Simple Techniques for Detecting Decomposability or Indecomposability of
Generalized Inverse Limits
by
Scott Varagona
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
May 7, 2012
Keywords: Continuum theory, Inverse Limits, upper semi-continuous functions,
indecomposability
Copyright 2012 by Scott Varagona
Approved by
Michel Smith, Chair, Professor of Mathematics
Gary Gruenhage, Professor of Mathematics
Stewart Baldwin, Professor of Mathematics
Abstract
A continuum X is said to be decomposable if it can be written as a union of two proper
subcontinua; otherwise, X is said to be indecomposable. For years, topologists have used
inverse limits with continuous bonding functions to study indecomposable continua. Now
that the topic of generalized inverse limits with upper semi-continuous (or \u.s.c.") bonding
functions has become popular, it is natural to consider how these new kinds of inverse limits
might be used to generate indecomposable (or decomposable) continua.
In this work, we build upon our past results (from \Inverse Limits with Upper Semi-
Continuous Bonding Functions and Indecomposability," [13]) to obtain new and more gen-
eral theorems about how to generate indecomposable (or decomposable) continua from u.s.c.
inverse limits. In particular, we seek su cient conditions for indecomposability (or decom-
posability) that are easily checked, just from a straightforward observation of the bonding
functions of the inverse limit. Our primary focus is the case of inverse limits whose factor
spaces are indexed by the positive integers, but we consider extensions to other cases as well.
ii
Acknowledgments
First and foremost, I would like to thank my advisor, Dr. Michel Smith, for everything
he has done for me. Dr. Smith introduced me to the wonderful world of inverse limits, and
he has been a constant source of helpful advice and encouragement ever since. Indeed, Dr.
Smith has done more than simply teach me about Topology; rather, by his own example, he
has taught me how to think like a mathematician. I look forward to passing that knowledge
on to a graduate student of my own someday.
I should also thank my other committee members, Dr. Gary Gruenhage and Dr. Stew-
art Baldwin, for the positive impact they have had on my mathematics career and research.
Because of the wonderful classes I took from Dr. Gruenhage when I was an undergraduate, I
chose to stick with the Mathematics major, and later, to become a topologist. On the other
hand, when I was giving an inverse limits presentation as a Master student, Dr. Baldwin
once asked me an interesting question about u.s.c. inverse limits. That question ultimately
led me to investigate the \two-sided" inverse limits featured in Chapter 9 of this dissertation.
Next, I must give a very special thanks to Dr. Tom Ingram, who has constantly
supported my e orts and given me wonderful feedback on my work. He has trusted me to
help proofread some of his publications; moreover, he has also cited my own work in those
publications. Along with Tom Ingram, I must thank Dr. Bill Mahavier, who rst brought
the topic of u.s.c. inverse limits into prominence. Without the work of Ingram and Mahavier
to build upon, this dissertation simply could not exist!
Finally, of course I must thank my family for their perennial love and support. My
academic career at Auburn has been a long journey, and my family has helped me every step
of the way. Also, just as importantly, I want to thank all the amazing friends I have made
iii
as a graduate student at Auburn. You have all enriched my life in so many ways: thank you
so much!
iv
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 General Topology and Classic Inverse Limits Results . . . . . . . . . . . . . . . 3
2.1 Background De nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2.2 Background Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3 Classic Inverse Limits De nitions and Theorems . . . . . . . . . . . . . . . . 11
3 Background on Inverse Limits with u.s.c. Bonding Functions . . . . . . . . . . . 13
4 Su cient Conditions For Decomposability of u.s.c. Inverse Limits . . . . . . . . 16
5 Su cient Conditions For Indecomposability of u.s.c. Inverse Limits . . . . . . . 21
6 Further Results on Indecomposability . . . . . . . . . . . . . . . . . . . . . . . . 29
7 A Generalization of the Two-Pass Condition . . . . . . . . . . . . . . . . . . . . 41
8 Inverse Limits on Initial Segments of the Ordinal Numbers . . . . . . . . . . . . 49
9 Two-Sided Inverse Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
10 Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
11 Possibilities For Future Research . . . . . . . . . . . . . . . . . . . . . . . . . . 69
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
v
List of Figures
4.1 A bonding function that gives rise to a degenerate inverse limit space . . . . . . 17
5.1 A bonding function that gives rise to an inverse limit without the full projection
property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
5.2 A u.s.c. function whose graph is topologically equivalent to a sin(1x) curve . . . 23
7.1 f1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
7.2 f2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
7.3 f3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
7.4 f4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
7.5 f5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
7.6 f6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
9.1 The spaces from Example 9.3. Left: one-sided inverse limit; right: two-sided. . . 56
9.2 The bonding function from Example 9.4 . . . . . . . . . . . . . . . . . . . . . . 57
9.3 The Henderson map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
9.4 The bonding function f from Example 9.7 . . . . . . . . . . . . . . . . . . . . . 59
10.1 The graph of the function f from Example 10.1 . . . . . . . . . . . . . . . . . . 61
vi
10.2 The graph of the function fj from Example 10.2 . . . . . . . . . . . . . . . . . . 62
10.3 The graph of the function f from Example 10.3 . . . . . . . . . . . . . . . . . . 63
10.4 The graph of the function f from Example 10.4 . . . . . . . . . . . . . . . . . . 63
10.5 The graph of the function f from Example 10.5 . . . . . . . . . . . . . . . . . . 64
10.6 The graph of the function f from Example 10.6 . . . . . . . . . . . . . . . . . . 64
10.7 The projection of the inverse limit from Example 10.7 onto its rst three factor
spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
10.8 The graph of the function f from Example 10.8 . . . . . . . . . . . . . . . . . . 66
10.9 The graph of the function f from Example 10.9 . . . . . . . . . . . . . . . . . . 66
10.10 Counter-clockwise from top: The graph of g, the short inverse limit, the long
inverse limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
10.11 Counter-clockwise from top: The graph of h, the short inverse limit, the long
inverse limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
vii
Chapter 1
Introduction
An inverse limit space is a valuable tool for topologists who wish to study indecom-
posable continua. Although a non-degenerate indecomposable continuum is a complicated
topological space by its very nature, it is often possible to represent such a space in a very
simple way|namely, as an inverse limit space with a single continuous bonding function. On
the other hand, by drawing a relatively simple bonding function f that satis es some special
properties, we can guarantee that the inverse limit space with the single bonding function f
is an indecomposable continuum. In this way, we may easily generate more indecomposable
continua as examples for further study.
A number of topologists have done research on the relationship between inverse limits
and indecomposable continua; see Chapter 1 of [4] for highlights from the history of this topic.
However, for many years, only inverse limits with continuous bonding functions had been
seriously considered in the literature. Now, after the work of Mahavier [8] and both Ingram
and Mahavier [3], generalized inverse limits with set-valued, upper semi-continuous (u.s.c.)
bonding functions have become popular. It is therefore a natural next step to consider how
these new kinds of inverse limits might be used to generate indecomposable continua.
After we gave presentations [15] and [16] addressing the issue of u.s.c. inverse limit
spaces and indecomposability, other mathematicians began publishing results on this topic
as well. In [5] and [4], Ingram extended some of his earlier results on indecomposability
in inverse limits with continuous bonding functions to the u.s.c. case. Also, in [17], Brian
Williams gave necessary and su cient conditions for an inverse limit to have the full pro-
jection property, a property that is vital for proving that some inverse limit spaces are
indecomposable continua.
1
Still, plenty of work remains to be done: we need to nd straightforward conditions
(stated in terms of the bonding functions fi : [0;1] ! 2[0;1]) to guarantee that the inverse
limit space is an indecomposable (or decomposable) continuum. The main goal of this work
is to provide many such conditions; moreover, we strive to give conditions that are simple
to check in practice. Ideally, by applying the theorems given here, one can tell from a quick
glance at the bonding functions whether or not an inverse limit is indecomposable.
In Chapter 2, we give basic topology de nitions and state theorems that may serve
as background. In Chapter 3, we de ne inverse limits with upper semi-continuous bonding
functions and cite important theorems that will be invoked repeatedly for the rest of this
work. Then, in Chapters 4 and 5, we recall our own past results (from [13]) on the main
problem before delving into new material. Chapter 6 features major generalizations of our
previous indecomposability theorems; using these new theorems, we may show that many
more u.s.c. functions with a structure akin to the sin(1x) curve give us indecomposable
inverse limits. In Chapter 7, we discuss a generalization of Ingram?s two-pass condition;
we name this new condition the \ -two-pass" condition, and we consider its impact on the
study of indecomposable inverse limits. Next, we look into inverse limits whose factor spaces
are indexed by sets other than the positive integers: Chapter 8 addresses inverse limits
indexed by large initial segments of the ordinals (i.e., \long" inverse limits), and Chapter 9
addresses inverse limits indexed by Z (i.e., \two-sided" inverse limits). Both of these topics
have implications for the study of indecomposability as well. Finally, although we give
speci c examples from time to time in the theory chapters, we set aside Chapter 10 solely
for additional examples. We then close with Chapter 11, a discussion of possible topics for
further research.
2
Chapter 2
General Topology and Classic Inverse Limits Results
2.1 Background De nitions
We begin with very basic topology de nitions, most of which should be covered in an
introductory topology course or may be found in an introductory text, such as [7] or [10].
For a detailed discussion of ordinal numbers, see, e.g., [6].
Let X be a set and let T be a collection of subsets of X with the following properties:
1. X2T;
2. ;2T;
3. If fOigi2 is a collection of members of T, then Si2 Oi2T;
4. If fOigni=1 is a nite collection of members of T, then Tni=1Oi2T.
Then the pair (X;T) is called a topological space with topology T. Such a topological
space will often be referred to simply as X when the associated topology T is understood.
The members of T are called open sets.
A subset K of a topological space X is closed if X K is open.
A topological space X is degenerate if it consists of only one point. Otherwise, X is
non-degenerate.
Suppose M is a subset of a topological space X. A point p2X is a limit point of M if
every open set containing p contains a point in M di erent from p.
3
Suppose M is a subset of a topological space X. The set of all limit points of M is
denoted by M0. The closure of M (denoted M) is M[M0.
Suppose a collection B of open sets of a space X satis es the following property:
If x2X and O is an open set containing x, then there exists a member b of B such
that x2b and b O.
Then B is a basis for the topology on X and a member b of B is called a basic open
set of X.
Suppose B is a collection of subsets of a set X such that
1. If x2X, there exists some b2B with x2b.
2. If b1 and b2 are members of B with x2b1\b2, then there exists some set b3 in B
with x2b3 (b1\b2).
Then the collection T = fSRjR Bg is a topology for X, and B is a basis for this
topology. It is said that the topology T is generated by the basis B.
A topological space X is called Hausdor if for every pair of distinct points p;q2X,
there exist disjoint open sets Op and Oq containing p and q respectively.
A space X is called regular if for every closed subset H of X and point p2X not in
H, there exist disjoint open sets OH and Op containing H and p, respectively.
A space X is called normal if for every pair of disjoint closed sets H and K in X, there
exist disjoint open sets OH and OK containing H and K, respectively.
If f : X !Y is a function from X to Y, and U is a subset of X, we de ne f(U) =
ff(u)ju2Ug.
4
Let X and Y be topological spaces, let f : X!Y be a function from X to Y, and let
x2X. Then f is said to be continuous at the point x if, whenever V is an open set in Y
containing f(x), there exists an open set U in X containing x such that f(U) V. If f is
continuous at each point x2X, we say f is continuous. A continuous function may also be
called a mapping.
A function f : X!Y is said to be surjective if for each y2Y, there exists some x2X
with f(x) = y.
A function f : X !Y is said to be 1-1 if for any pair of distinct points p, q in X,
f(p)6= f(q).
If f : X!Y is a function and y2Y, then the preimage of y via f (written as f 1(y))
is fx 2 Xj f(x) = yg. If A Y, then the preimage of A via f (written as f 1(A)) is
fx2Xjf(x)2Ag.
Suppose f : X!Y is a 1-1 surjective function. Then the function f 1 : Y !X given
by f 1(y) = x (where x is the unique point in X with the property that f(x) = y) is called
the inverse of f.
If X and Y are topological spaces and f : X!Y is 1-1, surjective, continuous, and has
a continuous inverse, then f is called a homeomorphism and the spaces X and Y are said to
be homeomorphic.
If f : X!X is a function, then we denote the composition f f by f2. More generally,
fn = f fn 1 for n 2.
5
If f : X!X is a function and A X, then we denote f 1(f 1(A)) by f 2(A). More
generally, f n(A) = f 1(f n+1(A)).
Let X be a topological space and let M X. A collection of sets fOigi2 in X is said
to be an open cover of M if each Oi is open in X and M Si2 Oi.
If fOigi2 is a cover of X, , and fOigi2 is also a cover of X, then fOigi2 is
called a subcover of the original cover fOigi2 . A subcover consisting of only nitely many
members is called a nite subcover.
A space X is compact if for every open coverfOigi2 of X, there exists a nite subcover
of X. (I.e., fOijgnj=1 for some positive integer n.)
A collection of subsets fGigi2 of a space X is called a monotonic collection if for each
pair of members Gj, Gk in the collection, either Gj Gk or Gk Gj.
For each i in some arbitrary index set , let Xi be a topological space. De ne X =
Q
i2 Xi to be the setf(xi)i2 jxi2Xi for each ig. De ne a topology on X as follows: a basic
open set containing (xi)i2 is given by Qi2 Oi, where Oi is open in Xi for each i, xi 2Oi
for each i, and Oi = Xi for all but nitely many i.
Then X together with the topology generated by this basis may be called a product
space (on the index set ).
In the case of a product space on the countably in nite index set consisting of the
positive integers (N), we denote the product space by Q1i=1Xi. Thus,
Q1
i=1Xi =f(x1;x2;x3;:::) jxi2Xi for each positive integer ig.
6
Let X = Qi2 Xi be a product space (with index set either nite or in nite). Let A
be a subset of . Then the function A : X !Qi2AXi de ned by A((xi)i2 ) = (xi)i2A
is called the projection map onto the set A. In the special case where A = fjg for some
j2 , we denote fjg simply by j and we call this function the projection map onto the j
coordinate.
If X;Y are topological spaces, f : X!Y is a function, and A X, then f restricted
to A (denoted by fjA) is the function given by fjA : A!Y, where fjA(a) = f(a) whenever
a2A.
A function f : X !Y is said to be open if for each open subset U of X, f(U) is an
open subset of Y.
Suppose X is a topological space with topology T and S X. Then the set S together
with the topology ^T = fS\OjO2Tg is called a subspace of X, where ^T is the subspace
topology.
Let be an ordinal. Then the set f j is an ordinal and < g is called an initial
segment of the ordinals. (Similarly, if < , then f j is an ordinal and < g is an
initial segment of .)
Suppose X is a topological space and d : X X ! R is a function satisfying the
following properties (for all x;y;z2X):
1. d(x;y) 0, and d(x;y) = 0 i x = y.
2. d(x;y) = d(y;x).
3. d(x;z) d(x;y) +d(y;z).
7
Then the function d is said to be a metric on X. For a given p2X and > 0, let
B(p; ) =fx2X jd(x;p) < g. If the collection fB(p; ) jp2X; > 0g is a basis for the
space X, then X is said to be a metric space.
Let X be a topological space. Two subsets H and K of X are called mutually separated
if neither set contains a point or a limit point of the other.
If X is a topological space and M X, then M is connected if M is not the union of
two mutually separated non-empty subsets of X.
A topological space X is a continuum if X is non-empty, compact, and connected.
A continuum that is Hausdor (but not necessarily metric) is called a Hausdor con-
tinuum.
A continuum that is metric is called a metric continuum.
If X is a continuum and A, a subset of X, is also a continuum, then A is called a
subcontinuum of X. If the subcontinuum A is a proper subset of X, then A is a proper
subcontinuum of X.
A continuum X is decomposable if it is the union of two proper subcontinua.
If X is a continuum but X is not decomposable, then X is said to be indecomposable.
If X is a continuum and p;q2X, then X is said to be irreducible between p and q if no
proper subcontinuum of X contains both p and q.
Let X be a connected set. If X fpg is not connected, then p is a cut point of X.
8
A continuum with exactly 2 non-cut points is called an arc.
A triod is a union of three arcs whose intersection is exactly one point.
A fan is a union of in nitely many arcs, all of which have exactly one point in common.
Let X be a topological space. Suppose that A, a subset of X, is an arc with the property
that whenever O X is an open set with O\A6= ;, there exists some point p2O with
p =2A. Then A is called a limit arc.
A subset A of a topological space X is said to be nowhere dense in X if every non-empty
open subset of X contains a non-empty open set that misses A.
A subset A of a topological space X is said to be dense in X if every non-empty open
subset of X contains a point of A.
2.2 Background Theorems
Most of the following basic theorems may be found in a standard topology text. The
proofs of these theorems are omitted, but most of the proofs may be found in one or more
of [7], [9], and [10]. A more general statement and proof of Theorem 2.26 may be found in
[4] (Lemma 221).
Theorem 2.1. A subset M of a topological space X is closed (in X) i M contains all of
its limit points in X.
Theorem 2.2. A subset M of a topological space X is closed (in X) i M = M.
Theorem 2.3. If A B, then A B.
9
Theorem 2.4. A closed subset of a compact space is compact.
Theorem 2.5. The continuous image of a compact set is compact.
Theorem 2.6. (Tychono ) If fXigi2 is a collection of compact topological spaces, then the
product space Qi2 Xi is compact.
Theorem 2.7. If fXigi2 is a collection of continua, then the product space Qi2 Xi is a
continuum.
Theorem 2.8. Let B be a basis for a topological space X. Then every open set of X is a
union of members of B.
Theorem 2.9. The following are equivalent for a function f : X!Y from topological space
X to topological space Y:
i. f is a continuous function.
ii. If O is a (basic) open set in Y, then f 1(O) is open in X.
Theorem 2.10. If Qi2 Xi is a product space and A , then the projection map A is
continuous.
Theorem 2.11. If X;Y are topological spaces, f : X!Y is continuous, and A X, then
fjA is also continuous.
Theorem 2.12. If X is a compact Hausdor space, then X is regular.
Theorem 2.13. If X is a compact Hausdor space, then X is normal.
Theorem 2.14. If X is regular, then X is Hausdor .
Theorem 2.15. If X is normal, then X is regular.
Theorem 2.16. The unit interval [0;1] is an arc.
Theorem 2.17. Suppose M is a subset of a topological space X. If M is closed and not
connected, then M is the union of two disjoint closed sets H and K.
10
Theorem 2.18. The continuous image of a connected set is connected.
Theorem 2.19. The continuous image of a continuum is a continuum.
Theorem 2.20. The common part of a monotonic collection of continua is a continuum.
Theorem 2.21. If X is a compact space, Y is a Hausdor space, and f : X!Y is a 1-1,
surjective, continuous function, then f 1 is continuous (and hence, f is a homeomorphism).
Theorem 2.22. A topological space X is connected i for any two distinct points p;q2X,
there exists a connected subset of X containing p and q.
Theorem 2.23. If fKig1i=1 is a collection of connected subsets of a space X such that
Ki\Ki+16=; for all i 1, then K = S1i=1Ki is connected.
Theorem 2.24. If K X is connected, then K is connected.
Theorem 2.25. If A is a dense subset of the topological space X, then A = X.
Theorem 2.26. Suppose T is an arc 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.
Theorem 2.27. A Hausdor continuum X is indecomposable i every proper subcontinuum
of X is nowhere dense in X.
2.3 Classic Inverse Limits De nitions and Theorems
As we stated in Chapter 1, the main focus of this work is inverse limits with upper
semi-continuous bonding functions. However, for the sake of introduction, here we present
the de nition of traditional inverse limits with continuous bonding maps rst. We also give
basic theorems about these inverse limits to contrast with the theorems about the upper
semi-continuous inverse limits given in Chapter 3. Proofs of the following theorems may be
found in one or both of [4], [14].
11
Suppose that, for each positive integer i, Xi is a topological space and fi is a continuous
function from Xi+1 to Xi. Let lim fXi;fig1i=1 be the set f(x1;x2;x3;:::) 2Q1i=1Xi jxi =
fi(xi+1) for all positive integers ig. Then we say lim fXi;fig1i=1 is an inverse limit space and
a basis for the topology on lim fXi;fig1i=1 isfO\lim fXi;fig1i=1jO is basic open in Q1i=1Xig.
The Xi?s are called the factor spaces of lim fXi;fig1i=1, and the fi?s are continuous bonding
maps. If fi;fi+1;:::;fj 1 are bonding maps, let us denote fi fi+1 ::: fj 1 by fi;j.
Theorem 2.28. Suppose X = lim fXi;fig1i=1 is an inverse limit space with continuous bond-
ing maps, fnig1i=1 is an increasing sequence of positive numbers, gi = fni;ni+1 for each i, and
Y = lim fXni;gig1i=1. Then X is homeomorphic to Y.
Theorem 2.29. Let X = lim fXi;fig1i=1 be an inverse limit space with continuous bonding
maps. If there is a natural number N so that fn is an onto homeomorphism for each n N,
then X is homeomorphic to XN.
Theorem 2.30. Let X = lim fXi;fig1i=1 be an inverse limit space with continuous bonding
maps and suppose Xi is non-empty and compact for each i. Then X is non-empty and
compact.
Theorem 2.31. Let X = lim fXi;fig1i=1 be an inverse limit space with continuous bonding
maps and suppose Xi is a continuum for each i. Then X is a continuum.
We note in advance that most of these theorems are false if upper semi-continuous
bonding functions are used instead of continuous bonding maps. Most importantly, as was
pointed out by Ingram and Mahavier in [3], some kind of extra hypothesis must be added
to Theorem 2.31 for that theorem to hold true in the upper semi-continuous case. (Contrast
Theorem 2.31 with, e.g., Theorem 3.3 or Theorem 3.4.)
12
Chapter 3
Background on Inverse Limits with u.s.c. Bonding Functions
Suppose X and Y are compact Hausdor spaces, and de ne 2Y to be the set of all
non-empty compact subsets of Y. A function f : X ! 2Y is called upper semi-continuous
(u.s.c.) if for any x2X and open V in Y containing f(x), there exists an open U in X
containing x so that f(u) V for all u2U. If f : X!2Y is u.s.c. and f(x) is connected
for each x2X, then f is a u.s.c. continuum-valued function; in this case, for emphasis, we
will sometimes write f : X!C(Y) instead, where C(Y) is the set of all subcontinua of Y.
If f : X ! 2Y is u.s.c. and f(x) = fyg for some x2X and y 2Y, then although f is
a set-valued function, we use the convention of writing simply f(x) = y. Therefore, in the
case where f : X ! 2Y is u.s.c. but f(x) is degenerate for all x2X, we may regard f as
the corresponding continuous function f : X!Y.
Again, let X;Y be compact Hausdor spaces and let f : X!2Y be a u.s.c. function.
If y2Y, then the preimage of y via f is f 1(y) = fx2Xjy2f(x)g. More generally, if
A Y, then the preimage of A via f is f 1(A) = fx2Xjf(x)\A6= ;g. We say f is
surjective if for each y2Y, f 1(y) is non-empty. Assuming that f : X!2Y is a surjective
u.s.c. function, the inverse of f, i.e., the set-valued function f 1 : Y ! 2X, is given by
f 1(y) = fx2Xjy 2f(x)g. It will later become evident that if f is a u.s.c. surjective
function, then its inverse, f 1, is also a u.s.c. surjective function.
If X;Y; and Z are compact Hausdor spaces and f : X!2Y and g : Y !2Z are u.s.c.
functions, then g f : X!2Z is the u.s.c. function given by (g f)(x) =fz2Z j9y2Y
such that y 2 f(x) and z 2 g(y)g. In the special case of a u.s.c. function f : X ! 2X,
we denote f f by f2; moreover, for any integer n 2, let us say fn = f fn 1. It will
also sometimes be helpful to use the following convention: whenever f : X!2X is a u.s.c.
13
function and A X, let us denote the preimage of the preimage of A, i.e., f 1(f 1(A)), by
f 2(A). More generally, for each integer n 2;f n(A) = f 1(f (n 1)(A)).
Given compact Hausdor spaces X;Y and a u.s.c. function f : X ! 2Y , the graph of
f, abbreviated G(f), is the set f(x;y) 2X Y jy2f(x)g. The inverse of the graph is
G(f) 1 =f(y;x)j(x;y)2G(f)g. If X1;X2;:::;Xn;Xn+1 are compact Hausdor spaces and
fi : Xi+1 ! 2Xi is u.s.c. for 1 i n, then G(f1;f2;:::;fn) = f(x1;x2;:::;xn;xn+1) 2
Qn+1
i=1 Xi jxi2fi(xi+1) for 1 i ng.
Now suppose that, for each positive integer i, Xi is a compact Hausdor space and
fi : Xi+1!2Xi is an upper semi-continuous function. We de ne lim fXi;fig1i=1 to be the set
f(x1;x2;x3;:::) 2Q1i=1Xi jxi 2fi(xi+1) for all positive integers ig. (For convenience, we
will denote a sequence (x1;x2;x3;:::) by the boldface x and denote the sequence of functions
(f1;f2;f3;:::) by the boldface f. Thus, we may abbreviate lim fXi;fig1i=1 by lim f.) Then we
say lim f is an inverse limit space with u.s.c. bonding functions, and a basis for the topology
on lim f is fO \ lim f jO is basic open inQ1i=1Xig. For brevity?s sake, we will sometimes
call an inverse limit space with u.s.c. bonding functions simply a u.s.c. inverse limit space.
Finally, in the special case where X is a compact Hausdor space, f : X!2X is u.s.c., and
f = (f;f;f;:::), we say lim f is the inverse limit with the single bonding function f. (If, in
the description of a particular inverse limit, only the single bonding function f : X!2X is
given, then it will be clear from context that lim f is the inverse limit with the single bonding
function f.)
As stated in Chapter 2, a continuum is a non-empty compact connected space; a con-
tinuum that is Hausdor (but not necessarily metric) will be called a Hausdor continuum.
We will usually assume that each factor space Xi is a non-degenerate Hausdor continuum.
In [3], Ingram and Mahavier prove various theorems about inverse limits with u.s.c.
bonding functions. There are four especially critical theorems (originally labeled 2.1, 3.2,
4.7, and 4.8 in [3]) that we will need as background in the next chapters, and so, we restate
them here:
14
Theorem 3.1. Suppose each of X and Y is a compact Hausdor 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).
Theorem 3.2. Suppose that for each positive integer i, Xi is a non-empty compact Hausdor
space and fi : Xi+1!2Xi is an upper semi-continuous bonding function. Then lim f is non-
empty and compact.
Theorem 3.3. Suppose that for each positive integer i, Xi is a Hausdor continuum, fi :
Xi+1!2Xi is an upper semi-continuous function, and for each x in Xi+1, fi(x) is connected.
Then lim f is a Hausdor continuum.
Theorem 3.4. Suppose that for each positive integer i, Xi is a Hausdor continuum, fi :
Xi+1 ! 2Xi is an upper semi-continuous function, and for each x2Xi, f 1i (x) is a non-
empty, connected set. Then lim f is a Hausdor continuum.
15
Chapter 4
Su cient Conditions For Decomposability of u.s.c. Inverse Limits
Let us recall once more that a Hausdor continuum X is decomposable if it is the union
of two proper subcontinua; if a Hausdor continuum X is not decomposable, X is said to
be indecomposable. Now, suppose for each positive integer i, Xi is a Hausdor continuum,
fi : Xi+1 ! 2Xi is a u.s.c. bonding function, and lim f is the resulting inverse limit. Our
rst major goal is to provide some simple means for recognizing when such a u.s.c. inverse
limit is a decomposable continuum. It would be especially convenient if we could infer
decomposability just from some easily-checked feature of some bonding function?s graph,
G(fi). In this chapter, we recall our previous results on this topic; these results and their
proofs may also be found in [13], but we include them here for the sake of completeness.
Theorem 4.1. Suppose that for each positive integer i, Xi is a non-degenerate Hausdor
continuum, fi : Xi+1 ! 2Xi is a surjective u.s.c. function, and fi(x) is connected for
each x 2 Xi+1. Suppose further that, for some positive integer j, there is an open set
U Xj+1 Xj intersecting G(fj) so that G(fj) nU is the graph of a u.s.c. function
h : Xj+1!2Xj satisfying the following conditions:
1) h(x) is connected for all x2Xj+1.
2) There is an open V Xj+1 Xj so that U\V =; and G(h)\V 6=;.
Then lim f is a decomposable continuum.
Proof. By Theorem 3.3, lim f is a continuum. To show lim f is decomposable, by Theorem
2.27, it su ces to nd a proper subcontinuum that is not nowhere dense. Fix some positive
integerj with fj : Xj+1!2Xj satisfying the hypothesis, so thatG(fj)nU is the graph of some
u.s.c. function h satisfying conditions 1 and 2. De neef = (f1;f2;:::;fj 1;h;fj+1;:::). Then
16
by Theorem 3.3, lim ef is a continuum. Moreover, because there is some (pj+1;pj)2G(fj)\U
and each fi is surjective, there is some point p = (p1;p2;:::;pj;pj+1;:::) with p2lim fnlim ef.
So lim ef is a proper subcontinuum of lim f.
Since there is a basic open subset Oj+1 Oj of V that also intersects G(h), (X1 X2
::: Xj 1 Oj Oj+1 Xj+2 :::)\lim f is a non-empty open subset of lim ef. Thus, lim ef
is not nowhere dense.
As we remarked in [13], each u.s.c. function fi must be surjective for this theorem to
succeed. There is a counterexample otherwise: Suppose that, for each positive integer i,
fi : [0;1]!2[0;1] is given by the graph consisting of the straight line segments from (0;0) to
(1;0) and from (1;0) to (1; 12). (See Figure 4.1.) Then lim f is a single point, (0;0;0;:::), and
is therefore an indecomposable continuum. To avoid trivial counterexamples such as this,
we will repeatedly assume that the bonding functions are surjective and the factor spaces
are non-degenerate. See Chapter 10 for examples of bonding functions for which Theorem
4.1 actually applies. In particular, Example 10.1 shows how the conditions in Theorem 4.1
are often easy to check in practice.
Figure 4.1: A bonding function that gives rise to a degenerate inverse limit space
Theorem 4.2. Suppose that for each positive integer i, Xi is a non-degenerate Hausdor
continuum, fi : Xi+1 ! 2Xi is a surjective u.s.c. function, and for each x 2 Xi, f 1i (x)
is connected. Suppose further that, for some positive integer j, there is an open set U
17
Xj+1 Xj intersecting G(fj) so that G(fj)nU is the graph of a u.s.c. function h : Xj+1!2Xj
satisfying the following conditions:
1) For all x2Xj, h 1(x) is a non-empty, connected set.
2) There is an open V Xj+1 Xj so that U\V =; and G(h)\V 6=;.
Then lim f is a decomposable continuum.
Proof. By Theorem 3.4, lim f is a continuum. Let ef = (f1;f2;:::;fj 1;h;fj+1;:::). Again,
by Theorem 3.4, lim ef is a continuum; the same argument as in Theorem 4.1 shows that lim ef
is a proper subcontinuum that is not nowhere dense.
Theorem 4.3. Suppose that for each positive integer i, Xi is a non-degenerate Hausdor
continuum and fi : Xi+1 !2Xi is a surjective u.s.c. function with fi(x) connected for each
x2Xi+1. Suppose there is some point x = (x1;x2;x3;:::) 2 lim f so that f 1i (xi) = xi+1
for i 2, and there is an open U X2 X1 so that G(f1)\U is a non-empty subset of
fx2g f1(x2). Then lim f is a decomposable continuum.
Proof. Theorem 3.3 implies that lim f is a continuum. Let O2 O1 be a basic open subset
of U with G(f1)\(O2 O1) a non-empty subset of fx2g f1(x2). We note that the proper
subcontinuum f1(x2) fx2g fx3g ::: of lim f contains the open set (O1 O2 X3 X4
:::)\lim f. Thus, there exists a proper subcontinuum that is not nowhere dense.
So far, we have seen decomposability arise in certain situations where lim f satis es
either Theorem 3.3 or Theorem 3.4. Now we turn to a situation where 3.3 or 3.4 no longer
apply, i.e., a situation where images (or preimages) of points need not be connected. In
general, if there exists some positive integer i and x 2 Xi+1 with fi(x) not connected,
then lim f need not be connected. However, in some special cases, lim f turns out to be a
continuum even if fi(x) is not connected for some x2Xi+1. One such special case is given
by the following theorem, a consequence of one of Ingram?s results (Theorem 3.3) in [5]: If
f : [0;1] ! 2[0;1] is a u.s.c. bonding function that is the union of two distinct continuous
functions g;h : [0;1] ! [0;1], at least one of which is surjective, then lim f is a continuum.
18
Our next theorem generalizes this result and also provides another su cient condition for
decomposability.
Theorem 4.4. Suppose for each positive integer i, Xi is a non-degenerate Hausdor con-
tinuum and fi : Xi+1 ! 2Xi is a u.s.c. function that is the union of two u.s.c. functions
gi;hi : Xi+1!2Xi satisfying the following properties:
1) At least one of gi;hi is surjective.
2) gi(x) and hi(x) are connected for all x2Xi+1.
3) G(gi)\G(hi)6=;.
Then lim f is a continuum. Moreover, if there is some positive integer j so that G(gj)6
G(hj) and G(hj)6 G(gj), then lim f is a decomposable continuum.
Proof. We rst show that lim f is a continuum. Since for each positive integer i, fi is u.s.c.
and each factor space is a continuum, it follows from Theorem 3.2 that lim f is compact. To
show that lim f is connected, we will show that, for any p;q2lim f, there exists a connected
subset of lim f that contains p and q.
Let p = (p1;p2;p3;:::) and q = (q1;q2;q3;:::) be in lim f. Then for each positive integer
i, (pi+1;pi)2G( i), where i2fgi;hig. Similarly, for each i, (qi+1;qi)2G( i), where i2
fgi;hig. Now de ne z1 = ( 1; 2; 3;:::), z2 = ( 1; 2; 3;:::), z3 = ( 1; 2; 3;:::); :::; zi =
( 1; 2;:::; i 1; i;:::), etc. By property 3, G(gi) and G(hi) must intersect at some point
(xi+1;xi). So, since each fi is surjective, there exists some point (x1;x2;:::;xi;xi+1;:::) in
lim f that lies in both lim zi and lim zi+1. That means, for all i, lim zi\lim zi+16=;.
By Theorem 3.3, lim zi is connected for each positive integeri. Since (lim zi)\(lim zi+1)6=
; for each i, it follows that K = S1i=1 lim zi is connected. Thus, the closure of K is also
connected. We already know p2lim z1 K. To show that q2K, we observe that, since
lim zi K for each i, points of the form (q1;:::);(q1;q2;:::); :::; (q1;q2;:::;qi;:::), etc.,
are all in K. Since q is a limit point of the set of these points, q2K. Thus, both p and q
lie in the connected set K. It follows that lim f is connected, and hence, is a continuum.
19
Now, if G(gj) 6 G(hj) and G(hj) 6 G(gj) for some positive integer j, lim f can be
decomposed into two proper subcontinua as follows. Let eg = (f1;f2;:::;fj 1; gj;fj+1;:::)
and eh = (f1;f2;:::;fj 1;hj;fj+1;:::). Then (by an argument similar to the one given above)
lim eg and lim eh are both proper subcontinua of lim f, and (lim eg)[(lim eh) = lim f.
Corollary 4.5. Suppose that, for each positive integer i, fi : [0;1]!2[0;1] is a u.s.c. function
that is the union of two distinct continuous functions gi;hi : [0;1] ! [0;1], at least one of
which is surjective. Then lim f is a decomposable continuum.
20
Chapter 5
Su cient Conditions For Indecomposability of u.s.c. Inverse Limits
Our next major goal is to give straightforward conditions on u.s.c. bonding functions
fi : Xi+1 ! 2Xi that guarantee that the inverse limit space lim f is a non-degenerate inde-
composable continuum. By way of introduction to the problem, at rst we will focus only
on the case where f : [0;1]!2[0;1] and f = (f;f;f;:::). That is, we will assume lim f is an
inverse limit with a single u.s.c. bonding function f : [0;1] ! 2[0;1]. Once again, here we
recall our earlier results from [13]; we expand our results to much more general cases in the
next chapter.
Let us begin by proving a lemma that gives us valuable information about the proper
subcontinua of lim f in some special cases. An inverse limit space lim f with factor spaces Xi
is said to have the full projection property if, whenever H is a proper subcontinuum of lim f,
there exists some positive integer N so that n(H)6= Xn for all n N.
Lemma 5.1. Let f : [0;1] ! 2[0;1] be a u.s.c. function with the property that lim f is a
continuum. Suppose that, for some A ( [0;1], fj[0;1]nA is a function, f([0;1]nA) = [0;1],
and P =f(p1;p2;:::)2lim f jpi =2A for all ig is a dense subset of lim f. Then lim f has
the full projection property.
Proof. Assume by way of contradiction that there is some proper subcontinuum H of lim f
so that, for each positive integer n, there exists some m n such that m(H) = [0;1]. For
any such m, we know that [0;1] = f([0;1]nA) = f( m(H)nA) m 1(H); from this it
follows that i(H) = [0;1] for all i m. Thus, since in nitely many positive integers m
with m(H) = [0;1] exist, we have that n(H) = [0;1] for each positive integer n. We will
now show that P H.
21
Let p = (p1;p2;p3;:::)2P. Then since 1(H) = [0;1], there exists some point in H of
form (p1;?;?;?;:::). Since 2(H) = [0;1], there exists some point in H of form (?;p2;?;?;:::).
However, p2 =2A and fj[0;1]nA is a function, so f(p2) is unique; therefore, f(p2) = p1. That
means some point of form (p1;p2;?;?;?;:::) lies in H. A similar argument shows that some
point of form (p1;p2;:::;pi 1;pi;?;?;:::) lies in H for all i; since p is a limit point of the set
of all such points, and H is closed, p 2H. Thus, P H. But then P H; because P is
dense, P = lim f, so lim f H. Therefore, lim f = H, contradicting the assumption that H
is a proper subcontinuum.
Figure 5.1: A bonding function that gives rise to an inverse limit without the full projection
property
A remark about the full projection property is in order. Any inverse limit lim f with
a continuous surjective bonding function f : [0;1] ! [0;1] has the full projection property
automatically. However, if f is u.s.c., then in general lim f need not have the full projection
property. Example 131 in [4] illustrates this point: For each positive integer i, let Xi = [0;1]
and let fi : [0;1]!2[0;1] be the graph consisting of the straight line segments joining (0;0) to
(1;0) and (0;0) to (1;1). (See Figure 5.1.) Then H =f(x;x;x;:::) jx2[0;1]g is a proper
subcontinuum of lim f, but i(H) = [0;1] for all i. Thus, some kind of additional hypotheses
(like those in Lemma 5.1) are required for the full projection property to hold.
With this lemma in hand, we may prove the following theorem. We note in advance that
the example motivating this theorem is the inverse limit generated using the u.s.c. function
22
f : [0;1] ! 2[0;1] whose graph is topologically equivalent to a sin(1x) curve (see Figure 5.2).
In this case, the set A is simply f0g.
Figure 5.2: A u.s.c. function whose graph is topologically equivalent to a sin(1x) curve
Theorem 5.2. Suppose f : [0;1] ! 2[0;1] is u.s.c. and there is some non-empty closed
nowhere dense set A [0;1] with the property that:
1) f(a) = [0;1] for all a2A.
2) fj[0;1]nA is an open continuous function.
3) For each a2A, y2[0;1] and > 0:
i. If 9b2[0;1] with b>a, then there exists some x12[0;1]nA such that x12(a;a+ )
and f(x1) = y.
ii. If9b2[0;1] with b 0 such that ((xn+1 ;xn+1 + )\[0;1]) On+1. By condition 3,
there exists some z2((xn+1 ;xn+1 + )\[0;1]) with z =2A and f(z) = xn. That means
(x1;x2;:::;xn;z;:::)2O.
In any case, there exists some point (x1;x2;:::;xn;xn+1;:::)2O with xn+1 =2A. Now
let [On+1 = On+1 nA. Since A is closed and xn+1 2 [On+1, it follows that [On+1 is open
and non-empty. Moreover, by condition 2, f([On+1) is open. Since f([On+1) contains xn,
which lies in On, we have that On\f([On+1) is open and non-empty. It follows that W =
(O1 O2 On 1 (On\f([On+1)) [0;1] )\lim f contains (x1;x2;:::;xn;xn+1;:::) and
is therefore a basic open set that satis es the inductive hypothesis. So, W contains a point
(p1;p2;:::;pn 1;pn;?;?;:::) 2P. Since pn 2f([On+1), there exists some pn+1 2On+1nA
such that f(pn+1) = pn. But (by condition 3) there exists pn+2 2 [0;1] nA such that
f(pn+2) = pn+1, there exists pn+3 2 [0;1]nA such that f(pn+3) = pn+2, etc. So, we have
shown that (p1;p2;:::;pn;pn+1;pn+2;:::), a point in P, lies in O. This means P is dense in
lim f.
By condition 2, fj[0;1]nA is a function. By condition 3, f([0;1]nA) = [0;1]. Thus, the
hypothesis of Lemma 5.1 is satis ed; this means that lim f has the full projection property.
24
Finally, suppose by way of contradiction that lim f is a union of two proper subcontinua
H and K. Because lim f has the full projection property, there exists some positive integer
N such that n(H) 6= [0;1] and n(K) 6= [0;1] for all n N. Since A is non-empty, there
exists some a2A lying in either N+1(H) or N+1(K); without loss of generality, assume
a 2 N+1(H). Since N+1(K) 6= [0;1], N+1(H) must be an arc; in particular, N+1(H)
must contain either [a ;a] or [a;a + ] for some > 0. In either case, by condition 3,
f( N+1(H)nA) = [0;1], which forces N(H) = [0;1]. (Contradiction.)
So lim f is indecomposable and the proof is complete.
We improve on this result in various ways in the next chapter. For now, let us turn to
a much di erent condition that also guarantees indecomposability. First, however, we must
de ne the \itinerary space" of an inverse limit. Suppose that, for each positive integer i, Pi
is a partition of [0;1]. That is, Pi is a collection of subsets Pi1;Pi2;:::;Pini of [0;1] so that
[0;1] = Snik=1Pik and Pij\Pil =;if j6= l. Suppose for each positive integer i, fi : [0;1]!2[0;1]
is u.s.c. and lim f is the associated inverse limit space. If x = (x1;x2;x3;:::)2lim f, and for
i 2, i is the unique member ofPi containing xi, then the itinerary representation of x is
(x) = (x1; 2; 3;:::). If for each positive integer i, Pi is assumed to have some topology
TPi, thenI = [0;1] Q1i=2Pi is a product space with the standard product topology. Thus,
: lim f!I is given by ((x1;x2;x3;:::)) = (x1; 2; 3;:::). That is, maps x2 lim f to
its unique itinerary representation (x). Then we call (lim f) the itinerary space of lim f.
Our approach toward itinerary spaces here is inspired by that of Stewart Baldwin in [1].
We intend to form the partitionsPi, each with respective topologyTPi, so that turns out to
be a homeomorphism between lim f and the itinerary space of lim f. Of course, our choice of
partition depends heavily on the nature of the bonding functions (fi), so we now introduce
the kind of bonding function that interests us here. Recall that if g is a u.s.c. bonding
function, then the inverse of the graph G(g) is given by G(g) 1 =f(y;x)j(x;y)2G(g)g. Let
the graph of f : [0;1]!2[0;1] be given by G(g) 1[G(h) 1, where, for some xed a2(0;1),
25
1) g : [0;1] ! [0;1] is a non-decreasing continuous function with g(0) = 0, g(1) = a,
and g((0;1)) = (0;a).
2) h : [0;1] ! [0;1] is a non-increasing continuous function with h(0) = 1, h(1) = a,
and h((0;1)) = (a;1).
Then we say f is a steeple with turning point a. Note that the graph of any steeple f
is closed, so that a steeple is automatically u.s.c. by Theorem 3.1. See Examples 10.5 and
10.6 in Chapter 10 for samples of bonding functions that are (and are not) steeples.
Now suppose that, for each positive integer i, fi : [0;1] ! 2[0;1] is a steeple with
turning point ai+1. Then for each i 2, we de ne the steeple partition with center ai to be
Pi =fLi;Ci;Rig, where Li = [0;ai);Ci =faig; and Ri = (ai;1]. For a given i, the topology
TPi onPi will befPi;;;fLig;fRig;fLi;Rigg. LetI = [0;1] Q1i=2Pi, so thatI is a product
space with the usual product topology. Direct inspection reveals that the itinerary space of
lim f is the subspace bI of I consisting of all points of the following forms:
1) (x1; 2; 3; 4;:::), where x12(0;1) and i2fLi;Rig for each i 2.
2) (0;L2;L3;L4;:::)
3) (0;L2;L3;:::;Lk;Rk+1;Ck+2; k+3; k+4;:::), where i2fLi;Rigfor each i k+ 3.
4) (0;R2;C3; 4; 5;:::), where i2fLi;Rig for each i 4.
5) (1;C2; 3; 4;:::), where i2fLi;Rig for i 3.
Because the only open set in Pi that contains Ci is Pi itself, Pi is not Hausdor , and
neither isI. However, as we will see in the following lemma (and as was rst pointed out in
[1]), bI turns out to be a Hausdor subspace of I.
Lemma 5.3. Suppose for each positive integer i, fi : [0;1]!2[0;1] is a steeple with turning
point ai+1. Then lim f is a Hausdor continuum. Moreover, if for i 2, Pi is the steeple
partition with center ai, then : lim f!I is a homeomorphism between lim f and bI.
Proof. For each positive integer i, the graph of each fi satis es the hypothesis of Theorem
3.1, so each fi is u.s.c.; indeed, by construction, fi(x) is connected for each positive integer
i and each x2[0;1]. Thus, by Theorem 3.3, lim f is a Hausdor continuum.
26
We have already observed that, since each fi is a steeple, (lim f) = bI; it remains to
verify that is a homeomorphism between lim f and bI. Of course maps onto its range.
If x;y 2 lim f and x 6= y, then either x1 6= y1 or, for some i 2, xi and yi lie in di erent
members of Pi, so that the itinerary representations of x and y must di er. So is 1-1.
To show that is continuous, let x = (x1;x2;x3;:::)2lim f so that (x) = (x1; 2; 3;:::),
and let O = O1 O2 ::: On Pn+1 ::: be a basic open set in I containing (x).
For convenience, shrink O to a smaller basic open set that also contains (x), i.e., eO =
O1 f 2g f 3g ::: f ng Pn+1 :::, where i = i if i = Li or Ri, and i =Pi if
i = Ci. Then let U = O1 2 3 ::: n [0;1] :::, where i = i if i = Li or Ri
and i = [0;1] if i =Pi. U contains x, and (U) eO O, so is continuous.
lim f is compact. Therefore, to show that 1 is continuous, it will su ce to show
that bI is Hausdor . Suppose that y = (y1;y2;:::) and z = (z1;z2;:::) with y;z 2 bI and
y6= z. If y1 6= z1, then there exist disjoint open sets Oy1;Oz1 [0;1] containing y1 and z1,
respectively, so that Oy1 P2 ::: and Oz1 P2 ::: are disjoint open sets inI containing
y and z. So, suppose y1 = z1. There are three subcases:
1) If y1 = z1 are not 0 or 1, then the remaining coordinates are all Li?s and Ri?s. Since
y 6= z, there must be some coordinate k 2 for which one of yk;zk is Lk and the other is
Rk. Thus, if Uk =fLkg, Vk =fRkg, and Ui = Vi =Pi for all i6= k, then [0;1] Q1i=2Ui and
[0;1] Q1i=2Vi are disjoint open sets, one containing y and the other containing z.
2) If y1 = z1 = 1, then y2 = z2 = C2 and the remaining coordinates of y and z are only
Li?s and Ri?s. Again, since y 6= z, there must be some coordinate k 3 for which one of
yk;zk is Lk and the other is Rk. Thus, if Uk = fLkg, Vk = fRkg, and Ui = Vi = Pi for all
i6= k, then [0;1] Q1i=2Ui and [0;1] Q1i=2Vi are disjoint open sets, one containing y and
the other containing z.
3) If y1 = z1 = 0, then there are two subcases:
A) y and z both have their rst R in the kth coordinate. So,
y = (0;L2;L3;:::;Lk 1;Rk;Ck+1;yk+2;yk+3;:::), and
27
z = (0;L2;L3;:::;Lk 1;Rk;Ck+1;zk+2;zk+3;:::). Since y6= z, for some j k + 2, one
of yj;zj is Lj and the other is Rj. Then two disjoint open sets containing y and z respectively
can be found in a manner similar to that of case 1 and 2.
B) y has its rst R in the kth coordinate but z either has no R?s at all or its rst R lies
in the jth coordinate, where (without loss of generality) k >j. If z has no R?s at all, then
y and z can be separated by open sets like those in case 1 and 2; if z has its rst R in the
jth coordinate, then y?s jth coordinate is Lj, and thus, it is again easy to separate y and z
with open sets.
All possible cases have been addressed, so (lim f) = bI is Hausdor . Hence, 1 is
continuous, and is a homeomorphism.
Theorem 5.4. Suppose for each positive integer i, fi : [0;1]!2[0;1] is a steeple with turning
point ai. Then lim f is homeomorphic to the bucket-handle continuum.
Proof. Let g : [0;1]![0;1] be the standard tent map, i.e., the function whose graph consists
of straight line segments from (0;0) to (12;1), and from (12;1) to (1;0). Then g is a steeple, so
lim g, the bucket-handle continuum, is homeomorphic to bI. Since lim f is also homeomorphic
to bI, lim f = lim g.
Thus, because the bucket-handle is indecomposable, so is any inverse limit space on unit
intervals with bonding functions that are steeples. It is possible to give an alternate argu-
ment for indecomposability here that does not involve itineraries; we do so, and generalize
the \steeple" construction a bit further, in Chapter 7. For a more detailed discussion of
itineraries, see, e.g., [1]. The author is indebted to Michel Smith and Tom Ingram for their
help in re ning our description of \steeple" bonding functions.
28
Chapter 6
Further Results on Indecomposability
Most of the results in the last two chapters have already been proven in the previous
work by the author, [13], but we have included them again here for the sake of completeness.
Now we seek new and more general results on detecting indecomposability in u.s.c. inverse
limits. The main results in this chapter are generalizations of the important Theorem 5.2.
However, we will begin with some indecomposability results that are more miscellaneous,
but still interesting and useful.
Theorem 6.1. Suppose that, for each positive integer i, fi : [0;1]!2[0;1] is u.s.c., surjective,
and continuum-valued. If lim f has the full projection property and for each i, either fi(0) =
[0;1] or fi(1) = [0;1], then lim f is an indecomposable continuum.
Proof. Assume by way of contradiction that lim f is a union of two proper subcontinua,
H and K. By the full projection property, there must exist some large enough integer
N so that for all n N, n(H) 6= [0;1] and n(K) 6= [0;1]. Let us consider the case
when fN(0) = [0;1]. One of N+1(H) or N+1(K), but not both, must contain 0; without
loss of generality, assume N+1(H) contains 0 but N+1(K) does not. Thus, the set R =
f(x1;x2;x3;:::) 2 lim f jxN+1 = 0g must be a subset of H. However, since fN(0) = [0;1],
it follows that N(R) = [0;1], so that N(H) = [0;1] also. But it was already stated that
N(H)6= [0;1], so we have a contradiction. The case when fN(1) = [0;1] is similar.
Theorem 6.2. Suppose for each positive integer i, Xi is a Hausdor continuum that is
irreducible between two of its points (ai and bi), and fi : Xi+1!2Xi is u.s.c., surjective, and
continuum-valued. If lim f has the full projection property and for all i, either fi(ai+1) = Xi
or fi(bi+1) = Xi, then lim f is an indecomposable continuum.
29
Proof. Assume by way of contradiction that lim f is a union of two proper subcontinua, H and
K. By the full projection property, there must exist some large enough integer N so that for
all n N, n(H)6= Xn and n(K)6= Xn. Now let us consider the case where fN(aN+1) =
XN. Since XN+1 is irreducible from aN+1 to bN+1, and XN+1 = N+1(H) [ N+1(K),
it follows that one of N+1(H) or N+1(K), but not both, must contain aN+1. Without
loss of generality, assume N+1(H) contains aN+1 but N+1(K) does not. Thus, the set
R = f(x1;x2;x3;:::) 2 lim f j xN+1 = aN+1g must be a subset of H. However, since
fN(aN+1) = XN, it follows that N(R) = XN, so that N(H) = XN also. But it was already
stated that N(H) 6= XN, so we have a contradiction. The case when fN(bN+1) = XN is
similar.
Theorem 6.3. Suppose f : [0;1]!2[0;1] is u.s.c. continuum-valued with f(0) = [0;1], and
lim f is an indecomposable continuum. Then f0g [0;1] is a limit arc of G(f).
Proof. Suppose by way of contradiction that f0g [0;1] is not a limit arc of G(f). Then
there is some point (0;y) that is not a limit point of G(f)n(f0g [0;1]). That means there
exist some small 1; 2 > 0 so that the open set of form [0; 1) (y 2;y + 2) contains no
element of G(f)n(f0g [0;1]). If [0; 1) (y 2;1] contains no element of G(f)n(f0g [0;1]),
then [0; 1) (y 2;1] may be used as the open set U in Theorem 4.1, contradicting that
lim f is indecomposable. A contradiction is similarly reached if [0; 1) [0;y + 2) contains
no element of G(f)n(f0g [0;1]). Thus, [0; 1) [y + 2;1] and [0; 1) [0;y 2] must
respectively contain points (a1;b1) and (a2;b2) from G(f)n(f0g [0;1]). Since G(f) is
connected, G(f) must contain either a point (a1;b3) in [0; 1) [0;y 2] or a point (a2;b4)
in [0; 1) [y + 2;1]. In either case, we have a contradiction because f has been shown not
to be continuum-valued.
Theorem 6.4. Suppose f : [0;1]!2[0;1] is u.s.c. continuum-valued with f(1) = [0;1], and
lim f is an indecomposable continuum. Then f1g [0;1] is a limit arc of G(f).
Proof. The argument is almost identical to that of the previous theorem.
30
The fact that 0 and 1 are endpoints of [0;1] was used strongly in the proof of the previous
two theorems. If instead f(a) = [0;1] for some a2(0;1), then there is a counterexample. Let
the graph of f : [0;1] ! 2[0;1] consist of the straight line segment between the points (0;0)
and (12;1), the straight line segment between the points (12;1) and (12;0), and the straight line
segment between the points (12;0) and (1;1). (This is Example 209 in [4].) Then f is u.s.c.
and continuum-valued, f(12) = [0;1], and (as proved by Ingram) lim f is indecomposable.
However, f12g [0;1] is not a limit arc of G(f)n(f12g [0;1]).
Theorem 6.5. (Michel Smith) Suppose that for each positive integer i, Xi is an indecom-
posable continuum and fi : Xi+1 !2Xi is a surjective u.s.c. bonding function. If lim f is a
continuum with the full projection property, then lim f is indecomposable.
Proof. Assume the hypothesis and suppose by way of contradiction that lim f is decompos-
able. Then lim f is a union of two proper subcontinua H and K. Since lim f has the full
projection property, there exists some positive integer N so that, for all n N, n(H) and
n(K) are proper subcontinua of Xn. However, since each bonding function fi is surjective,
it must be the case that N(lim f) = XN. From this it follows that N(H)[ N(K) = XN,
contradicting that XN is indecomposable.
Now, we seek to generalize Theorem 5.2. The next few theorems and lemmas may
be thought of as a warm-up before the most signi cant and useful result in this chapter,
Theorem 6.14.
Lemma 6.6. For each positive integer i, let Xi be a Hausdor continuum. Suppose that
fi : Xi+1!2Xi is a u.s.c. function for each i and lim f is a continuum. Suppose further that
there is a sequence of sets A2;A3;::: so that, for each i 2, Ai (Xi, fi 1jXinAi is a function,
and fi 1(XinAi) = Xi 1. Finally, suppose that P =f(p1;p2;:::)2lim f j8i 2; pi =2Aig
is a dense subset of lim f. Then lim f has the full projection property.
Proof. Assume by way of contradiction that there is some proper subcontinuum H of lim f
so that, for each positive integer n, there exists some m n such that m(H) = Xm. For
31
any such m 2, we know that Xm 1 = fm 1(XmnAm) = fm 1( m(H)nAm) m 1(H);
from this it follows that i(H) = Xi for each i m. Thus, since in nitely many positive
integers m with m(H) = Xm exist, we have that n(H) = Xn for each positive integer n.
We will now show that P H.
Let p = (p1;p2;p3;:::) 2P. Then since 1(H) = X1, there exists some point in H of
form (p1;?;?;?;:::). Since 2(H) = X2, there exists some point in H of form (?;p2;?;?;:::).
However, p2 =2A2 and f1jX2nA2 is a function, so f1(p2) is unique; therefore, f1(p2) = p1. That
means some point of form (p1;p2;?;?;?;:::) lies in H. A similar argument shows that some
point of form (p1;p2;:::;pi 1;pi;?;?;:::) lies in H for all i; since p is a limit point of the set
of all such points, and H is closed, p 2H. Thus, P H. But then P H; because P is
dense, P = lim f, so lim f H. Therefore, lim f = H, contradicting the assumption that H
is a proper subcontinuum.
Theorem 6.7. Suppose for each positive integer i, Xi is a Hausdor continuum and fi :
Xi+1 !2Xi is u.s.c. Suppose that, for each i 2, there is some non-empty closed nowhere
dense set Ai Xi with the property that:
1) fi 1(a) = Xi 1 for all a2Ai.
2) fjXinAi is an open continuous function.
3) For each a 2 Ai, y 2 Xi 1 and open Ua Xi containing a, there exists some
x2XinAi with x2Ua and fi 1(x) = y.
4) For each a2Ai, if H is a non-degenerate subcontinuum of Xi containing a, then H
must contain a subset eH of XinAi for which f(eH) = Xi.
Then lim f is an indecomposable continuum.
Proof. lim f is a continuum since fi is continuum-valued for each positive integer i. It remains
to show that lim f is indecomposable.
First, let P =f(p1;p2;p3;:::)2lim fjpi =2Ai8i 2g. We will show that P is dense in
lim f. Thus, we need to show that, for each positive integer n, if O1;O2;:::;On are arbitrary
32
opens subsets of X1;X2;:::;Xn respectively so that O = (O1 O2 On Xn+1 :::)\
lim f is non-empty, then O contains some point in P.
Proof by induction on n:
If n = 1, then O = (O1 X2 X3 )\lim f. O1 is open and non-empty, so O1
contains a point p1. By condition 3, there exists p2 2X2nA2 such that f1(p2) = p1, there
exists p3 2X3nA3 such that f2(p3) = p2, etc. It follows that p = (p1;p2;p3;:::) 2O, and
p2P also.
Now we assume the claim is true for n; we need to show it is true for n+1. So, suppose
O = (O1 O2 On On+1 Xn+2 )\lim f is non-empty. We need to show that
O contains a point in P.
We will begin by showing that there is some point (x1;x2;:::;xn;xn+1;:::) 2O with
xn+1 =2An+1. There is at least some (x1;x2;:::;xn;xn+1;:::) 2O, since O is non-empty;
if xn+1 =2An+1, we are done. So, suppose xn+1 2An+1. Then because xn+1 2On+1, which
is open, by condition 3, there exists some z2On+1 with z =2An+1 and fn(z) = xn. That
means (x1;x2;:::;xn;z;:::)2O.
In any case, there exists some point (x1;x2;:::;xn;xn+1;:::) 2O with xn+1 =2An+1.
Now let [On+1 = On+1nAn+1. Since An+1 is closed and xn+12 [On+1, it follows that [On+1 is
open and non-empty. Moreover, by condition 2, fn([On+1) is open. Since fn([On+1) contains
xn, which lies in On, we have that On\fn([On+1) is open and non-empty. It follows that W =
(O1 O2 On 1 (On\fn([On+1)) Xn+1 )\lim f contains (x1;x2;:::;xn;xn+1;:::)
and is therefore a basic open set that satis es the inductive hypothesis. So, W contains a
point (p1;p2;:::;pn 1;pn;?;?;:::) 2 P. Since pn 2 fn([On+1), there exists some pn+1 2
On+1nAn+1 such that fn(pn+1) = pn. But (by condition 3) there exists pn+22Xn+2nAn+2
such that fn+1(pn+2) = pn+1, there exists pn+3 2Xn+3nAn+3 such that fn+2(pn+3) = pn+2,
etc. So, we have shown that (p1;p2;:::;pn;pn+1;pn+2;:::), a point in P, lies in O. This
means P is dense in lim f.
33
By condition 2, for each i 2, fi 1jXinAi is a function. By condition 3, for each i 2,
fi 1(XinAi) = Xi 1. Thus, the hypothesis of Lemma 6.6 is satis ed; this means that lim f
has the full projection property.
Finally, suppose by way of contradiction that lim f is a union of two proper subcontinua
H and K. Because lim f has the full projection property, there exists some positive integer
N such that n(H) 6= Xn and n(K) 6= Xn for all n N. Since AN+1 is non-empty, there
exists some a2AN+1 lying in either N+1(H) or N+1(K); without loss of generality, assume
a2 N+1(H). Since N+1(H)6= XN+1 and N+1(K)6= XN+1, it follows that N+1(H) must
be a non-degenerate subcontinuum of XN+1 containing a. By condition 4, N+1(H) must
contain a subset eH of XN+1nAN+1 for which fN(eH) = XN. But fN(eH) N(H); since
N(H) is closed, it follows that fN(eH) = XN N(H). Thus, XN = N(H), which gives us
a contradiction.
So lim f is indecomposable and the proof is complete.
Next, we present a sequence of lemmas that will lead to the very powerful Theorem
6.14.
Lemma 6.8. Suppose f : X!Y is a continuous function and there is some closed nowhere
dense subset A of X such that fjXnA is open. Then if B is a nowhere dense subset of Y,
f 1(B) is a nowhere dense subset of X.
Proof. Let B be nowhere dense in Y. If f 1(B) is empty, we are done. So, suppose f 1(B)
is non-empty. Assume by way of contradiction that there exists a non-empty open U X
such that every non-empty open subset of U meets f 1(B). Since A is closed and nowhere
dense, UnA is a non-empty open subset of U. Hence, f(UnA) is open in Y, and therefore
f(UnA) contains an open set V that misses B. That means the (non-empty) open set
f 1(V) misses f 1(B). But then f 1(V)\(UnA) is a non-empty open subset of U that
misses f 1(B). (Contradiction.)
34
Lemma 6.9. If A X is nowhere dense in X and X is an open subset of Y, then A is
nowhere dense in Y.
Proof. Let O be open in Y. If O misses X, then O misses A. If O meets X, then O\X is
open in Y, so that O\X contains an open set U that misses A. Since U is open in X and
X is open in Y, U is open in Y.
Lemma 6.10. If A;B X are both nowhere dense in X, then A[B is nowhere dense in
X.
Proof. Let A;B be nowhere dense in X and assume O X is open. Then O contains an
open set U that misses A. In turn, U contains an open set V that misses B. Hence, O
contains the open set V which misses A[B.
Lemma 6.11. If A X is nowhere dense in X, then A is nowhere dense in X.
Proof. Suppose not, i.e., A X is nowhere dense but A is not. Then there exists an open
set O in X such that every open U O contains a point in A. Now since A is nowhere
dense, O contains an open set U that misses A. But U must contain a point in A, where of
course A = A[A0. Therefore, U must contain a point in A0. That means, by the de nition
of limit point, U must contain a point in A. (Contradiction.)
Lemma 6.12. Suppose f : [0;1] ! 2[0;1] is u.s.c. and there is some non-empty closed
nowhere dense set A [0;1] with the property that:
1) f(a) = [0;1] for all a2A.
2) fj[0;1]nA is a continuous function, and for some B [0;1]nA that is closed and
nowhere dense in [0;1]nA, fj[0;1]n(A[B) is open.
3) For each a2A, y2(0;1) and interval Ua of form (c;a) or (a;c) in [0;1], there exists
some x2UanA with f(x) = y.
Then if O = (O1 O2 On [0;1] )\lim f is a non-empty basic open set
in lim f, n(O) is a disjoint union V [Z, where V is a non-empty open set and Z is a set
35
so that if z2Z, then every non-empty subset of [0;1] of form (d;z) or (z;d) contains some
point of V.
Proof. We prove the lemma by induction on n. If n = 1, then since the bonding function
f is surjective, 1(O) is O1 itself. O1 may be written as V [Z, where V = O1 and Z = ;.
So assume the lemma is true for n; we must show it is true for n + 1. Let O = (O1 O2
On On+1 [0;1] )\lim f be basic open and non-empty. The inductive hypothesis
applies to the set eO = (O1 O2 On [0;1] )\lim f; so, n(eO) is a disjoint union
V[Z where V is open and non-empty, and if z2Z, then every non-empty subset of [0;1] of
form (d;z) or (z;d) contains some point in V. We must show that the analogous statement
holds true for n+1(O).
First, we note that n+1(O) is the intersection of On+1 with f 1( n(eO)). By the
inductive assumption, n(eO) = V[Z, as already described. So now, we consider f 1( n(eO)).
Since f is a continuous function on [0;1]nA and, by condition 3, (0;1) f([0;1]nA), we
have that the preimage of V via fj[0;1]nA is a non-empty open set U in [0;1]nA. (Moreover,
since [0;1]nA is open in [0;1], U is open in [0;1].) Since f(a) = [0;1] for each a2A, we
have that the preimage of V via fjA is A itself. Thus, f 1(V) = U[A. The preimage of
the set Z, f 1(Z), may be written as a disjoint union A[W for some W [0;1].
Now if a2A, we claim that every open interval in [0;1] of form (d;a) or (a;d) contains
a point of U. For, if we have (d;a), we may pick any y2V nf0;1g, and then, by condition
3, there exists some x2 (d;a)nA with f(x) = y. That means fj 1[0;1]nA(V) contains x, and
x2U. Since x2(d;a), we are done. A similar argument applies in the case of (a;d).
Next, suppose w2W; we intend to show that every open set in [0;1] of form (d;w)
and (w;d) contains a point in U. Since w2W, we know w62A; that implies that (since
A is closed and [0;1] is regular) there exists some > 0 so that (w ;w + ) misses A.
Therefore, on (w ;w + ), f is a continuous function. Let N be a large enough integer
so that (w 1N;w) (w ;w)\(d;w). We will show that (w 1N;w) contains a point
of U. First, we note that f((w 1N;w]) cannot be identically f(w); for, if it were, then
36
that would contradict the assumption that fj[0;1]n(A[B) is an open map (since B was assumed
to be nowhere dense). So, for some x2 (w 1N;w), either f(x) > f(w) or f(x) < f(w).
Either way, by the Intermediate Value Theorem, since f is a continuous map on [x;w], on
that interval f must achieve every value between f(x) and f(w). But f(w) is in Z. So, if
f(x) > f(w), the open set (f(w);f(x)) contains points in V; if f(x) < f(w), the open set
(f(x);f(w)) contains points in V. Either way, U = f 1(V)nA meets (x;w), and that shows
that every open set of form (d;w) contains a point in U. A similar argument shows that
every open set of form (w;d) contains a point in U.
Thus, we have shown that f 1( n(eO)) consists exactly of U[f 1(Z), where U is a
non-empty open set and each point t in f 1(Z) has the property that every non-empty
subset of [0;1] of form (d;t) or (t;d) contains some point in U. We note that, since V
and Z were disjoint, f 1(V) and f 1(Z) have only the points of A in common; thus, U =
f 1(V)nA and f 1(Z) are disjoint sets. Now, we consider f 1( n(eO))\On+1, which equals
(U\On+1)[(f 1(Z)\On+1). Note that U\On+1 and f 1(Z)\On+1 are disjoint sets, and
U\On+1 is open. Now, suppose t2f 1(Z)\On+1, and let (d;t) be some open interval
in [0;1]. Since t 2 On+1 and On+1 is open, we know that (for some small > 0) On+1
must contain an open interval (t ;t+ ). Hence, On+1 must contain (d;t)\(t ;t+ ),
which equals (maxfd;t g;t). But (maxfd;t g;t) must contain points in U, so that
(d;t) contains points in U\On+1. A similar argument may be given for an open interval
(t;d). Finally, the open set U\On+1 is non-empty; for, if U\On+1 = ;, then because O
was non-empty (and thus, f 1( n(eO))\On+1 was non-empty), f 1(Z)\On+1 must be non-
empty. However, by the above argument, since On+1 contains points in f 1(Z), it follows
that On+1 must contain points of U. Hence, U\On+1 is non-empty, which is a contradiction;
we conclude that U\On+1 was non-empty in the rst place.
We have therefore demonstrated that f 1( n(eO))\On+1 is a union of two disjoint
sets satisfying the condition given in the inductive hypothesis. However, n+1(O) equals
f 1( n(eO))\On+1, so the proof is complete.
37
Lemma 6.13. Let f : [0;1] ! 2[0;1] be a u.s.c. function with the property that lim f is a
continuum. Suppose that, for some A ( [0;1], fj[0;1]nA is a function, (0;1) f([0;1]nA),
and P = f(p1;p2;:::) 2 lim f jpi =2A[f0;1g for all ig is a dense subset of lim f. Then
lim f has the full projection property.
Proof. Assume by way of contradiction that there is some proper subcontinuum H of lim f
so that, for each positive integer n, there exists some m n such that m(H) = [0;1]. For
any such m, we know that (0;1) f([0;1]nA) = f( m(H)nA) m 1(H); since m 1(H)
is closed and contains (0;1), it follows that m 1(H) = [0;1]. Similarly, i(H) = [0;1] for all
i m. Thus, since in nitely many positive integers m with m(H) = [0;1] exist, we have
that n(H) = [0;1] for each positive integer n. We will now show that P H.
Let p = (p1;p2;p3;:::)2P. Then since 1(H) = [0;1], there exists some point in H of
form (p1;?;?;?;:::). Since 2(H) = [0;1], there exists some point in H of form (?;p2;?;?;:::).
However, p2 =2A and fj[0;1]nA is a function, so f(p2) is unique; therefore, f(p2) = p1. That
means some point of form (p1;p2;?;?;?;:::) lies in H. A similar argument shows that some
point of form (p1;p2;:::;pi 1;pi;?;?;:::) lies in H for all i; since p is a limit point of the set
of all such points, and H is closed, p 2H. Thus, P H. But then P H; because P is
dense, P = lim f, so lim f H. Therefore, lim f = H, contradicting the assumption that H
is a proper subcontinuum.
We are now ready to prove the main theorem of this chapter.
Theorem 6.14. Suppose f : [0;1] ! 2[0;1] is u.s.c. and there is some non-empty closed
nowhere dense set A [0;1] with the property that:
1) f(a) = [0;1] for all a2A.
2) fj[0;1]nA is a continuous function, and for some B [0;1]nA that is closed and
nowhere dense in [0;1]nA, fj[0;1]n(A[B) is open.
3) For each a2A, y2(0;1) and interval Ua of form (c;a) or (a;c) in [0;1], there exists
some x2UanA with f(x) = y.
38
Then lim f is an indecomposable continuum.
Proof. Since f(x) is connected for each x2[0;1], lim f is a continuum. Now we must show
it is indecomposable.
We will show that the set P = f(p1;p2;:::) 2 lim f j pi =2 A[f0;1g for all ig is
dense in lim f. To that end, let O = (O1 O2 On [0;1] ) \ lim f be a non-
empty basic open subset of lim f; we will show that O contains a point in P. We begin by
noting that, by Lemma 6.10, A[f0;1g= A1 is nowhere dense in [0;1]. Now f 1(A1) is the
union of the sets fj 1[0;1]nA(A1) and A; since (by Lemma 6.8) fj 1[0;1]nA(A1) is nowhere dense in
[0;1]nA, we conclude from Lemma 6.9 that fj 1[0;1]nA(A1) is also nowhere dense in [0;1]. We
already know A is nowhere dense, so we have shown that f 1(A1) is a union of two nowhere
dense sets and is therefore also nowhere dense. Let us call f 1(A1)[f0;1g = A2. Again,
by Lemma 6.10, A2 is nowhere dense. Next, by similar reasoning, f 1(A2)[f0;1g = A3
is nowhere dense. Continuing this way, we nd that An = f 1(An 1)[f0;1g is nowhere
dense. We note that An contains all the points of A;f 1(A);f 2(A);:::;f (n 1)(A), as well
as f0;1g;f 1(f0;1g);f 2(f0;1g);:::;f (n 1)(f0;1g).
Now, we note that (by Lemma 6.12) the projection of O onto the nth factor space,
n(O), contains a non-empty open set. Since An was nowhere dense, n(O)nAn is non-
empty. Thus, there exists a point xn 2 n(O)nAn such that xn 62 A[f0;1g;f(xn) =
xn 1 62 A[f0;1g;f2(xn) = xn 2 62 A[f0;1g, and so forth, so that xi =2 A[f0;1g for
each positive integer i n. We may also use condition 3 to select some element xn+1 of
f 1(xn)n(A[f0;1g), and then select some element xn+2 of f 1(xn+1)n(A[f0;1g), and so
forth. The sequence x = (x1;x2;:::;xn;xn+1;:::) is therefore an element of P. Since xi62A
for each positive integer i, f acts as a function on each xi; so, because xn2 n(O), x2O.
Thus, we have shown that P = f(p1;p2;:::) 2 lim f jpi =2A[f0;1g for all ig is dense in
lim f. From this, we conclude using Lemma 6.13 that lim f has the full projection property.
Finally, suppose by way of contradiction that lim f is the union of two proper subcon-
tinua H and K. Since lim f has the full projection property, there exists some large enough
39
integer N so that if n N, n(H) and n(K) are proper subcontinua of [0;1]. Since A is
non-empty, there is some a2A and either a2 N+1(H) or a2 N+1(K). Without loss of
generality, assume a2 N+1(H). Then for some small > 0, either (a ;a) N+1(H) or
(a;a + ) N+1(H). In either case, by condition 3, for any y2 (0;1) there exists some x
in (a ;a)nA or in (a;a + )nA with f(x) = y. That implies that N(H) must contain
(0;1). But then, since N(H) is closed, N(H) = [0;1]. This is a contradiction, so the proof
is complete.
With this grand theorem, we may detect indecomposability easily in a great many more
cases: See Example 10.8 in Chapter 10.
40
Chapter 7
A Generalization of the Two-Pass Condition
The two-pass condition, as described by Ingram in [5] and later in [4], is important to
the question of when indecomposability arises in inverse limits with u.s.c. bonding functions.
Supposef : [0;1]!2[0;1] is u.s.c. Thenf satis es the two-pass condition if there are mutually
exclusive connected open subsets U and V of [0;1] so that fjU and fjV are mappings and
f(U) = f(V) = [0;1]. A consequence of Ingram?s Theorem 4.3 from [5] is the following:
Theorem 7.1. (Ingram) Suppose f : [0;1] ! 2[0;1] is a u.s.c. function satisfying the two-
pass condition. Then if lim f is a continuum that has the full projection property, lim f is an
indecomposable continuum.
Our goal in this section is to further explore the relationship between the two-pass con-
dition, the full projection property, and indecomposability. In particular, we will introduce a
new generalization of the two-pass condition that applies to a wider variety of u.s.c. graphs.
The idea that such a generalization was possible arose in a discussion with Michel Smith,
when he noted that a certain u.s.c. graph comes within of satisfying the two-pass condi-
tion, for any choice of > 0. As it turns out, having a function f that \almost" satis es the
two-pass condition is enough to prove an indecomposability theorem analogous to Ingram?s.
Suppose f : [0;1] ! 2[0;1] is u.s.c. Then f satis es the -two-pass condition if 8 > 0
there exist mutually exclusive connected open sets U;V [0;1] so that, for somefa;bg U
and fc;dg V, fjfa;bg and fjfc;dg are mappings, f(a) and f(c) lie within of 0, and f(b)
and f(d) lie within of 1.
Theorem 7.2. Suppose the u.s.c. function f : [0;1]!2[0;1] satis es the -two-pass condition
and lim f is a continuum with the full projection property. Then lim f is indecomposable.
41
Proof. Suppose lim f = H[K, a union of two proper subcontinua. By the full projection
property, there is a positive integer N such that n(H) 6= [0;1] and n(K) 6= [0;1] for all
n N.
We consider the sets N(H) and N(K). Because these sets are both proper subcontinua
of [0;1] whose union is [0;1], one of them must contain 0 and the other must contain 1.
Without loss of generality, suppose N(H) contains 0 and N(K) contains 1, so that N(H) =
[0;h] for some 0 0 would have had the same e ect.) This example shows yet again the di culty in nding
a general \subsequence" theorem for u.s.c. inverse limits. (See also Example 3 in [3].)
Figure 10.2: The graph of the function fj from Example 10.2
Example 10.3. Let the graph of f : [0;1] ! 2[0;1] be given by the straight line segments
joining (0;0) to (13; 13), from (13; 13) to (13; 23), from (13; 23) to (23; 23), and from (23; 23) to (1;1).
(See Figure 10.3.)
Then (13; 13; 13;:::)2lim f, f 1(13) = 13, and U = [0;1] (13; 23) is an open subset of [0;1]
[0;1] with G(f)\U f13g f(13). Thus, Theorem 4.3 applies and lim f is a decomposable
continuum.
Also, let us note that neither Theorem 4.1 nor Theorem 4.2 applies in this case. For, if
U is any open subset of G(f), G(f)nU is either not the graph of a u.s.c. function from [0;1]
into 2[0;1] or it is the graph of a u.s.c. function that maps a point (13) to a disconnected set.
Example 10.4. Let f : [0;1] ! 2[0;1] be de ned by the graph that is the union of the
traditional tent map and the re ection of the tent map about the line y = 12. (See Figure
10.4.)
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Figure 10.3: The graph of the function f from Example 10.3
Then by Theorem 4.4 (or by Corollary 4.5), lim f is a decomposable continuum. This
space appears to contain a fan-like structure of bucket handle continua.
Figure 10.4: The graph of the function f from Example 10.4
Example 10.5. Let the graph of f : [0;1]!2[0;1] be given by drawing straight line segments
from (0;0) to (13; 13), (13; 13) to (13; 23), (13; 23) to (12;1), and then drawing the re ection of this
gure about the line x = 12. (See Figure 10.5.)
Then f is a steeple with turning point a = 12. So, by Theorem 5.4, lim f is homeomorphic
to the bucket handle and thus, lim f is indecomposable.
Example 10.6. Let the graph of f : [0;1]!2[0;1] be given by drawing straight line segments
from (0;0) to (0; 13), (0; 13) to (12;1) and then drawing the re ection of this gure about the
line x = 12. (See Figure 10.6.)
63
Figure 10.5: The graph of the function f from Example 10.5
Then lim f is decomposable by Theorem 4.1. (The open set U from Theorem 4.1 could
be [0; 14) [0; 14).) Note that f is not a steeple function because f(0) does not equal f0g.
Figure 10.6: The graph of the function f from Example 10.6
Example 10.7. Consider the u.s.c. function f : [0;1]!2[0;1] mentioned in Chapter 5 whose
graph is topologically equivalent to a sin(1x) curve. Speci cally, let the graph of f consist of
the following straight line segments:
i. For each odd integer n 1, the segment joining the points ( 12n;0) and ( 12n 1;1).
ii. For each even integer n 2, the segment joining the points ( 12n;1) and ( 12n 1;0).
iii. The vertical line segment joining the points (0;0) and (0;1).
(See Figure 5.2.)
64
Then, by Theorem 5.2, lim f is an indecomposable continuum. The projection of lim f
onto the rst three factor spaces (i.e., inside [0;1]3) is a countable sequence of sin(1x) curves
joined end to end and limiting to a sin(1x) curve on the back face of the cube. (See Fig-
ure 10.7.) We note that lim f has a structure reminiscent of the indecomposable continua
constructed by Michel Smith in [12].
Figure 10.7: The projection of the inverse limit from Example 10.7 onto its rst three factor
spaces
Example 10.8. Let the graph of f : [0;1] ! 2[0;1] be given by the u.s.c. function shown in
Figure 10.8.
We note that the set A = f0;1g is non-empty, closed, and nowhere dense; moreover,
f(0) = f(1) = [0;1]. Let B be the subset of [0;1]nf0;1gconsisting of points where fj[0;1]nf0;1g
is not di erentiable. Note that B is closed in [0;1]nf0;1g and also nowhere dense. Next,
note that fj[0;1]n(f0;1g[B) is an open mapping. Furthermore, for each interval of form (0;c) in
[0;1] and each y in the interval (0;1), there exists some x2 ([0;1]nf0;1g) with f(x) = y.
The same statement is true for each interval of form (c;1). Thus, the hypothesis of Theorem
6.14 is satis ed, and lim f is an indecomposable continuum.
Example 10.9. Let f : [0;1] ! 2[0;1] be given by squeezing two copies of the steeple graph
in Figure 10.5 into [0;1] [0;1], as shown in Figure 10.9.
If we take a1 = 0;a2 = 14;a3 = 12;a4 = 34; and a5 = 1, then the hypothesis of Theorem
7.4 is satis ed and thus, by Corollary 7.5, lim f is an indecomposable continuum.
65
Figure 10.8: The graph of the function f from Example 10.8
Figure 10.9: The graph of the function f from Example 10.9
Finally, we give some illustrative examples of inverse limits on various kinds of linearly
ordered sets.
Example 10.10. Let the graph of g : [0;1]!2[0;1] be the union of the line segment joining
the points (0;0) and (0;1) and the line segment joining the points (0;0) and (1;1).
We note that g2(x) = g(x) for each x2 [0;1]; that is, g g = g. So, if X = [0;1] for
each < !1, and g ; = g for each < < !1, we know that the collection of functions
fg ; : X !C(X )g <