Example 103 we actually show that lim g 2 produces this continuum, but in ←− light of Theorem 37 this is suﬃcient. Denote by p1 and p2 the projections of the plane to its ﬁrst and second coordinates, respectively. Then, p1 (S) = [0, 1] and p2 (S) = [−1, 1].

(1/4,1)

(1,1)

(1/2,1/2)

(0,0)

Fig. 1.21 The function f from Example 103

Example 103 Let f : I → → I be the map from I onto itself such that f (0) = 0, f (1/4) = 1, f (1/2) = 1/2, f (1) = 1, and f is linear on [0, 1/4], [1/4, 1/2], and [1/2, 1]. Then lim g and lim f = lim g 2 are homeomorphic to S. (See ←− ←− ←− Figures 1.21 and 1.22.) Proof. The proof is broken into several parts. First, we deﬁne the function h : lim f → S. Then we show that h is continuous, one-to-one, and surjective. ←− Because lim f is compact, this is suﬃcient for h to be a homeomorphism. ←− (The deﬁnition of h.) In deﬁning h we make use of two linear homeomorphisms, ϕ and ψ. Let ϕ : [1/2, 1] → → [−π/2, π/2] be the linear homeomorphism

1.13 An inverse limit homeomorphic to a sin(1/x)-curve

69

h(1,1,1/4,1/16,...) h(1/4,1/16,...) h(1,1,1,...)

h(1,1/4,1/16,...)

h(0,0,0,...)

h(1/2,1/2,...)

h(1/2,1/8,/1/32,...) h(1/2,1/2,1/8,1/32,...) h(1/2,1/2,1/2,1/8,1/32,...)

Fig. 1.22 Some values of the homeomorphism in Example 103

such that ϕ(1/2) = −π/2 and ϕ(1) = π/2. Let ψ : [0, 1/4] → → [1, π/2] be the linear homeomorphism such that ψ(0) = 1 and ψ(1/4) = π/2. We deﬁne h as follows. Let x ∈ lim f . If xi ∈ [1/2, 1] for each positive integer i, deﬁne h(x) = ←− 0, sin(ϕ(x1 )) . If x1 ∈ [0, 1/4), let h(x) = 1/ψ(x1 ), sin(ψ(x1 )) . If x1 ∈ [1/4, 1/2), let h(x) = 1/(π − ϕ(f (x1 ))), sin(ϕ(f (x1 ))) . If x1 ∈ [1/2, 1], but xi ∈ [1/2, 1] for some i > 1, there is a positive integer j such that xj ∈ [1/2, 1] and xj+1belongs to one of [1/8, 1/4) and [1/4, 1/2). If xj+1 ∈ [1/8, 1/4), let 1/(2jπ + ϕ(f (xj+1 ))), sin(ϕ(f (xj+1 ))) . If xj+1 ∈ [1/4, 1/2), let h(x) = 1/((2j + 1)π − ϕ(f (xj+1 ))), sin(ϕ(f (xj+1 ))) . This completes the deﬁnition of h. It may help the reader to observe that if xj ∈ [1/2, 1] and xj+1 ∈ [1/8, 1/4) then p1 (h(x)) ∈ 2/((4j + 1)π), 2/((4j − 1)π) whereas if xj ∈ [1/2, 1] and xj+1 ∈ [1/4, 1/2) then p1 (h(x)) ∈ 2/((4j + 3)π), 2/((4j + 1)π) . h(x) =

(The continuity of h.) Suppose x ∈ lim f . We consider two cases. ←− • There is a positive integer i such that xi ∈ [1/2, 1] • xi ∈ [1/2, 1] for each positive integer i. Suppose there is a positive integer i such that xi ∈ [1/2, 1]. If x1 < 1/2 and x1 = 1/4, because all of the functions involved in the deﬁnition of h are

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continuous, p1 ◦ h and p2 ◦ h are continuous.

It follows that h is continuous at x. Suppose x1 = 1/4. Then h(x) = 2/π, 1 . The continuity of h at x follows from the facts that if z ∈ lim f and z1 < 1/4, h(z) = 1/ψ(z1 ), sin(ψ(z1 )) ←− and as t → 1/4, ψ(t) → π/2.

Similarly, it follows directly from the deﬁnition of h and the continuity of the functions involved in its deﬁnition that h is continuous at x if 1/2 < xj ≤ 1 and either 1/8 < xj+1 < 1/4 or 1/4 < xj+1 < 1/2 for some positive integer j. Suppose j is a positive integer such that xi = 1/2 for 1 ≤ i ≤ j and xj+1 = 1/8. Then, h(x) = 2/((4j − 1)π), −1 . If z ∈ lim f and 1/16 ≤ ←− zj+1 < 1/8, then zj ∈ [1/4, 1/2) and zj−1 ∈ [1/2, 1]. Thus, h(z) = 1/((2j − 1)π − ϕ(f (zj ))), sin(ϕ(f (zj ))) . The continuity of h at x follows from the fact that as t → 1/8, ϕ(f (t)) → −π/2. Suppose j is a positive integer such that xi = 1 for 1 ≤ i ≤ j and xj+1 = 1/4. Then, h(x) = 2/((4j + 1)π), 1 . The continuity of h at x follows from the facts that if z ∈ lim f and zj+1 ∈ (1/8, 1/4) then h(z) = 1/(2jπ + ←− ϕ(f (zj+1 ))), sin(ϕ(f (zj+1 ))) and as t → 1/4, ϕ(f (t)) → π/2. This concludes the proof that h is continuous in the ﬁrst case. We now turn to the case that xi ∈ [1/2, 1] for each positive integer i. Suppose x1 , x2 , x3 , . . . is a sequence of points of lim f that converges to x. ←− Because of the continuity of the functions involved in the deﬁnition of h, we only need to consider the case that for each n there is a positive integer jn such that xnjn ∈ [1/2, 1] and xnjn +1 ∈ [1/2, 1]. Then, the continuity of h at x follows from the facts that as n → ∞ we have that jn → ∞ and xn1 → x1 . (h is one-to-one). Suppose x and y are in lim f and x = y. We consider ←− three cases. 1. xi ∈ [1/2, 1] for each positive integer i 2. There is a positive integer j such that xj ∈ [1/2, 1] and xj+1 ∈ [1/2, 1] 3. xi ∈ [0, 1/2) for each positive integer i. Suppose xi ∈ [1/2, 1] for each positive integer i. Then p1 (h(x)) = 0. If yi ∈ [1/2, 1] for each i then x1 = y1 , so sin(ϕ(x1 )) = sin(ϕ(y1 )) and, consequently, h(x) = h(y). If there is a nonnegative integer k such that yk+1 ∈ [0, 1/2) then p1 (h(y)) > 2/((4k + 3)π). Again, h(x) = h(y).

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71

Suppose case (2) holds. Then, p1 (h(x)) ∈ (2/((4j + 3)π), 2/((4j − 1)π)]. If yi ∈ [1/2, 1] for each i, then p1 (h(y)) = 0 and h(x) = h(y). If yi ∈ [0, 1/2) for each i then p1 (h(y)) > 2/(3π). Because j > 0, 2/((4j − 1)π) ≤ 2/(3π) and h(x) = h(y). If there is a positive integer k such that yk ∈ [1/2, 1] and yk+1 ∈ [1/8, 1/2), then p1 (h(y)) ∈ (2/((4k + 3)π), 2/((4k − 1)π)]. There are two possibilities: j = k and j = k. If j = k, then (2/((4j + 3)π), 2/((4j − 1)π)] ∩ (2/((4k + 3)π), 2/((4k − 1)π)] = ∅ and h(x) = h(y). If j = k, then xj+1 = yj+1 otherwise x = y. If one of xj+1 is in [1/8, 1/4) and the other is in [1/4, 1/2) then p1 (h(x)) = p1 (h(y)) and so h(x) = h(y). If both lie in [1/8, 1/4) or both lie in [1/4, 1/2) then sin(ϕ(f (xj+1 ))) = sin(ϕ(f (yj+1 ))) and again we have h(x) = h(y). Finally, suppose case (3) holds. Then p1 (h(x)) ∈ (2/(3π), 1]. If there is a positive integer i such that yi ∈ [1/2, 1] then p1 (h(y)) ≤ 2/((4i − 1)π) and h(x) = h(y). If yi ∈ [0, 1/2) for each i, then x1 = y1 . If one of x1 and y1 is in [0, 1/4) and the other is in [1/4, 1/2) then p1 (h(x)) = p1 (h(y)) and we have h(x) = h(y). If both x1 and y1 are in [0, 1/4) then ψ(x1 ) = ψ(y1 ). If both x1 and y1 are in [1/4, 1/2), ϕ(x1 ) = ϕ(y1 ). Either way we have h(x) = h(y). (h is surjective.) The image of (0, 0, 0, . . . ) under h is (1, sin(1)) and the image of (1, 1, 1, . . . ) under h is (0, 1). Because h(lim f ) is a subcontinuum of ←− S containing these two points and S is irreducible between them, h(lim f ) = ←− S.

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18. W. T. Ingram, Inverse limits on [0, 1] using tent maps and certain other piecewise linear bonding maps, 253–258, in: H. Cook, W. T. Ingram, K. T. Kuperberg, A. Lelek, P. Minc (Eds.), Continua with the Houston Problem Book, Marcel Dekker, New York, 1995. 19. W. T. Ingram, Inverse limits on [0, 1] using piecewise linear unimodal bonding maps, Proc. Amer. Math. Soc. 128 (2000), 279–286. 20. W. T. Ingram, Invariant sets and inverse limits, Topology Appl. 126 (2002), 393–408. 21. William T. Ingram and William S. Mahavier, Interesting dynamics in a family of one-dimensional maps, Amer. Math. Monthly 111 (2004), 198–215. 22. W. T. Ingram, A brief historical view of continuum theory, Topology Appl. 153 (2006), 1530–1539. 23. John L. Kelley, General Topology , Springer-Verlag, New York, 1975. 24. Judy Kennedy, A brief history of indecomposable continua, 103–126, in: H. Cook, W. T. Ingram, K. T. Kuperberg, A. Lelek, P. Minc (Eds.), Continua with the Houston Problem Book, Marcel Dekker, New York, 1995. 25. B. Knaster, Un continu dont tout sous-continu est ind´ ecomposable, Fund. Math. 3 (1922), 246-286. 26. C. Kuratowski, Th´ eorie des continus irr´ eductibles entre deux points I, Fund. Math. 3 (1922), 200–231. 27. C. Kuratowski, Topology. Vol. II, Academic Press, New York, 1968. 28. Tine-Yien Li and James A. Yorke, Period three implies chaos, Amer. Math. Monthly 82 (1975), 985–992. 29. W. S. Mahavier, A chainable continuum not homeomorphic to an inverse limit on [0, 1] with only one bonding map, Proc. Amer. Math. Soc. 18 (1967), 284-286. 30. W. S. Mahavier, Arcs in inverse limits on [0, 1] with only one bonding map, Proc. Amer. Math. Soc. 21 (1969), 587–590. 31. Dorothy S. Marsh, A chainable continuum not homeomorphic to an inverse limit on [0, 1] with only one bonding map, Colloq. Math. 43 (1980), 75–80. 32. John Milnor, Periodic orbits, external rays and the Mandelbrot set: an expository account, Ast´ erisque No. 261, xiii, (2000) 277–333. 33. Piotr Minc and W. R. R. Transue, Sarkovski˘ı’s theorem for hereditarily decomposable chainable continua, Trans. Amer. Math. Soc. 315 (1989), 173–188. 34. Piotr Minc and W. R. R. Transue, A transitive map on [0, 1] whose inverse limit is the pseudoarc, Proc. Amer. Math. Soc. 111 (1991), 1165–1170. 35. R. L. Moore, Foundations of Point Set Theory, Colloquium Publications, Vol. XIII, Amer. Math. Soc., Providence, RI, 1962. 36. Sam B. Nadler, Jr., Arc components of certain chainable continua, Canad. Math. Bull. 14 (1971), 183–189. 37. Sam B. Nadler, Jr., Continuum Theory, Marcel Dekker, New York, 1992. 38. Brian Raines, Orbits of turning points for maps of ﬁnite graphs and inverse limit spaces (electronic), Proc. Ninth Prague Top. Symp. 2001, Topol. Atlas, North Bay, ON (2002), 253–263. 39. William Thomas Watkins, Homeomorphic classiﬁcation of certain inverse limit spaces with open bonding maps, Paciﬁc J. Math. 103 (1982), 589–601.

Chapter 2

Inverse Limits in a General Setting

Abstract In this chapter we investigate inverse limits in a very general setting: over directed sets with factor spaces that are compact Hausdorﬀ spaces using upper semi-continuous closed set-valued bonding functions. Basic existence and connectedness theorems are proved and examples are provided that illustrate limitations to the generality of the theorems. One section is devoted to examples in the case where the factor spaces are all the interval [0, 1]. Basic theorems on mappings of inverse limits are included as well. As the chapter progresses additional hypotheses are added to the factor spaces (up to compact metric) and the bonding functions (continuous single-valued or unions of such). The chapter concludes with considerations of a few miscellaneous topics including dimension and a proof that a 2-cell is not an inverse limit with a single upper semi-continuous function on [0, 1].

2.1 Introduction Inverse limits are normally deﬁned for a pair of sequences X1 , X2 , X3 , . . . and f1 , f2 , f3 , . . . such that, for each i, Xi is a topological space and fi is a mapping (i.e., continuous function) from Xi+1 into Xi . Such a pair of sequences is often denoted {Xi , fi } and is called an inverse limit sequence and the mappings fi are called bonding maps and the spaces Xi are referred to as factor spaces. More generally an inverse limit system is deﬁned to be a triple that consists of a directed set D, a collection of topological spaces {Xα }α∈D , and a collection of mappings {fα β : Xβ → Xα | α, β ∈ D and α β}. Often this triple is shortened to {Xα , fα β , D}. This more general form has proved useful at times. See Howard Cook’s article [1] for a particularly nice construction that employs this type of inverse limit system. Throughout, we use the term inverse limit sequence when the underlying directed set is the

W.T. Ingram and W.S. Mahavier, Inverse Limits: From Continua to Chaos, Developments in Mathematics 25, DOI 10.1007/978-1-4614-1797-2_2, © Springer Science+Business Media, LLC 2012

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set of positive integers and the term inverse limit system when the underlying directed set may be a more general directed set. Inverse limit systems in which the bonding functions are mappings have been studied for quite some time, particularly those for which the underlying directed set is the set of positive integers. Recently, inverse limit sequences in which the bonding functions are upper semi-continuous set-valued functions were introduced in [8] as inverse limits of closed subsets of [0, 1] × [0, 1]. These were generalized in [7] to inverse limit sequences where the spaces were compact Hausdorﬀ spaces and the bonding functions were upper semicontinuous set-valued functions. A natural next step is to examine inverse limits in a setting that encompasses all of these scenarios. In Section 2.5 we prove the basic existence theorems in this very general setting over a directed set where the factor spaces are compact Hausdorﬀ spaces and the bonding functions are upper semi-continuous. In Sections 2.6 and 2.9 we address the connectedness of the inverse limit. In Section 2.7 we include some examples that demonstrate some of the variety of spaces that result as an inverse limit with upper semi-continuous bonding functions. In Section 2.8 we examine some basic theorems concerning mappings between inverse limits of inverse limit systems. In Section 2.10 of this chapter we contrast a couple of major diﬀerences between inverse limits of ordinary inverse limit sequences and these more general inverse limit systems through examples showing that certain basic theorems do not hold in this setting. Finally, in Section 2.11 we include some theorems requiring metric factor spaces. We recommend that the reader with little or no experience with inverse limits ﬁrst read Chapter 1 to get a better feel for inverse limits before reading the present chapter.

2.2 Deﬁnitions and a basic theorem A relation on a set D is a subset of D × D such that each member of D is a ﬁrst term of some pair in the relation. If is a relation on a set D and (x, y) is in then we write x y. A directed set is a pair (D, ) where is a relation on D such that (a) if α ∈ D then α α; (b) if α, β, and γ are in D and α β and β γ then α γ; and (c) if α and β are in D then there is a member γ of D such that α γ and β γ. If (D, ) is a directed set, for short, we usually say simply that D is a directed set. If D is a directed set and α and β are elements of D such that α β we say that α precedes β in D. A directed set is called totally ordered provided if α and β belong to D then α β or β α and in the case where α β and β α we have α = β.

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If Y is a topological space, then 2Y denotes the collection of nonempty closed subsets of Y whereas we denote by C(Y ) the elements of 2Y that are connected. Let each of X and Y be a topological space and let f be a function from X into 2Y . The function f : X → 2Y is upper semi-continuous at the point x ∈ X if and only if for each open set V in Y containing f (x) there is an open set U in X containing x such that if u is in U then f (u) ⊆ V ; f is called upper semi-continuous provided it is upper semi-continuous at each point of X. If A ⊆ X, f (A) denotes {y ∈ Y | y ∈ f (x) for some x ∈ A}. The graph of f is denoted by G(f ) and is the set of all points (x, y) ∈ X × Y such that y is in f (x). If f : X → 2Y and g : Y → 2Z , we denote by g ◦ f : X → 2Z the closed set-valued function given by (g ◦ f )(x) = {z ∈ Z | there is an element y ∈ Y such that y ∈ f (x) and z ∈ f (y)}. In the case where f is a singletonvalued upper semi-continuous function, we do not distinguish between f and the corresponding mapping associated with f . For example, if f : X → 2X is given by f (x) = {x} for each x ∈ X, we still refer to f as the identity on X. Suppose D is a directed set and, for each α in D, Xα is a topological space. Suppose further, for each α and β in D with α β, fα β : Xβ → Xα is an upper semi-continuous function from Xβ into 2Xα such that fα α is the identity on Xα and if α β γ then fα γ = fα β ◦ fβ γ . The triple {Xα , fα β , D} is called an inverse limit system. The spaces Xα are called The inverse factor spaces and the functions fα β are called bonding functions. limit of the system {Xα , fα β , D} is a subspace of Π = α∈D Xα with the product topology. We denote elements of Π using boldface type. If x ∈ Π, xα denotes the α-coordinate of x (i.e., x(α) = xα ). The points of the inverse limit are the elements x of Π such that if α β in D then xα ∈ fα β (xβ ). We denote the inverse limit by lim{Xα , fα β , D}. Consistent with our use of ←− boldface type to denote sequences of bonding maps in Chapter 1, we also use boldface type to denote collections of bonding functions fα β . Also consistent with the notation in Chapter 1, if f is the collection of all of the functions fα β in an inverse limit system {Xα , fα β , D} we normally denote the inverse limit of this inverse limit system by lim f . If, for each α β in D and each ←− point t of Xβ , fα β (t) is degenerate then this deﬁnition reduces to the usual one for systems over directed sets. If, in addition, the directed set is the set of positive integers, this deﬁnition reduces to the usual one for inverse limit sequences. The most commonly used directed set in inverse limits is the set of positive integers. Often, in the case where D is the set of positive integers, instead of specifying all of the bonding functions fi j in the system, only the terms of a sequence of functions are speciﬁed. This was the practice we used in Chapter 1 and is the way we present most of the examples in this chapter. The commonly used convention for expanding a sequence f1 , f2 , f3 , . . . of functions into the functions of an inverse sequence is to deﬁne fi j to be the composition fi ◦fi+1 ◦· · ·◦fj−1 for i < j and to let fi i be the identity for each

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i. In the case where the inverse limit system over the set of positive integers is speciﬁed by a sequence of mappings fi : Xi+1 → Xi , as mentioned in the introduction to this chapter, we may denote the inverse limit sequence by the pair, {Xi , fi }, and its inverse limit by lim{Xi , fi }, or simply lim f . In the ←− ←− speciﬁc instance where f : X → X is a function, Xi = X and fi = f for each positive integer i, we may denote the system by {X, f } and its inverse limit by lim f . Such systems determined by a single space and a single bonding ←− function are often called inverse limits with a single bonding function. The reader who is only interested in the proofs of these theorems for inverse sequences may assume throughout that D is the set of positive integers. Moreover, for the basic results on inverse limit sequences using single-valued continuous functions, the reader may also assume the bonding functions are mappings. α ∈ D} is a collection of topological spaces, we denote the proIf {Xα | jection of α∈D Xα onto the factor space Xα by pα (i.e., pα (x) = xα ). We are usually more interested in the inverse limit space than the product space, therefore we denote by πα the restriction of pα to the inverse limit space. The projection pα is an open mapping on the product space, but πα is not normally open. A useful feature of inverse limits lies in the interaction between the bonding functions and the projection mappings. The proof of the following analogue of Theorem 2 from Chapter 1 is an immediate consequence of the deﬁnitions and is left to the reader. Theorem 104 Suppose {Xα , fα β , D} is an inverse limit system and the inverse limit, M , of the system is nonempty. If x ∈ M and α β, πα (x) ∈ fα β (πβ (x)). If fα β is a mapping, πα (x) = fα β ◦ πβ (x).

2.3 Graphs of upper semi-continuous functions

That there is a close connection between closed subsets of product spaces and upper semi-continuous set-valued functions can be seen from the following theorem. Theorem 105 Suppose each of X and Y is a compact Hausdorﬀ space and M is a subset of X × Y such that if x is in X then there is a point y in Y such that (x, y) is in M . Then M is closed if and only if there is an upper semi-continuous function f : X → 2Y such that M = G(f ).

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79

Proof. We ﬁrst show that if f : X → 2Y is an upper semi-continuous function then G(f ) is closed. Let p = (p1 , p2 ) be a point of X × Y that is not in G(f ). Then, p2 ∈ / f (p1 ), so, because compact Hausdorﬀ spaces are regular, there are mutually exclusive open sets V and W in Y such that p2 ∈ V and f (p1 ) ⊆ W . Because f is upper semi-continuous, there is an open subset U of X containing p1 such that if t ∈ U then f (t) ⊆ W . Thus, U × V is an open subset of X × Y containing p that misses G(f ). It follows that G(f ) is closed. Assume that M is closed and, for each x in X, deﬁne f (x) to be {y ∈ Y | (x, y) ∈ M }. Because M is closed, f (x) is closed for each x in X. To see that f is upper semi-continuous, suppose x is in X and V is an open set in Y containing f (x). If f is not upper semi-continuous at x, then for each open set U containing x there exist points z of U and (z, y) of M such that y is not in V . For each open set U containing x, denote by MU the set of all points (p, q) of M such that p is in U and q is not in V . Observe that if U and U are open sets containing x and U ⊆ U then MU ⊆ MU . From this it follows that the collection M of all the closed sets MU has the ﬁnite intersection property. X × Y is compact, thus there is a point (a, b) common to all the sets in M. Each element of M is a subset of M , therefore (a, b) belongs to M so b ∈ f (a). Because x is the only point common to all the sets U , a = x. Furthermore, b is not in V . This contradicts the fact that b belongs to f (x).

2.4 Consistent systems In general, an inverse limit system with upper semi-continuous bonding functions over a directed set may fail to produce a nonempty inverse limit even if the factor spaces are compact. Consider the following example. In this example and hereafter we denote the identity from X into X by IdX . If no confusion should arise with respect to the domain, we may shorten this to Id. Example 106 Let D = {1, 2, 3, . . . } ∪ {a, b} where, if i and j are positive integers, then i j if and only if i ≤ j, a j if and only if j ≥ 2, b j if and only if j ≥ 3, 1 b, and a b (1 and a do not compare nor do 2 and b). Let Xα = {0, 1} for each α ∈ D. If 3 i j, let fi j = Id. Let f1 2 = fa 2 = fa b = Id as well, and let f1 b = 1 − Id. Let f2 3 (t) = {0, 1} for t ∈ {0, 1} and fb 3 = f2 3 . We expand this into a system by composition. Then, lim f = ∅. ←− Proof. Suppose x ∈ lim f . If x1 = 0, then x2 = 0 and xb = 1. Because x2 = 0, ←− xa = 0; but xa = 1 because xb = 1. This is a contradiction. Because x1 = 0,

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x1 = 1. Then, x2 = 1 and xb = 0. But, x2 = 1 yields xa = 1 whereas xb = 0 yields xa = 0, again a contradiction. So, lim f = ∅. ←− We call an inverse limit system {Xα , fα β , D} consistent provided for each η ∈ D and each t ∈ Xη there is a point x of Π such that xη = t and if α β η then xα ∈ fα β (xβ ). We now show that two important classes of inverse limit systems are consistent: those in which all the bonding functions are mappings and those for which the directed set is the set of positive integers. Theorem 107 Each inverse limit system {Xα , fα β , D} where each function fα β in the system is a mapping is consistent. Proof. Suppose η ∈ D and t ∈ Xη . Let z be a point of Π. For each γ such that γ η, let xγ = fγ η (t) and let xγ = zγ otherwise. Suppose α β η. Inasmuch as fα η (t) = fα β (fβ η (t)) we have xα = fα β (xβ ) and we see that the system is consistent. Theorem 108 Each inverse limit system {Xi , fi j , D} where D is the set of positive integers is consistent. Proof. Suppose n is a positive integer and t ∈ Xn . Let z ∈ Π. Let xk = zk if k > n. Let xn = t and x1 be a point belonging to f1 n (t). Inductively, suppose 1 ≤ j < n is an integer such that x1 , x2 , . . . , xj have been chosen so that xi ∈ fi n (t) for 1 ≤ i ≤ j and if l ≤ m ≤ j then xl ∈ fl m (xm ). Because fj n = fj j+1 ◦ fj+1 n there exists an element xj+1 of Xj+1 such that xj+1 ∈ fj+1 n (t) and xj ∈ fj j+1 (xj+1 ). It follows that the system {Xi , fi j } is consistent.

2.5 Compact inverse limits In this section we assume that {Xα , fα β , D} is a consistent inverse limit system over a directed set D with upper semi-continuous bonding functions and that Xα is a compact Hausdorﬀ space for each α ∈ D. Our goal in this section is to prove that under these conditions the inverse limit is nonempty and compact (Theorem 111). In the absence of assumptions of some sort, the inverse limit may be empty even if D is the set of positive integers, the factor spaces metric, and there is only one bonding map. Consider the following example.

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Example 109 Let D be the set of positive integers and, for each i, Xi = (0, 1), and fi = f where f : (0, 1) → (0, 1) is given by f (x) = x/2 for each x ∈ (0, 1). One can see that lim f = ∅ for if n is a positive integer no point ←− of lim f can have a point with ﬁrst coordinate greater than 1/2n−1 because in ←− that case its nth coordinate would have to be greater than 1. Recall that we use Π to denote α∈D Xα . It is convenient to introduce the following notation: if η is an element of D, Gη denotes the set of all points x of Π such that if α and β are elements of D and α β η then xα ∈ fα β (xβ ). The reader should note that this notation diﬀers slightly from the notation employed in Chapter 1 for D = N. Theorem 110 Suppose {Xα , fα β , D} is a consistent inverse limit system such that Xα is a compact Hausdorﬀ space for each α in D. Then, for each η ∈ D, Gη is a nonempty compact set. Proof. The system is consistent, therefore Gη is nonempty for each η ∈ D. Because Π is compact it suﬃces to show that Gη is closed. Let x be a point of Π that is not in Gη . There exist α and β in D with α β η such that xα is not in fα β (xβ ). By Theorem 105, the graph of fα β is closed, so there is an open set Uβ × Uα ⊆ Xβ × Xα such that (Uβ × Uα ) ∩ G(fα β ) = ∅. Let O = Uβ × Uα × ( γ∈D−{α,β} Xγ ). Then O contains x and if y ∈ O then / fα β (yβ ); that is, y ∈ / Gη . So Gη is closed and therefore compact. yα ∈ As an immediate consequence of Theorem 110 we have the result we sought in the following theorem. Theorem 111 Suppose {Xα , fα β , D} is a consistent inverse limit system such that Xα is a compact Hausdorﬀ space for each α in D. Then, K = lim f ←− is nonempty and compact. Proof. The collection of all the sets Gη for η ∈ D is a collection of nonempty compact sets in the compact Hausdorﬀ space Π. Note that if α and β are in D there is a member γ of D such that α γ and β γ. Because Gγ ⊆ Gα and Gγ ⊆ Gβ , it follows that the collection {Gη | η ∈ D} has the ﬁnite intersection Gη is nonempty and compact. Clearly property. From this we see that η∈D K = η∈D Gη . If f : X → 2Y is a set-valued function and A is a subset of X, recall that f (A) = {y ∈ Y | there exists x ∈ A such that y ∈ f (x)}. If f : X → 2Y , we call f surjective if f (X) = Y . This is consistent with the usual deﬁnition of surjective for mappings. In the case where D is the set of positive integers and there is a positive integer n such that, for each i ≥ n, fi is surjective then for each positive integer j ≥ n and for each point t of Xn there is a point

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x in the inverse limit with t ∈ πn (x). Thus, in the case where D is the set of positive integers, one does not need Theorem 111 to see that the inverse limit is nonempty. Indeed, we have the following theorem the proof of which is left to the reader. Theorem 112 Suppose X1 , X2 , X3 , . . . is a sequence of compact Hausdorﬀ spaces and fk : Xk+1 → 2Xk is an upper semi-continuous function for each positive integer k. If there is a positive integer n such that fk is surjective for each positive integer k ≥ n, i and j are positive integers not less than n with i < j, s is a point of Xj , and t ∈ fi j (s), then there is a point x of lim f such ←− that xi = t and xj = s. In the absence of compactness of the factor spaces or assumptions about the directed set such as D = {1, 2, 3, . . . }, even with surjective bonding functions, the inverse limit of a consistent system may be empty. An example of an inverse limit system with surjective bonding maps that has an empty inverse limit is given by Henkin in [3]. We conclude this section with a proof that the inverse limit of a consistent inverse limit system on compact Hausdorﬀ spaces is the inverse limit of an inverse limit system with surjective bonding functions. Theorem 113 Suppose {Xα , fα β , D} is a consistent inverse limit system such that Xα is a compact Hausdorﬀ space for each α in D. Then, lim f is ←− the inverse limit of an inverse limit system {Yα , gα β , D} where, for each α in D, Yα = αβ fα β (Xβ ), and if α β in D, gα β = fα β |Yβ is surjective. Proof. Choose α ∈ D. Observe that if α β γ, then fα γ = fα β ◦ fβ γ so fα γ (Xγ ) ⊆ fα β (fβ γ (Xγ )) ⊆ fα β (Xβ ). From this and the fact that D is a directed set it follows that the collection C = {fα β (Xβ ) | α β} has the ﬁnite intersection property. Because C is a collection of nonempty compact sets, Yα = αβ fα β (Xβ ) is a nonempty compact set. We now show that if α β then fα β (Yβ ) ⊆ Yα . Let x be an element of Yβ and suppose γ is an element of D such that α γ. There is an element η of D such that β η and γ η. Because x ∈ fβ η (Xη ), fα β (x) ⊆ fα β (fβ η (Xη )) = fα η (Xη ) ⊆ fα γ (Xγ ). Thus, fα β (x) ⊆ Yα . Next we show that if α β and y is in Yα then there is an element x of Yβ such that y ∈ fα β (x). Suppose α β and let y be an element of Yα . If γ is an element of D such that β γ, because y belongs to fα γ (Xγ ) and fα γ = fα β ◦ fβ γ , there is a point t of Xβ such that t ∈ fβ γ (Xγ ) and y ∈ fα β (t). For each γ ∈ D such that β γ, let Nγ = {t ∈ fβ γ (Xγ ) | y ∈ fα β (t)} and let Mγ = Nγ . We now show that if γ1 γ2 then Nγ2 ⊆ Nγ1 . To see this note that if t ∈ Nγ2 then t ∈ fβ γ1 (Xγ1 ) which follows from the fact that t ∈ fβ γ2 (Xγ2 ) and the deﬁnition of fβ γ1 ◦ fγ1 γ2 . So the collection of compact sets Mγ has

2.6 Connected inverse limits

83

the ﬁnite intersection property. Thus, βγ Mγ is a nonempty subset of Xβ . Let x be in βγ Mγ . To see that x is in Yβ , let U be an open set containing x and suppose β γ. Because x is a point of Mγ = N γ , there is a point t of Nγ in U . Because t is in Nγ , t ∈ fβ γ (Xγ ). It follows that x ∈ fβ γ (Xγ ). But, fβ γ (Xγ ) is closed so x ∈ fβ γ (Xγ ). Because x ∈ fβ γ (Xγ ) for each γ such that β γ, x ∈ Yβ . Finally, y ∈ fα β (x) for suppose y ∈ / fα β (x). There exist mutually exclusive open sets Oy and O containing y and fα β (x), respectively. There is an open set U containing x such that if z ∈ U then fα β (z) ⊆ O. If γ ∈ D and β γ, U contains a point t of Nγ so fα β (t) ⊆ O, a contradiction to the fact that y ∈ fα β (t). For α and β in D with α β, let gα β = fα β |Yβ . It is clear that lim g ⊆ ←− lim f . If x ∈ lim f and α ∈ D, xα ∈ fα β (xβ ) for each β such that α β. ←− ←− Thus, xα belongs to Yα for each α, and so x ∈ lim g. ←−

2.6 Connected inverse limits

We next turn our attention to conditions under which inverse limits are connected. One might suspect that imposing a natural condition such as all of the bonding functions have connected graphs would be suﬃcient to guarantee that the inverse limit is connected. That this condition is not suﬃcient even if D is the set of positive integers, each factor space is the interval [0, 1], and the sequence of bonding functions is constant may be seen from the following example. The reader will recall that Theorem 105 allows us to specify an upper semi-continuous closed set-valued function by identifying its graph as a closed subset of a product of two spaces that projects onto the ﬁrst factor space. In the examples of this chapter, I denotes the interval [0, 1] and Q denotes the Hilbert cube I ∞ . Example 114 (A function with a connected graph that yields an inverse limit that is not connected) Let G(f ) be the union of the four straight line intervals, I × {0}, {1} × I, the interval from (0, 0) to (1/4, 1/4), and the interval from (3/4, 1/4) to (1, 1) in I ×I (see Figure 2.1). Then, G(f ) is connected but lim f is not connected. ←− Proof. Let N be the set of all points p of K = lim f such that p1 = p2 = 1/4 ←− and p3 = 3/4. Note that N is closed. Let x be a point of N . Let R = R1 × R2 × R3 × Q be the region in Q where R1 = R2 = (1/8, 3/8) and R3 = (5/8, 7/8), and note that R contains x. Assume that the point y is in R ∩ K. Then y1 and y2 are in (1/8, 3/8). It follows that y2 ≤ 1/4. But if y2 < 1/4, inasmuch as y3 > 5/8 we have y3 = 1 and y is not in R. We

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conclude that y ∈ N , so N and K − N are closed and mutually exclusive, thus K is not connected.

(1,1)

(1/4,1/4) (3/4,1/4)

(0,0)

(1,0)

Fig. 2.1 The function from Example 114

Another natural condition one could impose with the expectation of obtaining a connected inverse limit would be that the bonding functions have connected values. However, even for inverse systems over directed sets in which every factor space is the interval [0, 1] this may not be the case as may be seen from the following example. Example 115 Let D be the directed set given in Example 106; that is, D = {1, 2, 3, . . . } ∪ {a, b} where, if i and j are positive integers, then i j if and only if i ≤ j, a j if and only if j ≥ 2, b j if and only if j ≥ 3, 1 b, and a b (1 and a do not compare nor do 2 and b). Let Xα = [0, 1] for each α ∈ D. Let fa 2 = f1 2 = fa b = Id. Let f1 b be the full tent map; that is, f1 b (t) = 2t for 0 ≤ t ≤ 1/2 and f1 b (t) = 2 − 2t for 1/2 < t ≤ 1. Let f2 3 be given by f2 3 (t) = [0, 1] for each t ∈ [0, 1] and fb 3 = f2 3 . Finally, let fi i+1 = Id for i ≥ 3. The inverse limit of the consistent system {Xα , fα β , D} is the union of two mutually exclusive arcs.

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85

Proof. We ﬁrst note that if x ∈ lim f and x1 = p then x2 = p. Thus, ←− xa = p and, as a consequence, xb = p. It follows that f1 b (p) = p. There are two ﬁxed points for f1 b so if x ∈ lim f , then x1 = x2 = xa = xb = 0 or ←− x1 = x2 = xa = xb = 2/3. In either case, x3 may be any element of [0, 1], but xi = x3 for all i ≥ 3. It may now be seen that the two arcs whose union is lim f are A = {x ∈ lim f | x1 = x2 = xa = xb = 0, x3 = t, xi = x3 for i ≥ ←− ←− 3 where t ∈ [0, 1]} and B = {x ∈ lim f | x1 = x2 = xa = xb = 2/3, x3 = ←− t, xi = x3 for i ≥ 3 where t ∈ [0, 1]}. Recall our notation from Section 2.5 that if η ∈ D then Gη = {x ∈ Π | if α β η then xα ∈ fα β (xβ )}. By a Hausdorﬀ continuum we mean a compact connected subset of a Hausdorﬀ space. This distinction allows us to continue to use the term continuum to mean a compact connected subset of a metric space. Our next theorem provides a suﬃcient condition that an inverse limit system in which the factor spaces are compact Hausdorﬀ spaces produces a connected inverse limit. Theorem 116 Suppose that {Xα , fα β , D} is a consistent inverse limit system such that Xα is a compact Hausdorﬀ space for each α in D. If, for each η in D, Gη is connected, then lim f is a Hausdorﬀ continuum. ←− Proof. By Theorem 110, for each η ∈ D, Gη is a nonempty compact set. This theorem is thus an immediate consequence of the observation that the collection {Gη | η ∈ D} of compact and connected subsets of the compact Hausdorﬀ space Π has the ﬁnite intersection property. We now turn to two important cases in which Theorem 116 applies. These are the case where all the bonding functions in the system are mappings and the case where the directed set is totally ordered.

2.6.1 Systems in which all of the bonding functions are mappings Theorem 117 Suppose that {Xα , fα β , D} is an inverse limit system such that Xα is a Hausdorﬀ continuum for each α ∈ D and each fα β is a mapping. Then lim f is a Hausdorﬀ continuum. ←− Proof. Suppose η ∈ D and let E = {η} ∪ {γ ∈ D | γ η}. Let ϕ : xα for each α ∈ E α∈E Xα → Gη be given by ϕ(x) = y where yα = and yα = fα η (xη ) otherwise. Suppose x and z are in α∈E Xα . If x = z there is an element γ ∈ E such that xγ = zγ . Then, ϕ(x) = ϕ(z), so ϕ is 1-1. We now show that ϕ is surjective and continuous. If y ∈ Gη , let x be the member of α∈E Xα such that xα = yα for each α ∈ E. Then, ϕ(x) = y,

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so ϕ is surjective. Finally, ϕ is continuous because, for each α ∈ D, πα ◦ ϕ is continuous. From the compactness of Π it follows that ϕ is a surjective homeomorphism, so Gη is connected being homeomorphic to the connected set α∈E Xα . The theorem now follows from Theorem 116.

2.6.2 Systems in which the directed set is totally ordered Next we consider systems in which the underlying directed set is totally ordered although some of the theorems of this subsection do not depend on this assumption. Theorem 118 Suppose that each of X and Y is a compact Hausdorﬀ space, X is connected, f is an upper semi-continuous function from X into 2Y and, for each x in X, f (x) is a Hausdorﬀ continuum. Then G(f ) is a Hausdorﬀ continuum. Proof. Recall that G(f ) = {(x, y) ∈ X × Y | y ∈ f (x)}. Note that G(f ) is closed by Theorem 105. Assume that G(f ) is not connected. There are then two nonempty mutually exclusive closed sets H and K whose union is G(f ). If x is in X, then {x} × f (x) is a connected subset of G(f ) and thus a subset of one of H and K. Let H1 be the set of all points x of X such that {x}×f (x) lies in H and let K1 be the points x of X such that {x} × f (x) lies in K. Because H1 and K1 are nonempty compact sets whose union is the connected set X, they have a common point z. But this is impossible because {z} × f (z) would then be a connected subset of both H and K. If M is a subset of the product X × Y of compact Hausdorﬀ spaces, then the inverse of M is the subset of Y × X consisting of all points (y, x) such that (x, y) is in M . We denote this inverse by M −1 . If f is an upper semicontinuous function, by the inverse of f , denoted f −1 , we mean the function from f (X) ⊆ Y into 2X such that if y ∈ f (X), f −1 (y) = {x ∈ X | y ∈ f (x)}. One of the consequences of Theorem 105 is that if f : X → Y is an upper semi-continuous function, then f −1 : f (X) → 2X is upper semi-continuous. We often use this observation in this section. Lemma 119 Suppose X and Y are compact Hausdorﬀ spaces, f : X → 2Y is an upper semi-continuous function, and M = G(f −1 ). Then, M −1 = G(f ). Proof. Note that (x, y) ∈ M −1 if and only if (y, x) ∈ M = G(f −1 ) if and only if x ∈ f −1 (y) if and only if y ∈ f (x) if and only if (x, y) ∈ G(f ). Using Lemma 119, the following theorem is an immediate consequence of Theorem 105.

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87

Theorem 120 If X and Y are compact Hausdorﬀ spaces and f : X → 2Y is an upper semi-continuous function then f −1 : f (X) → 2X is an upper semi-continuous function. Proof. Let M = G(f −1 ). By Lemma 119, M −1 = G(f ) which is closed by Theorem 105. Inasmuch as M −1 is closed, M is closed so f −1 is upper semicontinuous. Theorem 120 yields a corollary to Theorem 118 for certain upper semicontinuous functions whose inverses are Hausdorﬀ continuum-valued. Theorem 121 Suppose that X and Y are compact Hausdorﬀ spaces and f : X → 2Y is an upper semi-continuous function such that f (X) is connected and f −1 : f (X) → 2X is Hausdorﬀ continuum-valued. Then G(f ) is connected. Proof. Let M = G(f −1 ). By Theorem 120, f −1 is upper semi-continuous. By hypothesis, f −1 is Hausdorﬀ continuum-valued, so by Theorem 118, M is connected. Because M is connected, M −1 is connected. By Lemma 119, M −1 = G(f ), so G(f ) is connected. For the next few theorems it is convenient to generalize the deﬁnition of the graph G(f ) of an upper semi-continuous function. Suppose {Xα , fα β , D} is an inverse limit system with upper semi-continuous bonding functions and Xα is a compact Hausdorﬀ space for each α in D. Suppose {β1 , β2 , . . . , βn } with n ≥ 2 is a ﬁnite subset of D. Let G(β1 , β2 , . . . , βn ) = {x ∈ Xβ1 × Xβ2 × · · · Xβn | xβi ∈ fβi βj (xβj ) whenever βi βj }. Note that if β1 β2 then G(β1 , β2 )−1 is the graph of fβ1 β2 . This slight twist in the notation is convenient for our intended use of these sets in the proofs of theorems leading to Theorem 125. Theorem 122 Suppose {Xα , fα β , D} is a consistent inverse limit system such that, for each α ∈ D, Xα is a compact Hausdorﬀ space, n ≥ 2, and B = {β1 , β2 , . . . , βn } is a ﬁnite subset of D. Then G(β1 , β2 , . . . , βn ) is nonempty and compact. Proof. B is a ﬁnite subset of a directed set D, thus there is an element η of D such that βi η for i = 1, 2, 3, . . . , n. The system is consistent, therefore if t ∈ Xη , there is an element x of Π such that xη = t and if α β η then xα ∈ fα β (xβ ). Then, (xβ1 , xβ2 , . . . , xβn ) ∈ G(β1 , β2 , . . . , βn ) so this set is nonempty. Xβ1 × Xβ2 × · · · × Xβn is compact, therefore it is suﬃcient to show that G(β1 , β2 , . . . , βn ) is closed. If (xβ1 , xβ2 , . . . , xβn ) ∈ / G(β1 , β2 , . . . , βn ), there / fβi βj (xβj ). Because the exist integers i and j such that βi βj and xβi ∈

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graph of fβi βj is closed, there are open sets Uβj and Uβi containing xβj and xβi , respectively, such that Uβj × Uβi does not intersect the graph of fβi βj . Then, O = O1 × O2 × · · · × On , where Oi = Uβi , Oj = Uβj , and Ok = Xβk if k ∈ {1, 2, . . . , n} − {i, j}, is an open set containing (xβ1 , xβ2 , . . . , xβn ) and no point of G(β1 , β2 , . . . , βn ). Theorem 123 Suppose {Xα , fα β , D} is a consistent inverse limit system such that, for each α ∈ D, Xα is a Hausdorﬀ continuum. Further suppose n ≥ 2 and B = {β1 , β2 , . . . , βn } is a ﬁnite subset of D such that βi βj if and only if i ≤ j and if βi βj then fβi βj is Hausdorﬀ continuumvalued or fβi βj (Xβj ) is connected with fβ−1 : fβi βj (Xβj ) → 2Xβj Hausdorﬀ i βj continuum-valued. Then G(β1 , β2 , . . . , βn ) is a Hausdorﬀ continuum. Proof. By Theorem 122, G(β1 , β2 , . . . , βn ) is nonempty and compact so we only need to show that G(β1 , β2 , . . . , βn ) is connected. We proceed by induction on the number of elements in B. If there are only two elements of B, β1 and β2 with β1 β2 , G(β1 , β2 ) = G(fβ−1 ). The graph of fβ1 β2 is connected 1 β2 if and only if the graph of its inverse is connected. So it follows that G(β1 , β2 ) is connected by Theorem 118 if fβ−1 is Hausdorﬀ continuum-valued and by 1 β2 Theorem 121 if fβ1 β2 is Hausdorﬀ continuum-valued. Suppose the conclusion holds for any subset of D with n elements and let β1 , β2 , . . . , βn+1 be n + 1 elements of D such that βi βj if and only if i ≤ j. The proof reduces to two cases. Case (1): fβ1 β2 is Hausdorﬀ continuum-valued. By the inductive hypothesis, G(β2 , . . . , βn+1 ) is connected. Suppose H and K are closed sets whose union is G(β1 , . . . , βn+1 ) and let h : G(β1 , . . . , βn+1 ) → G(β2 , . . . , βn+1 ) be the mapping deﬁned by h((xβ1 , xβ2 , . . . , xβn+1 ))= (xβ2 , . . . , xβn+1 ). We shall show that h(H ∪ K) = G(β2 , . . . , βn+1 ). Suppose (xβ2 , xβ3 , . . . , xβn+1 ) is in G(β2 , . . . , βn+1 ) and let xβ1 be an element of fβ1 β2 (xβ2 ). If 2 ≤ i ≤ n + 1, xβ2 ∈ fβ2 βi (xβi ), so xβ1 ∈ fβ1 β2 (fβ2 βi (xβi )) = fβ1 βi (xβi ). Thus (xβ1 , xβ2 , . . . , xβn+1 ) ∈ G(β1 , β2 , . . . , βn+1 ) and h((xβ1 , . . . , xβn+1 )) = (xβ2 , . . . , xβn+1 ). Because h(H) and h(K) are closed and h(H) ∪ h(K) is connected, there is a point p belonging to h(H) and h(K). Then, C = {x ∈ G(β1 , . . . , βn+1 ) | xβ1 ∈ fβ1 β2 (pβ2 ), xβi = pβi for 2 ≤ i ≤ n + 1} is connected because fβ1 β2 (pβ2 ) is connected. There exist elements t of H and s of K such that h(t) = h(s) = p. Because tβ1 ∈ fβ1 β2 (pβ2 ) and tβk = pβk for 2 ≤ k ≤ n + 1, t is in C. Similarly, s is in C. Because C is a connected subset of H ∪ K intersecting both H and K, there is a point common to H and K. Consequently, H and K are not mutually separated. Case (2): fβ−1 is Hausdorﬀ continuum-valued. By the inductive hypoth1 β2 esis G(β1 , β3 , . . . , βn+1 ) is connected. Suppose H and K are closed sets whose union is G(β1 , . . . , βn+1 ) and let h : G(β1 , . . . , βn+1 ) → G(β1 , β3 , . . . , βn+1 ) be the mapping deﬁned by h((xβ1 , xβ2 , . . . , xβn+1 )) = (xβ1 , xβ3 , . . . , xβn+1 ). Because xβ1 ∈ fβ1 β3 (xβ3 ) there is an element xβ2 of Xβ2 such that

2.6 Connected inverse limits

89

xβ1 ∈ fβ1 β2 (xβ2 ) and xβ2 ∈ fβ2 β3 (xβ3 ). If 3 ≤ i ≤ n + 1, because xβ3 ∈ fβ3 βi (xβi ), we see that xβ2 ∈ fβ2 βi (xβi ) so (xβ1 , xβ2 , . . . , xβn+1 ) ∈ G(β1 , β2 , . . . , βn+1 ) and h((xβ1 , xβ2 , . . . , xβn+1 )) = (xβ1 , xβ3 , . . . , xβn+1 ). It follows that h(H ∪ K) = G(β1 , β3 , . . . , βn+1 ). Thus, there is a point p belonging to h(H) and h(K). Then, C = {x ∈ G(β1 , . . . , βn+1 ) | xβ2 ∈ fβ−1 (pβ1 ), xβi = 1 β2 (p ) is connected. pβi for 1 ≤ i ≤ n+1, i = 2} is connected inasmuch as fβ−1 β1 1 β2 There exist elements t of H and s of K such that h(t) = h(s) = p. Note that tβ1 ∈ fβ1 β2 (tβ2 ), so tβ2 ∈ fβ−1 (pβ1 ). Because tβk = pβk for k = 1, 3, . . . , n+1, 1 β2 t is in C. Similarly, s is in C. Because C is a connected subset of H ∪ K intersecting both H and K, there is a point common to H and K. Consequently, H and K are not mutually separated. Theorem 124 Suppose {Xα , fα β , D} is a consistent inverse limit system such that, for each α ∈ D, Xα is a Hausdorﬀ continuum and D is totally ordered. If, for each α and β in D such that α β, fα β is Hausdorﬀ continuum-valued or fα β (Xβ ) is connected with fα−1β : fα β (Xβ ) → 2Xβ Hausdorﬀ continuum-valued, then Gη is a Hausdorﬀ continuum for each η ∈ D. Proof. Suppose η ∈ D. If {β1 , β2 , . . . , βn } is a ﬁnite subset of D, because D is totally ordered, we may assume that βi βj if and only if i ≤ j. Let G (β1 , β2 , . . . , βn ) = {x ∈ Π | xβi ∈ fβi βj (xβj ) whenever i ≤ j}. Let D = D − {β1 , β2 , . . . , βn }. By Theorem 123, G(β1 , . . . , βn ) is connected. Therefore G (β1 , . . . , βn ), which is homeomorphic to G(β1 , . . . , βn ) × α ∈D Xα , is connected. Let G = {G (β1 , β2 , . . . , βn ) | {β1 , . . . , βn } is a ﬁnite subset of D such that βi η for 1 ≤ i ≤ n} and note that G is a collection of Hausdorﬀ continua by Theorem 123. If G (β1 , . . . , βn ) and G (γ1 , . . . , γm ) are in G, then G (ζ1 , . . . , ζk ) ⊆ G (β1 , . . . , βn ) ∩ G (γ1 , . . . , γm ) where {ζ1 , . . . , ζk } = {β1 , ldots, βn )} ∪ {γ1 , . . . , γm }. Consequently, G has the ﬁnite intersection property, so the common part C of all the elements of G is a Hausdorﬀ continuum. We now show that C = Gη . If x ∈ Gη and G (β1 , β2 , . . . , βn ) ∈ G then x ∈ G (β1 , β2 , . . . , βn ) so Gη ⊆ C. If x ∈ Gη then there exist α and β in D such that α β η and xα ∈ fα β (xβ ). Then, x ∈ G(α, β) so x ∈ C and we have that C ⊆ Gη . As a consequence of the preceding theorems, in the next theorem we have the result we sought in this subsection. Theorem 125 Suppose {Xα , fα β , D} is a consistent inverse limit system such that, for each α ∈ D, Xα is a Hausdorﬀ continuum and D is totally ordered. If, for each α and β in D such that α β, fα β is Hausdorﬀ continuum-valued or fα β (Xβ ) is connected with fα−1β : fα β (Xβ ) → 2Xβ Hausdorﬀ continuum-valued, then lim f is a Hausdorﬀ continuum. ←− Proof. This is immediate from Theorems 124 and 116.

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Because of its importance, we state the following special case of Theorem 125. Recall that inverse limit sequences are consistent (Theorem 108). Theorem 126 Suppose {Xi , fi } is an inverse limit sequence on Hausdorﬀ continua with upper semi-continuous bonding functions such that, if i is a positive integer, then fi is Hausdorﬀ continuum-valued or fi (Xi+1 ) is connected with fi−1 : f (Xi+1 ) → 2Xi+1 Hausdorﬀ continuum-valued. Then lim f ←− is a Hausdorﬀ continuum. If f : X → X is a mapping of a Hausdorﬀ continuum into itself, then f (X) is connected and f −1 : f (X) → 2X is an upper semi-continuous function whose inverse is continuum valued (in fact each value is degenerate). Thus, we have the following corollary to Theorem 126. Corollary 127 Suppose {X, f −1) is an inverse limit sequence where X is a Hausdorﬀ continuum and f is a mapping from X into X. Then lim f −1 is ←− a Hausdorﬀ continuum homeomorphic to f (X). That lim f −1 is a Hausdorﬀ continuum follows directly from Theorems 121 ←− and 126. Moreover, it is not diﬃcult to show that lim f −1 is homeomorphic ←− to the Hausdorﬀ continuum f (X) because h : f (X) → lim f −1 given by ←− h(x) = (x, f (x), f 2 (x), . . . ) is a homeomorphism. Theorem 125 gives a suﬃcient condition that inverse limits on Hausdorﬀ continua with upper semi-continuous bonding functions be Hausdorﬀ continua. However, in general the bonding functions do not have to satisfy the conditions of Theorem 125 in order to produce a connected inverse limit. For instance, see Example 132. Theorem 156 of Section 2.9 provides a diﬀerent suﬃcient condition for an inverse limit with upper semi-continuous bonding functions on [0, 1] to be connected. Deciding in general when an inverse limit system with upper semi-continuous bonding functions produces a connected inverse limit remains an interesting problem.

2.7 Examples in the special case that each factor space is [0, 1]

In this section we assume that the directed set is the set of positive integers and each factor space is [0, 1]. We give some examples of some upper semicontinuous bonding functions that illustrate additional aspects of the nature of inverse limits with upper semi-continuous bonding functions. Each example

2.7 Examples in the special case that each factor space is [0, 1]

91

is produced by a constant sequence of bonding functions. In most of these examples we specify the graph of the bonding function rather than a formula for the function. We denote the Hilbert cube [0, 1]∞ by Q. A point p of a Hausdorﬀ continuum M is a separating point of M provided M − {p} is not connected. Separating points are also called cut points. An arc is a continuum with only two nonseparating points. Our ﬁrst example shows that inverse limits with upper semi-continuous bonding functions can produce the Hilbert cube. Example 128 (The Hilbert cube) Let f : [0, 1] → 2[0,1] be given by f (x) = [0, 1] for each x in [0, 1]; that is, G(f ) = [0, 1] × [0, 1]. Then lim f is ←− the Hilbert cube. In our next example the inverse limit is a closed set in the Hilbert cube Q consisting of a convergent sequence of points. Example 129 (A convergent sequence) Let G(f ) be the union of the point {(1, 1)} and [0, 1] × {0} (see Figure 2.2). Proof. Let K = lim f . No point of K has a coordinate between 0 and 1 ←− inasmuch as the range of f is {0, 1}. If p is in K and pi = 1 then pi+1 = 1 whereas if pi = 0 then pi+1 is 0 or 1. Thus, the points of K are (0, 0, 0, . . .) and the set of all points p1 , p2 , p3 , . . . where, for each positive integer n, πi (pn ) is 1 if i ≥ n and 0 if i < n. A Cantor set can be the result of an inverse limit on [0, 1] with an upper semi-continuous bonding function. Example 130 (A Cantor set) Let G(f ) be the union of [0, 1] × {0} and [0, 1] × {1} (see Figure 2.3). Proof. Note that if x is in lim f and i is a positive integer xi cannot be strictly ←− between 0 and 1. Moreover, if xi ∈ {0, 1} then xi+1 ∈ {0, 1}. It follows that lim f is a Cantor set. ←− Example 114 showed that the inverse limit may fail to be connected even if the graph of the bonding function is connected. Theorem 125 provided a suﬃcient condition that the inverse limit be a Hausdorﬀ continuum and one of its consequences is that the inverse limit in the next example is a continuum. As we show in the remaining examples in this section, quite a variety of continua can be obtained as an inverse limit with upper semi-continuous bonding functions that generally do not satisfy the conditions from Theorem 125. The inverse limit in our next example is a fan (i.e., a continuum that is the union of a collection of arcs such that the intersection of each two of them is a point v, called the vertex, that is an endpoint of each of the arcs in the collection).

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(0,0)

(1,0)

Fig. 2.2 The function from Example 129

Example 131 (A simple fan) Let G(f ) be the union of the graph of the identity function and the interval I × {0} (see Figures 2.4 and 2.5). Proof. Inasmuch as f −1 is continuum-valued, lim f is connected by Theorem ←− 126. Let v = (0, 0, 0, . . . ). Note that if p ∈ lim f and for some positive integer ←− n, pn > 0, then pj = pn for all j > n. For each positive integer n, let Kn be the set of all points p ∈ lim f such that pj = 0 for j < n and pj = pn > 0 for ←− j ≥ n. The closure of Kn is an arc of length 1/2n−1 having oneendpoint v. Moreover no two of these arcs intersect except at v and lim f = i>0 Ki . ←− In the following example we see that a simple upper semi-continuous function can produce a Cantor fan. The reader will also note that the inverse limit is a continuum even though the bonding function does not satisfy the hypothesis of Theorem 125. Although here we show directly that the inverse limit in this example is a continuum, this fact is also a consequence of Theorem 156 of Section 2.9. Example 132 (The Cantor fan) Let G(f ) be the union of the identity, Id, and the map 1 − Id on [0, 1] (see Figures 2.6 and 2.7).

2.7 Examples in the special case that each factor space is [0, 1]

93

(0,1)

(1,1)

(0,0)

(1,0)

Fig. 2.3 The function from Example 130

Proof. Let K = lim f . The vertex v of K is the point (1/2, 1/2, 1/2, . . .) and ←− the Cantor set C at the base of the fan is lim{{0, 1}, f |{0, 1}}. If c is a point ←− of C, the arc joining v and c is the inverse limit of the inverse sequence {Ji , gi } where, for each i, Ji is the interval joining ci and 1/2 and gi is a homeomorphism that ﬁxes 1/2 and whose graph is a subset of the graph of f . (It could be of help for the reader to observe that the graph of f is the union of four intervals having only the point (1/2, 1/2) in common. For each i, the graph of gi is one of these four intervals.) Next we consider three examples produced by similar upper semi-continuous functions. Each graph consists of the union of the horizontal line I × {0} and a vertical line. We produce the examples by choosing the vertical line that intersects this horizontal line in three places: (1, 0), (0, 0), and (1/2, 0). The resulting inverse limits are, respectively, an arc (i.e., a continuum with only two nonseparating points), inﬁnite-dimensional, and an arc with a sequence of stickers. Example 133 (An arc) Let G(f ) be the union of [0, 1]×{0} and {1}×[0, 1] (see Figure 2.8). Proof. Let M = lim f . We show that if x ∈ M − {(0, 0, 0, . . . ), (1, 1, 1, . . . )} ←− then x is a separating point of M thus showing that M is an arc. If x ∈

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2 Inverse Limits in a General Setting (1,1)

(1,0)

(0,0)

Fig. 2.4 The function from Example 131

K2 K3

.

.

. v

K1

Fig. 2.5 A depiction of the inverse limit from Example 131

M − {(0, 0, 0, . . . ), (1, 1, 1, . . . )} then there is a positive integer k such that xk = 0 but xk−1 = 0 if k > 1. There are two cases: xk = 1 and xk < 1. Suppose xk = 1. Then, k ≥ 2 so xk−1 = 0. Let A = {y ∈ M | yk−1 ∈ (0, 1]} and B = {y ∈ M | yk ∈ [0, 1)}. Then A and B are open in M . If z ∈ M − {x} and zk = 1 then z ∈ B whereas if zk = 1 then zk−1 = 0 so z ∈ A. Thus, M − {x} = A ∪ B. If z ∈ B, then zk ∈ [0, 1) and zk−1 = 0. Thus, z ∈ / A so A ∩ B = ∅. Therefore, x is a separating point of M .

2.7 Examples in the special case that each factor space is [0, 1] (0,1)

95 (1,1)

(1/2,1/2)

(1,0)

(0,0)

Fig. 2.6 The function from Example 132 v

...

...

...

...

Fig. 2.7 A depiction of the inverse limit from Example 132

Suppose xk < 1. Let A = {y ∈ M | yk < xk } and B = {y ∈ M | yk > xk } and note that A and B are mutually exclusive open sets in M . If y ∈ M and yk = xk it follows that y = x so M − {x} = A ∪ B and, again, we have that x is a separating point of M . In either case, we have shown M has only two nonseparating points so M is an arc.

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(0,0)

(1,0)

Fig. 2.8 The function from Example 133

Let f be the function from Example 133 and M = lim f . The following ←− is another way to look at M . Let A1 = {x ∈ M | x2 = 1}. Note that if x ∈ A1 then xk = 1 for k ≥ 2. For each positive integer i > 1, let Ai = {x ∈ lim f | xk = 0 for k i}. Then, lim f = A1 ∪ A2 ∪ ←− ←− A3 ∪ · · · ∪ {(0, 0, 0, . . . )}. To see that A is an arc observe that Ai ∩ Aj = ∅ if |i − j| > 1 and Ai ∩ Ai+1 = {Pi } where the ﬁrst i coordinates of Pi are 0 and all of the remaining coordinates are 1. This arc plays a crucial role later in analyzing Examples 136 and 137. Example 134 (An inﬁnite-dimensional continuum) Let G(f ) be the union of [0, 1] × {0} and {0} × [0, 1] (see Figure 2.9). Let M = lim f . Observe that M contains [0, 1] × {0} × [0, 1] × {0} × ←− [0, 1] × {0} × . . . and this set is homeomorphic to the Hilbert cube, so M is inﬁnite-dimensional. Example 135 (An arc with stickers) Let G(f ) be the union of [0, 1]×{0} and {1/2} × [0, 1] (see Figure 2.10). Proof. Let M = lim f . For each positive integer n, let An = {x ∈ M | xn ∈ ←− [0, 1], xj = 1/2 for j > n and if n > 1 then xi = 0 for i < n}. Note that for

2.7 Examples in the special case that each factor space is [0, 1]

97

(0,1)

(0,0)

(1,0)

Fig. 2.9 The function from Example 134

each positive integer i, Ai+1 ∩ Ai = {(0, 0, . . . , 0, 1/2, 1/2, 1/2, . . .)} where the last 0 occurs in the ith coordinate. Moreover, Ai ∩ Aj = ∅ if |i − j| > 1. Finally, (0, 0, 0, . . . ) is a limit point of A1 ∪A2 ∪A3 ∪· · · = M − {(0, 0, 0, . . . )}. For a picture of M see Figure 2.11.

Example 136 (An arc with a fan and spines) Let G(f ) be the union of [0, 1] × {0} and {1} × [0, 1] together with the portion of the identity above [0, 1/4] (see Figures 2.12 and 2.13).

Let M = lim f . Let A = A1 ∪ A2 ∪ A3 ∪ · · · ∪ {(0, 0, 0, . . . )} where A1 = ←− {x ∈ M | x2 = 1} and if i is an integer greater than one then Ai = {x ∈ lim f | xk = 0 for k i}. Then, A is an arc lying in M . ←− The arc A is the inverse limit in Example 133. Let B1 = {x ∈ M | xj ∈ [0, 1/4] for all j} and, for i > 1, let Bi = {x ∈ M | xj = 0 for j < i, xi ∈ [0, 1/4] and xj = xi for j ≥ i}. Let B = B1 ∪ B2 ∪ B3 ∪ · · · ∪ {(0, 0, 0, . . . )} and note that B is a fan lying in M with vertex (0, 0, 0, . . . ). Suppose each of i and j is a positive integer. Let C1 j = {x ∈ M | xk = 1 for k ≥ j + 1 and xk ∈ [0, 1/4] for 1 ≤ k ≤ j} and, for i > 1, let Ci j = {x ∈ M | xk = 1 for k ≥ i + j, xk = 0

98

2 Inverse Limits in a General Setting (1/2,1)

(1/2,1/2)

(0,0)

(1/2,0)

(1,0)

Fig. 2.10 The function from Example 135 (1,1/2,1/2, . . .)

(0,1,1/2, . . .) (0,0,1,1/2, . . .) (0,0,0, . . .) (0,0,1/2, . . .) (0,1/2,1/2, . . .)

Fig. 2.11 A depiction of the inverse limit from Example 135

(1/2,1/2,1/2, . . .)

2.7 Examples in the special case that each factor space is [0, 1]

99 (1,1)

(1/4, 1/4)

(1,0)

(0,0)

Fig. 2.12 The function from Example 136 P0

B1 B2 B3

.. .

(0,0,0,...)

C1 4 A1 P4

A4

C2 2

C1 3

C3 1 P3

A3

C2 1

P2

C1 2

A2

P1

Fig. 2.13 A depiction of the inverse limit from Example 136

100

2 Inverse Limits in a General Setting

for k < i, and xk ∈ [0, 1/4] for i ≤ k < i + j}. Note that if x is a point of M not in A ∪ B then x ∈ Ci j for some i and j. Also, C1 1 ⊂ A1 . Because A ∪ B is connected and Ci j intersects A for each i and j with j ≤ i, M is connected. In Figure 2.13, we depict M . In the picture, P0 = (1, 1, 1, . . . ) and, if i is a positive integer, Pi = (0, . . . , 0, 1, 1, 1, . . . ) has its ﬁrst i coordinates 0 followed by all 1s. Example 137 (An arc plus a sequence of fans) Let G(f ) be the union of [0, 1] × {0} and {1} × [0, 1] together with the straight line interval from (3/4, 1/4) to (1, 1) (see Figures 2.14 and 2.15). Let M = lim f . Let A = A1 ∪ A2 ∪ A3 ∪ . . . ∪ {(0, 0, 0, . . . )} where ←− A1 = {x ∈ M | x2 = 1} and if i is an integer greater than one then Ai = {x ∈ lim f | xk = 0 for k i}. Then, A is an arc ←− lying in M . Let D0 = {x ∈ M | x1 ∈ [0, 1] and for each positive integer k, xk+1 = (xk + 2)/3} and, for i ≥ 1, let Di = {x ∈ M | xk = 0 for k ≤ i, xi+1 ∈ [1/4, 1], and xk+1 = (xk + 2)/3 for each k > i}. For each positive integer j, let E0 j = {x ∈ M | x1 ∈ [1/4, 1], xk = 1 if k > j , and if 1 < k < j then xk+1 = (xk + 2)/3}. If each of i and j is a positive integer, let Ei j = {x ∈ M | xk = 1 for k > j, xk = 0 for k < i, xi ∈ [1/4, 1] and if i ≤ k < j then xk+1 = (xk + 2)/3}. For each nonnegative integer i let Fi = Ei 1 ∪ Ei 2 ∪ Ei 3 ∪ · · · ∪ Di . Note that each Fi is a fan with vertex Pi where P0 = (1, 1, 1, . . . ) and, for i > 0, Pi is the point of M having kth coordinate 0 for k ≤ i and all remaining coordinates 1. Furthermore, if x ∈ M and x ∈ / A then x ∈ Fi for some i ≥ 0. It follows that M is a continuum. In Figure 2.15 we depict the continuum M . At this point we observe that the set-theoretic union of the graphs of the functions from Examples 136 and 137 is the graph of the function from Example 114. Although the inverse limit in Example 136 is a continuum and the inverse limit in Example 137 is a continuum and the arc A lies in the intersection of these two continua, the inverse limit of the union of the two graphs is not connected. In Section 2.9 we study inverse limits of upper semicontinuous functions that are graphs of unions of mappings. Among other things we show for maps of intervals that if at least one of the mappings is surjective then the inverse limit is connected. A consequence of Theorem 166 of Section 2.10 is that if f is a mapping from I into I, then lim f is homeomorphic to lim f 2 . In the case where ←− ←− an upper semi-continuous bonding function f is not a mapping, lim f and ←− lim f ◦ f may not be homeomorphic. The following example shows this. In ←− this example we specify the function instead of the graph of the function. This example is a minor modiﬁcation of an example in [7].

2.7 Examples in the special case that each factor space is [0, 1]

101 (1,1)

(3/4, 1/4)

(0,0)

(1,0)

Fig. 2.14 The function from Example 137 D0

.. .

P0

E0 3 E0 2

(0,0,0,...) E0 1

A1 P3 E3 1

A3 .. .

D3

P2 A2 .. .

E2 1 E2 2

D2 P1 E1 1

Fig. 2.15 A depiction of the inverse limit from Example 137

.. .

E1 2

E1 3

D1

102

2 Inverse Limits in a General Setting

Example 138 (lim f and lim f ◦ f are not necessarily homeomor←− ←− phic) Let f : [0, 1] → 2[0,1] be given by f (x) = {1/2, 1 − x} for 0 ≤ x < 1/2, f (x) = {1/2} for 1/2 ≤ x < 1 and f (1) = [0, 1/2] (see Figure 2.16). Proof. Note that K = lim f contains a triod which is the union of the ←− three arcs described below. Let A1 be the set of all points of K whose ﬁrst coordinate is in the half-open interval (1/2, 1]. The closure of A1 is an arc from (1, 0, 1, 0, . . .) to (1/2, 1/2, 1, 0, 1, 0, . . .). Let A2 be the set of all points of K whose ﬁrst two coordinates are 1/2 and whose third coordinate is in the half-open interval (1/2, 1]. The closure of A2 is an arc from (1/2, 1/2, 1/2, 1/2, 1, 0, 1, 0, . . .) to (1/2, 1/2, 1, 0, 1, 0, . . .). Finally let A3 be the set of all points of K whose ﬁrst coordinate is 1/2 and whose second coordinate is in the half-open interval [0, 1/2). The closure of A3 is an arc from (1/2, 0, 1, 0, 1, 0, . . .) to (1/2, 1/2, 1, 0, 1, 0, . . .). The union of the closures of A1 , A2 , and A3 is a simple triod contained in K. (0,1)

(0,1/2)

(1/2,1/2)

(1,1/2)

(1,0)

Fig. 2.16 The function from Example 138

On the other hand, f ◦ f is the union of the three straight line intervals, {0} × [0, 1/2], I × {1/2}, and {1} × [1/2, 1] (see Figure 2.17). H = lim f ◦ f ←− is a continuum by Theorem 126. We show that H is an arc with endpoints a = (0, 0, 0, . . .) and b = (1, 1, 1, . . .). Let p be a point of H diﬀerent from

2.7 Examples in the special case that each factor space is [0, 1]

103

a and b. There is an n such that pn is neither 0 nor 1. If pn = 1/2 then H ∩ πn−1 (pn ) is degenerate and separates H into the two mutually separated sets H ∩ πn−1 ([0, pn )) and H ∩ πn−1 ((pn , 1]). If pn = 1/2 and pn+1 ∈ {0, 1} then H ∩ πn−1 (pn ) is degenerate and a separating point (i.e., cut point) of H. Thus we may assume that pn+1 is neither 0 nor 1 and again conclude −1 (pn+1 ) is a separating point unless pn+1 = 1/2. Continuing that H ∩ πn+1 this process, the only point remaining to consider is the constant sequence (1/2, 1/2, 1/2, . . .). But this point also is clearly a separating point of H. Thus H is a continuum having at most two nonseparating points and is an arc. See [4, Theorem 2-1, p. 49] and [4, Theorem 2-27, p. 54].

(1,1)

(0,1/2)

(1,1/2)

(0,0)

Fig. 2.17 The graph of f 2 where f is the function from Example 138

The next example provides an upper semi-continuous function whose inverse limit is the union of a 2-cell and an arc with a single point in common. With a sequence of upper semi-continuous functions it is possible to get a 2-cell as the inverse limit. One only needs to use the sequence f1 , f2 , f3 , . . . where G(f1 ) = [0, 1] × [0, 1] and G(fi ) is the identity on [0, 1] for i > 1. However, Van Nall has shown [10] that if f : [0, 1] → 2[0,1] is an upper semicontinuous function then lim f is not a 2-cell (see Theorem 186 of Section ←− 2.12).

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2 Inverse Limits in a General Setting

Example 139 (A two-dimensional example) Let G(f ) consist of the union of the four straight line intervals, [0, 1/2] × {0}, {1/2} × [0, 1/2], [1/2, 1] × {1/2}, and {1} × [1/2, 1] (see Figures 2.18 and 2.19). Proof. Let K = lim f . Here K is the union of a 2-cell D and an arc A. To ←− identify D, let i and j be positive integers with j > i + 1 and let Di,j be the 2-cell, {p ∈ K | pi ∈ [0, 1/2], pj ∈ [1/2, 1], pk = 0 if k < i, pk = 1/2 if i < k < j, pk = 1 if k > j}. In Figure 2.19 we provide a schematic picture to assist the reader. In this picture we have labeled a few of the disks Di,j . In the ﬁgure the disks Di,j and Di,j+1 share a common horizontal border and the disks Di,j and Di+1,j share a common vertical border as long as i + 1 < j − 1. Let D be the closure of the union of all the disks Di,j where i ≥ 1 and j > i + 1. Suppose k is a positive integer. Let αk = {p ∈ K | pk ∈ [1/2, 1], pm = 1/2 for m < k, and pm = 1 for m > k} and βk = {p ∈ K | pk ∈ [0, 1/2], pm = 0 for m < k, and pm = 1/2 for m > k}. The arc αk forms the right-hand vertical edge of the disk Dk−2,k for k = 3, 4, 5, . . . in the ﬁgure and the arc βk lies directly below all of the disks Dk,k+n for n = 2, 3, 4, . . . . The closure of β1 ∪ β2 ∪ β3 ∪ · · · is an arc from (0, 0, 0, . . . ) to (1/2, 1/2, 1/2, . . . ) forming the bottom edge of the disk D and the closure of α3 ∪ α4 ∪ α5 ∪ · · · is an arc from (1/2, 1/2, 1/2, . . . ) to (1/2, 1/2, 1, 1, . . . ) forming the right-hand edge of D. Let A = α1 ∪ α2 . Then, A is an arc, D is a 2-cell, K = D ∪ A, and D ∩ A = {(1/2, 1/2, 1, 1, . . .)}. One can modify Example 139 to produce an inverse limit of dimension n for any choice of n. For example, to produce an inverse limit of dimension 3 add a second stairstep between 1/4 and 1/2. That is, let M be the union of the intervals [0, 1/4] × {0}, {1/4} × [0, 1/4], [1/4, 1/2] × {1/4}, {1/2} × [1/4, 1/2], [1/2, 1] × {1/2}, and {1} × [1/2, 1]. Additional stairsteps can be added to produce higher-dimensional inverse limits.

2.8 Mapping theorems The inverse limit of an inverse limit system {Xα , fα β , D} is a subset of the product of the factor spaces, therefore a number of mapping theorems for inverse limits derive from mapping theorems for product spaces. One fundamental mapping property of product spaces is that a function into a product

2.8 Mapping theorems

105 (1,1)

(1,1/2) (1/2,1/2)

(0,0)

(1/2,0)

Fig. 2.18 The function from Example 139

is continuous if and only if the composition of the function with each of the projection maps is continuous. A mapping f is called one-to-one (or reversible or 1-1) provided if f (x) = f (y) then x = y. Suppose D and E are sets and σ : E → D is a one-to-one function from E into D. If {Xα | α ∈ D} and {Yβ | β ∈ E} are collections of sets and for each β ∈ E there is a function ϕβ : Xσ(β) → Yβ then the collection {ϕβ | β ∈ E} induces a function Φ : α∈D Xα → β∈E Yβ deﬁned by πβ (Φ(x)) = ϕβ (πσ(β) (x)) (i.e., πβ (Φ(x)) = ϕβ (xσ(β) )) for each β in E. Theorem 140 Suppose {Xα , fα β , D} and {Yα , gα β , E} are inverse limit systems with upper semi-continuous bonding functions and σ : E → D is a one-to-one function such that if α β in E then σ(α) σ(β) in D. Suppose further that for each α ∈ E there is a function ϕα : Xσ(α) → Yα such that if α β in E then ϕα ◦ fσ(α) σ(β) = gα β ◦ ϕβ . Then, if x is in lim f , ←− Φ(x) is in lim g. ←− Proof. Suppose x ∈ lim f . If α and β are in E and α β, then σ(α) σ(β), ←− so xσ(α) ∈ fσ(α) σ(β) (xσ(β) ). Thus, ϕα (xσ(α) ) ∈ ϕα (fσ(α) σ(β) (xσ(β) )). Because πα (Φ(x)) = ϕα (xσ(α) ) and ϕα ◦ fσ(α) σ(β) = gα β ◦ ϕβ , we have πα (Φ(x)) ∈ gα β (ϕβ (xσ(β) )). But, ϕβ (xσ(β) ) = πβ (Φ(x)). It follows that πα (Φ(x)) ∈ gα β (πβ (Φ(x))) and thus Φ(x) ∈ lim g. ←−

106

2 Inverse Limits in a General Setting third coord in [0,1/2]

second coord in [0,1/2]

first coord in [0,1/2] (1,1,1, . . . )

first coord in [1/2,1] (1/2,1,1, . . . ) second coord in [1/2,1] (0,1/2,1, . . . ) (1/2,1/2,1, . . . ) D 1,3

third coord in [1/2,1] (0,0,1/2, . . . )

fourth coord in [1/2,1]

(0, 1/2,1/2, . . . )

D 2,4

D1,4

D

D1,5

(1/2,1/2,1/2,1, . . . )

(0,0,0,1/2, . . . )

fifth coord in [1/2,1]

D

3,5

2,5

(1/2,1/2,1/2, . . . ) (0,0,0, . . . )

(0,0,1/2, . . . )

(0,1/2,1/2, . . . )

Fig. 2.19 A schematic representation of the inverse limit from Example 139

In the case where ϕβ is a mapping for each β in E, the function induced by {ϕβ | β ∈ E} is a mapping. This observation and Theorem 140 lead directly to the following theorem. Theorem 141 Suppose {Xα , fα β , D} and {Yβ , gα β , E} are inverse limit systems with upper semi-continuous bonding functions such that lim f is ←− nonempty and σ : E → D is a one-to-one function such that if α β in E then σ(α) σ(β) in D. Suppose further that for each β ∈ E there is a map ϕβ : Xσ(β) → Yβ such that if α β in E then ϕα ◦ fσ(α) σ(β) = gα β ◦ ϕβ . Then, lim g is nonempty and ϕ = Φ| lim f is a mapping of lim f into lim g. ←− ←− ←− ←− As in the previous theorem, in the following we denote Φ| lim f by ϕ. If D ←− is a directed set, a subset E of D is said to be coﬁnal in D provided if α ∈ D

2.8 Mapping theorems

107

there is an element β of E such that α β. It is not diﬃcult to show that if E is coﬁnal in D then E is a directed set. If {Xα , fα β , D} is an inverse limit system and E is a coﬁnal subset of D, there is an associated inverse limit system {Yβ , gα β , E} such that Yβ = Xβ for each β ∈ E and gα β = fα β for each α and β in E such that α β. We refer to this system over E as the restriction of {Xα , fα β , D} to E. Theorem 142 Suppose D = {1, 2, 3, . . . } and {Xn , fn m , D} is an inverse limit system with upper semi-continuous bonding functions such that lim f ←− is nonempty, E is a coﬁnal subset of D, and {Ym , gn m , E} is the restriction of {Xn , fn m , D} to E. For each m ∈ E, let ϕm denote the identity on Xm . Then, the function ϕ induced by {ϕm | m ∈ E} is a mapping from lim f onto ←− lim g. ←− Proof. By Theorem 141, ϕ : lim f → lim g is a mapping. Suppose y is in ←− ←− lim g. Let x be a point of lim f chosen as follows. For m ∈ E let xm = ym . ←− ←− For n and m in E with n < m − 1 and no element of E between them, choose xn+1 ∈ fn+1 m (xm ) such that xn ∈ fn n+1 (xn+1 ). (Such a choice is possible because fn m = fn n+1 ◦ fn+1 m .) Next, choose xn+2 ∈ fn+2 m (xm ) such that xn+1 ∈ fn+1 n+2 (xn+2 ). Continuing, we may choose xi for each i, n < i < m, so that xi ∈ fi k (xk ) for i ≤ k ≤ m and xn ∈ fn i (xi ). This process determines a point x of lim f such that ϕ(x) = y. ←− In the case where the bonding functions are mappings, we can draw the stronger conclusion in Theorem 142 that the induced mapping is a homeomorphism even for directed sets D other than the set of positive integers. (See Theorem 165 of Section 2.10.) One crucial diﬀerence between inverse limits with upper semi-continuous bonding functions and those with (ordinary) bonding maps is that the map induced by identities (i.e., each map in the collection {ϕβ | β ∈ E} is the identity) need not be a homeomorphism. In Example 138 we saw that lim f and lim f ◦ f may not be homeomor←− ←− phic. Example 143 not only produces another example demonstrating this phenomenon but also shows that the induced map need not be a homeomorphism when the bonding functions have values that are closed sets. In this example, E = {1, 3, 5, . . . } and ϕn = Id for each n ∈ E. Example 143 (The Hurewicz continuum) Let G(f ) be the union of the four straight line intervals joining the points (0, 1/2) to (1/2, 1), (1/2, 1) to (1, 1/2),(1, 1/2) to (1/2, 0), and (1/2, 0) to (0, 1/2). Then, for g = f ◦f , lim f ←− and lim g are not homeomorphic (see Figure 2.20). ←− Proof. Note that the four arcs whose union is G(f ) form a diamond in I 2 . Label these arcs Ai for i ∈ {1, 2, 3, 4} in a clockwise direction so that A1 ⊆ [0, 1/2] × [1/2, 1] and A4 ⊆ [0, 1/2] × [0/1/2]. Let K = lim f . The set K ←− contains a simple closed curve that is the union of the four arcs Bi for i ∈ {1, 2, 3, 4} determined as follows. If i = 2 or i = 4, Bi is the set of all points

108

2 Inverse Limits in a General Setting

(1/2,1)

(1,1/2)

(0,1/2)

(1/2,0)

Fig. 2.20 The function from Example 143

p ∈ K such that for each n, (pn+1 , pn ) ∈ Ai . If i = 1 or 3, then Bi is the set of all points p ∈ K such that, for each odd n, (pn+1 , pn ) ∈ Ai and, for each even n, (pn+1 , pn ) ∈ Ai+2(mod4) . On the other hand, the graph of g = f ◦f is the union of two arcs, one from (0, 0) to (1, 1) and the other from (0, 1) to (1, 0) and lim g is homeomorphic ←− to the cone over a Cantor set (see Example 132) so lim f and lim g are ←− ←− not homeomorphic. (See also Example 138 in Section 2.7 for another such example.) Finally, let D denote the set of positive integers, E the set of odd positive integers, and ϕn the identity on [0, 1] for each n ∈ E. The surjective induced map ϕ : lim f → → lim f 2 from Theorem 142 cannot be a homeomorphism. ←− ←− Perhaps also of interest is the continuum K = lim f . K contains two ←− mutually exclusive Cantor sets: C0 consisting of all points p of K such that pn = 1/2 if n is even and C1 consisting of all points p of K such that pn = 1/2 if n is odd. If a is a point of C0 and b is a point of C1 , then, for each n, there is an integer in where 1 ≤ in ≤ 4 such that (an+1 , an ) and (bn+1 , bn ) are endpoints of the arc Ain . It can be shown that the set of all points x of K

2.8 Mapping theorems

109

such that (xn+1 , xn ) is in Ain is an arc joining a and b and K is the union of all these arcs, no two of which have a point in common that is not an endpoint. This is the example of a universal continuum given by Hurewicz in [5]. In fact, Hurewicz showed that if C is a (metric) continuum then there exist a subcontinuum H of K and a monotone map of H onto C. In the case where D = E, we are able to draw some stronger conclusions regarding the nature of the induced map from Theorem 141. Theorem 144 Suppose {Xα , fα β , D} and {Yα , gα β , D} are inverse limit systems with upper semi-continuous bonding functions such that lim f is ←− nonempty. Suppose further that for each α ∈ D there is a one-to-one mapping ϕα : Xα → Yα such that if α β in D then ϕα ◦ fα β = gα β ◦ ϕβ . Then, ϕ = Φ| lim f is a one-to-one mapping of lim f into lim g. ←− ←− ←− Proof. By Theorem 141, ϕ is a mapping from lim f into lim g so we need ←− ←− only show that ϕ is one-to-one. Suppose ϕ(x) = ϕ(t). Then, if α ∈ D, πα (ϕ(x)) = πα (ϕ(t)). From this we conclude that ϕα (xα ) = ϕα (tα ) for each α ∈ D. Each ϕα is 1-1, thus xα = tα for each α ∈ D, consequently x = t. Theorem 145 Suppose {Xα , fα β , D} and {Yα , gα β , D} are inverse limit systems with upper semi-continuous bonding functions such that lim f is ←− nonempty. Suppose further that, for each α ∈ D, ϕα : Xα → Yα is a one-to-one and surjective mapping such that if β is in D and α β then ϕα ◦ fα β = gα β ◦ ϕβ and ϕ = Φ| lim f . Then, ϕ : lim f → lim g is one-to-one ←− ←− ←− and surjective. Proof. By Theorem 141 with E = D and for each α ∈ D, σ(α) = α, lim g is ←− nonempty. By Theorem 144 ϕ is a 1-1 mapping of lim f into lim g, so that ←− ←− we only have to check that ϕ is surjective. Let ψ be the function from lim g ←− to lim f induced by {ϕ−1 α | α ∈ D} as given by Theorem 140. Suppose that y ←− is in lim g and let x = ψ(y). Then x ∈ lim f and ϕ(x) = y because if α ∈ D, ←− ←− πα (x) = ϕα (xα ) = ϕα (ϕ−1 α (yα )). Corollary 146 Suppose {Xα , fα β , D} and {Yα , gα β , D} are inverse limit systems on compact Hausdorﬀ spaces with upper semi-continuous bonding functions. Suppose further that, for each α ∈ D, ϕα : Xα → Yα is a homeomorphism and if β is in D and α β then ϕα ◦ fα β = gα β ◦ ϕβ and ϕ = Φ| lim f . Then, ϕ : lim f → lim g is a homeomorphism. ←− ←− ←− As an application of Corollary 146 we present the following example. We begin with a lemma. Lemma 147 Let g be the mapping from [0, 1] onto [0, 1] whose graph consists of two straight line intervals, one from (0, 0) to (1/2, 1) and the other from (1/2, 1) to (1, 1/2). Suppose f : [a, b] → → [a, b] is a mapping of the interval

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2 Inverse Limits in a General Setting

[a, b] onto itself and c is a point of the open interval (a, b) such that f (a) = a, f (c) = b, f (b) = c, and h = f |[a, c] and k = f |[c, b] are homeomorphisms. If ϕ is a homeomorphism of [0, 1] onto [a, b] such that ϕ(0) = a, ϕ(1/2) = c, and ϕ(1) = b then there is a homeomorphism ψ : [0, 1] → → [a, b] such that ψ(0) = a, ψ(1/2) = c, ψ(1) = b, and f ◦ ψ = ϕ ◦ g. Proof. Let ψ(x) = h−1 (ϕ(g(x))) if 0 ≤ x ≤ 1/2 and ψ(x) = k −1 (ϕ(g(x))) if 1/2 ≤ x ≤ 1. Because g(1/2) = 1, ϕ(1) = 1, and h−1 (1) = k −1 (1) = 1/2, we see that ψ is a mapping. Suppose x and y are in [0, 1] and x = y. If one of x and y is in [0, 1/2] and the other is in (1/2, 1], then by deﬁnition ψ(x) = ψ(y). If both are in [0, 1/2] or both are in (1/2, 1] then again ψ(x) = ψ(y) inasmuch as g(x) = g(y). Thus, ψ is one-to-one and therefore is a homeomorphism. Example 148 Suppose f : [a, b] → [a, b] is a mapping of the interval [a, b] onto itself and c is a point of the open interval (a, b) such that f (a) = a, f (c) = b, and f (b) = c and f |[a, c] is a homeomorphism as is f |[c, b]. Then, lim f is ←− homeomorphic to the closure of the graph of y = sin(1/x) on (0, 1]. Proof. Let g : [0, 1] → [0, 1] be the map whose graph consists of two straight line intervals, one from (0, 0) to (1/2, 1) and the other from (1/2, 1) to (1, 1/2). In Chapter 1 we showed that lim g is homeomorphic to the closure of the ←− graph of y = sin(1/x) on (0, 1]. We now show that lim f and lim g are home←− ←− omorphic. Let ϕ1 be the homeomorphism of [0, 1] onto [a, b] such that the graph of ϕ consists of two straight line intervals, one from (0, a) to (1/2, c) and the other from (1/2, c) to (1, b). Inductively, suppose ϕ1 , ϕ2 , ϕ3 , . . . , ϕn have been deﬁned so that for 1 ≤ i ≤ n we have ϕi (0) = a, ϕi (1/2) = c, ϕi (1) = b, and, if i > 1, ϕi−1 ◦ g = f ◦ ϕi . By Lemma 147 there is a homeomorphism ϕn+1 such that ϕn+1 (0) = a, ϕn+1 (1/2) = c, ϕn+1 (1) = b, and ϕn ◦ g = f ◦ ϕn+1 . With the sequence ϕ thus deﬁned it follows from Corollary 146 with D as the set of positive integers that lim f and lim g are homeomorphic. ←− ←− Suppose X is a compact Hausdorﬀ space. If f : X → 2X and g : X → 2 are upper semi-continuous functions, f and g are topologically conjugate provided there is a homeomorphism h such that h(X) = X and h ◦ f = g ◦ h. We conclude this section with a theorem that provides suﬃcient conditions under which inverse limit sequences with topologically conjugate bonding functions produce homeomorphic inverse limits. X

Theorem 149 Suppose D is the set of positive integers and X is a compact Hausdorﬀ space such that for each i in D, Xi = X. If f : X → 2X and g : X → 2X are topologically conjugate upper semi-continuous functions, then lim f is homeomorphic to lim g. ←− ←−

2.9 Upper semi-continuous functions that are unions of functions

111

Proof. X is a compact Hausdorﬀ space and D is the set of positive integers, thus Theorem 111 gives that lim f and lim g are not empty. There is a home←− ←− omorphism h : X → X such that h(X) = X and h ◦ f = g ◦ h. Let each map ϕi = h and let ϕ : lim f → lim g be the mapping induced by ϕ1 , ϕ2 , ϕ3 , . . .. ←− ←− By Theorem 145, ϕ is 1-1 and surjective. Because ϕ is a 1-1 mapping from a compact space onto a Hausdorﬀ space, ϕ is a homeomorphism.

2.9 Upper semi-continuous functions that are unions of functions

In this section we consider an interesting class of upper semi-continuous functions, those whose graphs are unions of the graphs of (set-valued) functions. We are primarily concerned with conditions that ensure that inverse limits of inverse limit sequences with such functions as bonding functions are connected. Of course, without some conditions on the functions, the inverse limit may not be connected because the union of the mapping that is identically 0 on [0, 1] with the mapping that is identically 1 on [0, 1] yields a Cantor set for its inverse limit; see Example 2.3. Although the subject of this section is of interest in and of itself, some who are looking at applications of inverse limits in economics have asked about the nature of inverse limits with upper semi-continuous functions that are unions of mappings. If f : X → 2Y and g : X → 2Y are set-valued functions, we say that f and g have a coincidence point provided there is a point x of X such that f (x) ∩ g(x) = ∅. Lemma 150 Suppose X1 , X2 , X3 , . . . is a sequence of compact Hausdorﬀ spaces and fi : Xi+1 → 2Xi is an upper semi-continuous function for each positive integer i. If n is a positive integer, g : Xn+1 → 2Xn is an upper semi-continuous function such that fn and g have a coincidence point, fi is surjective for each i ≥ n, and ϕ is a sequence of functions such that ϕi = fi for i = n and ϕn = g, then lim f and lim ϕ have a point in common. ←− ←− Proof. Inasmuch as fn and g have a coincidence point, there are points t of Xn+1 and z of Xn such that z ∈ fn (t) ∩ g(t). Because fi is surjective for each i > n, by Theorem 112 there is a point x of lim f such that xn+1 = t and ←− xn = z. Because z ∈ g(t), x is in lim ϕ. ←− Suppose X and Y are compact Hausdorﬀ spaces and F is a collection of set-valued functions from X into 2Y . A function f ∈ F is said to be universal with respect to F provided f has a coincidence point with each member of F . Recall that C(X) denotes the connected elements of 2X .

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Theorem 151 If F is a collection of upper semi-continuous functions of a continuum X into C(X) one of which is surjective and universal with respect to F and f is a closed subset of X × X that is the set-theoretic union of the graphs of the functions in the collection F , then f : X → 2X is an upper semi-continuous function such that, if fi = f for each positive integer i, then lim f is a continuum. ←− Proof. f is a closed subset of X × X and each point of X is a ﬁrst coordinate of some point of f , therefore f is upper semi-continuous. Because lim f is ←− compact, we only need to show that this inverse limit is connected. Suppose f1 ∈ F and f1 is surjective and universal with respect to F . Choose a point x ∈ lim f1 and let y ∈ lim f . There exists a sequence ϕ1 , ϕ2 , ϕ3 , . . . such ←− ←− that ϕi ∈ F and yi ∈ ϕi (yi+1 ) for each positive integer i. Let C1 = lim f 1 , ←− and, if n is an integer with n > 1, let Cn be the inverse limit of the sequence ϕ1 , ϕ2 , . . . , ϕn−1 , f1 , f1 , f1 , . . . . For each n, Cnis a continuum by Theorem 126 and, by Lemma 150, Cn ∩ Cn+1 = ∅. Thus, i>0 Ci is connected. Moreover, for each n, because f1 is surjective there is a point pn of Cn such that πi (pn ) = yi for i ≤ n. It follows that y ∈ C and because x ∈ C1 , lim f is the ←− union of a collection of continua all containing x. Thus lim f is connected. ←− The set-theoretic union of a ﬁnite collection of mappings of [0, 1] into itself is a closed subset of [0, 1] × [0, 1], thus we have the following corollary to Theorem 151. In Section 2.12 we show that the continuum that results in Theorem 152 is one-dimensional (see Theorem 185). Theorem 152 If F is a ﬁnite collection of mappings from [0, 1] into itself one of which is surjective, f is the set-theoretic union of the maps in F , and fi = f for each positive integer i, then lim f is a one-dimensional continuum. ←− Just after Example 137 we observed that the preceding theorem does not hold if F is allowed to contain upper semi-continuous functions. In fact, as we show in our next example, there is a two-element collection F consisting of one upper semi-continuous function having a connected inverse limit and one mapping that has a union with a nonconnected inverse limit. Example 153 (An upper semi-continuous function and a map whose union produces a nonconnected inverse limit) Let g1 be the function from Example 136 and g2 be the piecewise linear mapping passing through (0, 1), (3/4, 1/4), (7/8, 1/2), and (1, 0) and let F = {g1 , g2 }. If f = G(g1 ) ∪ g2 , lim f is not connected (see Figure 2.21 for the graph of f ). ←− By an argument virtually identical to that provided in Example 114 it may be shown that if f is the upper semi-continuous function whose graph is the

2.9 Upper semi-continuous functions that are unions of functions (0,1)

113 (1,1)

(7/8,1/2)

(1/4,1/4)

(3/4,1/4)

(0,0)

(1,0)

Fig. 2.21 The function from Example 153

set-theoretic union of the graphs of g1 and g2 , then lim f is not connected. ←− In fact {x ∈ lim f | x1 = x2 = 1/4, x3 = 3/4} is both open and closed in the ←− inverse limit. This inverse limit contains (I × {1} × {0})∞ . For the remainder of this section we consider unions of mappings. We are interested in inverse limits with upper semi-continuous functions on [0, 1] that are unions of ﬁnitely many mappings that are not necessarily surjective such as the function in Example 143 (Figure 2.20). First we prove a lemma that is of use in the proof of Theorem 155. Lemma 154 If f : [0, 1] → [0, 1] is a mapping from [0, 1] into itself such that f 2 ([0, 1]) = f ([0, 1]), t ∈ f ([0, 1]), and fi = f for each positive integer i, then there is a point x ∈ lim f such that x1 = t. ←− Proof. Let t be a point of f ([0, 1]) and let x1 = t. Because f (f ([0, 1])) = f ([0, 1]) and t ∈ f ([0, 1]) there is a point x2 of f ([0, 1]) such that f (x2 ) = x1 . Similarly, because x2 is in f ([0, 1]) = f 2 ([0, 1]), there is a point x3 ∈ f ([0, 1]) such that f (x3 ) = x2 . Continuing in this manner we obtain a point x ∈ lim f ←− such that x1 = t. Theorem 155 Suppose F is a ﬁnite collection of mappings from [0, 1] into itself that contains a mapping f1 with the following properties.

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1. f1 ([0, 1]) is nondegenerate 2. If g ∈ F there is a point pg ∈ f1 ([0, 1]) such that f1 (pg ) = g(pg ) 3. If g ∈ F then g(f1 ([0, 1])) = g([0, 1]). If f is the set-theoretic union of all the elements of F and fi = f for each positive integer i, then lim f is a one-dimensional continuum. ←− Proof. Choose a point y in lim f . There exists a sequence ϕ1 , ϕ2 , ϕ3 , . . . such ←− that ϕi is in F and ϕi (yi+1 ) = yi for each positive integer i. Let C1 be the inverse limit of the sequence f1 , f1 , f1 , . . . and if n is an integer greater than one, let Cn be the inverse limit of the sequence ϕ1 , ϕ2 , . . . , ϕn−1 , f1 , f1 , . . . . Using condition (2), it follows from Lemma 150 that Ci and Ci+1 have a point in common for each positive integer i. Thus, C1 ∪ C2 ∪ C3 ∪ · · · is connected. Using Lemma 154 with t = yn there is a point x of lim f1 such that x1 = yn . ←− The point pn = (y1 , y2 , . . . , yn , x2 , x3 , . . . ) belongs to Cn and the distance from y to pn is less than 1/2n . Thus, y belongs to Cl(C1 ∪ C2 ∪ C3 ∪ · · · ). Each point of lim f belongs to a continuum lying in lim f that contains the ←− ←− continuum lim f1 , therefore lim f is a continuum. ←− ←− From condition (1) and Lemma 154 it follows that lim f1 is nondegenerate. ←− Thus, the dimension of lim f is one by Theorem 185. ←− Theorem 156 If f : [0, 1] → 2[0,1] is an upper semi-continuous function that is the union of a ﬁnite collection F of mappings from [0, 1] into itself one of which is surjective and fi = f for each positive integer i, then lim f ←− is a one-dimensional continuum that contains a copy of every inverse limit lim g where gi ∈ F for each i. ←− Proof. Theorem 152 yields that lim f is a continuum. It is easy to see that ←− lim f contains a copy of every inverse limit lim g where gi ∈ F for each i. ←− ←− That the dimension of the inverse limit is one follows from Theorem 185 from the ﬁnal section of this chapter. Richard M. Schori [11] constructed a chainable continuum that contains a copy of every chainable continuum. Although Schori’s result is stronger, we still observe the following. Corollary 157 There exists an upper semi-continuous function f such that if fi = f for each positive integer i, then lim f is a one-dimensional contin←− uum that contains a copy of every chainable continuum. Proof. There exist two mappings ϕ : [0, 1] → → [0, 1] and ψ : [0, 1] → [0, 1] such that if M is a chainable continuum then there exists a sequence k such that ki ∈ {ϕ, ψ} for each i and M is homeomorphic to lim k, [2] or [12]. Let ←− f = ϕ ∪ ψ and apply Theorem 156.

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115

2.10 Theorems for inverse limit systems with bonding functions that are mappings

In this section we consider inverse limit systems in which the bonding functions are mappings and contrast some of these results with the fact that they fail for systems with upper semi-continuous bonding functions. Earlier in this chapter we established the inverse limit is a nonempty compact Hausdorﬀ space for any inverse limit system {Xα , fα β , D} where each Xα is a nonempty compact Hausdorﬀ space and each fα is a mapping; see Theorems 107 and 111. Furthermore, in Theorem 117 we showed that the inverse limit of a system with mappings is a Hausdorﬀ continuum when each factor space is a Hausdorﬀ continuum.

2.10.1 A basis for the topology One useful feature of inverse limits of systems of mappings lies in the fact that a collection of open sets that looks as if it were only a subbasis for the topology of the inverse limit is, in fact, a basis for the topology of the inverse limit. We see this in our next theorem. In its proof we employ the convention that the domain of πα is the inverse limit space. Theorem 158 Suppose {Xα , fα β , D} is an inverse limit system where each fα β is a mapping and M = lim f is nonempty. Then, B = {πα−1 (O) | α ∈ ←− D and O is an open subset of Xα } is a basis for the topology for lim f . ←− Proof. Choose an element R of the usual basis for the topology of the product space Π and let x be an element of R ∩ M . Then R = α∈D Uα where each factor is open and Uα = Xα except for ﬁnitely many elements of D, say α1 , α2 , . . . , αn . There is a member α of D such that αi α for 1 ≤ i ≤ n. Because fαi α is a mapping for each i, there is an open subset O of Xα containing xα such that fαi α (O) ⊆ Uαi for 1 ≤ i ≤ n. Then, πα−1 (O) contains x and is a subset of R ∩ M . It follows that B is a basis for the topology of lim f . ←−

2.10.2 Closed subsets Suppose {Xα , fα β , D} is an inverse limit system and M = lim f is nonempty. ←− If H is a closed subset of M , we denote πα (H) by Hα . For convenience, if

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f : X → 2Y and A ⊂ X and no confusion should arise, we may denote f |A by f . Theorem 159 Suppose {Xα , fα β , D} is an inverse limit system such that each fα β is a mapping and M = lim f is nonempty. Then, if H is a closed ←− subset of M , H is the inverse limit of the inverse limit system {Hα , gα β , D} where gα β = fα β | Hβ . Proof. It is immediate that H ⊆ lim{Hα , gα β , D}. On the other hand, sup←− pose x ∈ lim{Hα , gα β , D}. Then, xα ∈ Hα for each α ∈ D. Therefore, if ←− α ∈ D and Oα is an open subset of Xα such that x ∈ πα−1 (Oα ), there is a point p of H such that pα = xα . Thus, p ∈ πα−1 (Oα ). Each basis element containing x contains a point of H, therefore x ∈ H. Because H is closed, x ∈ H. Corollary 160 Suppose {Xα , fα β , D} is an inverse limit system such that each fα β is a mapping and M = lim f is nonempty. Then, if H is a closed ←− subset of M such that Hα = Xα for each α ∈ D, we have H = M .

(1,1)

(0,0)

Fig. 2.22 The function from Example 161

(1,0)

2.10 Inverse limit systems with mappings

117

2.10.3 Closed subsets of a system with upper semi-continuous bonding functions That each of Theorem 158, Theorem 159, and Corollary 160 fail to hold for inverse limit sequences using upper semi-continuous bonding functions may be seen from the following example. We specify the upper semi-continuous bonding function by means of its graph. Example 161 Let D be the set of positive integers and let G(f ) be the subset of [0, 1] × [0, 1] containing the point (1, 0) and the line joining (0, 0) and (1, 1) (see Figure 2.22). Proof. Let K = lim f . Without proof, we note that K is the union of an arc ←− A = {(t, t, t, . . . ) | t ∈ [0, 1]} and a sequence of points p1 , p2 , p3 , . . . not in A converging to (0, 0, 0, . . . ) where pi has its ﬁrst i coordinates 0 and all other coordinates 1. To see that Theorem 158 fails for K, one needs only to note that the point (0, 1, 1, 1, . . . ) of K belongs to R = [0, 1/4) × (3/4, 1] × Q but R fails to contain any set of the form πi−1 (O) where O is open in [0, 1]. Indeed, if O is an open subset of [0, 1], O contains a point t of [0, 1] not in both [0, 1/4) and (3/4, 1]. If i is a positive integer (t, t, t, . . . ) is a point of πi−1 (O) that is not in R. Furthermore, H = {(t, t, t, . . .) ∈ M | t ∈ [0, 1]} is a closed proper subset of K such that Hi = [0, 1] for each positive integer i so Theorem 159 and Corollary 160 fail for f .

2.10.4 Intersections of closed subsets of the inverse limit Another interesting property of inverse limit systems with mappings is found in the following theorem. As before, if H is a closed subset of lim f , we denote ←− πα (H) by Hα . Theorem 162 Suppose {Xα , fα β , D} is an inverse limit system such that each fα β is a mapping, and H and K are closed subsets of lim f . Then, ←− H ∩ K = lim{Hα ∩ Kα , fα β |(Hβ ∩ Kβ ), D}. ←− Proof. Because πα (H ∩ K) ⊆ Hα ∩ Kα for each α ∈ D, H ∩ K ⊆ lim{Hα ∩ ←− Kα , fα β |(Hβ ∩ Kβ ), D}. On the other hand, if x ∈ lim{Hα ∩ Kα , fα β |(Hβ ∩ ←− Kβ ), D} then xα ∈ Hα ∩Kα for each α ∈ D so x ∈ lim{Hα , fα β |Hβ , D} = H. ←− Similarly, lim{Hα ∩ Kα , fα β |(Hβ ∩ Kβ ), D} ⊆ K. ←−

118

2 Inverse Limits in a General Setting (1,1)

(1/4,5/8)

(3/4,3/8)

(0,0)

Fig. 2.23 The mapping from Example 163

In the previous theorem, πα (H ∩ K) is not necessarily equal to Hα ∩ Kα for any α ∈ D even if D is the set of positive integers. An example of this phenomenon follows. Example 163 Let f : [0, 1] → [0, 1] be the piecewise linear map passing through the points (0, 0), (1/4, 5/8), (3/4, 3/8), and (1, 1). Then, lim f is the ←− union of two arcs H = lim{[0, 5/8], f |[0, 5/8]} and K = lim{[3/8, 1], f |[3/8, 1]}. ←− ←− Then, H ∩K = (1/2, 1/2, 1/2, . . .) but Hi ∩Ki = [7/16, 9/16] for each positive integer i (see Figure 2.23). Proof. The maps f |[0, 5/8] and f |[3/8, 1] are topologically conjugate, so H and K are homeomorphic by Theorem 149. To see that H is an arc, we employ Theorem 39 from Chapter 1 to observe that H is the closure of a ray with remainder lim{[1/4, 5/8], f |[1/4, 5/8]}. The slope of f on [1/4, 5/8] ←− is −1/2, thus n>0 f n ([1/4, 5/8]) = {1/2}. It follows from Theorem 113 that the remainder is a single point.

2.10 Inverse limit systems with mappings

119

2.10.5 The subsequence theorem One of the fundamental tools in analyzing the nature of inverse limits of inverse limit systems using mappings as bonding functions is the subsequence theorem, Theorem 166. Unfortunately, this theorem does not hold for general inverse limit systems with upper semi-continuous bonding functions under study earlier in the present chapter (see Example 143). Here we state and prove a general version of the subsequence theorem for inverse limit systems in which the bonding functions are mappings. Theorem 164 Suppose {Xα , fα β , D} is an inverse limit system over a directed set D such that the inverse limit is nonempty, each fα β is a mapping, and E is a coﬁnal subset of D. If {Yβ , fα β , E} is the restriction of {Xα , fα β , E} to E, then there is a one-to-one mapping from lim f onto lim g. ←− ←− Proof. For each α in E, let ϕα denote the identity on Xα . By Theorem 141 the function ϕ induced by {ϕα | α ∈ E} is a mapping of lim f into lim g. ←− ←− Suppose y ∈ lim g and γ ∈ D. We now construct a point x of lim f so that ←− ←− ϕ(x) = y. If γ ∈ E let xγ = yγ . Suppose γ is not in E. There exists an element δ of E such that γ δ. Let xγ = fγ δ (yδ ). Every bonding function is a mapping, thus the choice of xγ is independent of the choice of δ and it is not diﬃcult to check that x is in lim f so ϕ is surjective. To see that ϕ is ←− 1-1, suppose each of x and t is in lim f and ϕ(x) = ϕ(t). Let α be a member ←− of D. There is a member β of E such that α β. Because β ∈ E, xβ = tβ , so fα β (xβ ) = fα β (tβ ). Thus, xα = tα for each α ∈ D, so x = t. Theorem 165 Suppose D is a directed set and Xα is a compact Hausdorﬀ space for each α ∈ D. Suppose further {Xα , fα β , D} is an inverse limit system where each fα β is a mapping, and E is a coﬁnal subset of D. Then lim{Xα , fα β , D} and lim{Xα , fα β , E} are homeomorphic. ←− ←− Proof. The 1-1 mapping ϕ from Theorem 164 is a homeomorphism because lim{Xα , fα β , D} is compact and lim{Xα , fα β , E} is Hausdorﬀ. ←− ←− Of course, if D is the set of positive integers, any increasing sequence of positive integers n1 , n2 , n3 , . . . is coﬁnal in D. This observation leads to a restatement of Theorem 165. Although it is merely a restatement of the previous theorem in the speciﬁc case that D is the set of positive integers, it is of enough value to the theory of inverse limits to merit a separate statement. We recall the following standard notation. If X is a sequence of spaces, f is a sequence of mappings such that fi : Xi+1 → Xi , and i < j, then fi j : Xj → Xi = fi ◦ fi+1 ◦ · · · ◦ fj−1 and fi i is the identity on Xi . Theorem 166 (The subsequence theorem) Suppose D is the set of positive integers and n1 , n2 , n3 , . . . is an increasing sequence of positive integers. Suppose further that X is a sequence of compact Hausdorﬀ spaces and f is

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2 Inverse Limits in a General Setting

a sequence of mappings such that fi : Xi+1 → Xi for each i. Let g be the sequence of maps such that gi = fni ni+1 for each i. Then, lim f is homeo←− morphic to lim g. ←− We end this subsection with an application of Theorem 166. We employ the following terminology in the next theorem. If f : X → Y is a mapping, we say that f factors through Z if there are maps g : X → Z and h : Z → Y such that f = h ◦ g. Theorem 167 Suppose {Xi , fi } is an inverse limit sequence such that, for each positive integer i, fi is a mapping and n1 , n2 , n3 , . . . is an increasing sequence of positive integers such that fni ni+1 factors through [0, 1] for each i. Then M = lim f is homeomorphic to an inverse limit on [0, 1]; that is, M ←− is a chainable continuum. Proof. By Theorem 166, M is homeomorphic to lim g where gi = fni ni+1 for ←− each i. Because fni ni+1 factors through [0, 1], there exist maps ψi : [0, 1] → Xni and ϕi : Xni+1 → [0, 1] such that fni ni+1 = ψi ◦ ϕi . It follows from Theorem 166 that M is homeomorphic to lim{Yi , hi } where Yi = Xn(i+1)/2 ←− and hi = ψ(i+1)/2 if i is odd, and Yi = [0, 1] and hi = ϕi/2 if i is even. One ﬁnal application of Theorem 166 yields that lim{Yi , hi } is homeomorphic to ←− lim k where ki = ϕi ◦ ψi+1 , a map from [0, 1] to [0, 1]. ←− Of course, a more general theorem than the one stated here for [0, 1] holds, but this theorem illustrates a way that the subsequence theorem can be used.

2.10.6 Other induced homeomorphisms Other important consequences of Theorem 165 are found in the next two theorems. Theorem 168 (The shift homeomorphism) Suppose {Xi , fi , D} is an inverse limit sequence over the set of positive integers where, for each i, Xi is a compact Hausdorﬀ space and fi is a mapping. Let E = D − {1}. Then, h : lim{Xi , fi , D} → lim{Xi , fi , E} given by h(x) = (x2 , x3 , x4 , . . . ) is a ←− ←− homeomorphism. The homeomorphism h from Theorem 168 is called the shift homeomorphism. In dynamics, one reason for interest in inverse limits is that by passing to the inverse limit, one is able to replace a dynamical system consisting of a topological space and a continuous function with a (possibly more complicated) space (the inverse limit) and a homeomorphism (the shift). In the case

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121

where each factor space Xi is the same topological space X and each bonding map fi is the same map f , the inverse of the shift homeomorphism is given by h−1 (x) = (f (x1 ), x1 , x2 , . . .). The map h−1 is also called the shift homeomorphism by many authors (consequently, in each instance we have tried to make it clear which of these “shifts” we are using as we did in Theorem 20 in Chapter 1). It is interesting to note that, in this case, h−1 is induced by a sequence of mappings that are not necessarily 1-1, ϕi = f for each i. Our next theorem is another valuable tool in analyzing inverse limits with a constant sequence of factor spaces and a constant sequence of bonding maps, the so-called inverse limits with a single bonding map. In the case where we have an inverse limit with a single bonding map, we denote the inverse limit sequence by {X, f } and the inverse limit by lim f . If f : X → X ←− is a mapping, f n : X → X denotes the n-fold composition of f with itself. Theorem 169 Suppose X is a compact Hausdorﬀ space and f : X → X is a mapping. If n is a positive integer, lim f is homeomorphic to lim f n . ←− ←− Proof. Let n1 = 1 and nk+1 = nk + n for k = 1, 2, 3, . . . and apply Theorem 166.

2.10.7 Inverse limits as sequential limiting sets Suppose M is a sequence of sets in a topological space. By the limiting set (or lim sup) of the sequence is meant the set to which the point P belongs if and only if it is true that if U is an open set containing P then U contains a point of Mi for inﬁnitely many integers i. By the sequential limiting set of the sequence is meant the set that is the limiting set of every subsequence of M. Theorem 170 Suppose D is the set of positive integers and {Xi , fi j , D} is an inverse limit sequence where each Xi is a compact Hausdorﬀ space and each map fi j is surjective. Let p be a point of Π = i>0 Xi and, for each positive integer n, let hn : Xn → Π be given by πi (hn (x)) = fi n (x) if i ≤ n and πi (hn (x)) = pi if n < i. Let Yn = hn (Xn ). Then, for each n, hn is a homeomorphism and lim f is the sequential limiting set of the sequence Y . ←− Proof. That each hn is a homeomorphism is an immediate consequence of the fact that hn is 1-1 and πi ◦ hn is continuous for each i. Let K = lim f ←− and let x be a point of K. If Yn1 , Yn2 , Yn3 , . . . is a subsequence of Y and O = i>0 Oi is a basic open set containing x, then, inasmuch as there is a positive integer j such that Oi = Xi for i ≥ j, O contains a point of Yni for each ni ≥ j so x is in the limiting set of Yn1 , Yn2 , Yn3 , . . .. On the other hand

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if x is in the limiting set of Yn1 , Yn2 , Yn3 , . . . and O is a basic open set in the product space containing x there is a positive integer n such that if j ≥ n then Oj = Xj . Because O contains a point y of Ynk for some nk > n and there is a point t of K such that ti = yi for i ≤ nk , O contains a point of K. Because K is closed, x is in K.

2.10.8 Inverse limits as intersections of closed sets We close Section 2.10 with a theorem relating inverse limits and intersections of monotonic collections of closed sets. Theorem 171 Suppose D is a directed set and {Xα | α ∈ D} is a collection of topological spaces such that if α β in D then Xβ is a subset of Xα . Furthermore, if α β in D, let fα β be the identity on X . Then, β α∈D Xα is homeomorphic to lim f . Moreover, α∈D Xα = ∅ if and only if lim f = ∅. ←− ←− lim f if and only if thereis a point p Proof. The point x of α∈D Xα is in ← − of α∈D Xα such that xα = p for each α ∈ D. Let h : lim f → α∈D Xα be ←− given by h(x) is the point p such that xα = p for each α ∈ D. That h−1 is a homeomorphism follows from the observation that the composition of h−1 with each projection is the identity on α∈D Xα . It is clear that α∈D Xα = ∅ if and only if lim f = ∅. ←−

2.11 Some theorems for inverse limit systems with metric factor spaces

In this section we present some theorems for inverse limits that require a metric on the factor spaces. Every metric space has an equivalent metric bounded by 1, thus we assume that all of our metric spaces have a metric bounded by 1. If (X1 , d1 ), (X2 , d2 ), (X3 , d3 ), . . . is a sequence of metric spaces each with a metric bounded by 1, then a metric for i>0 Xi is given by d(x, y) = i>0 di (xi , yi )/2i . This is the metric that we use for the inverse limit. Theorem 172 Suppose {Xα , fα β , D} is an inverse limit system such that Xα is a compact metric space for each α ∈ D and each bonding function is a mapping. If D has a countable coﬁnal subset then lim f is a metric space. ←−

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123

Proof. Suppose E is a countable coﬁnal subset of D. By Theorem 165, lim f ←− is homeomorphic to lim{Xα , fα β , E}. The latter is a metric space because it ←− is a subset of the metric space α∈E Xα . For the remainder of this chapter we deal only with inverse limit sequences in metric spaces. Our next theorem, although easy to prove, is of fundamental importance to inverse limits in continuum theory. By the diameter of a subset A of a metric space with metric d we mean the least upper bound of {d(x, y) | x, y ∈ A}. We denote the diameter of A by diam A. Theorem 173 Suppose {Xi , fi } is an inverse limit sequence where, for each i, Xi is a compact metric space. If ε > 0 there exist a positive integer n and a positive number δ such that if A is a subset of Xn and diam A < δ then diam πn−1 (A) < ε. Proof. Let ε > 0. There is a positive integer n such that j≥n 2−j < ε/3. For each i < n, fi n is uniformly continuous so there is a positive number δ < ε/3 such that if p and q are points of Xn and dn (p, q) < δ then di (fi n (p), fi n (q)) < ε/3. Suppose A is a subset of Xn and the diameter of A is less than δ. If x and y are points of πn−1 (A) then dn (xn , yn ) < δ so d(x, y) < 2ε/3. One consequence of Theorem 173 is the following theorem that generalizes the theorem from the ﬁrst chapter that inverse limits on [0, 1] are chainable. Theorem 174 If M1 , M2 , M3 , . . . is a sequence of chainable continua and f1 , f2 , f3 , . . . is a sequence of mappings such that fi : Mi+1 → Mi , then lim f ←− is a chainable continuum. If f : X → → Y is a mapping of a metric space X onto a topological space Y , then f is called an ε-map provided diam f −1 (y) < ε for each point y of Y . We end this section with a theorem similar to Theorem 173. Its proof is left to the reader. Theorem 175 Suppose {Xi , fi } is an inverse limit sequence where Xi is a compact metric space and fi is a mapping for each positive integer i. Then, if ε > 0, there is a positive integer n such that if i ≥ n then πi is an ε-map. Mardeˇsi´c and Segal, [9, Theorem 1∗ , p. 148], have shown that a converse of Theorem 175 holds in the case where the factor spaces are connected polyhedra (triangulable continua). We state this result without proof, instead referring the reader to their paper. If Π is a class of polyhedra, a compact metric space is said to be Π-like provided for each ε > 0 there exist a polyhedron P ∈ Π and an ε-mapping f : X → → P from X onto P .

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Theorem 176 (Mardeˇ si´ c and Segal) Let Π be a class of connected polyhedra. Then, the class of Π-like continua coincides with the class of inverse limits of inverse sequences {Pi , fi } where Pi ∈ Π and fi is surjective for each positive integer i.

2.12 Dimension

If G is a ﬁnite collection of sets and n is a positive integer, we say that the order of G is n provided n is the largest of the integers i such that there are i + 1 members of G with a common element. Recall that the mesh of a ﬁnite collection G of sets is the largest of the diameters of the elements of G. If G and H are collections of sets we say that H reﬁnes G provided for each element h of H there is an element g of G such that h ⊆ g. If n is a positive integer, the compact metric space X is said to have dimension not greater than n, written dim(X) ≤ n, provided, for each positive number ε, there is a ﬁnite collection of open sets covering X that has mesh less than ε and order not greater than n. We say the dimension of X is n, written dim(X) = n, provided dim(X) ≤ n and dim(X) ≤ n − 1. It is convenient to use this deﬁnition of dimension (sometimes called covering dimension) in the study of inverse limits. For compact metric spaces the property of having dimension n (respectively, not greater than n) is equivalent to the usual deﬁnition of having small inductive dimension n (respectively, not greater than n) [6, Theorem V 8, p. 67]. In [6, Theorem V 1, p. 54] it is shown that if X is a compact metric space with dim(X) ≤ n and G is a ﬁnite collection of open sets covering X then there is a ﬁnite collection H of open sets covering X that reﬁnes G having order not greater than n. We begin our look at dimension in inverse limits by presenting a theorem of Nall, Theorem 181, on the dimension of an inverse limit when the bonding functions have zero-dimensional values. Suppose X1 , X2 , X3 , . . . , Xn is a ﬁnite collection of compact metric spaces and f1 , f2 , . . . , fn−1 is a ﬁnite collection of upper semi-continuous functions such that fi : Xi+1 → 2Xi for 1 ≤ i < n. Let Gn = {(x1 , x2 , . . . , xn ) ∈ X1 × X2 × · · · × Xn | xi ∈ fi (xi+1 ) for 1 ≤ i < n}. Our ﬁrst lemma is a special case of Theorem 110. By using the directed set D in that theorem to be the set of integers {1, 2, . . . , n} we have the following. Lemma 177 Suppose X1 , X2 , X3 , . . . , Xn is a ﬁnite collection of compact Hausdorﬀ spaces and fi : Xi+1 → 2Xi is an upper semi-continuous function for 1 ≤ i < n. Then Gn is compact.

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125

Suppose X1 , X2 , X3 , . . . , Xn is a ﬁnite collection of compact metric spaces and f1 , f2 , . . . , fn−1 is a ﬁnite collection of upper semi-continuous functions such that fi : Xi+1 → 2Xi for 1 ≤ i < n. Let Y = X1 × X2 × · · · × Xn−1 and deﬁne Fn : Xn → 2Y by Fn (x) = {(x1 , x2 , . . . , xn−1 ) ∈ Gn−1 | xn−1 ∈ fn (x)}. In [10], Nall makes the following useful observation. Theorem 178 Suppose X1 , X2 , X3 , . . . , Xn is a ﬁnite collection of compact metric spaces and fi : Xi+1 → 2Xi is an upper semi-continuous function for 1 ≤ i < n. Then, Fn is upper semi-continuous. Proof. The graph of Fn is homeomorphic to Gn which is compact by Lemma 177. Theorem 105 yields that Fn is upper semi-continuous. Lemma 179 Suppose X1 , X2 , X3 , . . . , Xn is a ﬁnite collection of compact metric spaces and fi : Xi+1 → 2Xi is an upper semi-continuous function for 1 ≤ i < n. If x is a point of Xn such that dim(Fn (x)) > 0 then there exist an integer j, 1 ≤ j < n, and a point z of Xj+1 such that dim(fj (z)) > 0. Proof. dim(Fn (x)) > 0, therefore it contains a nondegenerate continuum K [6, Theorem D, p.22]. Some projection of K into one of the factor spaces Xi is nondegenerate. Let j be the largest integer i so that the projection of K into Xi is nondegenerate. If j = n − 1, let z = x and it follows that fj (z) contains a nondegenerate continuum so dim(fj (z)) > 0. If j < n − 1 then the projection of K into Xj+1 is a single point z. It follows that dim(fj (z)) > 0. If (X1 , d1 ), (X2 , d2 ), . . . , (Xn , dn ) is a ﬁnite collection of compact metric spaces, there are numerous metrics that are compatible with the product topology on X1 ×X2 ×· · ·×Xn . One that is particularly convenient is d(x, y) = i=n i i=1 di (xi , yi )/2 . Lemma 180 Suppose X1 , X2 , X3 , . . . , Xn is a ﬁnite collection of compact metric spaces, fi : Xi+1 → 2Xi is an upper semi-continuous function for 1 ≤ i < n, and m is a positive integer. If dim(Xn ) ≤ m and for each i, 1 ≤ i < n and each point x of Xi+1 dim(fi (x)) = 0 then dim(Gn ) ≤ m. Proof. Suppose ε > 0 and x is a point of Xn . It follows from Lemma 179 that dim(Fn (x)) = 0 so there exists a ﬁnite collection Vx of mutually exclusive open sets covering Fn (x) such that the mesh of Vx is less than ε/3. Because Fn is upper semi-continuous, there is an open set ux containing x of diameter less than ε such that Fn (ux ) ⊆ Vx ∗ (where Vx ∗ denotes the union of all the sets in Vx ). The collection of open sets U = {ux | x ∈ Xn } covers the compact set Xn so there is a ﬁnite subcollection U of U that covers Xn . The dimension of Xn is not greater than m so there is a ﬁnite collection W of open sets covering Xn such that the order of W is not greater than m and W reﬁnes U . The mesh of W is less than ε/2. For each w ∈ W choose a point xw of

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Xn in w such that wxw ⊆ uxw ∈ U . Because U ⊆ U, Fn (wxw ) ⊆ Vx∗w . Then, {v × w | w ∈ W and there is a point x ∈ Xn such that w = wx and v ∈ Vx } is a collection of open sets covering Gn of order not greater than m and mesh less than ε. Thus, dim(Gn ) ≤ m. Theorem 181 (Nall) Suppose {Xi , fi } is an inverse limit sequence with upper semi-continuous bonding functions such that Xi is a compact metric space for each positive integer i and m is a positive integer. Suppose further that dim(fi (x)) = 0 for each positive integer i and each point x of Xi+1 . If there is an increasing sequence n1 , n2 , n3 , . . . of positive integers such that dim(Xni ) ≤ m for i = 1, 2, 3, . . . , then dim(lim f ) ≤ m. ←− Proof. Recall that lim f = n>2 Gn where G n = {x ∈ i>0 Xi | xi ∈ ←− fi (xi+1 ) for 1 ≤ i < n}. ObserveGn = Gn × i>n Xi . Let ε > 0. There is a positive integer N such that i≥N 2−i < ε/2. Let i be an integer such that ni > N . By Lemma 180 dim(Gni ) ≤ m. Let U be a collection of open sets of order not greater than m and mesh less than ε/2 that covers Gni . Then, {π −1 (u) | u ∈ U} is a collection of open sets of mesh less than ε and order not greater than m that covers lim f . ←− If each of X and Y is a compact metric space and f : X → 2Y is a function that is the union of ﬁnitely many mappings of X into Y , then f is upper semi-continuous and dim(f (x)) = 0 for each x in X. As a consequence, we have the following corollary to Theorem 181. Corollary 182 Suppose n is a positive integer and {Xi , fi } is an inverse limit sequence in which each Xi is a compact metric space and each fi is the union of ﬁnitely many mappings. If for each j there is a positive integer i ≥ j such that dim(Xi ) ≤ n then dim(lim f ) ≤ n. ←− Of course one consequence of Corollary 182 is that ordinary inverse limits do not raise dimension. On the other hand, dimension may be lowered by the ordinary inverse limit construction even if the bonding maps are surjective as may be seen from the following example. Example 183 Let ϕ denote the projection of the unit square C = [0, 1]×[0, 1] onto the interval I = [0, 1]. Let ψ denote a map of I onto C. Let Xi = C and fi = ψ ◦ ϕ for each i. Let Yi = I and gi = ϕ ◦ ψ for each i. Let Zi = C for odd integers i and Zi = I for even integers i. Let ki = ψ for odd i and ki = ϕ for even i. Using n1 = 1, n2 = 3, n3 = 5, . . . in the subsequence theorem we see that lim f is homeomorphic to lim k. Using n1 = 2, n2 = 4, n3 = 6, . . . in the ←− ←− subsequence theorem we see that lim k is homeomorphic to lim g. Therefore, ←− ←− lim f is homeomorphic to lim g so dim (lim f ) ≤ 1 even though each factor ←− ←− ←− space is two-dimensional and each bonding map is surjective.

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By substituting an n-cell or the Hilbert cube for C in Example 183, we see that an inverse limit of n-dimensional or even inﬁnite-dimensional continua can have dimension one. Actually, the dimension of lim f in Example 183 is one as shown by the ←− following theorem. Theorem 184 Suppose {Xi , fi } is an inverse limit sequence where Xi is a continuum of dimension one and fi is a surjective mapping for each i. Then, dim (lim f ) = 1. ←− Proof. Let M = lim f . Because dim(X1 ) = 1, X1 is nondegenerate. Let p ←− and q be two diﬀerent points of X1 . Each bonding map is surjective, thus there are points x and y of M such that x1 = p and y1 = q. Because M is a nondegenerate continuum, dim (M ) ≤ 0. By Theorem 181, dim (M ) ≤ 1, so dim (M ) = 1. In the case that the bonding functions are upper semi-continuous functions each of which is the union of ﬁnitely many mappings, the dimension of the inverse limit is not greater than one. Theorem 185 If fi : I → 2I is an upper semi-continuous function that is the union of ﬁnitely many mappings f1i , f2i , . . . , fki i of I = [0, 1] into itself for each i, then the dimension of lim f is not greater than one. Moreover, if there ←− is a sequence g such that gi ∈ {f1i ,f2i ,. . . , fki i } for each positive integer i and lim g is nondegenerate then the dimension of lim f is one. ←− ←− Proof. Inasmuch as f is the union of ﬁnitely many mappings, dim(f (t)) = 0 for each t ∈ [0, 1]. By Nall’s theorem (Theorem 181), the dimension of lim f ←− is not greater than one. If there is a sequence g such that gi ∈ {f1i ,f2i ,. . . ,fki i } for each positive integer i and lim g is nondegenerate, then lim f contains a ←− ←− nondegenerate continuum so its dimension is one. We close with a proof that one cannot get a 2-cell as an inverse limit with a single upper semi-continuous bonding function from [0, 1] into 2[0,1] . Recall that, unless otherwise noted, if f : [0, 1] → 2[0,1] is an upper semicontinuous function, we consider the domain of the projection πi to be the inverse limit space, lim f , whereas fˆ denotes the shift map on lim f given ←− ←− by fˆ(x) = (x2 , x3 , x4 , . . . ). In general, the shift map on an inverse limit with upper semi-continuous bonding functions is not a homeomorphism. However, when it is restricted to a compact set on which it is one-to-one, its restriction is a homeomorphism. The following proof is based on work of Nall [10]. Theorem 186 (Nall) Suppose f : [0, 1] → 2[0,1] is an upper semi-continuous function such that if y ∈ [0, 1] there exists a point x ∈ [0, 1] such that y ∈ f (x). Then lim f is not a 2-cell. ←−

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Proof. Suppose M = lim f is a 2-cell. If 0 < t < 1, then π1−1 (t) separates M . ←− Because M is a 2-cell, π1−1 (t) is not zero-dimensional [6, Corollary 2, p. 48], so it contains a nondegenerate continuum H. There is a positive integer m ≥ 2 such that πm (H) is nondegenerate but πi (H) is a single point for 1 ≤ i < m. Suppose J is an interval such that πm (H) = J and if 1 ≤ i < m, let πi (H) = {ti } where t1 = t. Let K = {x ∈ M | xi = ti for 1 ≤ i < m and xm ∈ J}. Note that H ⊆ K ⊆ π1−1 (t). By [6, Theorem IV 3, p. 44], π1−1 (J) is 2-dimensional being a closed set with interior lying in a 2-cell. Let z be a point of π1−1 (J). Because z1 ∈ J and J = πm (H), there is a point w of H such that wm = z1 . Let y be the point of [0, 1]∞ such that yi = ti for 1 ≤ i < m and ym+i = zi+1 for i = 0, 1, 2, . . . . Because w ∈ H ⊆ M and z ∈ M , it follows that y ∈ K. Moreover, fˆm−1 (y) = z. Thus, z ∈ fˆm−1 (K) and we have established that π1−1 (J) ⊆ fˆm−1 (K). Note that fˆ is 1-1 on K because π1 (K) is degenerate so fˆ is a homeomorphism on K. In fact, fˆm−1 is a homeomorphism on K and fˆm−1 (K) contains a two-dimensional subset so K contains a two-dimensional subset. But, K is a subset of π1−1 (t) so it follows that π1−1 (t) contains an open set. Thus, we have for each t in (0, 1), π1−1 (t) contains an open set. But, if s = t and 0 < s, t < 1 then π1−1 (s) and π1−1 (t) have no point in common so the 2-cell M contains uncountably many mutually exclusive open sets, a contradiction. Nall actually proves more than we state in Theorem 186. He shows that a continuum that is the union of a countable collection of n-cells and compact n-dimensional manifolds is not homeomorphic to an inverse limit on [0, 1] with a single upper semi-continuous bonding function. His proof is similar to the one we present for Theorem 186.

References

1. Howard Cook, Continua which admit only the identity mapping onto non-degenerate subcontinua, Fund. Math. 60 (1967), 241–249. 2. Howard Cook and W. T. Ingram, Obtaining AR-like continua as inverse limits with only two bonding maps, Glasnik Mat. 4(24) (1969), 309–312. 3. L. Henkin, A problem on inverse mapping systems, Proc. Amer. Math. Soc. 1 (1950), 224–225. 4. John G. Hocking and Gail S. Young, Topology, Dover, New York, 1988. ¨ 5. Witold Hurewicz, Uber oberhalb-stetige Zerlegungen von Punktmengen in Kontinua, Fund. Math. 15 (1930), 57–60. 6. Witold Hurewicz and Henry Wallman, Dimension Theory, Princeton University Press, Princeton, NJ, 1941. 7. W. T. Ingram and William S. Mahavier, Inverse limits of upper semi-continuous set valued functions, Houston J. Math. 32 (2006), 119–130. 8. William S. Mahavier, Inverse limits with subsets of [0, 1] × [0, 1], Topology and Its Applications 141/1-3 (2004), 225–231. 9. Sibe Mardeˇsi´ c and Jack Segal, ε-mappings onto polyhedra, Trans. Amer. Math. Soc. 109 (1963), 146—164. 10. Van Nall, Inverse limits with set valued functions, Houston J. Math. 37 (4) (2011). 11. Richard M. Schori, A universal snake-like continuum, Proc. Amer. Math. Soc. 16 (1965), 1313–1316. 12. Sam W. Young, The representation of chainable continua with only two bonding maps, Proc. Amer. Math. Soc. 23 (1969), 653–654.

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Chapter 3

Inverse Limits in Continuum Theory

Abstract This chapter is devoted to some topics in the theory of continua. We look at the eﬀect of imposing atriodicity or unicoherence on the factor spaces or monotonicity on the bonding mappings. We conclude the chapter with characterizations of irreducibility and indecomposability of inverse limits followed by a brief discussion of selected miscellaneous topics from continuum theory.

3.1 Introduction In this chapter we investigate the interaction of several continuum-theoretic properties with the inverse limit construction. These properties include atriodicity, unicoherence, irreducibility, chainability, and indecomposability. In Section 3.3 we include the Capel theorems on inverse limits on arcs and simple closed curves with monotone bonding maps. Section 3.4 is devoted to a look at indecomposability including Kuykendall’s characterizations of irreducibility and indecomposability. This is followed by a section on indecomposability in inverse limits with set-valued functions. We close the chapter with a brief discussion of a few selected topics from continuum theory not covered elsewhere in this book including span in inverse limits.

3.2 Indecomposability, atriodicity, and unicoherence

Recall that by a Hausdorﬀ continuum we mean a compact, connected Hausdorﬀ space and by a continuum we mean a compact, connected metric space. W.T. Ingram and W.S. Mahavier, Inverse Limits: From Continua to Chaos, Developments in Mathematics 25, DOI 10.1007/978-1-4614-1797-2_3, © Springer Science+Business Media, LLC 2012

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A (Hausdorﬀ) continuum is decomposable provided it is the union of two of its proper subcontinua and is indecomposable otherwise. If {Xα , fα β , D} is an inverse limit system over a directed set D where each Xα is a compact Hausdorﬀ space such that M = lim f = ∅ (this condition is assured if the ←− system is consistent) and H ⊆ M , for convenience of notation we denote πα (H) by Hα .

3.2.1 Indecomposability The following theorem provides a simple suﬃcient condition that an inverse limit of an inverse limit system with (single-valued) bonding maps on Hausdorﬀ continua be indecomposable. In Section 3.4 of this chapter we give an example (Example 207) showing that the converse of this theorem is false. Theorem 187 Suppose {Xα , fα β , D} is an inverse limit system over a directed set D where each Xα is a Hausdorﬀ continuum, each fα β is a mapping, and E is a coﬁnal subset of D. If for each α and β in E, α = β, such that α β and for each two subcontinua A and B of Xβ such that Xβ = A ∪ B, then fα β (A) = Xα or fα β (B) = Xα , then lim f is an indecomposable Haus←− dorﬀ continuum. If, in addition, E is countable and Xα is metric for each α in E, then lim f is an indecomposable continuum. ←− Proof. Let M = lim f . That M is a Hausdorﬀ continuum is a result of The←− orem 117 from Chapter 2. Suppose H and K are proper subcontinua of M such that M = H ∪ K. It follows from Corollary 160 of Chapter 2 that there is an element α of D such that if α β then Hβ = Xβ and Kβ = Xβ . Choose an element β of E, β = α, such that α β. Observe that Xβ = Hβ ∪ Kβ so Hα = fα β (Hβ ) = Xα or Kα = fα β (Kβ ) = Xα . This contradicts the choice of α. If E is countable and Xα is metric for each α ∈ E then, by Theorem 172 of Chapter 2, lim{Xα , fα β , E} is a metric space so the Hausdorﬀ continuum ←− M is a continuum. By Theorem 165 of Chapter 2, because E is coﬁnal in D, lim{Xα , fα β , D} is homeomorphic to lim{Xα , fα β , E}. ←− ←− Corollary 188 Suppose {Xα , fα β , D} is an inverse limit system over a directed set D where each Xα is a Hausdorﬀ continuum and each fα β is a mapping. If E is a coﬁnal subset of D such that Xα is indecomposable for each α in E, then lim f is an indecomposable Hausdorﬀ continuum. If E is ←− a countable coﬁnal subset of D and Xα is an indecomposable continuum for each α in E, then lim f is an indecomposable continuum. ←−

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That Theorem 187 does not always hold for inverse limit systems with upper semi-continuous bonding functions may be seen from the following example. Example 189 Let f : [0, 1] → 2[0,1] be given by f (x) = [0, 1] for each x in [0, 1]. Then, lim f is the Hilbert cube, yet if A and B are subintervals of [0, 1] ←− such that [0, 1] = A ∪ B, then f (A) = f (B) = [0, 1].

3.2.2 Triods and atriodicity A (Hausdorﬀ) continuum M is a triod provided M contains a subcontinuum C such that M − C has at least three components (or, equivalently, M is the union of three continua with a common point such that the common part of each two of them is the common part of all three of them and is a proper subcontinuum of each one of them). A (Hausdorﬀ) continuum is atriodic provided it contains no triod. We recall Sorgenfrey’s triod theorem, [31, Theorem 1.8, p. 443]. Theorem 190 (Sorgenfrey) If a Hausdorﬀ continuum contains three subcontinua with a common point so that no one of them is a subset of the union of the other two then it contains a triod. Sorgenfrey’s proof is given in a metric setting, but his proof carries directly over to Hausdorﬀ continua. We use this result in the proof of the following theorem. Theorem 191 Suppose {Xα , fα β , D} is an inverse limit system such that each Xα is an atriodic Hausdorﬀ continuum and each fα β is a mapping. Then, lim f is atriodic. ←− Proof. Let M = lim f . By Theorem 159 of Chapter 2, each subcontinuum of ←− M is the inverse limit of its projections, so it is suﬃcient to show that M is not a triod. Suppose M is the union of three continua H, K, and L such that the common part of each two of them is the common part of all three of them and is a proper subcontinuum of each one of them. Let p be a point common to H, K, and L. There exist points x ∈ H, y ∈ K, and z ∈ L such that x ∈ / K ∪ L, y ∈ / H ∪ L, and z ∈ / H ∪ K. There is a member β of D such that if α ∈ D and β α then Hα , Kα , and Lα are three subcontinua of / πα (K ∪ L) = Kα ∪ Lα , yα ∈ / Hα ∪ Lα , and Xα each containing pα and xα ∈ zα ∈ / Hα ∪ Kα . By Theorem 190, Xα contains a triod for any α such that β α.

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The inverse limit may be a triod even if each factor space is not a triod. This may be seen from the following example. By a theta curve we mean a continuum homeomorphic to the union of the unit circle in the plane and the interval [−1, 1], that is, a continuum that is the union of three mutually exclusive copies of [0, 1] with the points corresponding to 0 identiﬁed and the points corresponding to 1 identiﬁed. We call the map T of [0, 1] onto itself given by T (x) = 2x for 0 ≤ x ≤ 1/2 and T (x) = 2 − 2x for 1/2 ≤ x ≤ 1 the full tent map. The inverse limit, lim T , is the BJK horseshoe. It has a single ←− endpoint, (0, 0, 0, . . .). Example 192 Let Θ be a theta curve and f : Θ → Θ the mapping that restricted to each copy of [0, 1] is a full tent map. Then lim f is the union of ←− three copies of the BJK horseshoe intersecting only at a common endpoint. The inverse limit is a triod even though each factor space is not a triod.

3.2.3 Unicoherence We turn our attention to the preservation of unicoherence by the inverse limit construction. Our proof relies on the following lemma that identiﬁes an important property of the projections in an inverse limit system. Lemma 193 Suppose {Xα , fα β , D} is an inverse limit system such that each Xα is a (Hausdorﬀ ) continuum and each fα β is a mapping. If C1 and C2 are mutually exclusive closed subsets of lim f , there is a member α of D such ←− that if α β then πβ (C1 ) and πβ (C2 ) are mutually exclusive. Proof. Suppose C1 and C2 are mutually exclusive closed subsets of lim f such ←− that if α is in D there is an element β of D such that α β and πβ (C1 ) ∩ πβ (C2 ) = ∅. It follows that πα (C1 ) ∩ πα (C2 ) = ∅ for each α ∈ D. If gα β = fα β |(πβ (C1 ) ∩ πβ (C2 )), then gα β : πβ (C1 ) ∩ πβ (C2 ) → πα (C1 ) ∩ πα (C2 ). By Theorem 162 of Chapter 2, C1 ∩ C2 = lim{πα (C1 ) ∩ πα (C2 ), gα β , D}, but ←− this inverse limit is nonempty inasmuch as each factor space is nonempty and compact. A (Hausdorﬀ) continuum M is unicoherent provided it is true that if M is the union of two subcontinua H and K then H ∩ K is connected. A (Hausdorﬀ) continuum is hereditarily unicoherent provided each subcontinuum of it is unicoherent. The following theorem ﬁrst appeared in [24, Corollary 1, p. 228]. Theorem 194 Suppose {Xα , fα β , D} is an inverse limit system such that each Xα is a unicoherent Hausdorﬀ continuum and each fα β is a surjective mapping. Then, lim f is unicoherent. ←−

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Proof. Suppose M = lim f is not unicoherent. Then, there exist subcontinua ←− H and K of M such that M = H ∪ K and H ∩ K is not connected. So, H ∩ K is the union of two mutually exclusive closed sets A and B. There exist open sets U and V such that A ⊆ U , B ⊆ V and U and V are mutually exlcusive. Let H = H − H ∩ (U ∪ V ) and K = K − K ∩ (U ∪ V ). Because H and K are continua, H and K are nonempty, thus H and K are mutually exclusive closed sets. There exist open sets O and R such that H ⊆ O, K ⊆ R, and O and R are mutually exclusive. Using Lemma 193 twice (once with U ∩ M and V ∩ M and once with O ∩ M and R ∩ M ) we obtain a member γ of D such that πγ (U ∩ M ) and πγ (V ∩ M ) are mutually exclusive as are πγ (O ∩ M ) and πγ (R ∩ M ). Next, we show that Hγ ∩ Kγ is a subset of πγ (U ) ∪ πγ (V ). To see this, let p be a point of Hγ ∩ Kγ . There exist points x of H and y of K such that xγ = p = yγ . If p is not in πγ (U ) ∪ πγ (V ), then neither x nor y is in U ∪ V . Thus, x is in H and y is in K , and so p is in πγ (O) ∩ πγ (R), contrary to the choice of γ. Observe that Hγ ∩ Kγ is not connected inasmuch as it intersects both of the mutually exclusive closed sets πγ (U ) and πγ (V ) and is a subset of their union. The bonding maps are surjective, thus Xγ = Hγ ∪ Kγ , but this contradicts the fact that Xγ is unicoherent. Surjectivity is necessary in the hypothesis of Theorem 194 as may be seen from the following example of a nonunicoherent continuum that is the inverse limit of unicoherent continua using nonsurjective bonding maps. We denote by S 1 the unit circle in the plane. Example 195 Let X be the closure of a ray with remainder S 1 in the plane. Formally, let X = R where R = {(ρ, θ) | ρ = 1 + 1/θ, 2π ≤ θ} (in polar coordinates). Note that X is unicoherent. Let f : X → X be given by f (t) = t for t ∈ S 1 , f ((ρ, θ)) = (1 + (1/(θ + 2π)), θ) for 2π ≤ θ. Then, lim f = S 1 . ←−

3.2.4 Irreducibility For the theorems of this subsection, we assume the factor spaces are metric because results that generally do not hold for compact Hausdorﬀ spaces are used. If M is a (Hausdorﬀ) continuum and H is a closed subset of M , M is irreducible about H provided no proper subcontinuum of M contains H. If M is a (Hausdorﬀ) continuum, a collection G of subsets of M is called a proper decomposition of M provided M is the union of all of the elements of G and each element of G contains a point belonging to no other element of G. We make use of the following theorem of Sorgenfrey [32, p. 667].

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Theorem 196 (Sorgenfrey) Suppose M is a continuum and k ≥ 2 is an integer. Then, M is irreducible about some k of its points if and only if for each proper decomposition of M into k + 1 subcontinua the union of some k elements of the decomposition is not connected. Theorem 197 Suppose k is a positive integer and k ≥ 2. If {Xi , fi } is an inverse limit sequence such that, for each i, Xi is a continuum irreducible about k points, then lim f is a continuum irreducible about k points. ←− Proof. Suppose M = lim f is not irreducible about any k of its points. Then, ←− by Theorem 196, M is the union of k +1 of its subcontinua H 1 , H 2 , . . . , H k+1 such that the union of any k of them is a proper subcontinuum of M . There exists a positive integer N such that if n ≥ N then the nth projection of each union of k of the continua H 1 , H 2 , . . . , H k+1 is a proper subcontinuum of Xn . Thus, by Sorgenfrey’s theorem, for n ≥ N , Xn is not irreducible about any k of its points inasmuch as it is the union of k + 1 of its subcontinua πn (H 1 ), πn (H 2 ), . . . , πn (H k+1 ) and the union of any k of these is a proper subcontinuum of Xn . A (Hausdorﬀ) continuum M is irreducible provided it is irreducible about some two-point subset; that is, there exist points p and q of M such that no proper subcontinuum of M contains both p and q. If M is irreducible about {p, q}, we say that M is irreducible from p to q. One consequence of Theorem 197 is that an inverse limit on irreducible continua is irreducible. Unfortunately, this theorem only yields that the inverse limit is irreducible without producing two points such that the inverse limit is irreducible between them. Producing points of irreducibility in an inverse limit can be quite diﬃcult, even for inverse limits on [0, 1]. David Ryden solved this problem for inverse limits on [0, 1] with sequences of bonding maps in [27]. However, for inverse limits on [0, 1] with a single bonding map, it is not diﬃcult to produce points of irreducibility as we see from the following theorem due to Kuykendall [15]. Theorem 198 (Kuykendall) Suppose f : [0, 1] → → [0, 1] is a mapping from [0, 1] onto [0, 1], p is the ﬁrst ﬁxed point for f 2 on [0, 1], and q is the last ﬁxed point for f 2 on [0, 1]. Then, lim f 2 is irreducible between (p, p, p, . . . ) ←− and (q, q, q, . . . ) and therefore lim f is irreducible between (p, f (p), p, f (p), . . . ) ←− and (q, f (q), q, f (q), . . . ). Proof. Let a and b be points of [0, 1] such that f (a) = 0 and f (b) = 1 and let J be the interval with endpoints a and b. Let c be the ﬁrst point of J such that f (c) = a and d be the last point of J such that f (d) = b. By considering cases a < b and b < a, it is easy to see that c < d. Because f 2 (c) = 0, f 2 has a ﬁxed point x, 0 ≤ x ≤ c, so p ≤ c. Because f 2 (d) = 1, f 2 has a ﬁxed point y, d ≤ y ≤ 1, so d ≤ q. Then f 2 ([p, q]) = [0, 1], so if H is a subcontinuum of lim f 2 containing (p, p, p, . . . ) and (q, q, q, . . . ) then πi (H) = [0, 1] for each ←− positive integer i. It follows that H = lim f 2 . ←−

3.3 Monotone bonding maps

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3.3 Monotone bonding maps In this section we present the Capel theorems [3] that inverse limits of sequences of monotone surjective maps of arcs are arcs and inverse limits of sequences of monotone surjective maps of simple closed curves are simple closed curves. Recall that a mapping f : X → Y is monotone provided f −1 (y) is connected for each y ∈ f (X); a point p of a continuum M is a separating point of M provided M − {p} is not connected (separating points are also called cut points). Every continuum has at least two nonseparating points (Theorem 262 of the Appendix, see also [6, Theorem 2-18, p. 48]). An arc is a continuum with only two nonseparating points. Each arc is homeomorphic to [0, 1] [6, Theorem 2-27, p. 54] and is thereby endowed with an order that we shall denote by <. If A and B are arcs and f : A → B is a mapping, we say that f is order-preserving (respectively, order-reversing) provided f (x) ≤ f (y) (respectively, f (x) ≥ f (y)) whenever x < y. If f : A → B is a monotone mapping of the arc A to the arc B, then f is either orderpreserving or order-reversing. The following theorem is not diﬃcult to prove and its proof is left to the reader. Theorem 199 If f : A → B and g : B → C are order-preserving (orderreversing) mappings of arcs to arcs, then g ◦ f is order-preserving. Theorem 200 (Capel) If {Xi , fi } is an inverse limit sequence such that Xi is an arc and fi is monotone and surjective for each i then lim f is an ←− arc. Proof. Let M = lim f . If there is a positive integer n such that fi is ←− order-preserving for each i ≥ n, M is homeomorphic to lim{Yi , gi } where ←− Yi = Xn+i−2 and gi = fn+i−1 for each i. If inﬁnitely many bonding maps are order-reversing, by employing Theorem 199 we may obtain an increasing sequence n1 , n2 , n3 , . . . so that fni ni+1 is order-preserving for each i. By the subsequence theorem, M is homeomorphic to lim{Xni , fni ni+1 }. Conse←− quently, we assume that each bonding map is order-preserving. For each i, denote by ai and bi the endpoints of Xi and assume ai < bi . If xis in M and x is neither(a1 , a2 , a3 , . . .) nor (b1 , b2 , b3 , . . .), then M − {x} = ( n≥1 πn−1 ([an , xn ))) ∪ ( n≥1 πn−1 (xn , bn ])) because if y = x then there is a positive integer n such that yn = xn so yn belongs to one of the open sets [an , x) and (x, bn ]. Because M − {x} is the union of two mutually exclusive open sets, it is not connected. M has only two nonseparating points, therefore it is an arc. By a simple closed curve we mean a continuum such that the complement of each two-point subset of it is not connected.

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Lemma 201 Suppose each of X and Y is a simple closed curve and f : X → → Y is a monotone mapping from X onto Y . Suppose further that p and q are points of X such that f (p) = f (q) and U and V are the open sets whose union is Y − {f (p), f (q)}. Then, there exist open sets R and S such that X − {p, q} = R ∪ S and f (R) ⊆ U and f (S) ⊆ U. Proof. Inasmuch as f is monotone f −1 (f (p)) and f −1 (f (q)) are mutually exclusive arcs lying in X, so there exist mutually exclusive connected open sets A and B such that X − {f −1 (f (p)), f −1 (f (q))} = A ∪ B. Because f (A) and f (B) are connected and do not contain either f (p) or f (q), neither f (A) nor f (B) intersects both U and V . Suppose f (A) ⊆ U . Then, f (B) ⊆ V . There exist mutually exclusive open sets R and S such that X −{p, q} = R∪S with A ⊆ R and B ⊆ S. It follows that f (R) ⊆ U and f (S) ⊆ V .

Theorem 202 (Capel) If {Xi , fi } is an inverse limit sequence such that Xi is a simple closed curve and fi is monotone and surjective for each positive integer i then lim f is a simple closed curve. ←− Proof. Let M = lim f . Suppose each of x and y is a point of M and x = y. ←− There is a positive integer N such that if i ≥ N then xi = yi . For convenience of notation we assume N = 1. Because x1 and y1 are diﬀerent points of the simple closed curve X1 , X1 − {x1 , y1 } = U1 ∪ V1 where U1 and V1 are mutually exclusive open sets. Inductively, using Lemma 201, there exist sequences U1 , U2 , U3 , . . . and V1 , V2 , V3 , . . . such that Xi − {xi , yi } = Ui ∪ Vi and f i (Ui+1 ) ⊆ U i and fi (V i+1 ) ⊆ V i for each positive integer i. Let U = i>0 πi−1 (Ui ) and V = i>0 πi−1 (Vi ). Note that U ⊆ M − {x, y} for if p ∈ U then pk ∈ πk−1 (Uk ) for some k so pk = xk and we have p = x. Similarly, p = y. Using a similar argument, we obtain that V ⊆ M − {x, y}. On the other hand, if p ∈ M − {x, y} then there is an integer k such that pk = xk and pk = yk . Thus, pk ∈ Uk ∪ Vk from which it follows that p ∈ U or p ∈ V . Because X − {x, y} is the union of two mutually exclusive open sets U and V , it is not connected.

3.4 Characterizations of irreducibility and indecomposability

Irreducibility and indecomposability are closely linked through the theorem that a continuum is indecomposable if and only if it contains three points such that it is irreducible between each two of them [6, Theorem 3-51, p. 141]. In

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this section we present characterizations of irreducibility and indecomposability of a continuum in terms of its inverse limit representation. We begin with a couple of theorems that require neither a metric nor a countable directed set. Recall that we often denote the closure of a set A by Cl(A). Theorem 203 Suppose {Xα , fα β , D} is an inverse system such that each Xα is a compact Hausdorﬀ space and each fα β is a mapping. If p is a point of lim f , E is a coﬁnal subset of D, Kα is a subcontinuum of Xα containing ←− pα for each α in E, and Yα = Cl( β∈E fα β (Kβ )) for each α in D, then Yα αβ

is a subcontinuum of Xα and if α β then fα β (Yβ ) ⊆ Yα for each α ∈ D. Proof. Let α be an element of D. Each element of the collection Cα = {fα β (Kβ ) | β ∈ E, α β} is a subcontinuum of Xα that contains pα so Yα is a continuum. Suppose α β. Note that {γ ∈ E | β γ} ⊆ {γ ∈ E | α γ} and for each γ ∈ E such that β γ we have fα γ = fα β ◦ fβ γ . It consequently follows using the continuity of fα β that fα β (Yβ ) = fα β (Cl( γ∈E fβ γ (Kγ ))) ⊆ βγ Cl( γ∈E fα β (fβ γ (Kγ ))) ⊆ Yα . βγ

Theorem 204 Suppose {Xα , fα β , D} is an inverse system such that each Xα is a Hausdorﬀ continuum and each fα β is a surjective mapping. If lim f ←− is irreducible about the closed set H and there is a coﬁnal subset E of D such that Hη is a subset of a subcontinuum Kη of Xη for each η ∈ E, then Xα = Cl( αβ fα β (Kβ )) for each α ∈ D. β∈E

Proof. Let M = lim f and suppose that p is a point of the closed subset ←− H of M . Suppose E is a coﬁnal subset of D and Kη is a subcontinuum of Xη containing Hη for each η ∈ E. Then pη ∈ Hη for each η ∈ E. For each α ∈ D, let Yα = Cl( αη fα η (Kη )). From Theorem 203 it follows that Yα is η∈E

a subcontinuum of Xα and if α and β are in D with α β, fα β (Yβ ) ⊆ Yα . Therefore, L = lim{Yα , fα β |Yβ , D} is a subcontinuum of M containing H. ←− Suppose there exists an element α of D such that Yα = Xα . Then, because each fα β is surjective, it follows that there is a point z of M such that zα ∈ Xα but zα ∈ / Yα so z ∈ / L. Thus, L is a proper subcontinuum of M that contains H, a contradiction. Theorem 205 (Kuykendall)[16] Suppose {Xi , fi } is an inverse sequence such that, for each i, Xi is a continuum and fi is a surjective mapping. If p and q are points of M = lim f , then M is irreducible between p and q if and ←− only if for each positive integer n and each positive number ε there exists a positive integer N ≥ n such that if m ≥ N and K is a subcontinuum of Xm containing pm and qm then dn (x, fn m (K)) < ε for each x in Xn .

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Proof. Suppose M is irreducible between p and q but there exist a positive integer n and a positive number ε such that if N ≥ n is an integer then there exist an integer m ≥ N and a subcontinuum K m of Xm containing pm and qm such that dn (xm , fn m (K m )) ≥ ε for some xm in Xn . It follows that there exist an increasing sequence m1 , m2 , m3 , . . . of positive integers with n ≤ m1 , a sequence K m1 , K m2 , K m3 , . . . of continua such that K mi ⊆ Xmi with {pmi , qmi } ⊆ K mi , and a sequence xm1 , xm2 , xm3 , . . . of points of Xn such that dn (xmi , fn mi (K mi )) ≥ ε for each i. Some subsequence of xm1 , xm2 , . . . converges. For notational convenience, we may assume that xm1 , xm2 , . . . converges to x0 ∈ Xn . There is a positive integer j such that if i ≥ j, dn (x0 , xmi ) < ε/2. Let n1 , n2 , n3 , . . . be the subse. . such that n1 = mj , n2 = mj+1 , n3 = mj+2 , . . .. For quence of m1 , m2 , . each i, let Yi = Cl( i≤nk fi nk (K nk )). It follows from Theorem 203, Yi is a subcontinuum of Xi and fi (Yi+1 ) ⊆ Yi for each i. So, lim{Yi , fi |Yi+1 } is a ←− subcontinuum of M containing p and q. Because M is irreducible between p and q, by Theorem 204 we have that Yi = Xi for each i. In particular, x0 ∈ Yn = Cl( n≤nk fn nk (K nk )) so there exist a positive integer k and a point z of fn nk (K nk ) so that dn (x0 , z) < ε/2. Then, dn (xnk , z) < ε, a contradiction. On the other hand, if M is not irreducible between p and q, there is a proper subcontinuum K of M that contains p and q. There is a positive integer n such that if j ≥ n then Kj = Xj . There is a point x of Xn −Kn and, consequently, a positive number ε such that dn (x, Kn ) ≥ ε. If m ≥ n then Km is a subcontinuum of Xm containing pm and qm and dn (x, fn m (Km )) ≥ ε. Theorem 206 (Kuykendall)[16] Suppose {Xi , fi } is an inverse sequence such that for each i, Xi is a continuum and fi is a surjective mapping. Then, M = lim f is indecomposable if and only if for each positive integer n and ←− each positive number ε there exist a positive integer m > n and three points of Xm such that if K is a subcontinuum of Xm containing two of them then dn (x, fn m (K)) < ε for each x in Xn . Proof. Suppose M is indecomposable, n is a positive integer, and ε > 0. There exist three points of M such that M is irreducible between each two of them. Applying Theorem 205 to each two of these three points and taking n to be the largest of the three integers obtained yields the desired conclusion. On the other hand, if M is decomposable then M = H ∪ K where H and K are proper subcontinua of M . There is a positive integer n such that if m ≥ n then Hm and Km are proper subcontinua of Xm . There exist a positive number ε and points x of Hn and y of Kn such that dn (x, Kn ) ≥ ε and dn (y, Hn ) ≥ ε. If m > n and a, b, and c are three points of Xm then two of them belong to Hm or two of them belong to Km . Either possibility leads to a contradiction.

3.4 Irreducibility and indecomposability

141 (7/8,15/16)

(1,1)

(1/4,7/8)

(3/4,1/8) (0,0)

(1/8,1/16)

Fig. 3.1 The map f in Example 207

That the converse of Theorem 187 is not true may be seen from the following example. Example 207 Let Xi be the interval [0, 1] and fi = f for each i where f denotes the piecewise linear map whose graph is the union of ﬁve straight line intervals from (0, 0) to (1/8, 1/16) to (1/4, 7/8) to (3/4, 1/8) to (7/8, 15/16) to (1, 1) as shown in Figure 3.1. Then, lim f is indecomposable by Theorem ←− 206 for if J is a subinterval of [0, 1] containing two of the three points 0, 1/2, and 1 then d(x, f n (J)) < 1/2n+2 for each positive integer n and each x in [0, 1]. However, if J is a proper subinterval of [0, 1], f n (J) = [0, 1] for any positive integer n.

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3.5 Closed subsets of inverse limits with upper semi-continuous bonding functions Earlier in this book we proved that each closed subset of an inverse limit with mappings is the inverse limit of its projections; see Theorem 159 of Chapter 2. This theorem fails to hold for inverse limits with upper semicontinuous bonding functions. In Example 189 of Chapter 2 we observed that the Hilbert cube is the inverse limit on [0, 1] using the single bonding function f : [0, 1] → 2[0,1] given by f (x) = [0, 1] for each x ∈ [0, 1]. However, the arc α = {x ∈ lim f | xi = x1 for each positive integer i} is a proper ←− subcontinuum such that πn (α) = [0, 1] for each positive integer n. In this section we consider closed subsets of inverse limits with upper semi-continuous functions. Our emphasis is on subcontinua of inverse limits with set-valued functions. However, before we restrict our attention to subcontinua, we can make one observation about closed subsets of an inverse limit with upper semi-continuous bonding functions in the form of the following theorem. We leave its proof to the reader. Theorem 208 Suppose X is a compact Hausdorﬀ space and f : X → 2X is an upper semi-continuous function. If Y is a closed subset of X and g : Y → 2Y is an upper semi-continuous function such that G(g) ⊆ G(f ), then lim g ←− is a closed subset of lim f . ←− Alexander N. Cornelius presents a very nice characterization of the compact subsets of an inverse limit with set-valued functions that are inverse limits of their projections in [4]. However, his characterization requires rather detailed knowledge of the inverse limit in order for it to be used. Consequently, next we discuss a speciﬁc variation of this phenomenon that is somewhat easier to verify and, as we show in Section 3.6, turns out to be useful in proving indecomposability of some inverse limits with set-valued functions.

3.5.1 The full projection property One useful application of the fact that closed subsets of inverse limits with mappings are the inverse limit of their projections allows us to conclude that a subcontinuum of the inverse limit that projects onto the entire factor space in inﬁnitely many factors is the entire inverse limit; see Corollary 160 of Chapter 2. This suggests the following deﬁnition. We say that an inverse limit M on continua X1 , X2 , X3 , . . . with upper semi-continuous bonding functions has the full projection property provided if H is a subcontinuum of M such that πi (H) = Xi for inﬁnitely many integers i then H = M . The full

3.5 Closed subsets

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projection property is not just a property of continua that are inverse limits with mappings as may be seen by the following example. Example 209 Let f : [0, 1] → 2[0,1] be the function whose graph consists of the union of three straight line segments, one from (0, 0) to (1/2, 1), one from (1/2, 1) to (1/2, 0), and one from (1/2, 0) to (1, 1). Then, lim f has the full ←− projection property (and is an indecomposable continuum). (See Figure 3.2 for the graph of this function.) Proof. f (x) is connected for each x ∈ [0, 1], thus lim f is a continuum. Let n ←− be a positive integer and Gn = {(x1 , x2 , . . . , xn+1 ) ∈ [0, 1]n+1 | xi ∈ f (xi+1 ) for 1 ≤ i ≤ n}. We ﬁrst show inductively that if (p1 , p2 , . . . , pn+1 ) ∈ Gn and pn+1 ∈ / {0, 1} then Gn − {(p1 , . . . , pn+1 )} is the union of two mutually separated sets: one containing (0, 0, . . . , 0) and the other containing (1, 1, . . . , 1); that is, Gn is an arc with endpoints (0, 0, . . . , 0) and (1, 1, . . . , 1). Because G1 = (G(f ))−1 this is true for n = 1. Assume the statement is true for the positive integer k and let p = (p1 , p2 , . . . , pk+2 ) be a point of Gk+1 such that pk+2 ∈ / {0, 1}. We consider cases: pk+2 = 1/2 and pk+2 = 1/2. If 0 < pk+2 < 1/2, then pk+1 ∈ / {0, 1} so Gk − {(p1 , . . . , pk+1 )} = Ak,0 ∪ Ak,1 where Ak,0 and Ak,1 are mutually separated sets with (0, 0, . . . , 0) ∈ Ak,0 and (1, 1, . . . , 1) ∈ Ak,1 . Let Ak+1,0 = {(x1 , x2 , . . . , xk+2 ) ∈ Gk+1 | xk+2 < 1/2 and (x1 , x2 ,. . . , xk+1 ) ∈ Ak,0 } and let Ak+1,1 = {(x1 , x2 , . . . , xk+2 ) ∈ Gk+1 | (1) xk+2 ≥ 1/2 or (2) xk+2 < 1/2 and (x1 , x2 , . . . , xk+1 ) ∈ Ak,1 }. Then, (0, 0, . . . , 0) is in Ak+1,0 , (1, 1, . . . , 1) is in Ak+1,1 , Ak+1,0 and Ak+1,1 are mutually separated with Gk+1 − {p} = Ak+1,0 ∪ Ak+1,1 . If 1/2 < pk+2 < 1, we obtain the desired separation by letting Ak+1,0 = {(x1 , x2 , . . . , xk+2 ) ∈ Gk+1 | (1) xk+2 ≤ 1/2 or (2) xk+2 > 1/2 and (x1 , x2 , . . . , xk+1 ) ∈ Ak,0 } and Ak+1,1 = {(x1 ,x2 ,. . . , xk+2 ) ∈ Gk+1 | xk+2 > 1/2 and (x1 ,x2 ,. . . ,xk+1 ) ∈ Ak,1 }. If pk+2 = 1/2 there are three possibilities: / {0, 1}. Suppose pk+1 = 0. Note that p = pk+1 = 0, pk+1 = 1, and pk+1 ∈ (0, 0, . . . , 0, 1/2). Let Ak+1,0 = {(x1 ,x2 ,. . . , xk+2 ) ∈ Gk+1 | xk+2 ≤ 1/2} − {p} and Ak+1,1 = {(x1 ,x2 ,. . . , xk+2 ) ∈ Gk+1 | xk+2 > 1/2}. From the observation that p is the only limit point of Ak+1,1 having last coordinate 1/2 it follows that Ak,0 and Ak,1 are mutually separated. The case that pk+1 = 1 is similar. If pk+1 ∈ / {0, 1}, then Gk = Ak,0 ∪ Ak,1 where (0, 0, . . . , 0) ∈ Ak,0 , (1, 1, . . . , 1) ∈ Ak,1 and Ak,0 and Ak,1 are mutually separated. We obtain the desired separation by letting Ak+1,0 = {(x1 ,x2 , . . . ,xk+2 ) ∈ Gk+1 | (1) xk+2 < 1/2 or (2) xk+2 = 1/2 and (x1 , x2 , . . . , xk+1 ) ∈ Ak,1 } and Ak+1,1 = {(x1 , x2 , . . . , xk+2 ) ∈ Gk+1 | (1) xk+2 > 1/2 or (2) xk+2 = 1/2 and (x1 , x2 , . . . , xk+1 ) ∈ Ak,0 }. Now suppose H is a subcontinuum of lim f such that πi (H) = [0, 1] ←− for inﬁnitely many positive integers i. Let p be a point of lim f such that ←− p ∈ / {(0, 0, 0, . . . ), (1, 1, 1, . . . )}. Suppose n is a positive integer. There is a positive integer m ≥ n such that pm+1 ∈ / {0, 1} and πm+1 (H) = [0, 1]. Then, (p1 , . . . , pm+1 ) is in Gm and Gm − {(p1 , . . . , pm+1 )} = Am,0 ∪ Am,1 where (0,

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0, . . . , 0) ∈ Am,0 and (1, 1, . . . , 1) ∈ Am,1 . Thus, H intersects the two mutually separated sets Am,0 × [0, 1]∞ and Am,1 × [0, 1]∞ so H contains a point in the boundary of each of them. Consequently, H contains a point q such that (q1 , q2 , . . . , qm+1 ) = (p1 , p2 , . . . , pm+1 ). Therefore d(q, p) < 2−(m+1) < 2−n . It follows that p ∈ H so H = lim f . ←− The indecomposability of lim f is a consequence of Theorem 212 from the ←− next section.

(1/2,1)

(0,0)

(1,1)

(1/2,0)

Fig. 3.2 The graph of the upper semi-continuous bonding function in Example 209

A second continuum having the full projection property produced by an upper semi-continuous function that is not a mapping is a continuum ﬁrst studied by Scott Varagona. In his analysis of this example Varagona showed that it has the full projection property although the property was not named until later. We include his example without proof, referring the reader to his paper [33] for the details. However, after one shows that the example has the full projection property, its indecomposability is a consequence of Theorem 212 from the next section. Example 210 (Varagona) Let f : [0, 1] → 2[0,1] be the function whose graph is the union of straight line intervals joining (1/2n , 0) and (1/2n−1 , 1)

3.5 Closed subsets

145

for all odd positive integers, straight line intervals joining (1/2n−1 , 0) and (1/2n , 1) for all even positive integers, and the straight line interval joining (0, 0) and (1, 0) (a graph homeomorphic to a sin(1/x)-curve); see Figure 3.3. Then, M = lim f has the full projection property (and is an indecomposable ←− continuum).

(0,1)

(0,0)

(1/4,1)

(1,1)

(1/2,0)

Fig. 3.3 The graph of the upper semi-continuous bonding function in Example 210

As deﬁned here, the full projection property only requires consideration of subcontinua of the inverse limit. If the scope is extended to include all closed subsets of an inverse limit with surjective set-valued bonding functions, one obtains a property that Brian Williams, in a study subsequent to that of Cornelius mentioned early in this section, calls the weak full projection property [34]. Making use of the weak full projection property, Williams characterizes the full projection property [34, Lemma 3.2.4, p.48]. Later in that same section of his dissertation, Williams addresses the general question of when closed subsets of an inverse limit with surjective upper semi-continuous bonding functions are the inverse limit of their projections [34, Corollary 3.2.9]. Perhaps quite useful is a suﬃcient condition for a single bonding function to produce an inverse limit with the weak full projection property [34, Theorem 3.2.11, p. 48].

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3.6 Indecomposability of inverse limits with upper semi-continuous bonding functions

Not much is known about indecomposability of inverse limits when the bonding functions are not mappings. Varagona has shown that if f : [0, 1] → C([0, 1]) is a function such that G(f ) is the sin(1/x)-curve “squeezed” into [0, 1] × [0, 1] (see Example 210), then lim f is indecomposable [33, Theorem ←− 3.2, p. 1024]. In that same article, he also shows that sequences of functions he calls “steeple” functions produce inverse limits homeomorphic to the BJK horseshoe and, consequently, are indecomposable continua [33, Theorem 3.4, p. 1028]. In this section we provide suﬃcient conditions for indecomposability of the inverse limit with a sequence of upper semi-continuous bonding functions. Unfortunately, the full projection property (deﬁned in the previous section) is not easily checked for upper semi-continuous functions that are not mappings (although that condition is satisﬁed by mappings). Varagona provides a suﬃcient condition on a single upper semi-continuous bonding function on [0, 1] so that its inverse limit has the full projection property [33, Lemma 3.1, p. 1022]. If X and Y are Hausdorﬀ continua and f : X → 2Y is an upper semicontinuous function, we say that f satisﬁes the two-pass condition if there are mutually exclusive connected open subsets U and V of X so that f | U and f | V are mappings and f (U ) = f (V ) = Y . If n ≥ 3 is an integer, we say that the continuum T is a simple n-od provided there is a point J of T such that T is the union of n arcs each two of which intersect only at J. We call J the junction point of T and the other endpoints of the arcs that make up T the endpoints of T . Lemma 211 Suppose T is an arc or a simple n-od for some integer n ≥ 3 and T is the union of two proper subcontinua H and K. If U and V are mutually exclusive connected open subsets of T , then one of U and V is a subset of one of H and K. Proof. If T is an arc let J denote a separating point of T and let J be the junction point of T if T is an n-od. The point J cannot belong to both U and V . Suppose J ∈ / U and A is the endpoint of T such that U is a subset of the arc [J, A]. Assume A ∈ H. If J ∈ H then U ⊆ H. If J ∈ / H and U is not a subset of either H or K, then H ∩ K ⊆ U . If A ∈ U then T − U ⊆ K so V ⊆ K. If A ∈ / U then T − U is the union of two mutually exclusive closed sets C and D with A ∈ C. Then V ⊆ C or V ⊆ D. Because C ⊆ H and D ⊆ K, V ⊆ H or V ⊆ K.

3.7 Continua that cannot be obtained with one bonding function

147

One consequence of our next theorem is that the inverse limit from Example 209 is indecomposable. This theorem generalizes a theorem on inverse limits with mappings found in [14, Theorem 3.4, p. 4]. Theorem 212 Suppose T1 , T2 , T3 , . . . is a sequence such that if i is a positive integer then Ti is an arc or there is a positive integer ni so that Ti is a simple ni -od. If fi : Ti+1 → 2Ti is an upper semi-continuous function satisfying the two-pass condition for each positive integer i and lim f is a continuum with ←− the full projection property, then lim f is indecomposable. ←− Proof. Suppose M = lim f is the union of two proper subcontinua H and ←− K. Because M has the full projection property, there is a positive integer n such that if m ≥ n then πm (H) = Tm and πm (K) = Tm . Because fn satisﬁes the two-pass condition there are mutually exclusive connected open subsets U and V of Tn+1 so that fn | U and fn | V are mappings and fn (U ) = fn (V ) = Tn . Because πn+1 (H) and πn+1 (K) are two subcontinua whose union is Tn+1 , by Lemma 211, one of U and V is a subset of one of πn+1 (H) and πn+1 (K). Suppose U ⊆ πn+1 (H). If t ∈ fn (U ) there is a point s of U such that fn (s) = t. Because U ⊆ πn+1 (H), there is a point x of H such that xn+1 = s. Then, xn = t so t ∈ πn (H) and it follows that fn (U ) ⊆ πn (H). However, fn (U ) = Tn , contradicting the fact that πn (H) = Tn . The other possibilities similarly lead to a contradiction. On the other hand, there are inverse limits satisfying the two-pass condition that do not have the full projection property. The following example is due to Varagona. Example 213 Let f : [0, 1] → 2[0,1] be the upper semi-continuous function whose graph consists of three straight line intervals, one from (0, 1) to (0, 0), one from (0, 0) to (1/2, 1), and one from (1/2, 1) to (1, 0). Then, lim f sat←− isﬁes the two-pass condition but it does not have the full projection property (see Figure 3.4). Proof. Let g : [0, 1] → [0, 1] be the full tent map, that is, the map whose graph consists of two straight line intervals, one from (0, 0) to (1/2, 1) and the other from (1/2, 1) to (1, 0). Then, by Theorem 208, lim g is a subcontinuum of ←− lim f . Moreover, lim g projects onto [0, 1] for each positive integer n. However, ←− ←− the point (1, 0, 0, 0, . . . ) is a point of lim f that is not in lim g. ←− ←−

3.7 Continua that cannot be obtained as an inverse limit on [0, 1] using a single set-valued function In Chapter 2, Theorem 186, we included Nall’s proof that [0, 1]2 is not homeomorphic to an inverse limit with a single set-valued function on [0, 1]. Although the square disk is not a ﬁnite graph, this result stimulated interest

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3 Inverse Limits in Continuum Theory (0,1)

(1/2,1)

(0,0)

(1,0)

Fig. 3.4 The upper semi-continuous function f in Example 213

in the question of which ﬁnite graphs can be represented by an inverse limit on [0, 1] with a single upper semi-continuous bonding function. Speciﬁcally, the question of whether a simple closed curve or a simple triod could be so obtained attracted attention. Illanes recently showed that S 1 is not homeomorphic to an inverse limit on [0, 1] using a single bonding function [8]. As this manuscript was being prepared for print, Nall announced that no ﬁnite graph other than an arc can be obtained as such an inverse limit [26].

3.8 Additional topics There are numerous additional topics in continuum theory where inverse limits have been employed that have not been included elsewhere in this monograph. We touch on a few of these in this section.

3.8 Additional topics

149

3.8.1 Span Andrew Lelek’s notion of span played a crucial role in the construction of the ﬁrst published example of an atriodic tree-like continuum that is not chainable, [10]. The continuum constructed there has positive span and, inasmuch as Lelek had earlier shown that chainable continua have span zero, the continuum is not chainable. Subsequent work of Piotr Minc has provided additional means of producing nonchainable tree-like continua. Recently, Logan Hoehn has shown that there exist nonchainable continua whose span is zero [7]. There are several variants of the notion of span and a couple of equivalent means of deﬁning them. Here we choose to deﬁne only one. If f : X → Y is a mapping of continua, by the span of f , denoted σ(f ), we mean the least upper bound of {ε ≥ 0 | there is a subcontinuum Z of X × X such that p1 (Z) = p2 (Z) and d(f (x), f (y)) ≥ ε for each (x, y) ∈ Z} where d is the metric for Y and p1 and p2 are the two projections of X × X onto X. By the span of a space X, denoted σ(X), we mean the span of the identity map on X. The following theorem appears in [17]. Theorem 214 (Lelek) If M is a chainable continuum then σ(M ) = 0. There are many examples of inverse limits in which the factor spaces have positive span but the inverse limit has span zero. The following is one such example. Example 215 Let p : S 1 → [−1, 1] be the projection of S 1 onto [−1, 1] and h : [−1, 1] → [0, 1] be the map given by h(t) = −1/2(t − 1). Let ϕ : [0, 1] → S 1 be the map given by ϕ(t) = e2πit and f : S 1 → S 1 be the map given by f = ϕ ◦ h ◦ p. Then, M = lim f is homeomorphic to an inverse limit on [0, 1] ←− so σ(M ) = 0. Proof. The bonding map f is deﬁned as a map that factors through [0, 1] so, by Theorem 167 from Chapter 2, M is chainable. Thus, σ(M ) = 0. Recall that by the limiting set (or lim sup) of a sequence X of subcontinua of a continuum M , we mean {x ∈ M | if O is an open set containing x and j is a positive integer then there is an integer i ≥ j such that O contains a point of Xi }. It is well known that the limiting set of a sequence of subcontinua of a continuum is a continuum. Lelek made the following observation in [17]. Theorem 216 If X is a sequence of subcontinua of a continuum M with limiting set H and ε is a positive number such that σ(Xi ) ≥ ε for each i, then σ(H) ≥ ε.

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Suppose X is a sequence of continua and f is a sequence of mappings such that fi : Xi+1 → Xi for each i. Recall the notation for i < n, fi n : Xn → Xi where fi n = fi ◦ fi+1 ◦ · · · ◦ fn−1 and fn n denotes the identity on Xn .. Theorem 217 Suppose {Xi , fi } is an inverse limit sequence such that Xi is a continuum and fi is a mapping for each positive integer i. If M = lim f ←− and ε is a positive number such that σ(f1 n ) ≥ ε for each positive integer n, then σ(M ) > 0. Proof. Let q be a point of M and n be a positive integer. The function hn : Xn → i>0 Xi determined by πi (hn (x)) = fi n (x) for i ≤ n and πi (hn (x)) = qi for i > n is a homeomorphism as observed in Theorem 170 of Chapter 2, and, by that same theorem, M is the (sequential) limiting set of the sequence h1 (X1 ), h2 (X2 ), h3 (X3 ), . . . . Because σ(f1 n ) ≥ ε, there is a subcontinuum Zn of Xn × Xn such that p1 (Zn ) = p2 (Zn ) and d1 (f1 n (x), f1 n (y)) ≥ ε for each (x, y) ∈ Zn (where d1 is the metric for X1 ). Then, (hn × hn )(Zn ) is a subcontinuum of hn (Xn ) × hn (Xn ) such that p1 ((hn × hn )(Zn )) = p2 ((hn × hn )(Zn )). If (s, t) ∈ (hn × hn )(Zn ), there is a point (x, y) ∈ Zn such that hn (x) = s and hn (y) = t. But, ρ(s, t) ≥ d1 (f1 n (x), f1 n (y))/2 ≥ ε/2 where ρ is the metric for i>0 Xi . Thus, σ(hn (Xn )) ≥ ε/2. That σ(M ) > 0 now follows from Theorem 216.

3.8.2 Property of Kelley The Property of Kelley is a continuum approximation property that has found wide application in the theory of hyperspaces. In [9] this property is simply called Kelley’s Property or property (κ). There is some literature involving this property and inverse limits. In [30] Dorothy Sherling was concerned with a question of Nadler whether the example of [10] has the Property of Kelley. In a subsequent paper [11] it was shown that the example of [10] has the Property of Kelley. An alternate proof that the example of [10] has the Property of Kelley appears in [12] where a study of permutation maps was initiated. Suppose σ is a permutation on {1, 2, . . . , n}. Partition [0, 1] by a1 , a2 , . . . , an where ai = (i − 1)/(n − 1) for i ∈ {1, 2, . . . , n}. Deﬁne a map fσ : [0, 1] → [0, 1] by letting fσ (ai ) = aσ(i) and extending linearly. Each such map is called a permutation map. A follow-up article, [13], to [12] continued the study of permutation maps as they relate to the Property of Kelley. In [1, Theorem 2.2] Robbie Beane showed that every continuum that is an inverse limit on intervals with a single permutation map has the Property of Kelley. Before stating the theorems, we provide the deﬁnition of the Property of Kelley.

3.8 Additional topics

151

Suppose M is a continuum with metric d. If C is a closed subset of M and ε > 0, we deﬁne U (C, ε) to be {x ∈ M | there is a point y ∈ C such that d(x, y) < ε}. If H and K are subcontinua of M , the Hausdorﬀ distance from H to K, denoted H(H, K), is inf{ε > 0 | H ⊆ U (K, ε) and K ⊆ U (H, ε)}. It is well known that H is a metric for the space of closed sets of M , 2M , as well as the space of subcontinua of M , C(M ) [9]. The continuum M is said to have the Property of Kelley provided if ε > 0 there is a positive number δ such that if p, q ∈ M with d(p, q) < δ and H is a subcontinuum of M containing p then there is a subcontinuum K of M containing q such that H(H, K) < ε. The following theorems are representative of some of the known results about the Property of Kelley in inverse limits. We present them without proof and refer the interested reader to the literature, including [12] and [13]. If f : X → X is a mapping and n is a positive integer, we say that a point x ∈ X is periodic of period n provided f n (x) = x whereas f k (x) = x if 1 ≤ k < n. Theorem 218 [12] Suppose f : [a, b] → → [a, b] is a unimodal map from the interval [a, b] onto [a, b] with critical point c and n is a positive integer such that c is periodic of period n,f (c) = b and f (b) = a. If k is an integer, 1 ≤ k < n, such that f k (a) is the ﬁrst element (in the usual order on [a, b]) of the orbit of c greater than a and n and k are relatively prime, then lim f ←− has the Property of Kelley. Theorem 219 [13] Suppose X1 , X2 , X3 , . . . is a sequence of continua each with the Property of Kelley and f1 , f2 , f3 , . . . is a sequence of conﬂuent mappings such that fi is a mapping from Xi+1 onto Xi . Then lim f has the ←− Property of Kelley. Theorem 220 [1] If f : [0, 1] → → [0, 1] is a permutation map from [0, 1] onto [0, 1] then lim f has the Property of Kelley. ←−

3.8.3 Fixed point property We have previously mentioned the ﬁrst example of a tree-like continuum without the ﬁxed point property provided by David Bellamy [2]. Based on Bellamy’s example, an inverse limit technique due to Brauch Fugate and Lee Mohler provides an example of a tree-like continuum having a homeomorphism without a ﬁxed point [5]. Subsequently, Minc provided both a weakly chainable continuum without the ﬁxed point property [22] and an hereditarily indecomposable tree-like continuum without the ﬁxed point property [23]. (A continuum is said to be weakly chainable provided it is a continuous

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image of the pseudo-arc.) M. M. Marsh has used inverse limit techniques extensively in numerous articles in his studies of the ﬁxed point property. Some samples of his work include [18] through [21].

3.8.4 Hyperspaces Jack Segal [28] showed that the hyperspace of the inverse limit of an inverse limit sequence {Xi , fi } where Xi is a continuum and fi is a mapping for each positive integer i is homeomorphic to the inverse limit on the hyperspaces of subcontinua of the factor spaces with induced mappings as bonding maps. In [25, Theorem 1.169, p. 171] Sam Nadler simpliﬁed Segal’s proof and observed that the theorem holds for Hausdorﬀ continua in inverse limit systems over directed sets. We state this result without providing any details of its proof, instead referring the reader to Nadler’s book. If X and Y are Hausdorﬀ continua and f : X → Y is a mapping, then the induced map, fˆ : C(X) → C(Y ), is given by fˆ(H) = {f (x) | x ∈ H} for H ∈ C(X). Nadler also observes that the following theorem holds if the hyperspace of subcontinua C(X) is replaced by the hyperspace of closed sets 2X . Theorem 221 Suppose {Xα , fα β , D} is an inverse limit system such that Xα is a Hausdorﬀ continuum for each α in the directed set D and each bonding function is a mapping and M = lim f . Then, C(M ) is homeomorphic ←− to the inverse limit of the inverse limit system {C(Xα ), fˆα β , D}. There is some research that involves the topic of this subsection with that of the previous one. Speciﬁcally, Segal and Watanabe use approximate inverse limit systems (a topic not covered in this book) to show that the hyperspace C(X) of an arc-like or circle-like Hausdorﬀ continuum X has the ﬁxed point property [29]. They also show that the hyperspaes C(X) and 2X of a locally connected Hausdorﬀ continuum X have the ﬁxed point property.

References

1. Robbie A. Beane, Inverse Limits of Permutation Maps, Ph.D. dissertation, Missouri University of Science and Technology, 2008. 2. David P. Bellamy, A tree-like continuum without the ﬁxed-point property, Houston J. Math. 6 (1980), 1–13. 3. C. E. Capel, Inverse limit spaces, Duke J. Math. 21 (1954), 233–245. 4. Alexander N. Cornelius, Inverse Limits of Set-Valued Functions, Ph.D. dissertation, Baylor University, 2008. 5. J. B. Fugate and L. Mohler, A note on ﬁxed points in tree-like continua, Topology Proc. 2 (1977), 457–460. 6. J. G. Hocking and G. S. Young, Topology, Addison-Wesley, Reading, MA 1961 (now available from Dover Publications, New York). 7. L. C. Hoehn, A non-chainable plane continuum with span zero, Fund. Math. 211 (2011), 149–174. 8. Alejandro Illanes, A circle is not the generalized inverse limit of a subset of [0, 1]2 , Proc. Amer. Math. Soc. 139 (2011), 2987–2993. 9. Alejandro Illanes and Sam B. Nadler, Jr., Hyperspaces, Marcel Dekker, New York, 1999. 10. W. T. Ingram, An atriodic tree-like continuum with positive span, Fund. Math. 77 (1972), 99–107. 11. W. T. Ingram and D. D. Sherling, Two continua having a property of J. L. Kelley, Canad. Math. Bull. 34 (1991), 351-356. 12. W. T. Ingram, Inverse limits and a property of J. L. Kelley, I, Bol. Soc. Mat. Mexicana 8 (2002), 83–91. 13. W. T. Ingram, Inverse limits and a property of J. L. Kelley, II, Bol. Soc. Mat. Mexicana 9 (2003), 135–150. 14. W. T. Ingram, Two-pass maps and indecomposability of inverse limits of graphs, Topology Proc. 29 (2005), 175–183. 15. Daniel P. Kuykendall, Irreducibility in Inverse Limits of Intervals, Master’s Thesis, University of Houston, 1969. 16. Daniel P. Kuykendall, Irreducibility and indecomposability in inverse limits, Fund. Math. 80 (1973), 265–270. 17. Andrew Lelek, Disjoint mappings and the span of spaces, Fund. Math. 55 (1964), 199–214.

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References

18. M. M. Marsh, A ﬁxed point theorem for inverse limits of simple n-ods, Topology Appl. 24 (1986), 213–216. 19. M. M. Marsh, Mappings of trees and the ﬁxed point property, Trans. Amer. Math. Soc. 315 (1989), 799–810. 20. M. M. Marsh, Covering spaces, inverse limits, and induced coincidence producing mappings, Houston J. Math. 29 (2003), 983–992. 21. M. M. Marsh, Inverse limits of projective spaces and inverse limits, Houston J. Math. 36 (2010), 945–957. 22. Piotr Minc, A weakly chainable continuum without the ﬁxed point property, Trans. Amer. Math. Soc. 351 (1999), 1109–1121. 23. Piotr Minc, A hereditarily indecomposable tree-like continuum without the ﬁxed point property, Trans. Amer. Math. Soc. 352 (2000), 643–654. 24. Sam B. Nadler, Jr., Multicoherence techniques applied to inverse limits, Trans. Amer. Math. Soc. 157 (1971), 227–234. 25. Sam B. Nadler, Jr., Hyperspaces of Sets, Marcel Dekker, New York, 1978 (also available as Aportaciones Matematic´ as: Textos, vol. 33, Sociedad Matem´ atica Mexicana). 26. Van Nall, No ﬁnite graph diﬀerent from an arc is a generalized inverse limit with a closed subset of [0, 1]2 , preprint. 27. David J. Ryden, Irreducibility in inverse limits on intervals, Fund. Math. 165 (2000), 29–53. 28. Jack Segal, Hyperspaces of the inverse limit space, Proc. Amer. Math. Soc. 10 (1959), 706–709. 29. J. Segal and T. Watanabe, Cosmic approximate inverse limits and ﬁxed points, Trans. Amer. Math. Soc. 333 (1992), 1–61. 30. D. D. Sherling, Concerning the cone=hyperspace property, Canad. J. Math. 35 (1983), 1030–1048. 31. R. H. Sorgenfrey, Concerning triodic continua, Amer. J. Math. 66 (1944), 439–460. 32. R. H. Sorgenfrey, Concerning continua irreducible about n points, Amer. J. Math. 68 (1946), 667–671. 33. Scott Varagona, Inverse limits with upper semi-continuous bonding functions and indecomposability, Houston J. Math. 37 (2011), 1017–1034. 34. Brian R. Williams, Indecomposability in Inverse Limits, Ph.D. dissertation, Baylor University, 2010.

Chapter 4

Brown’s Approximation Theorem

Abstract This chapter is devoted to an approximation theorem that allows certain adjustments of the bonding mappings in an inverse limit sequence without altering the topology of the inverse limit. We apply this fundamental result to show that if f is a mapping of an interval J onto itself that is the union of two monotone maps from nonoverlapping subintervals onto J, then the inverse limit using f as a single bonding map is the familiar BJK horseshoe.

4.1 Introduction A valuable tool in the theory of inverse limits is a theorem proved by Morton Brown [1, Theorem 3, p. 481]. This theorem allows certain adjustments in the sequence of bonding maps without changing the topology of the inverse limit. For example, as we show in an application of Brown’s theorem in the second section of this chapter, in inverse limits on intervals, this theorem allows certain alterations in the bonding maps such as the removal of “ﬂat spots” while maintaining a homeomorphic inverse limit. Sarah Holte [3] makes use of Brown’s theorem in showing that, for unimodal interval maps with ﬁnite kneading sequences, the kneading sequence and the dynamics of the left endpoint determine the topology of the inverse limit. In this chapter we provide Brown’s proof of Theorem 231 along with an application, Theorem 234, of Brown’s theorem.

4.2 Brown’s theorem Throughout this section we assume X = X1 , X2 , X3 , . . . is a sequence of compact metric spaces. All inverse limit sequences in this chapter are based W.T. Ingram and W.S. Mahavier, Inverse Limits: From Continua to Chaos, Developments in Mathematics 25, DOI 10.1007/978-1-4614-1797-2_4, © Springer Science+Business Media, LLC 2012

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on this sequence of factor spaces; that is, if we say f = f1 , f2 , f3 , . . . is an inverse limit sequence we mean that fi : Xi+1 → Xi is a mapping of Xi+1 into Xi for each positive integer i. We denote by dn a metric bounded by 1 for the factor space Xn and by d the usual metric on i>0 Xi given by d(x, y) = i i≥1 di (xi , yi )/2 . If f is an inverse limit sequence and i and j are positive integers with i < j, we denote by fi j the composition fi ◦ fi+1 ◦ · · · ◦ fj−1 mapping Xj into Xi . It is convenient to let fi i denote the identity on Xi , IdXi . Most of the lemmas and theorems in this section as well as their proofs are adaptations of theorems and proofs found in [1]. A sequence δ of positive numbers is called a Lebesgue sequence for the inverse limit sequence f provided if i and j are positive integers with i < j and x and y are points of Xj such that dj (x, y) < δj then di (fi j (x), fi j (y)) < 2−j . The following lemma is an immediate consequence of the uniform continuity of the bonding maps deriving from the compactness of the factor spaces. Lemma 222 Each inverse limit sequence has a Lebesgue sequence. Suppose X and Y are compact metric spaces and d is the metric for Y . If f : X → Y and g : X → Y are functions, let d(f, g) = l.u.b. {d(f (x), g(x)) | x ∈ X}. If f : X → Y and g : Y → Z are functions, for convenience we denote g ◦ f by gf . Observe that if f : X → Y , g : X → Y , and ϕ : Z → X are functions, then d(f ϕ, gϕ) ≤ d(f, g). We make use of this near the end of the proof of our next lemma as well as near the end of the proof of Lemma 226. Lemma 223 Suppose f and g are inverse limit sequences, and, for each positive integer k, πk is the projection of lim f into Xk . If 1 ≤ n ≤ i < j then ←− dn (gn i πi , gn j πj ) ≤

j−1

dn (gn r fr , gn r gr ).

r=i

Proof. Suppose i, j, and n are integers such that 1 ≤ n ≤ i < j. Note that dn (gn i πi , gn j πj ) ≤ dn (gn i gi i πi , gn i gi i+1 πi+1 ) + dn (gn i gi i+1 πi+1 , gn i gi j πj ) ≤ dn (gn i gi i πi , gn i gi i+1 πi+1 ) + dn (gn i gi i+1 πi+1 , gn i gi i+2 πi+2 ) + dn (gn i gi i+2 πi+2 , gn i gi j πj ). Continuing inductively, we have dn (gn i πi , gn j πj ) ≤

j−1 r=i

dn (gn i gi r πr , gn i gi r+1 πr+1 ).

4.2 Brown’s theorem

157

Because πk denotes the projection of lim f into Xk , if i ≤ r < j, πr = ←− fr πr+1 . Because gn r = gn i gi r , we have gn i gi r πr = gn r fr πr+1 . Furthermore gn i gi r+1 πr+1 = gn r gr πr+1 and dn (gn r fr πr+1 , gn r gr πr+1 ) ≤ dn (gn r fr , gn r gr ) from which the conclusion of the lemma follows.

Lemma 224 Suppose f and g are inverse limit sequences, δ is a Lebesgue sequence for g, and di (fi , gi ) < δi for each positive integer i. Denote by πk the projection of lim f into Xk . Then, if n is a positive integer, the sequence ←− gn n πn , gn n+1 πn+1 , gn n+2 πn+2 , . . . is a Cauchy sequence. Proof. Suppose n is a positive integer and ε > 0. There is a positive integer N , n ≤ N , such that r≥N 2−r < ε. Suppose i, j are positive integers with N ≤ i < j. Then, from Lemma 223 we have dn (gn i πi , gn j πj ) ≤

j−1

dn (gn r fr , gn r gr ).

r=i

Inasmuch as δ is a Lebesgue sequence for g and dk (fk , gk ) < δk for each positive integer k, if i ≤ r < j then dn (gn r fr , gn r gr ) ≤ 2−r . Thus, dn (gn i πi , gn j πj ) < ε. Theorem 225 Suppose f and g are inverse limit sequences, δ is a Lebesgue sequence for g and di (fi , gi ) < δi for each positive integer i. Let M = lim f ←− and K = lim g, and denote by πi the projection of M into Xi . Then for each ←− positive integer n the sequence gn n πn , gn n+1 πn+1 , gn n+2 πn+2 , . . . converges uniformly to a mapping F n : M → Xn . Moreover, if i < j then gi j Fj = Fi so the function F : M → i>0 Xi given by F (x) = (F1 (x), F2 (x), F3 (x), . . .) is a mapping of M into K. Proof. Suppose n is a positive integer. That the sequence gn n πn , gn n+1 πn+1 , gn n+2 πn+2 , . . . converges uniformly for each positive integer n is a consequence of Lemma 224. Denote the limit function by Fn . The uniform limit of a sequence of continuous functions is continuous, so Fn is continuous. F is coordinatewise continuous, therefore it is a mapping. Suppose i and j are positive integers with i < j. To see that gi j Fj = Fi , let x be a point of M . The sequence gj j πj (x), gj j+1 πj+1 (x), gj j+2 πj+2 (x), . . . converges to Fj (x) from which it follows that the sequence gi j gj j πj (x), gi j gj j+1 πj+1 (x), gi j gj j+2 πj+2 (x), . . . converges to gi j Fj (x). But, for k ≥ j, gi j (gj k (πk (x))) = gi k (πk (x)). The sequence gi i (πi (x)), gi i+1 (πi+1 (x)), gi i+2 (πi+2 (x)), . . . converges to Fi (x), thus it follows that gi j (Fj (x)) = Fi (x). Next, we show that the mapping deﬁned in Theorem 225 is surjective. However, in the proof we need a lemma that is similar to Lemma 223.

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Lemma 226 Suppose f and g are inverse limit sequences. If 1 ≤ n ≤ i < j then j−1 dn (gn r fr , gn r gr ). dn (gn i fi j , gn i gi j ) ≤ r=i

Proof. Proceeding in a manner similar to that used in the proof of Lemma 223, if i, j, and n are integers such that 1 ≤ n ≤ i < j then dn (gn i fi j , gn i gi j ) ≤

j−1

dn (gn i gi r fr j , gn i gi r+1 fr+1 j ).

r=i

For i ≤ r < j we have gn i gi r fr j = gn r fr fr+1 j , gn i gi r+1 fr+1 j = gn r gr fr+1 j , and dn (gn r fr fr+1 j , gn r gr fr+1 j ) ≤ dn (gn r fr , gn r gr ), therefore the conclusion of the lemma follows.

Theorem 227 Suppose f and g are inverse limit sequences and δ is a Lebesgue sequence for g. Let M = lim f , K = lim g, and denote by πi ←− ←− the projection of M into Xi . If di (fi , gi ) < δi for each positive integer i, the mapping F : M → K as deﬁned in Theorem 225, given by F (x) = (F1 (x), F2 (x), F3 (x), . . .) is surjective. Proof. Suppose y is in K and n is a positive integer. We ﬁrst show that there is a sequence p1 , p2 , p3 , . . . of points of M such that if k is a positive integer then dn (Fn (pk ), yn ) < 1/k. To that end let k be a positive integer. There is a positive integer i > n such that dn (gn i πi , Fn ) < 1/(3k) and we may assume that i is large enough that −r < 1/(3k) as well. Suppose j is a positive integer greater than r≥i 2 i. Because δ is a Lebesgue sequence for g and, by hypothesis, dr (fr , gr ) < δr , we have dn (gn r fr , gn r gr ) ≤ 2−r for each positive integer r with i ≤ r ≤ j − 1. j−1 Using Lemma 226, dn (gn i fi j , gn i gi j ) ≤ r=i 2−r < 1/(3k). It follows from Theorem 113 in Chapter 2, πi (M ) = j>i fi j (Xj ), so there is a positive integer j > i such that if t is a point of fi j (Xj ) then there is a point s of πi (M ) such that di (s, t) < δi . Because yj is in Xj , there is a point p of M such that di (pi , fi j (yj )) < δi . Because δ is a Lebesgue sequence for g, dn (gn i (pi ), gn i fi j (yj )) < 2−i . Then, because yn = gn i (gi j (yj )), dn (Fn (p), yn ) ≤ dn (Fn (p), gn i (πi (p))) + dn (gn i (πi (p)), gn i fi j (yj )) + dn (gn i fi j (yj ), gn i gi j (yj )) < 1/(3k) + 2−i + 1/(3k) < 1/k. Let pk = p. We next show that there is a point xn of M such that Fn (xn ) = yn . Some subsequence of the sequence p1 , p2 , p3 , . . . converges. For convenience of notation we assume p1 , p2 , p3 , . . . converges. Let xn be its sequential limit point. Suppose ε > 0. Because Fn is continuous, there is a positive number δ such that if q belongs to M and d(xn , q) < δ then dn (Fn (xn ), Fn (q)) < ε/2.

4.2 Brown’s theorem

159

There is a positive integer N such that if k ≥ N then dn (pk , xn ) < δ, so for k ≥ N , dn (Fn (xn , Fn (pk ))) < ε/2. There is a positive integer k ≥ N such that 1/k < ε/2. For such an integer k, we have dn (Fn (xn ), yn ) < 1/k. Thus, dn (Fn (xn ), yn ) ≤ dn (Fn (xn ), Fn (pk )) + dn (Fn (pk ), yn ) < ε/2 + 1/k < ε. It follows that Fn (xn ) = yn . We now have a sequence x1 , x2 , x3 , . . . of points of M such that Fi (xi ) = yi for each i. Suppose i is a positive integer. Observe that, by Theorem 225, if j > i, gi j Fj = Fi , so Fi (xj ) = gi j Fj (xj ) = gi j (yj ) = yi . Some subsequence of the sequence x1 , x2 , x3 , . . . converges to a point x of M . For convenience of notation we may assume x1 , x2 , x3 , . . . converges to x. Because Fi is continuous, Fi (x1 ), Fi (x2 ), Fi (x3 ), . . . converges to Fi (x). But, for j > i, Fi (xj ) = yi so we have that Fi (x) = yi . It follows that F (x) = y. Under additional conditions on the sequence g, the surjective mapping F from Theorem 225 is actually a homeomorphism. To show this we ﬁrst introduce Brown’s notion of a measure for an inverse limit sequence. We say that a sequence c of positive numbers is a measure for the inverse limit sequence f provided (1) i≥n+1 ci < cn /2 for n = 1, 2, 3, . . . and (2) if x and y are in lim f and x = y, then there is a positive integer n such that ←− dn+1 (xn+1 , yn+1 ) > cn . Lemma 228 Each inverse limit sequence has a measure. Proof. Let δ be a Lebesgue sequence for the inverse limit sequence f . Let for each positive integer i, let ci+1 = (1/2) min {ci /2, δi+2 }. c1 = δ2 /2 and, We thus have i≥n+1 ci < cn /2 for n = 1, 2, 3, . . .. Suppose x and y are points of lim f and x = y. There is a positive integer k such that xk = yk . ←− There is a positive integer n > k such that dk (xk , yk ) > 2−(n+1) . Then, dn+1 (xn+1 , yn+1 ) > cn for if dn+1 (xn+1 , yn+1 ) ≤ cn then dn+1 (xn+1 , yn+1 ) < δn+1 . Because δ is a Lebesgue sequence for f , dk (xk , yk ) < 2−(n+1) . This is a contradiction. Suppose (X, d) and (Y, ρ) are compact metric spaces. If f : X → Y is a mapping and ε > 0, let L(ε, f ) = l.u.b. {δ ≤ diam(X) | if x, y ∈ X and d(x, y) < δ then ρ(f (x), f (y)) < ε}. Lemma 229 Suppose f is an inverse limit sequence with measure c, and g is an inverse limit sequence. If n and i are positive integers with n < i and dk (fk , gk ) < L(ck , gn k ) for n ≤ k ≤ i − 1, then dn (fn i , gn i ) ≤ i−1 j=n cj . Proof. Recalling that fn i = fn i−1 fi−1 and gn i = gn i−1 gi−1 we have dn (gn i , fn i ) ≤ dn (gn i−1 gi−1 , gn i−1 fi−1 ) + dn (gn i−1 fi−1 , fn i−1 fi−1 ).

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4 Brown’s Approximation Theorem

Inasmuch as di−1 (fi−1 , gi−1 ) < L(ci−1 , gn i−1 ), dn (gn i−1 gi−1 , gn i−1 fi−1 ) ≤ ci−1 . This and the fact that dn (gn i−1 fi−1 , fn i−1 fi−1 ) ≤ dn (gn i−1 , fn i−1 ) yields dn (gn i , fn i ) ≤ ci−1 + dn (gn i−1 , fn i−1 ). In a similar manner we obtain dn (gn i−1 , fn i−1 ) ≤ ci−2 + dn (gn i−2 , fn i−2 ), and therefore dn (gn i , fn i ) ≤ ci−1 + ci−2 + dn (gn i−2 , fn i−2 ). Proceeding inductively and recalling that gn n = fn n = IdXn , the lemma follows. Theorem 230 Suppose f is an inverse limit sequence with measure c, g is an inverse limit sequence with Lebesgue sequence δ, and di (fi , gi ) < δi for each positive integer i. If, moreover, for each two positive integers n and i with n < i, dk (fk , gk ) < L(ck , gn k ) for each k satisfying n ≤ k ≤ i − 1, then the function F as deﬁned in Theorem 225 is one-to-one and, thus, is a homeomorphism from lim f onto lim g. ←− ←− Proof. Beause δ is a Lebesgue sequence for g and di (fi , gi ) < δi for each i, the function F deﬁned in Theorem 225 is continuous, and, by Theorem 227, F is surjective. To see that F is 1-1, suppose t and x are points of lim f , x = t, ←− and F (t) = F (x). Because c is a measure for f there is an integer n > 1 such that dn (xn , tn ) > cn−1 . Suppose i > n. Then, xn = fn i (xi ) and tn = fn i (ti ), therefore dn (xn , tn ) ≤ dn (fn i (xi ), gn i (xi )) + dn (gn i (xi ), gn i (ti )) + dn (gn i (ti ), fn i (ti )). Using the fact that F (x) = F (t), dn (gn i (xi ), gn i (ti )) ≤ dn (gn i (xi ), Fn (x)) + dn (Fn (t), gn i (ti )). Using this and Lemma 229 we obtain dn (xn , tn ) ≤ cj + dn (gn i (xi ), Fn (x)) + dn (Fn (t), gn i (ti )) + cj . j≥n

j≥n

Let ε = cn−1 /2 − j≥n cj . By the deﬁnition of F , we may choose i large enough that d n (gn i πi (x), F (x)) < ε and dn (F (t), gn i πi (t)) < ε. Thus, dn (xn , tn ) ≤ 2 j≥n cj + 2ε = cn−1 contrary to the choice of n. Theorem 231 (Brown) Suppose f is an inverse limit sequence with each factor space a compact metric space. For each positive integer i let Ki denote a collection of mappings from Xi+1 into Xi such that if ε > 0 then there is a map g ∈ Ki such that di (fi , g) < ε. Then there is a sequence g of mappings such that, for each i, gi ∈ Ki and lim f is homeomorphic to lim g. ←− ←−

4.3 An application of Brown’s theorem

161

Proof. Let c be a measure for f . We construct sequences g and δ such that 1. 2. 3. 4.

gi ∈ Ki for each positive integer i di (fi , gi ) < δi for each positive integer i δ is a Lebesgue sequence for g For each positive integer i, δi ≤ min{L(ci , gp i ) | p ≤ i}.

Let δ1 = L(c1 , IdX1 ). There is a mapping g1 ∈ K1 such that d1 (f1 , g1 ) < δ1 . There is a positive number δ2 < min{L(c2 , IdX2 ), L(c2 , g1 )} such that if x and y are in X2 and d2 (x, y) < δ2 then d1 (g1 (x), g1 (y)) < 2−2 . Inductively, suppose m ≥ 1 and g1 , g2 , . . . , gm and δ1 , δ2 , . . . , δm+1 have been deﬁned satisfying • gi ∈ Ki for 1 ≤ i ≤ m • di (fi , gi ) < δi for 1 ≤ i ≤ m • If j is a positive integer, j ≤ m+1, x and y belong to Xj , and dj (x, y) < δj , then di (gi gi+1 · · · gj−1 (x), gi gi+1 · · · gj−1 (y)) < 2−j for 1 ≤ i < j • δi < min{L(ci , gk gk+1 · · · gi−1 ) | 1 ≤ k < i} for each integer i, 1 ≤ i ≤ m + 1. There exists a map gm+1 ∈ Km+1 such that dm+1 (fm+1 , gm+1 ) < δm+1 . There is a positive number δm+2 < min{L(cm+2 , IdXm+2 ), L(cm+2 , gm+1 ), L(cm+2 , gm gm+1 ), . . . , L(cm+2 , g1 g2 . . . gm+1 }) such that for x and y in Xm+2 with dm+2 (x, y) < δm+2 we have di (gi gi+1 . . . gm+1 (x), gi gi+1 . . . gm+1 (y)) < 2−(m+2) for 1 ≤ i < m+2. The ﬁrst two conditions of the inductive hypothesis hold for m + 1 by the choice of gm+1 whereas the choice of δm+2 ensures that the other two conditions hold to complete the induction. Note that if n and i are positive integers with n < i and k is an integer such that n ≤ k < i then condition (4) yields that δk < L(ck , gp k ) for p ≤ k. Because n < k we have δk < L(ck , gn k ) and, consequently, from condition (2) that dk (fk , gk ) < L(ck , gn k ). Thus we have constructed an inverse limit sequence g with gi ∈ Ki for each i such that f with measure c and g with Lebesgue sequence δ satisfy the hypotheses of Theorem 230, therefore lim f ←− is homeomorphic to lim g under the homeomorphism from Theorem 230. ←−

4.3 An application of Brown’s theorem In this section we present an application of Theorem 231. Let T : [0, 1] → → [0, 1] be the map of the interval [0, 1] onto itself given by 2x, 0 ≤ x ≤ 1/2, T (x) = 2 − 2x, 1/2 ≤ x ≤ 1.

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4 Brown’s Approximation Theorem

The inverse limit, lim T , is the BJK horseshoe. In this section we show that if ←− f is a sequence of maps from an interval [a, b] onto itself each term of which is a union of two homeomorphisms from nonoverlapping subintervals onto [a, b], then lim f is homeomorphic to lim T . We then show that if f : [a, b] → → [a, b] ←− ←− is a mapping that is the union of two monotone mappings, then lim f is also ←− homeomorphic to lim T . For a picture representing the BJK horseshoe, see ←− Figure 1.5 in Chapter 1. Theorem 232 Suppose f1 , f2 , f3 , . . . is a sequence of mappings of the interval [a, b] onto itself such that for each i there exist a number ci , a < ci < b, and homeomorphisms hi : [a, ci ] → [a, b] and ki : [ci , b] → [a, b] such that hi (x), a ≤ x ≤ ci , fi (x) = ki (x), ci ≤ x ≤ b. Then, lim f is a BJK horseshoe. ←− → [a, b] be the linear homeomorphism given by ϕ1 (x) = Proof. Let ϕ1 : [0, 1] → (b − a)x + a. Inductively, let ϕi+1 : [0, 1] → → [a, b] be the homeomorphism given by 0 ≤ x ≤ 1/2, h−1 i (ϕi (T (x))), ϕi+1 (x) = −1 ki (ϕi (T (x))), 1/2 ≤ x ≤ 1. ϕn ◦ T = fn ◦ ϕn+1 for each n, thus the sequence ϕ1 , ϕ2 , ϕ3 , . . . induces a homeomorphism of lim T onto lim f . ←− ←− A map f from a continuum X onto a continuum Y is conﬂuent provided f (H) = K for each subcontinuum K of Y and each component H of f −1 (K). For a conﬂuent mapping f of an interval [a, b] onto itself, there exist ﬁnitely many surjective monotone maps m1 , m2 , . . . , mk such that f = m1 ∪ m2 ∪ · · · ∪ mk . The integer k is sometimes called the degree of f and is denoted deg(f ). Theorem 233 Suppose f : [a, b] → → [a, b] is a conﬂuent mapping and m1 : [a, c] → → [a, b] and m2 : [c, b] → → [a, b] are surjective monotone mappings such that m1 (x), a ≤ x ≤ c, f (x) = m2 (x), c ≤ x ≤ b. There exists a sequence g1 , g2 , g3 , . . . of mappings of [a, b] onto itself such that for each i there are surjective homeomorphisms hi : [a, c] → → [a, b] and → [a, b] with ki [c, b] → hi (x), a ≤ x ≤ c, gi (x) = ki (x), c ≤ x ≤ b,

4.3 An application of Brown’s theorem

163

and lim f is homeomorphic to lim g. ←− ←− Proof. For convenience, we assume f (a) = f (b) = a and f (c) = b. Let g be the open mapping of [a, b] onto itself given by g(x) = (b−a)(x−a)/(c−a) + a for a ≤ x ≤ c and g(x) = (b−a)(x−b)/(c−b) + a for c ≤ x ≤ b. For each number ε, 0 < ε < 1, let gε : [a, b] → [a, b] be given by gε (x) = (1−ε)f (x) + εg(x) and let K = {gε | 0 < ε < 1}. Note that d(f, gε ) ≤ ε d(f, g) where d is determined by the usual metric for [a, b]. Thus we may apply Theorem 231 to obtain a sequence g1 , g2 , g3 , . . . of elements of K such that lim f is homeomorphic to ←− lim g by Corollary 146 from Chapter 2. ←− The following corollary generalizes Theorem 35 from Chapter 1. Corollary 234 (Davis) If f : [a, b] → → [a, b] is a conﬂuent mapping that is the union of two monotone mappings, then lim f is homeomorphic to the ←− BJK horseshoe. The techniques used in this section could be used to show that if f is a sequence of conﬂuent mappings from [0, 1] onto [0, 1] and g is a sequence of open mappings from [0, 1] onto [0, 1] such that deg(fi ) = deg(gi ) for each i, then lim f and lim g are homeomorphic. This general result may be found in ←− ←− [2].

References

1. Morton Brown, Some applications of an approximation theorem for inverse limits, Proc. Amer. Math. Soc. 11 (1960), 478–483. 2. James F. Davis, Conﬂuent mappings on [0, 1] and inverse limits, Topology Proc. 15 (1990), 1–9. 3. Sarah Holte, Inverse limits of Markov interval maps, Topology Appl. 123 (2002), 421–427.

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Chapter 5

Appendix: An Introduction to the Hilbert Cube

Abstract This appendix is devoted to the topological background for the ﬁrst chapter of this book

5.1 Introduction In this appendix, we include the background material that an advanced undergraduate or beginning graduate student needs to read the ﬁrst chapter of this book. Throughout the book we assume that the reader has some background in topology. A typical undergraduate course in topology or even a senior-level course in analysis should be suﬃcient. There are many good books that contain much of the required material. One such book is Fred H. Croom’s book, Principles of Topology [1]. Sam Nadler’s book, Continuum Theory [5], is a more advanced source that contains much of the needed background as well. In an eﬀort to make this book as self-contained as possible, however, we have included in this appendix the deﬁnitions and theorems needed to read our ﬁrst chapter.

5.2 A brief introduction to topology If X is a set, a collection T of subsets of X is a topology for X provided 1. Both X and the empty set, ∅, belong to T 2. If G ⊆ T then the union of all of the sets in G is in T 3. If H is a ﬁnite subcollection of T then the common part (intersection) of all of the elements of H is in T . W.T. Ingram and W.S. Mahavier, Inverse Limits: From Continua to Chaos, Developments in Mathematics 25, DOI 10.1007/978-1-4614-1797-2_5, © Springer Science+Business Media, LLC 2012

167

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5 Appendix: An Introduction to the Hilbert Cube

The pair (X, T ) is called a topological space, the elements of X are often called points, and the elements of T are called open sets. Normally, when it is clear from the context which topology we intend for X we simply call X a topological space. A point x of X is said to be a limit point of a subset M of X provided each open set containing x contains a point of M distinct from x. A subset M of X is closed provided if x is a limit point of M then x ∈ M (or equivalently, X − M is open). If M is a subset of a topological space X, M = {x ∈ X | x ∈ M or x is a limit point of M }. A topological space X is called a Hausdorﬀ space provided if x and y are in X there exist mutually exclusive open sets one containing x and the other containing y. All spaces we consider in this book, including this appendix, are Hausdorﬀ spaces. It is often more convenient to deﬁne a topology on a set by specifying a basis for the topology. A collection B of subsets of X is called a basis for a topology for X provided 1. Each element of X belongs to some element of B 2. If U and V belong to B and x ∈ U ∩ V then there is an element W of B such that x ∈ W ⊆ U ∩ V . It is not diﬃcult to show that if B is a basis for a topology for X then T = {O | there is a subcollection C of B such that O is the union of all of the elements of C} is a topology for X. Neither is it diﬃcult to show that if B is a basis for a topology for X and T is the topology for X generated in this manner, then a point p is a limit point of a set M if and only if each element of B containing p contains a point of M distinct from p. A fundamental property of topological spaces is found in the following theorem. Theorem 235 If p is a limit point of the union A ∪ B of two subsets of a topological space, then p is a limit point of A or p is a limit point of B. Proof. If p is not a limit point of A then there is an open set U containing p such that U ∩ A = ∅. If p is not a limit point of B then there is an open set V containing p such that V ∩ B = ∅. Then, W = U ∩ V is an open set containing p such that W ∩ (A ∪ B) = ∅ so p is not a limit point of A ∪ B. A function is a set of ordered pairs such that no two pairs in the set have the same ﬁrst term. If f is a function, the set of all ﬁrst terms of pairs in f is called the domain of f and the set of all second terms of pairs in f is called the range of f . If f is a function and (x, y) ∈ f , we often write y = f (x). If f is a function with domain X and range a subset of Y we often write f : X → Y whereas if Y is the set of all second terms of pairs in f we say that f is a function from X onto Y and write f : X → → Y . In the case where f :X→ → Y is a function from X onto Y , we sometimes denote Y by f (X). If X is a set, a function d : X × X → R where R is the set of real numbers is called a metric for X provided for each x and y belonging to X,

5.3 The Hilbert cube

1. 2. 3. 4.

169

d(x, y) ≥ 0 d(x, y) = 0 if and only if x = y d(x, y) = d(y, x) d(x, y) ≤ d(x, z) + d(z, y) for each z ∈ X.

Suppose X is a set and d is a metric for X. If x ∈ X and ε > 0 we deﬁne B(x, ε) = {y ∈ X | d(x, y) < ε}. It is not diﬃcult to show that B = {B(x, ε) | x ∈ X and ε > 0} is a basis for a topology for X. This topology is called the metric topology for X generated by d and the pair (X, d) is sometimes called a metric space. A sequence is a function deﬁned on the set of positive integers N, although sometimes it is convenient for the domain of a sequence to be N ∪ {0}. Normally we specify a sequence s more simply by listing its terms (i.e., its range) s(1), s(2), s(3), . . . , or s1 , s2 , s3 , . . . . In this appendix we employ the convention used throughout the book of denoting sequences in boldface type and the terms of a sequence in italics. A sequence x of points of X is said to converge to a point p of X provided if O is an open set containing p then there is a positive integer N such that if n ≥ N then xn ∈ O.

5.3 The Hilbert cube

All of the inverse limits studied in the ﬁrst few sections of our ﬁrst chapter are subsets of the Hilbert cube Q. Those from the entire chapter that are not directly embedded in Q are subsets of a product of intervals that, in turn, is homeomorphic to Q under a natural homeomorphism. As a consequence some understanding of the topology of Q is crucial to understanding inverse limits on intervals, so in this section we develop some basic topological properties of Q. The points of Q are sequences of real numbers from the interval [0, 1]; that is, a point of Q is a sequence x = x1 , x2 , x3 , . . . where xi ∈ [0, 1] for each positive integer i. We normally denote a point x of Q by (x1 , x2 , x3 , . . . ). There are two equivalent ways normally employed to endow Q with a topology. One way is to introduce a metric for Q and give Q the metric topology. An alternative is to endow Q with the product topology. Each of these has its advantages. The reader who may need to read parts of this appendix may more likely be familiar with metric spaces, therefore we begin with a metric on Q. Then we develop the product topology for Q and ﬁnally we close this section with a proof that the topologies are identical.

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5.3.1 The metric topology for Q If each of x and y is a point of Q, let d(x, y) =

|xi − yi | i>0

2i

.

Thus deﬁned, d is a metric for Q, which we demonstrate in our next theorem. Theorem 236 d is a metric for Q. Proof. Only the triangle inequality needs any argument. To that end suppose that each of x, y, and z is a point of Q. Suppose integer. If n n is a positive n i − z | ≤ |x − y | + |y − z | so |x − z |/2 ≤ 1 ≤ i ≤ n, |x i i i i i i i i i=1 i=1 |xi − n yi |/2i + i=1 |yi − zi |/2i ≤ d(x, y) + d(y, z). The triangle inequality now follows. If x ∈ Q and ε > 0, let B(x, ε) = {y ∈ Q | d(x, y) < ε}. As discussed earlier, a basis for the metric topology for Q is the collection of all sets B(x, ε) where x ∈ Q and ε > 0. We refer to this basis as the usual basis for the metric topology for Q.

5.3.2 The product topology for Q If X1 , X2 , X3 , . . . is a sequence of sets, the product of the sets, denoted xi ∈ Xi i>0 Xi , is the collection of all sequences x1 , x2 , x3 , . . . such that for each positive integer i. As with the points of the cube, if x ∈ i>0 Xi , we normally denote x by (x1 , x2 , x3 , . . . ). If x ∈ Q, a basis element for the product topology for Q containing x is i>0 Oi where Oi is an open subset of [0, 1] containing xi for each positive integer i and there is a positive integer n such that Oi = [0, 1] for i > n (the reader should note that this does not preclude that some of the open sets Oi are [0, 1] for i ≤ n). A basis for the product topology for Q is the collection of all such basis elements. The reader who has not done so should construct an argument that this collection is a basis for a topology for Q. We refer to this basis as the usual basis for the product topology for Q.

5.4 A countable basis for the topology of Q

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5.3.3 The metric topology and the product topology for Q are identical By means of the next two theorems, we conclude that a subset of Q is open in the metric topology if and only if it is open in the product topology. This is accomplished by showing that each B(x, ε) is open in the product topology and each element of the usual basis for the product topology is open in the metric topology. The next two theorems are suﬃcient to show this. Theorem 237 If O is an element of the usual basis for the product topology for Q and p is a point of O, there is a positive number ε such that B(p, ε) ⊆ O. Proof. Suppose p ∈ O = i>0 Oi where Oi is open in [0, 1] for each positive integer i and n is a positive integer such that Oi = [0, 1] for i > n. For each integer i, 1 ≤ i ≤ n, let εi be a positive number such that if t ∈ I and |t − pi | < εi then t ∈ Oi . Let ε be the least of ε1 /2, ε2 /4, . . . , εn /2n . Suppose x ∈ Q. If x ∈ / O, there is an integer j, 1 ≤ j ≤ n, such that xj ∈ / Oj . Then, |xj − pj | ≥ εj so |xj − pj |/2j ≥ ε and we have d(p, x) ≥ ε. It follows that B(p, ε) ⊆ O. Theorem 238 If x ∈ Q and ε > 0 and y ∈ B(x, ε) there exists an element O of the usual basis for the product topology such that y ∈ O and O ⊆ B(x, ε). Proof. Suppose x ∈ Q and ε > 0. There is a positive integer n such that −i < ε/2. For each integer i, 1 ≤ i ≤ n, let Oi = [0, 1] ∩ (xi − ε/2, xi + i>n 2 O= i>0 Oi . If y ∈ O, d(x, y) = ε/2) and Oi = [0, 1] n for i > n. Let n i i i i=1 |xi − yi |/2 + i=1 (ε/2)/2 + ε/2 < ε. Thus, O ⊆ i>n |xi − yi |/2 < B(x, ε). Theorem 239 The metric topology for Q is the same as the product topology for Q. Henceforth, we simply refer to the metric topology as the topology for Q.

5.4 A countable basis for the topology of Q Let n be a positive integer and R n = {p ∈ Q | pi is rational for 1 ≤ i ≤ n and pi = 0 for i ≥ n}. Let R = n>0 Rn . Then, R is a countable subset of Q such that every point of Q is a limit point of R. We establish this in our next theorem.

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Theorem 240 Every point of Q is a limit point of R. Proof. Suppose x ∈ Q and ε > 0. It is suﬃcient to show that there is a point p of R such that d(x, p) < ε. To that end, there is a positive integer k such that i>k 2−i < ε/2. For each integer i, 1 ≤ i ≤ k, there is a rational number ri ∈ [0, 1] such that 0 < |ri − xi | < ε/2. Let p be the point of Q such 0 for i > k and pi = ri for 1 ≤ i ≤ k.Then, p ∈ R, p = x, and that pi = k k d(p, x) = i=1 |pi − xi |/2i + i>k |pi − xi |/2i < i=1 (ε/2)/2i + ε/2 < ε. One very useful feature of the topology for Q is that there are countably many open sets that form a basis for the topology. Theorem 241 The topology for Q has a countable basis. Proof. Let B = {B(p, ε) | p ∈ R and ε is a positive rational number}. Note that B is countable. We next show that B is a basis for the topology for Q. Suppose O is open in Q and x ∈ O. There exists a positive number ε such that B(x, ε) ⊆ O. Let p be a point of R such that d(p, x) < ε/2. Let r be a rational number such that d(p, x) < r < ε/2. Then B(p, r) ∈ B and x ∈ B(p, r). If y ∈ B(p, r), then d(x, y) ≤ d(x, p) + d(p, y) < ε so y ∈ B(x, ε) ⊆ O. Thus B(p, r) ⊆ O. Alternatively, we could have shown that the product topology has a countable basis by recalling that there is a countable basis B0 for the topology of [0, 1] and then proving that the countable set { i>0 Oi | Oi ∈ B0 for ﬁnitely many integers i and Oi = [0, 1] otherwise} is a basis for the product topology for Q.

5.5 Q is compact A subset M of a topological space X is called compact provided if G is a collection of open sets covering M then there is a ﬁnite subcollection H of G that covers M (or equivalently in a space whose topology is determined by a basis, if G is a collection of basis elements covering M then some ﬁnite subcollection H of G covers M ). To show that Q is compact, we ﬁrst show that every inﬁnite subset of Q has a limit point. Our proof relies on the fact that [0, 1] has the property that every sequence of points of [0, 1] has a convergent subsequence. In the following proof, in order to avoid certain multiple subscripting, if x = x1 , x2 , x3 , . . . is a sequence, we also denote xi by pi (x).

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Theorem 242 Every inﬁnite subset of Q has a limit point. Proof. Suppose M is an inﬁnite subset of Q. There exists a sequence x1 , x2 , x3 , . . . of points of M such that xi = xj if i = j. Then, p1 (x1 ), p1 (x2 ), p1 (x3 ), . . . is a sequence of numbers in [0, 1] so some subsequence converges to a number y1 ∈ [0, 1]; that is, there is a subsequence s1 of x1 , x2 , x3 , . . . such that p1 (s11 ), p1 (s12 ), p1 (s13 ), . . . converges to y1 . (In conventional notation, there is an increasing sequence n1 , n2 , n3 , . . . of positive integers such that s1 = xn1 , xn2 , xn3 , . . . ; i.e., s11 = xn1 , s12 = xn2 , s13 = xn3 , . . . .) Then, p2 (s11 ), p2 (s12 ), p2 ((s13 ), . . . is a sequence of numbers in [0, 1] so there is a subsequence s2 of s1 such that p2 (s21 ), p2 (s22 ), p2 (s23 ), . . . converges to a number y2 ∈ [0, 1]. Observe that inasmuch as s2 is a subsequence of s1 , p1 (s21 ), p1 (s22 ), p1 (s23 ), . . . converges to y1 . Continuing this process, we obtain sequences s1 , s2 , s3 , . . . of points of Q and y1 , y2 , y3 , . . . of numbers in [0, 1] such that si+1 is a subsequence of si and pk (si1 ), pk (si2 ), pk (si3 ), . . . converges to yk for each positive integer i and each integer k, 1 ≤ k ≤ i. We now show that y = (y1 , y2 , y3 , . . . ) is a limit point of M . Let O = O1 × O2 × O3 × · · · × Oj × Q be an open set containing y. For k ≤ j, pk (sj1 ), pk (sj2 ), pk (sj3 ), . . . converges to yk , so there is an integer Nk such that if i ≥ Nk then pk (sji ) ∈ Ok . Let N denote the largest of {N1 , N2 , . . . , Nj }. If i ≥ N and k ≤ j, then pk (sji ) ∈ Ok . Thus, O contains inﬁnitely many terms of sj and consequently inﬁnitely many points of M and we have that y is a limit point of M . Suppose x1 , x2 , x3 , . . . is a sequence of points of Q and x ∈ Q. The statement that x1 , x2 , x3 , . . . converges to x means if O is an open set containing x then a positive integer n such that if m is an integer not less than n then xm ∈ O. The statement that the sequence x1 , x2 , x3 , . . . converges means there is a point x ∈ Q such that x1 , x2 , x3 , . . . converges to x. If n1 , n2 , n3 , . . . is an increasing sequence of positive integers, we say that xn1 , xn2 , xn3 , . . . is a subsequence of x1 , x2 , x3 , . . . . Theorem 243 If M is a closed subset of Q and x1 , x2 , x3 , . . . is a sequence of points of M , then there exist a point x ∈ M and a subsequence of x1 , x2 , x3 , . . . that converges to x. Proof. Let X = {x1 , x2 , x3 , . . . }. If X is ﬁnite then there exist a point x ∈ M and an increasing sequence n1 , n2 , n3 , . . . of positive integers such that xni = x for i = 1, 2, 3, . . . . It is clear that xn1 , xn2 , xn3 , . . . converges to x. If X is inﬁnite, then, by Theorem 242, X has a limit point. Let x be a limit point of X. Because X ⊆ M and M is closed, x ∈ M . For each positive integer n, let Un = B(x, 1/n). Let xn1 be a point of X in U1 . There is a point xn2 of U2 in X − {x1 , x2 , . . . , xn1 } so n1 < n2 . Inductively, there exist an increasing

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sequence n1 , n2 , n3 , . . . of positive integers and points xn1 , xn2 , xn3 , . . . of X such that xni ∈ Ui . Let O be an open set containing x. There is a positive integer n such that Um ⊆ O for m ≥ n. Thus, xnm ∈ O for m ≥ n and we have that xn1 , xn2 , xn3 , . . . converges to x. Let B = {B(p, ε) | p ∈ R and ε is a positive rational number} (R was deﬁned in Section 5.4). In the proof of Theorem 241 it was shown that B is a basis for the topology for Q. The following lemma leads to a proof that Q is compact. Lemma 244 Suppose M is a closed subset of Q and every inﬁnite subset of M has a limit point. If G is a collection of elements of B that covers M then some ﬁnite subcollection of G covers M . Proof. Because B is countable, denote the elements of G by U1 , U2 , U3 , . . . . Suppose no ﬁnite subcollection of G covers M . For each positive integer n, let xn be a point of M not in U1 ∪ U2 ∪ U3 ∪ . . . ∪ Un . The set K = {x1 , x2 , x3 , . . . } is inﬁnite otherwise there is a point x of M such that x = xi for inﬁnitely many integers i, but such a point could not be in any element of G. By hypothesis, K has a limit point x. Because x is a limit point of K and K ⊆ M , x is a limit point of the closed set M . Thus, x ∈ M so x belongs to Uk for some k. Then Uk contains a point xi of K with i > k. However, inasmuch as i > k, xi is not in Uk . Theorem 245 If M is a closed subset of Q such that every inﬁnite subset of M has a limit point, then M is compact. Proof. If G is a collection of open sets covering M , then because B is a basis for the topology for Q, if x ∈ M and O is an element of G that contains x, there is an element of B that contains x and is a subset of O. Let H = {U ∈ B | there is an element O of G such that U ⊆ O}. Then, H is a collection of elements of B covering M so, by Lemma 244 some ﬁnite subcollection H0 of H covers M . For each element U of H0 choose an element O of G that contains U . The collection of all of these open sets so chosen is a ﬁnite subcollection of G that covers M . The following theorem is an immediate consequence of Theorems 242 and 245. Theorem 246 Q is compact.

5.6 Q is connected

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5.6 Q is connected Two nonempty sets are said to be mutually separated provided they have no point in common and neither contains a limit point of the other. A set is said to be connected provided it is not the union of two mutually separated sets. We now turn to a proof that Q is connected. We accomplish this by means of a sequence of lemmas. Lemmas 247 through 249 are all normally proved in an introductory course in topology. We include their proofs for the sake of completeness. Lemma 247 If a connected set in a topological space is a subset of the union of two mutually separated sets then it is a subset of one of them. Proof. Suppose C is a connected subset of the union of two mutually separated sets A and B. If C intersects both A and B, then C ∩ A and C ∩ B are mutually separated sets whose union is C, a contradiction. Lemma 248 If G is a collection of connected subsets of a topological space with a point in common, then the union of all of the sets in G is connected. Proof. Let p denote a point common to all of the elements of G and let C denote the union of all of the elements of G. Suppose C = A ∪ B where A and B are mutually separated with p ∈ A. Suppose g is an element of G. By Lemma 247, g is a subset of A because g contains the point p of A. It follows that C = A. Lemma 249 If M is a connected set subset of a topological space and L is a set such that M ⊆ L ⊆ M then L is connected. Proof. If L is the union of two mutually separated sets A and B, then, by Lemma 247, M is a subset of one of them, say M ⊆ A. Consequently, no point of B is a limit point of M so L is a subset of A. We now turn to some lemmas speciﬁc to Q. We employ the following notation in a couple of these lemmas. If n is a positive integer, let Cn = {x ∈ Q | xi = 0 for i = n}. Lemma 250 For each positive integer n, Cn is connected. Proof. Suppose n is a positive integer such that Cn = A ∪ B where A ∩ B = ∅. Let A = {t ∈ I | there is a point x ∈ A such that xn = t} and B = {t ∈ I | there is a point x ∈ B such that xn = t}. Then A ∩ B = ∅ and A ∪ B = I. Suppose 1 ∈ B . Let a = l.u.b A and let p be the point of Cn such that pn = a. If a ∈ A then every open interval containing a contains a point of B . Let O = i>0 Oi be a member of the usual basis for the product topology for Q containing p. Because On contains a, On contains a point of B , so O

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contains a point of B. It follows that A contains a limit point of B. In the case where a ∈ B , in a similar manner we obtain that B contains a limit point of A. Thus, A and B are not mutually separated so Cn is connected. Lemma 251 Suppose n is a positive integer and K is a connected subset of Q such that, for each x ∈ K, xi = 0 for i > n. If t ∈ [0, 1] and Kt = {x ∈ Q | xn+1 = t, xi = 0 for i > n + 1, and (x1 , x2 , . . . , xn , 0, 0, . . . ) ∈ K}, then Kt is connected. Proof. Suppose t ∈ [0, 1] and Kt = A ∪ B with A ∩ B = ∅. Let A = {x ∈ K | (x1 , x2 , . . . , xn , t, 0, 0, . . . ) ∈ A} and B = {x ∈ K | (x1 , x2 , . . . , xn , t, 0, 0, . . . ) ∈ B}. Then A ∩ B = ∅ and K = A ∪ B . Because K is connected, one of A and B contains a limit point of the other. Suppose p ∈ A and p is a limit point of B . Let p = (p1 , p2 , . . . , pn , t, 0, 0, . . . ) and suppose O = i>0 Oi is a member of the usual basis for the product topology for Q that contains p. Then, O = O1 × O2 × · · · × On × On+2 × On+3 × · · · is an open set containing p so O contains a point q of B . Then, q = (q1 , q2 , . . . , qn , t, 0, 0, . . . ) is a point of B in O so p is a limit point of B. In a similar manner it can be shown that if B contains a limit point of A then B contains a limit point of A. It follows that A and B are not mutually separated so Kt is connected. In Lemma 252 and Theorem 253 we also employ the following notation. For each positive integer n let In = {x ∈ Q | xi = 0 for i > n}. Lemma 252 For each positive integer n, In is connected. Proof. We proceed by induction. Note that I1 = C1 so by Lemma 250, I1 is connected. Suppose k is a positive integer such that Ik is connected. Let t be an element of I and let Ik, t = {x ∈ Q | xk+1 = t and xi = 0 for i > k + 1}. Inasmuch as Ik is connected, by Lemma 251, Ik, t is connected and Ck+1 is connected by Lemma 250. The point y ∈ Q such that yi = 0 for i = k + 1 and yk+1 = t is in both Ik, t and Ck+1 , thus we have that Ik, t ∪ Ck+1 is connected by Lemma 248. Let G = {Ik, t ∪ Ck+1 | t ∈ I}. Then G is a collection of connected sets all containing (0, 0, 0, . . . ), so again by Lemma 248 the union of all the sets of G is connected. But, the union of all the sets of G is Ik+1 . Theorem 253 Q is connected. Proof. Let M = I1 ∪ I2 ∪ I3 ∪ · · · . By Lemma 248, In is connected for each n, thus M is connected because (0, 0, 0, . . . ) belongs to all of the sets whose union is M . By Lemma 249, M is connected. But, M = Q.

5.7 Consequences of compactness

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In this section we examine some consequences of compactness although compactness is not required in our ﬁrst theorem. Theorem 254 If G is a collection of closed subsets of Q with a common point, then the set of all points common to all the elements of G is closed. Proof. Let C be the common part of all of the elements of G. If p is a limit point of C then p is a limit point of each element of G inasmuch as C is a subset of each element of G. Thus, p is in each element of G, so p ∈ C. Theorem 255 M is a closed subset of Q if and only if M is compact. Proof. Suppose M is a closed subset of Q and M = Q. Let G be a collection of open sets covering M such that each element of G intersects M . Then G0 = G ∪ {Q − M } is a collection of open sets covering Q. By Theorem 246 there is a ﬁnite subcollection H0 of G0 that covers Q. Let H = H0 − {Q − M }. Then, H is a ﬁnite subcollection of G that covers M . On the other hand, suppose M is a compact subset of Q and p is a point that is not in M . Then G = {B(x, εx ) | x ∈ M and εx = d(x, p)/2)} is a collection of open sets covering M . There is a ﬁnite subcollection H = {B(x1 , εx1 ), B(x2 , εx1 ), . . . , B(xn , εx1 )} of G that covers M . Let ε be the least of {εx1 , εx2 , . . . , εxn }. Then, B(p, ε) contains p and no point of M so p is not a limit point of M . Theorem 256 If M1 , M2 , M3 , . . . is a sequence of compact subsets of Q such that Mi+1 ⊆ Mi for each positive integer i, then the common part of all the sets in the sequence is a nonempty compact set. Proof. Suppose no point of Q belongs to every term of M1 , M2 , M3 , . . . . Let i be a positive integer and let Oi = Q − Mi . By Theorem 255, Mi is closed for each positive integer i so Oi is a nonempty open set. Then, Oi ⊆ Oi+1 for each positive integer i, and O = {O1 , O2 , O3 , . . . } is a collection of open sets covering Q. By Theorem 246 there is a positive integer n such that {O1 , O2 , . . . , On } covers Q. Because Oi ⊆ On for 1 ≤ i ≤ n, On = Q. This involves a contradiction so there is a point common to all of the terms of the sequence M1 , M2 , M3 , . . . . Let C denote the common part of all the sets M1 , M2 , M3 , . . . . Then C is closed by Theorem 254 and thus C is compact by Theorem 255.

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5.8 Continuity We now turn to continuity and show that compactness and connectedness are preserved under continuous transformation. If X and Y are topological spaces, M is a subset of X, and f : M → Y is a function, we say that f is continuous at the point x ∈ M provided if V is an open set containing f (x) then there is an open set U containing x such that if t ∈ U ∩ M then f (t) ∈ V . A function f : M → Y is called continuous provided it is continuous at each point of M . If f is a function, we denote by f −1 the set of all pairs (y, x) such that (x, y) ∈ f . If f is a function such that f −1 is also a function, we say that f is 1-1. If f is a 1-1 continuous function such that f −1 is continuous, f is called a homeomorphism. By a mapping we mean a continuous function. Theorem 257 Suppose M and K are subsets of Q. If M is connected and f :M → → K is a mapping of M onto K, then K is connected. Proof. Suppose K is the union of two mutually separated sets A and B. Let A = {x ∈ M | f (x) ∈ A} and B = {x ∈ M | f (x) ∈ B}. Then M = A ∪ B and A ∩ B = ∅. If p ∈ A there is an open set V containing f (p) such that V ∩ B = ∅. By the continuity of f , there is an open set U containing p such that if x ∈ U ∩ M then f (x) ∈ V . Thus, U ∩ B = ∅ so p is not a limit point of B . Similarly, B does not contain a limit point of A , so A and B are mutually separated, a contradiction. Theorem 258 Suppose M and K are subsets of Q. If M is compact and f :M → → K is a mapping of M onto K, then K is compact. Proof. Let G be a collection of open sets covering K. For each x ∈ M there is an element gx ∈ G such that f (x) ∈ gx . By the continuity of f there is an open set hx containing x such that f (hx ∩ M ) ⊆ gx . The collection of all such open sets hx covers M so there is a ﬁnite subset {x1 , x2 , . . . , xn } of M such that {hx1 , . . . , hxn } covers M . For each i, 1 ≤ i ≤ n, let gxi be an element of G such that f (hxi ∩ M ) ⊆ gxi . The collection {gx1 , gx2 , . . . , gxn } is a ﬁnite subcollection of G that covers K. It follows that K is compact. Theorem 259 Suppose M and K are subsets of Q. If M is compact and f :M → → K is a 1-1 mapping of M onto K, then h is a homeomorphism. Proof. We only need to show that f −1 is a mapping. Let y be a point of K. There is a point x ∈ M such that f (x) = y. Let V be an open set containing x that does not contain M . Then M − (M ∩ V ) is a closed subset of M so it is compact. By Theorem 258, f (M − (M ∩ V )) is compact and thus closed by Theorem 255. Then U = Q − f (M − (M ∩ V )) is an open set containing y. If z ∈ U ∩ f (M ) then f −1 (z) ∈ V so f is continuous at y. Because f −1 is continuous at each point of K, it is a homeomorphism.

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5.9 Arcs A point p of a set M is said to be a separating point of M provided M − {p} is not connected. Some authors call separating points cut points. An arc is a compact connected set that has only two nonseparating points. A subset D of M is said to be dense in M if every point of M − D is a limit point of D. In Theorem 240 we showed that Q contains a countable dense set. We now show that every compact subset of Q has this property. Lemma 260 If M is a compact subset of Q then M contains a countable dense subset. Proof. Suppose n is a positive integer. The collection Gn = {B(x, 1/n) | x is in M } covers M so there is a ﬁnite subcollection Hn = {B(xn 1 , 1/n), n n n , 1/n), . . . , B(x , 1/n)} of G that covers M . Let D = {x B(xn n n 2 kn 1 , x2 , n xn , . . . , x } and let D = D . If p ∈ M − D and ε > 0 there is n>0 n 3 kn a positive integer n such that 1/n < ε. Because Hn covers M one of its n elements contains p. If B(xn i , 1/n) contains p then d(p, xi ) < ε so B(p, ε) contains a point of D. It follows that p is a limit point of D. The following lemma is useful in the proof of Theorem 262. Lemma 261 Suppose M is a compact connected subset of Q and p is a point of M . If M − {p} is the union of two mutually separated sets H and K then H ∪ {p} and K ∪ {p} are connected. Proof. Suppose H ∪ {p} is the union of two mutually separated sets A and B with p ∈ A. Then M = (A ∪ B) ∪ K = (A ∪ K) ∪ B. However, by Theorem 235, B does not contain a limit point of A ∪ K. But, A ∪ K does not contain a limit point of B thereby contradicting the fact that M is connected. So, H ∪ {p} is connected. Similarly, K ∪ {p} is connected. Theorem 262 If M is a compact connected subset of Q, then M has at least two nonseparating points. Proof. Suppose M is a compact connected subset of Q and D = {x1 , x2 , x3 , . . . } is a countable dense subset of M . If M has no more than one nonseparating point, we may assume every point of D separates M . Then, M −{x1} is the union of two mutually separated sets, H1 and K1 . We show that H1 contains a nonseparating point of M . Let xn2 be the ﬁrst term of D in H1 . Then M − {xn2 } is the union of two mutually separated sets H2 and K2 with x1 ∈ K2 . By Lemma 261, H2 ∪ {xn2 } and K2 ∪ {xn2 } are connected. Then, H2 ∪ {xn2 } ⊆ H1 ∪ {x1 } and K1 ∪{x1 } ⊆ K2 ∪{xn2 }. Note that {x1 , x2 , . . . , xn2 } ⊆ K2 ∪ {xn2 }. Let xn3 be the ﬁrst term of D in H2 . Then, M −{xn3 } = H3 ∪ K3 where H3 and

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K3 are mutually separated with xn2 ∈ K3 . Furthermore, H3 ∪ {xn3 } ⊆ H2 ∪ {xn2 } and K2 ∪ {xn2 } ⊆ K3 ∪ {xn3 } and {x1 , x2 , . . . , xn3 } ⊆ K3 ∪ {xn3 }. Continuing this process we obtain two sequences of closed and connected sets, H1 ∪ {x1 } ⊇ H2 ∪ {xn2 } ⊇ H3 ∪ {xn3 } ⊇ · · · and K1 ∪ {x1 } ⊆ K2 ∪ {xn2 } ⊆ K3 ∪ {xn3 } ⊆ · · · such that D ⊆ (K1 ∪ {x1 }) ∪ (K2 ∪ {xn2 }) ∪ (K3 ∪ {xn3 }) . . . . By Theorem 256 there is a point p common to all the sets H1 ∪ {x1 }, H2 ∪ {x2 }, H3 ∪ {x3 }, . . . . Suppose M − {p} is the union of two mutually separated sets A and B with x1 ∈ B. Then, by Lemma 247, the connected set K ∪ {x1 } ⊆ B. It follows that Ki ∪ {xni } ⊆ B for each i and thus that D ⊆ B. Hence, p is not a limit point of D, a contradiction. So p is a nonseparating point of M that belongs to H1 . By a similar argument, K1 contains a nonseparating point of M .

Theorem 263 If M is a subset of Q and h : I → → M is a homeomorphism, then M is an arc. Proof. Let x be a point of M − {h(0), h(1)}. To show that M is an arc, it is suﬃcient to show that x is a separating point of M . Let s be h−1 (x) and J1 = [0, s) and J2 = (s, 1]. Then, M − {x} = h(J1 ) ∪ h(J2 ) and h(J1 ) ∩ h(J2 ) = ∅ because h is 1-1. We now show that h(J1 ) does not contain a limit point of h(J2 ). Let p be a point of h(J1 ). There is a point t ∈ J1 such that h(t) = p. There is an open interval J containing t such that J ∩ J2 = ∅. Because h−1 is continuous there is an open set O containing p such that h−1 (O ∩ M ) ⊆ J. Then, O ∩ h(J2 ) = ∅ so p is not a limit point of h(J2 ). Similarly, h(J2 ) does not contain a limit point of h(J1 ) so h(J1 ) and h(J2 ) are mutually separated and we have M − {x} is not connected. Although we do not need it as background for Chapter 1, we should remark that the converse of Theorem 263 is also true. A proof of this may be found in [2, Section 6.38, p. 375] or [5, Theorem 6.17, p. 96]. Consequently, our use of the term arc is consistent with that of authors who deﬁne an arc as a homeomorphic image of [0, 1].

5.10 Boundary bumping

This section is somewhat more advanced than the previous sections mainly because a consequence of the axiom of choice, Zorn’s lemma, is used in a proof but also because we step outside the more familiar setting of metric spaces and into the realm of compact Hausdorﬀ spaces. Our goal in this section is to provide a proof of Theorem 276. Although this theorem is used only once in Chapter 1, it is also employed elsewhere in the book. Consequently, we give its

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proof in a more general setting. We begin by proving some useful properties of compact Hausdorﬀ spaces, Theorem 268, starting with the following lemma. Lemma 264 If p is a point of a Hausdorﬀ space X and C is a compact subset of X not containing p then there exist mutually exclusive open sets U and V such that p ∈ U and C ⊆ V . Proof. From the deﬁnition of a Hausdorﬀ space, it follows that G = {O | O is open, O contains a point of C, and O contains no point of some open set RO containing p} is a collection of open sets covering C. Because C is compact there is a ﬁnite subcollection H of G that covers C. For each O ∈ H there is an open set RO containing p such that O ∩ RO = ∅. If V is the union of all the sets in H and U = O∈H RO , then U and V are mutually exclusive open sets such that p ∈ U and C ⊆ V . One immediate consequence of Lemma 264 is the following theorem. Theorem 265 Each compact subset of a Hausdorﬀ space is closed. Proof. Suppose C is compact. By Lemma 264, if p is a point not in C there is an open set containing p that contains no point of C, so p is not a limit point of C. Thus, C is closed. The proof of the following theorem is similar to the proof that a closed subset of Q is compact (see Theorem 255) and is left to the reader. Theorem 266 Each closed subset of a compact Hausdorﬀ space is compact. In the following lemma, we strengthen the conclusion of Lemma 264 in compact Hausdorﬀ spaces. Lemma 267 If p is a point of a compact Hausdorﬀ space X and C is a closed subset of X not containing p then there exist open sets U and V such that p ∈ U , C ⊆ V , and U ∩ V = ∅. Proof. C is a closed subset of a compact Hausdorﬀ space, therefore C is compact by Theorem 266. By Lemma 264 there exist mutually exclusive open sets U and V such that p ∈ U and C ⊆ V . Because X − U is closed, it is compact by Theorem 266, so again by Lemma 264, there exist mutually exclusive open sets U and V such that p ∈ U and X − U ⊆ V . Because U and V are mutually exclusive, U ∩ V = ∅. Theorem 268 If H and K are mutually exclusive closed subsets of a compact Hausdorﬀ space, there exist open sets U and V one containing H and the other K such that U ∩ V = ∅.

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Proof. By Lemma 266, H and K are compact. If x ∈ H, by Lemma 267 there exist open sets Ox and Rx such that x ∈ Ox , K ⊆ Rx , and O x ∩ Rx = ∅. Thus, the collection G = {O | O is an open set containing a point of M and there is an open set R such that K ⊆ R and O ∩ R = ∅} covers H. Because H is compact, there exists a ﬁnite subcollection H of G that covers H. For each O ∈ H choose an open set RO such that K ⊆ RO and O ∩ RO = ∅. If U is the union of all the sets in H and V = O∈H RO , then H ⊆ U , K ⊆ V , and U ∩ V = ∅. We return brieﬂy to the Hilbert cube for one additional theorem used in Chapter 1. Theorem 269 If M1 , M2 , M3 , . . . is a sequence of compact and connected subsets of Q such that Mi+1 ⊆ Mi for each positive integer i, then the common part of all the sets in the sequence is a nonempty, compact, and connected set. Proof. In light of Theorem 256, we only have to prove that the common part M is connected. Suppose M is the union of two mutually exclusive closed sets A and B. By Theorem 268 there exist mutually exclusive open sets U and V with A ⊆ U and B ⊆ V . For each positive integer i, Mi is connected and intersects both U and V so Ni = Mi − (Mi ∩ (U ∪ V )) = ∅. Then, N1 , N2 , N3 , . . . is a sequence of closed sets such that Ni+1 ⊆ Ni for each positive integer i. By Theorem 256, there is a point p that belongs to / U ∪ V . This involves a contradiction. i>0 Ni . Then, p ∈ M but p ∈ By a Hausdorﬀ continuum we mean a compact connected subset of a Hausdorﬀ space. A collection G of sets is said to be monotonic provided if g and h are elements of G then g ⊆ h or h ⊆ g. The ﬁrst part of our next theorem may be compared with Theorem 256. Theorem 270 Suppose G is a monotonic collection of closed subsets of a compact Hausdorﬀ space X. Then there is a point common to all the elements of G and the set of all points common to all the elements of G is closed. Moreover, if all the elements of G are Hausdorﬀ continua, the common part is a Hausdorﬀ continuum. Proof. The elements of G are compact by Theorem 266. The proof that the common part of all the elements of G is a nonempty closed set is similar to the proof of Theorem 256 and is left to the reader. Let g0 denote the closed set that is the common part of all the sets in G. If O is an open set containing g0 then there is an element g ∈ G such that g ⊆ O for otherwise each element of G intersects the closed set X −O. Then the collection {g ∩ (X −O) | g ∈ G} is a monotonic collection of closed sets so it has a common point in X − O and thus not in g0 , a contradiction. If g0 is not connected, it is the union of

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two mutually exclusive closed sets h and k. There exist mutually exclusive open sets U and V containing h and k, respectively. Some element of G is a subset of U ∪ V . However, it is connected and cannot intersect both of U and V . It follows that g0 is connected. Theorem 271 Suppose H and K are mutually exclusive closed subsets of a compact Hausdorﬀ space and G is a monotonic collection of closed sets such that if g ∈ G then g is not the union of two mutually exclusive closed sets one intersecting H and the other intersecting K. Then, the common part of all the elements of G is not the union of two mutually exclusive closed sets one intersecting H and the other intersecting K. Proof. There is a point common to all of the elements of G by Theorem 270. Let C denote the common part of all the elements of G. By Theorem 270, C is a nonempty closed set. If C is the union of two mutually exclusive closed sets CH and CK such that CH ∩ H = ∅ and CK ∩ K = ∅, then by Theorem 268 there exist open sets D1 and D2 containing CH and CK , respectively, such that D1 ∩D2 = ∅. If g ∈ G then g is not a subset of D1 ∪D2 for otherwise g is the union of two mutually exclusive closed sets one intersecting H and the other intersecting K. Let G = {g ∩ (X − (D1 ∪ D2 )) | g ∈ G}. Then G is a monotonic collection of closed sets so, by Theorem 270, there is a point x common to all the elements of G . Thus, x ∈ C but x ∈ X − (D1 ∪ D2 ), a contradiction. We now make use of one of the equivalent forms of the axiom of choice. A particularly convenient form is Zorn’s lemma. Proof of its equivalence to the axiom of choice is available in numerous sources, for example, [3]. That Zorn’s lemma follows from the axiom of choice is Theorem 39 on page 14 of [4]. The equivalence of Zorn’s lemma and the axiom of choice is discussed in [3]. Zorn’s lemma If G is a collection of sets such that if H is a monotonic subcollection of G then there is an element of G that is a subset of every element of H, then there is an element of G that contains no other set of the collection G. By a component of a subset C of a Hausdorﬀ space X we mean a connected subset of C such that each connected subset of C that intersects it is a subset of it. Lemma 272 Suppose H and K are mutually exclusive closed subsets of a compact Hausdorﬀ space and C is a closed set intersecting both H and K but no component of C intersects both H and K. If p ∈ C ∩ H and q ∈ C ∩ K then C is the union of two mutually exclusive closed sets one containing p and the other containing q.

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Proof. By Theorem 266, both H and K are compact. Suppose C is not the union of two mutually exclusive closed sets one containing p and the other containing q and let G = {g ⊆ C | g is a closed set intersecting both H and K such that g is not the union of two mutually exclusive closed sets one containing p and the other containing q}. If H is a monotonic subcollection of G, by Theorem 271 (with {p} and {q} as the closed sets in the hypothesis), the common part of all of the elements of H is a closed set that is not the union of two mutually exclusive closed sets one containing p and the other containing q. By Zorn’s lemma, there is an element g0 ∈ G that contains no other element of G. If g0 is not connected then g0 = g1 ∪ g2 where g1 and g2 are mutually exclusive closed sets. One of g1 and g2 contains both p and q; suppose it is g1 . Because g1 ⊆ g0 and g1 = g0 , g1 = gp ∪ gq where gp and gq are mutually exclusive closed sets containing p and q, respectively. However, g0 = gp ∪ (gq ∪ g2 ) and gp and gq ∪ g2 are mutually exclusive closed sets containing p and q, respectively, a contradiction. Thus, g0 is connected and contains both p and q. Because g0 is a subset of a component of C, this involves a contradiction, so C = Cp ∪ Cq , two mutually exclusive closed sets one containing p and the other containing q. From here, our treatment closely follows that of the ﬁrst chapter of [4] leading up to Moore’s theorem 50. We include this in order to make this appendix virtually complete while demonstrating that Theorem 276 holds in compact Hausdorﬀ spaces. Theorem 273 Suppose H and K are mutually exclusive closed subsets of a compact Hausdorﬀ space and C is a closed set intersecting both H and K but no component of C intersects both H and K. Then C is the union of two mutually exclusive closed sets one containing C ∩ H and the other containing C ∩ K. Proof. Suppose p is a point of C ∩ H. If q is a point of C ∩ K, by Lemma 272, C is the union of two mutually exclusive closed sets Cp and Cq with p ∈ Cp and q ∈ Cq . There exist open sets Up,q and Vp,q containing Cp and Cq , respectively, such that Up,q ∩ Vp,q = ∅. The collection {Vp,q | q ∈ C ∩ K} is a collection of open sets covering the compact set C ∩ K, so there exists a ﬁnite subcollection, say Vp,q1 , Vp,q2 , . . . , Vp,qn , covering C ∩ K. Let i=n i=n Up,K = i=1 Up,qi and Vp,K = i=1 Vp,qi . Then, p ∈ Up,K , C ∩ K ⊆ Vp,K , C ⊆ (Up,K ∪ Vp,K ), and Up,K ∩ Vp,K = ∅. The collection {Up,K | p ∈ H} is a collection of open sets covering the compact set C ∩ H so there exists a ﬁnite subcollection, say Up1 ,K , Up2 ,K , . . . , Upm ,K i=m i=m that covers C ∩ H. Let U = i=1 Upi ,K and V = i=1 Vpi ,K . Note that C ∩ H ⊆ U , C ∩ K ⊆ V , C ⊆ U ∪ V , and U ∩ V = ∅. Then C ∩ U and C ∩ V are mutually exclusive closed sets whose union is C and contain C ∩ H and C ∩ K, respectively.

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If H and K are mutually exclusive closed sets and C is a Hausdorﬀ continuum that intersects both H and K, then C is said to be irreducible from H to K provided no proper subcontinuum of C intersects both H and K. Theorem 274 If M is a Hausdorﬀ continuum that intersects both of the mutually exclusive closed point sets H and K then M contains a subcontinuum irreducible from H to K. Proof. Let G = {C | C is a subcontinuum of M intersecting both H and K}. Note that M ∈ G. If G is a monotonic subcollection of G and h is the common part of all the elements of G , then by Theorem 270, h is a Hausdorﬀ continuum that intersects both H and K, so h ∈ G. By Zorn’s lemma, there is an element g ∈ G that contains no other element of G. Then, g is irreducible from H to K. Theorem 275 If H and K are mutually exclusive closed sets and C is a Hausdorﬀ continuum irreducible from H to K then C − (C ∩ (H ∪ K)) and C − (C ∩ H) are connected and every point of C ∩ H is a limit point of C − (C ∩ H). Proof. Suppose L = C−(C ∩ (H ∪ K)) is the union of two mutually separated sets A and B. Then, A ∪ (C ∩ (H ∪ K)) is closed and it intersects both H and K. By Theorem 273, A ∪ (C ∩ (H ∪ K)) is the union of two mutually exclusive closed sets AH and AK containing C ∩ H and C ∩ K, respectively. Similarly B ∪ (C ∩ (H ∪ K)) is the union of two mutually exclusive closed sets BH and BK containing C ∩ H and C ∩ K, respectively. Then, AH ∪ BH and AK ∪BK are mutually exclusive closed sets whose union is the connected set C, a contradiction. Observe that L has a limit point in H for otherwise C is the union of two mutually exclusive closed sets C ∩ H and (C ∩ K) ∪ L. Similarly, L has a limit point in K. Thus, L is a subcontinuum of C intersecting both H and K. Because C is irreducible from H to K, C = L. Thus, every point of C ∩ H is a limit point of L as is every point of C ∩ K. Then, L ∪ (C ∩ K) is connected and its closure is C. Because C − (C ∩ H) = L ∪ (C ∩ K), we have C − (C ∩ H) is connected and every point of C ∩ H is a limit point of C − (C ∩ H). A point p is said to be a boundary point of a point set M provided every open set containing p contains a point in M and a point not in M . Theorem 276 If p is a point of the Hausdorﬀ continuum M in a compact Hausdorﬀ space X and D is an open set containing p that does not contain M , then the component C of D containing p has a limit point in the boundary of D. Proof. The connected set M intersects both D and X − D, therefore there is a point common to D and X − D. Such a point is in the boundary of D. Let

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B denote the boundary of D. Then {p} and B are mutually exclusive closed sets, so by Theorem 274, there is a subcontinuum K of M irreducible from {p} to B. By Theorem 275, K − (K ∩ B) is a connected subset of D that contains p so K − (K ∩ B) ⊆ C. By Theorem 275, K − (K ∩ B) has a limit point in B, so C does.

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575. 576. 577. 578. 579. 580. 581.

582.

583.

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588.

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Index

Cl(set), 10 Gη , 81 Gn , 3 I (the interval [0, 1]), 2 Id, 79 lim sup limiting set, 121 N, 2 R, 21 Q, 2, 169 G ∗ , 60 set, 10 sin(1/x)-curve, 10, 67 ε-chain, 59 ε-map, 123 f (set), 2 f : set → → set, 22 f : set → set, 22 fa b , 48 2-cell not an inverse limit on [0, 1], 127 2-cell with sticker, 104 Anderson, vii Andrews, 14 Arc, 5, 179 Atriodic, 133 inverse limit is, 133 Barge, viii Basic open set basis, 168 Basis, 168 Bennett, 11, 17 BJK horseshoe, 16, 162 three copies joined at a common endpoint, 134

two copies joined at a common endpoint, 44, 51, 53 Bonding maps, 2 monotone, 137 Brouwer, 15 Brown, 155, 160 Buckethandle, see BJK horseshoe Cantor fan, 92 Cantor set, 91 Capel, 137 Chain, 59 ε-chain, 59 reﬁne, 60 strongly reﬁne, 60 taut, 60 Chainable, 59 deﬁning sequence, 61 inverse limit is, 120 inverse limit of chainable continua is, 123 Choquet, vii Closure, 10 Compact inverse limit is, 81 inverse limit on I is, 4 Conﬂuent, 162 produces BJK horseshoe, 163 Conjugate, 23, 110 Connected inverse limit is, 85 inverse limit on I is, 5 Consistent system, 80 Continuum, 5, 131 decomposable, 14 Hausdorﬀ, 85 hereditarily indecomposable, 18

215

216 Hurewicz, 107 indecomposable, 14 inverse limit is, 111 inverse limit on I is, 5, 9 Cook, vii, 75 Core indecomposable, 31 of a map, 31 of an inverse limit, 31 tent family, 53 Cut point, see separating point Davis, 163 Decomposable continuum, 14 Hausdorﬀ continuum, 131 inverse limit on intervals is, 26 Deﬁning sequence, 61 Degenerate, 14 Dense, 179 Dimension, 124 Directed set, 76 Endpoint of a continuum, 43 of a topological ray, 10 of an arc, 5 three endpoint indecomposable continuum, 47 Finite graphs not an inverse limit on [0, 1], 147 Fixed point, 2 Fixed point property, 151 Full projection property, 142 Graphs of upper semi-continuous functions equivalent to closed subset of the product, 78 Hausdorﬀ continuum, 85, 131 inverse limit is, 85, 86, 89, 90 Henderson, vii, 18 Hereditarily decomposable, 21 Hereditarily indecomposable, 18 Hereditarily unicoherent, 134 Hilbert cube, 169 Holte, 155 Homeomorphism, 178 Hyperspaces, 152 Identity map, 2 Id, 79 Indecomposable

Index Hausdorﬀ continuum, 131 inverse limit contains, 21 inverse limit is characterization, 140 inverse limit on I is, 16, 17 inverse limit on intervals is, 24 inverse limit with unimodal maps is, 29 two-pass condition, 147 Indecomposable continuum, 14 Induced function, 105 Induced homeomorphism, 109 Induced mapping, 106 Ingram, viii, 12 Injection, 2 Into, 2 Inverse limit decomposable, 26 dimension, 125 is indecomposable Hausdorﬀ continuum, 132 is nonempty, 81 is one dimensional, 127 logistic map, 31 Feigenbaum limit, 46 on I, 2 on I is nonempty, 4 one bonding map on interval, 22 sequence, 75 single bonding map on I, 13 surjective bonding maps, 35 system, 77 basis, 115 closed subsets, 116 consistent, 80 induced homeomorphism, 109 induced mapping, 106 Subsequence Theorem, 119 surjective bonding functions, 82 upper semi-continuous bonding functions, 77 Inverse limit sequence Lebesgue sequence, 156 measure, 159 Irreducible, 136 inverse limit is, 136, 139 Irreducible about a closed set, 135 Janiszewski, 15 Kennedy, viii Knaster, 15 Knaster continuum, see BJK horseshoe Kuratowski, 15

Index Lebesgue sequence, 65, 156 Limiting set, 121 Logistic family, 31 Logistic map, 31 Mahavier, vii, 14, 17 Mapping, 2, 75, 178 Mardeˇsi´ c, 123 Markov maps, 57 Marsh, D., 14 Marsh, M., 152 Martin, viii Measure, 159 Mesh, 60 Minc, 18, 151 Monotone, 137 Monotone map, 12 Onto, 2 Orbit, 30 Period, 20 Period n, 151 Periodic, 151 Periodic point, 20 Permutation map, 58 Prime period, see period Projection, 2, 78 Property of Kelley, 150 Pseudo-arc, 18 Raines, viii Ray, 10 Reﬁne, 60 strongly, 60 Region, 3 Remainder, 12 Sarkovskii’s Theorem, 21 Schwarzian derivative, 32

217 Segal, 123, 152 Separating point, 5, 90, 137 Sequential limiting set, 121 set-valued, see upper semi-continuous function Shift homeomorphism, 13, 120 Shift map, 13 Simple closed curve, 137 Smale horseshoe, see BJK horseshoe Span, 149 Stockman, viii Strictly unimodal, 32 Strongly reﬁne, 60 Subsequence Theorem, 119 Surjection, 2 Surjective, 81 Taut chain, 60 Tent family, 52 Theta curve, 134 Topological ray, 10 Topologically conjugate, 23 Totally ordered, 76 Transitive, 18 Transue, 18 Triod, 133 Two-pass condition, 146 Unicoherent, 134 inverse limit is, 134 Unimodal, 27 strictly, 32 Union of functions, 111 Upper semi-continuous function, 77 Weakly chainable, 151 Williams, viii Yorke, viii