Inverse Limit Spaces
Except where reference is made to the work of others, the work described in this thesis is
my own or was done in collaboration with my advisory committee. This thesis does not
include proprietary or classifled information.
Scott Varagona
Certiflcate of Approval:
Gary Gruenhage
Professor
Mathematics & Statistics
Michel Smith, Chair
Professor
Mathematics & Statistics
Stewart Baldwin
Professor
Mathematics & Statistics
Thomas Pate
Professor
Mathematics & Statistics
George T. Flowers
Dean
Graduate School
Inverse Limit Spaces
Scott Varagona
A Thesis
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Master of Science
Auburn, Alabama
December 19, 2008
Inverse Limit Spaces
Scott Varagona
Permission is granted to Auburn University to make copies of this thesis at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Scott Michael Varagona, son of Michael and Janet Varagona, was born in Birmingham,
Alabama on August 15, 1983. After graduating from John Carroll Catholic High School in
2001, he entered Auburn University as a mathematics major. In 2005, he graduated with
a dual degree in Mathematics and Philosophy. He maintained a 4.0 GPA throughout his
entire undergraduate career, and also received the Andy Connor Award for Outstanding
Creativity and Originality in Mathematics. He remained at Auburn to pursue a Master?s
Degree in Mathematics. In 2008, he received the Emily Haynsworth Fellowship and won
the College of Science and Mathematics Dean?s Master Research Award.
Scott is also an avid chessplayer and a successful chess tournament competitor. After
winning the Alabama High School Chess Championship in the year 2000, he went on to
become Alabama State Chess Champion in 2002, 2005, 2007, and 2008.
iv
Thesis Abstract
Inverse Limit Spaces
Scott Varagona
Master of Science, December 19, 2008
(B.S., Auburn University, 2005)
64 Typed Pages
Directed by Michel Smith
This paper is a vast survey of inverse limit spaces. After deflning an inverse limit on
continuous bonding functions, we prove important theorems about inverse limits, provide
examples, and explore various generalizations of traditional inverse limits. In particular,
we present original proofs of theorems given by Ingram and Mahavier in \Inverse Limits of
Upper Semi-Continuous Set Valued Functions." We then use this new sort of inverse limit
to enliven the notion of a \two-sided" inverse limit; flnally, we use inverse limits on u.s.c.
functions to produce an indecomposable continuum.
v
Acknowledgments
Many thanks to my advisor, Dr. Michel Smith, for guiding and encouraging me
throughout my graduate career. I have been truly fortunate to have such a kind and
patient advisor, not to mention one who shares my love for inverse limit spaces. I must also
thank the rest of my committee members: Dr. Stewart Baldwin, whose insightful comments
inspired Chapter 5 of this thesis; Dr. Gary Gruenhage, who flrst inspired me to become a
topologist when I was still an undergraduate; and Dr. Thomas Pate, whose Matrices class
taught me to persevere, even when I felt I couldn?t prove a thing!
I would like to thank my family for their constant love and support. Thanks especially
to my parents for convincing me to stay in mathematics when, as a young undergraduate,
I felt uncertain about my choice of major. Were it not for their in uence, I would have
missed out on so much.
Finally, I would like to thank my fellow graduate students in the Math Department
at Auburn University. They have been more than just colleagues: they have been the most
wonderful friends I have ever had.
vi
Style manual or journal used Journal of Approximation Theory (together with the style
known as \aums"). Bibliography follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (speciflcally LATEX)
together with the departmental style-flle aums.sty.
vii
Table of Contents
1 Introduction 1
2 Background Definitions and Theorems 2
3 Inverse Limits with Continuous Bonding Maps 14
4 Inverse Limits of Upper Semi-Continuous Set Valued Functions 28
5 An Extension of the Inverse Limit with u.s.c. Bonding Functions 47
6 An Indecomposable Continuum Produced by an Inverse Limit on u.s.c.
Functions 51
Bibliography 55
Appendix 56
viii
Chapter 1
Introduction
An inverse limit space is a powerful topological tool. Inverse limits not only help us
generate complicated continua with interesting properties, but also allow us to represent
such continua in a simple and elegant way. Now, a new generalization of inverse limit
spaces has opened up more possibilities for topologists to explore. As we will see, this new
generalization even breathes new life into a difierent kind of inverse limit.
This paper provides a vast survey of inverse limits. In Chapter 2, we list basic def-
initions and theorems that will serve as background material. In Chapter 3, we begin by
deflning an inverse limit with continuous bonding maps and proving some important pre-
liminary theorems. We demonstrate the power of inverse limits by using them to prove the
formidable Tychonofi Theorem; then, we use inverse limits to represent some complicated
continua in a simple, straightforward way. Next, in Chapter 4, we consider a generalization
of inverse limits from Ingram and Mahavier?s \Inverse Limits of Upper Semi-Continuous
Set Valued Functions" [1]. After giving original proofs of the theorems from that paper, in
Chapter 5 we show how this new notion of inverse limit can revitalize the formerly redun-
dant notion of the \two-sided" inverse limit. Finally, in Chapter 6, we use inverse limits
with upper semi-continuous bonding functions to produce an indecomposable continuum,
and raise a few questions open for further research.
1
Chapter 2
Background Definitions
and Theorems
Let X be a set and let T be a collection of subsets of X with the following properties:
1. X 2 T;
2. ;2 T;
3. If fOigi2? is a collection of members of T, then Si2? Oi 2 T;
4. If fOigni=1 is a flnite collection of members of T, then Tni=1 Oi 2 T.
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.
Suppose M is a subset of a topological space X. A point p 2 X is a limit point of M
if every open set containing p contains a point in M difierent from p.
Suppose M is a subset of a topological space X. The closure of M (denoted M) is the
union of M with the set of all limit points of M.
Suppose a collection B of open sets of a space X satisfles the following property:
2
If x 2 X and O is an open set containing x, then there exists a member b of B such
that x 2 b 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 x 2 X, there exists some b 2 B with x 2 b.
2. If b1 and b2 are members of B with x 2 b1 \b2, then there exists some set b3 in B
with x 2 b3 (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 Hausdorfi if for every pair of distinct points p;q 2 X,
there exist disjoint open sets Op and Oq containing p and q respectively.
A space X is called regular if for every closed set H ? X and point p 2 X 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 the set X to Y, and U is a subset of X, we deflne
f(U) = ff(u)j u 2 Ug.
3
Let X and Y be topological spaces and let f : X ! Y be a function from X to Y.
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 x 2 X, we say f is continuous.
A function f : X ! Y is said to be onto if for each y 2 Y, there exists some x 2 X
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 y 2 Y, then the preimage of y (written as f?1(y)) is
fx 2 Xj f(x) = yg.
Suppose f : X ! Y is a 1-1 onto 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, onto, continous, and has a
continous inverse, then f is called a homeomorphism and the spaces X and Y are said to
be homeomorphic.
4
Let X be a topological space. A collection B of open sets of X is a local basis at the
point x 2 X if
1. For each member b 2 B, x 2 b;
2. If O is an open set in X containing x, then there exists a member b of B with
x 2 b O.
A space X is called flrst countable if for each x 2 X, there exists a countable local
basis at x.
A space X is called second countable if X has a basis that is countable.
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 flnitely many
members is called a flnite subcover.
A space X is compact if for every open coverfOigi2? of X, there exists a flnite subcover
of X. (I.e., fOijgnj=1 for some natural number 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.
5
A space X is perfectly compact if whenever fGigi2? is a monotonic collection of subsets
of X, there exists a point p in X that is either a point or a limit point of each Gi.
For each i, 1 ? i ? n, let Xi be a topological space. Deflne X = Qni=1 Xi = X1?X2?
????Xn to be the set f(x1;x2;:::;xn)j xi 2 Xi for 1 ? i ? ng. Deflne a topology on X as
follows: a basic open set containing (x1;x2;:::;xn) is given by Qni=1 Oi, where (for each i)
Oi is open in Xi and xi 2 Oi.
Then X together with the topology generated by this basis is called a (flnite) product
space.
For each positive integer i, let Xi be a topological space. Deflne X = Q1i=1 Xi to be
the set f(x1;x2;:::)j xi 2 Xi for each positive integer ig. Deflne a topology on X as follows:
a basic open set containing (x1;x2;:::) is given by Q1i=1 Oi, where (for each i) Oi is open
in Xi, xi 2 Oi, and for some positive integer N, On = Xn if n ? N.
Then X together with the topology generated by this basis is called a (countably
inflnite) product space.
For each i in some arbitrary index set ?, let Xi be a topological space. Deflne X =
Q
i2? Xi to be the set f(xi)i2?j xi 2 Xi for each ig. Deflne a topology on X as follows: a
basic open set containing (xi)i2? is given by Qi2? Oi, where (for each i) Oi is open in Xi,
xi 2 Oi, and for all but flnitely many i, Oi = Xi.
6
Then X together with the topology generated by this basis may be called a product
space on the index set ?.
Let X = Qi2? Xi be a product space (with index set ? either flnite or inflnite). Then
the function ?j : X ! Xj deflned by ?j((xi)i2?) = xj is called the projection map on the
jth coordinate.
Suppose X is a topological space with topology T and S ? X. Then the set S together
with the topology ^T = fS \Oj O 2 Tg is called a subspace of X, where ^T is the subspace
topology.
Let X be a set. Then the relation < on X is a linear ordering on X (and X is said to
be ordered with respect to <) if for any a;b;c 2 X,
1. If a 6= b, either a < b or b < a,
2. If a < b then b 6< a,
3. If a < b and b < c, then a < c.
Let X be a set with a linear ordering <. Let B be the collection of all subsets of X of
the following form:
1. fxj x < pg for some p 2 X,
2. fxj p < xg for some p 2 X,
3. fxj p < x < qg for some p,q 2 X, p < q.
7
Then the topology generated by B is called the order topology on X.
Let X and Y be two spaces with linear orderings 0, let
B(p;?) = fx 2 X j d(x;p) < ?g. If the collection fB(p;?) j p 2 X; ? > 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 Hausdorfi (but not necessarily metric) is called a Hausdorfi
continuum.
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 A is a proper subset of X, then A is a proper subcontinuum.
A point p of a space X is called an isolated point if there exists an open set O X
such that O = fpg.
Let X be a connected set. If X ?fpg is not connected, then p is a cut point of X.
9
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 inflnitely 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 \ A 6= ;, there exists some point
p 2 O with p =2 A. Then A is called a limit arc.
Background Theorems
Most of the following basic theorems may be found in [5]. The proofs of these theorems
are omitted, but may be found in one or more of [2], [3], and [4].
2.1. Let X be a topological space with M X. If M is compact, M is perfectly
compact.
2.2. Let X be a topological space with M X. If M is closed and perfectly compact,
M is compact.
2.3. A closed subset of a compact space is compact.
10
2.4. Projection maps are continuous.
2.5. The continuous image of a compact set is compact.
2.6. Any flnite product of compact sets is compact.
2.7. Let B be a basis for a topological space X. Then every open set of X is a union
of members of B.
2.8. The following are equivalent:
i. f : X ! Y is a continuous function from topological space X to topological space
Y.
ii. If O is a (basic) open set in Y, then f?1(O) is open in X.
2.9. Suppose X and Y are both well-ordered with respect to the order relations j,
then pj = fnkj (pnk) = fnkj (qnk) = qj. This is a contradiction, so h is 1-1.
To show h is continuous, let ??O be basic open in Y. So O is open in some Xnj, where
nj 2fnig1i=1. Thus, h?1(??O) = ??O, which is open in X, so h is continuous.
To show h?1 is continuous, suppose ??O is basic open in X containing (xi)1i=1. If O
is open in some Xni, then (h?1)?1(??O) = h(??O) = ??O is open in Y. On the other hand,
suppose O is open in some Xj, where j 6= ni for all i. Then if nk is the flrst ni with ni > j,
A = (fnkj )?1(O) is an open set in Xnk with ??A containing (xni)1i=1; moreover, h?1(??A) ??O.
So, in either case, h?1 is continuous.
Thus, h is a homeomorphism and the proof is complete. ?
Theorem 3.3. Let X = lim??fXi;fig1i=1 be an inverse limit space. If there is a natural
number N so that fn is an onto homeomorphism for each n ? N, then X is homeomorphic
to XN.
Proof: Deflne a function h : X ! XN by h((xi)1i=1) = xN. We must show that h is a
homeomorphism.
i) h is onto:
16
Because fn is an onto homeomorphism for n ? N, for each xN 2 XN and each n > N,
(fnN)?1(xN) = xn for some xn 2 Xn. Also, for each n < N, fNn (xN) = xn for some xn 2 Xn.
It follows that for each xN 2 XN, there exists a sequence x = (x1;x2;:::;xN;xN+1;:::) 2 X
with h(x) = xN: So h is onto.
ii) h is 1-1:
Suppose h(x) = h(y); we must show that x = y. Since h(x) = h(y), we know xN = yN,
so that xn = fNn (xN) = fNn (yN) = yn for all n < N. Moreover, since fn is a homeomorphism
for n ? N, xn = (fnN)?1(xN) = (fnN)?1(yN) = yn for all n > N.
So xn = yn for each positive integer n. That is, x = y, and h is 1-1.
iii) h is continuous:
Let O be open in XN. Then h?1(O) = ??O, which is open in X; thus, h is continuous.
iv) h?1 is continuous:
Let ??O be a basic open set in X containing x = (xi)1i=1. We must show there exists
an open set G in XN containing h(x) = xN with h?1(G) ? ??O. If O ? XN, then clearly
h?1(O) ???O, and xN 2 O, so G = O. Similarly, if O ? Xn with n > N, fnN(O) = G is open
in XN containing xN, and h?1(G) ? ??O. Finally, if O ? Xn with n < N, then (since fNn is
continuous, and O contains xn) there is an open G ? XN containing xN with fNn (G) ? O,
so that ??G ??O. It follows that h?1(G) ???O.
All cases are accounted for, so h?1 is continuous.
17
Therefore, h is a homeomorphism and the proof is complete. ?
It may be shown that, if each Xi is compact, the inverse limit space X = lim??fXi;fig1i=1
is a closed subspace of the product spaceQ1i=1 Xi. Thus, by the Tychonofi Theorem, if each
Xi is compact, then X (a closed subspace of a compact space) is compact. However, we
will prove directly that an inverse limit on compact spaces is compact (Theorem 3.4); we
will then use this result to prove the Tychonofi Theorem for countable products Q1i=1 Xi.
Finally, wewillgeneralizethenotionofinverselimitinordertoprovetheTychonofiTheorem
in its full generality.
Theorem 3.4. Let X = lim??fXi;fig1i=1 be an inverse limit space with Xi non-empty
and compact for each i. Then X is non-empty and compact.
Proof: Since each Xi is compact, each Xi is perfectly compact. We intend to show that
X is perfectly compact.
Let fGigi2? be a monotonic collection of non-empty subsets of X. (Need to show:
there is a point p = (pi)1i=1 in X such that p is either a point or limit point of every Gi.)
Deflne Gij = ?j(Gi).
18
Since fGi1gi2? is a monotonic collection of non-empty subsets in X1, a perfectly com-
pact space, there exists p1 2Ti2? Gi1, i.e., a point p1 in X1 that is a point or limit point of
every Gi1. For convenience, let us say f1 = f.
(We need to show that there exists some element in f?1(p1) that is also in Ti2? Gi2.)
Because ?1(Gi) = f ??2(Gi), clearly Gi1 = f(Gi2) for each i. Thus,
p1 2Ti2? Gi1 = Ti2? f(Gi2) Ti2? f(Gi2).
But the continuous image of Gi2, a compact set, is compact (and hence, closed); there-
fore:
p1 2Ti2? Gi1 = Ti2? f(Gi2) Ti2? f(Gi2) = Ti2? f(Gi2).
(We need to show that Ti2? f(Gi2) f(Ti2? Gi2).)
Let a 2 Ti2? f(Gi2). Since f(Ti2? Gi2) is closed, it will su?ce to show that a is a
point or limit point of f(Ti2? Gi2).
Proof by Contradiction: Suppose O X1 is an open set containing a but missing
f(Ti2? Gi2). Then f?1(O) is open, contains f?1(a), and misses Ti2? Gi2. In particular,
f?1(a) misses Ti2? Gi2, i.e., f?1(a)\(Ti2? Gi2) = ;.
19
Claim: There exists some Gj2 for which f?1(a)\Gj2 = ;.
Justiflcation: Suppose not. Then f?1(a)\Gi2 6= ; for all i.
Thus, since ff?1(a)\Gi2gi2? is a monotonic collection of non-empty closed sets, by
perfect compactness, we have that Ti2?(f?1(a) \ Gi2) 6= ;. However, by set theory, we
have Ti2?(f?1(a)\Gi2) = f?1(a)\(Ti2? Gi2), so that f?1(a)\(Ti2? Gi2) 6= ;. This is a
contradiction.
So, there exists some Gj2 for which f?1(a) \ Gj2 = ;. But a 2 Ti2? f(Gi2), so
a 2 f(Gi2) for all i. That means for each i, there exists yi 2 f?1(a) such that yi 2 Gi2.
In particular, for j, there exists some yj 2 f?1(a) such that yj 2 Gj2. But f(yj) = a,
so f?1(a)\Gj2 6= ;. (Contradiction.)
Thus, if a 2 Ti2? f(Gi2), then a 2 f(Ti2? Gi2). That is, Ti2? f(Gi2) f(Ti2? Gi2),
and that means Ti2? Gi1 f(Ti2? Gi2).
So, since there exists some point p1 2 Ti2? Gi1, there exists some p2 2 Ti2? Gi2 with
f(p2) = f1(p2) = p1. The same argument shows that there is a point p3 2 Ti2? Gi3 with
f2(p3) = p2, etc.
20
It is therefore easy to see that (p1;p2;p3;:::) is a point in X that is a point or limit
point of each Gi. Thus, X is non-empty and perfectly compact. Since X is closed in itself
and perfectly compact, X is compact. This completes the proof. ?
Theorem 3.5. Suppose X = Q1i=1 Xi is a product space, Yn = Qni=1 Xi, and fn :
Yn+1 ! Yn is the continuous function deflned by fn(x1;x2;:::;xn+1) = (x1;x2;:::;xn).
Then Y = lim??fYi;fig1i=1 is homeomorphic to X.
Proof: Let F : Y ! X be deflned by
F((x1);(x1;x2);(x1;x2;x3);:::;(x1;x2;x3;:::;xn);:::) = (x1;x2;x3;:::;xn;:::).
Clearly F is 1-1 and onto; now we must show F is continuous. Let O = Q1i=1 Oi
be basic open in X, so that for some positive integer k, Oi = Xi for all i > k. Then
F?1(O) = f((x1);(x1;x2);(x1;x2;x3);:::) 2 Y j xi 2 Oi for 1 ? i ? kg, i.e., the set of all
points in Y whose kth coordinate (x1;x2;:::;xk) lies in Qki=1 Oi. Since Qki=1 Oi is open in
Yk, F?1(O) =
??????Qk
i=1 Oi is open in Y, and F is continuous.
To show F?1 is continuous at a given point x = (x1;x2;x3;:::) in X, let ??O be basic
open in Y (where O is open in some Yk) so that F?1(x) 2 ??O. Since O is open in Yk,
there exists a basic open set Qki=1 Oi in Yk that is a subset of O and contains the point
(x1;x2;:::;xk). It follows that (Qki=1 Oi)?(Q1i=k+1 Xi), which is open in X, contains x.
However, F?1[(Qki=1 Oi)?(Q1i=k+1 Xi)] ???O; F?1 is therefore continuous.
21
So F is a homeomorphism. ?
Tychonofi Theorem for Countable Products. Let X = Q1i=1 Xi be a topological
product space with Xi compact for each i. Then X is compact.
Proof: By Theorem 3.5, if Yn = Qni=1 Xi, Y = lim??fYi;fig1i=1 is homeomorphic to X.
Any flnite product of compact spaces is compact, so each Yi is compact; thus, by Theorem
3.4, Y is compact. It follows that X is compact also. ?
Thus, we have used inverse limits to prove the Tychonofi Theorem for countable prod-
ucts, i.e., that a product of countably many compact spaces is compact. We will now
introduce a more general form of inverse limit that, among other things, will allow us to
prove the general Tychonofi Theorem, i.e., that any product of compact spaces is compact.
A directed set I is a set with a partial order < such that for each pair fi;fl 2 I, there
exists some 2 I such that fi < and fl < . Suppose that for each i 2 I, Xi is a
topological space; also suppose there exists a collection of functions ffji gi h, let fkh : Yk ! Yh be the continuous function deflned by fkh((xi)i?k) = (xi)i?h.
Then the inverse limit Y = lim??fYi;fkj gi2I; jfi Xi)] ???O, and F?1 is continuous. Thus, F?1
is a homeomorphism. ?
Lemma 3.8. Let I be an arbitrary index set, and let ~I be the set I with a well-ordering
placed upon it. Then X = Qi2I Xi is homeomorphic to ~X = Qi2~I Xi.
Proof: Let F : X ! ~X be deflned by F((xi)i2I) = (xi)i2~I. Clearly F is onto and 1-1;
it remains to show that F and F?1 are continuous.
Let O be basic open in ~X, so that O = Qi2~I Oi where Oi = Xi for all but flnitely many
i. Then F?1(O) = Qi2I Oi, which is open in X. So F is continuous, and the argument is
easily altered to show that F?1 is also continuous. So F is a homeomorphism. ?
With these two lemmas in hand, we are flnally prepared to prove the general Tychonofi
Theorem.
25
Tychonofi Theorem: Suppose for each i in some index set I, Xi is a compact topo-
logical space. Then Qi2I Xi is compact.
Proof: The result is already known if the index set I is flnite, so assume I is inflnite.
By Lemma 3.8, without loss of generality we may assume that the index set I is well-
ordered. Since I is easily re-ordered in a way that keeps I well-ordered without having a
last element, we may also assume without loss of generality that I is a directed set. We will
use transflnite induction: Suppose that, for any given fi 2 I, Qi?fi Xi is compact. We will
show that X = Qi2I Xi is compact.
For each i 2 I, let Yi = Qj?i Xj. Deflne fkh for h < k 2 I as in Lemma 3.7. Then
Y = lim??fYi;fkhgi2I; h
<
>:
3
2t;0 ? t ?
2
3
5
3 ?t;
2
3 ? t ? 1
9>
=
>;
26
Let fi = f for each i. Then lim??fXi;fig1i=1 = lim??f[0;1];fg1i=1 is homeomorphic to the
topologist?s sine curve, i.e., f(x;sin(1x)) j x 2 [?1;0)g[f(0;x) j x 2 [?1;1]g.
2) For each i, let Xi = [0;1]. Suppose f : [0;1] ! [0;1] is given by
f(t) =
8>
<
>:
2t; 0 ? t ? 12
2?2t; 12 ? t ? 1
9>
=
>;
Let fi = f for each i. Then lim??fXi;fig1i=1 = lim??f[0;1];fg1i=1 is homeomorphic to the
Knaster continuum, a.k.a., \the bucket handle."
27
Chapter 4
Inverse Limits of Upper Semi-Continuous
Set Valued Functions
Suppose X and Y are compact Hausdorfi spaces, and deflne 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 x 2 X and open V in Y containing f(x), there exists an open U in X
containing x so that f(u) ? V for all u 2 U. Upper semi-continuity is a generalization
of continuity; hence, using upper semi-continuous bonding functions instead of continuous
bonding maps provides us with a more generalized notion of an inverse limit.
Suppose that, for each positive integer i, Xi is a compact Hausdorfi space and fi :
Xi+1 ! 2Xi is an upper semi-continuous function. We deflne lim??fXi;fig1i=1 to be the set
of all points in Q1i=1 Xi with xi 2 fi(xi+1) for all i. (For convenience, we shall 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 j O is basic open inQ1i=1 Xig.
As in Chapter 3, if Oi is a subset of Xi, we deflne ??Oi = fx 2 lim??f j xi 2 Oig; if Oi is open
in Xi, then ??Oi is open in lim??f.
Remark: However, unlike the inverse limit spaces seen in Chapter 3, in general, the
collection f??Oj O is open in some Xig is not a basis for lim??f. An example is given in the
appendix to explain why this is so.?
28
In [1], Ingram and Mahavier not only prove generalizations of the sorts of theorems
already seen in Chapter 3, but also provide examples to show when such results do not
generalize. In this chapter, I give my own proofs of their theorems and explain their coun-
terexamples in detail. (Note that the theorems are numbered here in a way that is consistent
with the original numbering in [1]; for example, Theorem 2.1 from [1] has been relabeled
4.2.1, etc.)
First, Ingram and Mahavier introduce the useful notion of the graph of an upper semi-
continuous function. If X and Y are compact Hausdorfi spaces and f : X ! 2Y is u.s.c.,
the graph of f (abbreviated G(f)) is the set f(x;y) 2 X ?Y j y 2 f(x)g.
Theorem 4.2.1. Suppose each of X and Y is a compact Hausdorfi 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).
Proof: Assume the hypothesis.
(() Suppose there is an upper semi-continuous function f : X ! 2Y such that M =
G(f). We need to show that M is closed.
29
Proof by contradiction: Let (x;y) be a limit point of M with (x;y) =2 M. We know
the set f(x) is compact, and hence closed; moreover, fxg?f(x) is a subset of M. Because
(x;y) =2 M, we have y =2 f(x).
Y is a compact Hausdorfi space, so Y is regular. Thus, there exist disjoint open O1,
O2 in Y with f(x) ? O1 and y 2 O2. O1 contains f(x), so by u.s.c. there exists an open U
in X containing x so that f(U) ? O1.
U ? O2 is open in X ? Y and contains (x;y), a limit point of M, so U ? O2 must
contain some other point (x0;y0) 2 M. We note that not every point in U ?O2 can have x
as its flrst coordinate, for otherwise, U ?O2 would have no points in M. (For, each point
would be of form (x;z) where z =2 f(x), so that (x;z) =2 G(f) = M.) Thus, there is some
(x0;y0) 2 U ?O2 with x0 6= x.
However, x0 2 U, so f(x0) O1. But (x0;y0) 2 M, so y0 2 f(x0). It follows that a
point in f(x0) (namely, y0) lies in O2, which was disjoint from O1. This is a contradiction,
so M is closed.
()) Suppose that M is closed. We must show that there is an upper semi-continuous
function f : X ! 2Y such that M = G(f).
For each x 2 X, consider fxg ? Y. This set is closed in X ? Y, so that Kx =
(fxg?Y) \ M is also closed and non-empty. A closed subset of X ?Y is compact, so Kx
is compact. Thus, ?2(Kx) is compact in Y.
30
Deflne f : X ! 2Y by f(x) = ?2(Kx); we must show f is an upper semi-continuous
function.
Let V be open in Y with ?2(Kx) = f(x) V. We need to show there exists an open
set U in X with x 2 U such that f(U) ? V.
Proof by contradiction: Suppose no such U exists. Let fufigfi2? be the set of all points
ufi in X with f(ufi) 6? V. Then every open set in X containing x must contain inflnitely
many ufi?s. (For, suppose not. Then some open O containing x contains only flnitely
many ufi?s, say, ufi1;ufi2;:::;ufik. Thus, because X is regular, there exists an open set R
containing x that misses (X ?O)[fufiigki=1, and hence, misses all ufi?s.)
Claim: the collection of points W = f(ufi;yfi)jfi 2 ?;yfi 2 f(ufi);yfi =2 Vg has a limit
point (x;z) with z =2 V.
For, suppose not. Then fxg?(Y ?V) and W are disjoint closed sets in X ?Y. So,
since X?Y is normal, there exist disjoint open O1 and O2 containing fxg?(Y ?V) and W
respectively. Hence, for each (x;z) 2fxg?(Y ?V), we may flnd a basic open set (A?B)(x;z)
about (x;z) lying in O1. By the compactness of X ?Y, a flnite number (say, n) of these
open sets covers fxg?(Y ?V), so that f(Tni=1 Ai)?Bjgnj=1 also covers fxg?(Y ?V). We
note that W misses the union of the members of this flnite open cover.
31
Tn
i=1 Ai is open in X, contains x, and contains no ufi such that f(ufi) 6? V. This
means there does exist an open set (namely, Tni=1 Ai) in X with f(Tni=1 Ai) ? V. This is a
contradiction, so there does exist a point (x;z) 2fxg?Y, with z =2 V, that is a limit point
of W. But W M, which is closed, so any limit point of W is an element of M. That is,
(x;z) 2 M, and z 2 f(x). But f(x) ? V, and z =2 V. A contradiction has been reached, so
the proof is complete. ?
For the next two theorems, let the following be a standing hypothesis: suppose that for
each positive integer n, Xn is a non-empty compact Hausdorfi space and fn : Xn+1 ! 2Xn is
an upper semi-continous bonding function. IfQ = Q1i=1 Xi, let Gn = fx 2Qjxi 2 fi(xi+1)
for i ? ng:
Theorem 4.3.1. For each positive integer n, Gn is a non-empty compact set.
Proof: We flrst show that Gn is non-empty for each positive integer n. Pick any
point xn+1 2 Xn+1. fn(xn+1) is compact (and non-empty) in Xn; we may pick a point
xn 2 fn(xn+1), so that fn?1(xn) is compact (and non-empty) in Xn?1; next, pick a point
xn?1 2 fn?1(xn), and so forth. By flnishing this process at x1 and then (for i > n + 1)
choosing xi from Xi arbitrarily, we flnd that fx1;x2;:::;xn;xn+1;xn+2;:::g 2 Gn. So
Gn 6= ;.
Now we must show that Gn is compact. Since Gn is a subspace ofQ, which is compact,
it will su?ce to show that Gn is closed.
32
Proof by contradiction: Let p = fp1;p2;:::;pn;pn+1;pn+2;:::g be a limit point of Gn
in Q, with p =2 Gn. Since p =2 Gn, it follows that pi =2 fi(pi+1) for some i, 1 ? i ? n.
Xi is regular, so (in Xi) there exist disjoint open sets Opi and Ofi(pi+1) containing pi
and fi(pi+1) respectively. By the upper semi-continuity of fi, there exists an open U in
Xi+1 containing pi+1 so that f(U) ? Ofi(pi+1). That is, pi =2 f(U). Hence, for all u 2 U,
f(U) ? Ofi(pi+1) and f(u)\Opi = ;.
Thus, X1?X2?????Xi?1?Opi ?U?Xi+2???? is open in Q, contains p, but misses
Gn. However, p was a limit point of Gn, so this is a contradiction.
So Gn is closed in Q, and therefore, Gn is compact. ?
Theorem 4.3.2. K = lim??f is non-empty and compact.
Proof: K = lim??f = T1n=1 Gn. But fGng1n=1 is a monotonic collection of non-empty
closed (compact) subsets of Q; so, since Q is perfectly compact, T1n=1 Gn is non-empty.
Moreover, any intersection of closed sets is closed, so T1n=1 Gn is also closed, and therefore
compact. ?
Having dealt with the issue of compactness, we now turn to theorems about connect-
edness.
33
Theorem 4.4.1. Suppose X, Y are compact Hausdorfi spaces, X is connected, f :
X ! 2Y is u.s.c., and for each x in X, f(x) is connected. Then the graph G(f) is connected.
Proof: Suppose by way of contradiction that G(f) is not connected. Then, since
G(f) is closed, G(f) = H [ K, a union of disjoint closed sets. For a given x, deflne
(x;f(x)) = f(x;y)j y 2 f(x)g. We note that K \ (x;f(x)) and H \ (x;f(x)) cannot both
be non-empty. (For, if they were, then (x;f(x)) = [K \ (x;f(x))] [ [H \ (x;f(x))],
a union of two disjoint closed point sets. But f(x) was connected.) Thus, for all x 2 X,
either K \ (x;f(x)) = ; or H \ (x;f(x)) = ;. That is, for each x, (x;f(x)) must lie
either in H or K but not both.
Because X?Y is compact and Hausdorfi, X?Y is normal. So there exist disjoint open
sets OH and OK containing H and K respectively. Without loss of generality, consider a
given set (x;f(x)) that is a subset of K. We may flnd a union of basic open sets of form
Bi = Aij ? Rij in X ? Y that contains (x;f(x)) and lies in OK. By the compactness of
(x;f(x)), only flnitely many (say, n) Bi?s cover (x;f(x)). Thus, f(Tnj=1 Aij)?Ritgnt=1 is an
open cover of (x;f(x)).
Since R = Snt=1 Rit is open in Y and contains f(x), by u.s.c. there exists an open U
in X containing x so that f(U) ? R. Then V = U \Tnj=1 Aij is also open in X, contains
x, and clearly f(V) ? R. (Indeed, f(x;f(x)) j x 2 Vg?Snt=1((Tnj=1 Aij)?Rit).).
34
Hence, no points z in V can be such that (z;f(z)) ? H. (For that would contradict
the fact that (z;f(z)) ? OK, where OK \H = ;.) So, we have found an open set V = Vx
in X containing x so that Vx \ ?1(H) = ;. Such an open Vx can be found for each x
with (x;f(x)) ? K, so that the union of all such Vx?s is open in X and contains the set
fx j (x;f(x)) ? Kg.
But such an open set (disjoint from the union of the Vx?s) can also be found containing
the set fx j (x;f(x)) ? Hg. So, we have disjoint (non-empty) open sets in X whose union
equals X itself, and this contradicts the fact that X was connected. ?
Theorem 4.4.2. Suppose that X and Y are compact Hausdorfi spaces, Y is connected,
and f is an upper semi-continuous function from X into 2Y such that for each y in Y,
fx 2 X j y 2 f(x)g is a non-empty, connected set. Then G(f) is connected.
Proof: Suppose by way of contradiction that G(f) = H [K, a union of disjoint closed
sets. For each y 2 Y, let Ay = fx 2 X j y 2 f(x)g, and let (Ay;y) = fAyg?fyg. Each
Ay is connected, so for each y, either (Ay;y) ? H or (Ay;y) ? K but not both. We
know that H = S(Ay;y)?Hf(Ay;y)g and K = S(Ay;y)?Kf(Ay;y)g; thus, the sets ?2(H) =
fy j (Ay;y) ? Hg and ?2(K) = fy j (Ay;y) ? Kg are disjoint closed sets whose union is Y.
However, Y is connected, so this is a contradiction. ?
Next, it will be useful to extend the notion of the graph of one function, G(f), to
the graph of a flnite sequence of functions. If for 1 ? i ? n, Xi is a compact Hausdorfi
35
space and fi : Xi+1 ! 2Xi is u.s.c., we deflne G(f1;f2;:::;fn) = f(x1;x2;:::;xn;xn+1) 2
Qn+1
i=1 Xi j xi 2 fi(xi+1) for 1 ? i ? ng.
Theorem 4.4.3. Suppose X1;X2;:::;Xn+1 is a flnite collection of Hausdorfi continua
and f1;f2;:::;fn is a flnite collection of upper semi-continuous functions such that fi :
Xi+1 ! 2Xi for 1 ? i ? n. If fi(x) is connected for each x in Xi+1 and each i;1 ? i ? n,
then G(f1;f2;:::;fn) is connected.
Proof: We will use induction on the number of spaces, n. For the base case, suppose
X1;X2 are Hausdorfi continua, f1 : X2 ! 2X1 is an upper-semi continuous function, and
f1(x) is connected for each x in X2. Then G(f1) is connected by Theorem 4.4.2.
Now suppose the theorem is true for a graph on n spaces; we must show that the
theorem also holds for n+1 spaces. That is, we must show G(f1;f2;:::;fn) is connected.
By the inductive hypothesis, the graph G(f2;f3;:::;fn) is connected. Deflne an upper
semi-continuous function f? : G(f2;f3;:::;fn) ! 2X1 by f?(x2;x3;:::;xn+1) = f1(x2). To
show that f? is indeed upper semi-continuous, let (x2;x3;:::;xn+1) be in G(f2;f3;:::;fn),
so that f?(x2;x3;:::;xn+1) = f1(x2). Let V be an open set in X1 that contains f1(x2).
We need to flnd an open set in G(f2;f3;:::;fn) containing x whose image lies in V. Since
f1 is u.s.c., there exists some open U in X2 (with x2 2 V) so that f1(U) is a subset of V.
Thus, O = (U ?X3?X4?:::?Xn+1)\G(f2;f3;:::;fn) is an open set in G(f2;f3;:::;fn)
containing (x2;x3;:::;xn+1) such that f?(O) V.
36
Thus, f? is u.s.c. Moreover, f?(x2;x3;:::;xn+1) is connected for all (x2;x3;:::;xn+1) 2
G(f2;f3;:::;fn). (For, f?(x2;x3;:::;xn+1) = f1(x2), which was assumed to be connected.)
Thus, by Theorem 4.4.2, the graph of f? is connected. However, the graph of f? is precisely
the set of all ordered pairs (x1;x2;x3;:::;xn+1) with xi 2 fi(xi+1) for each i. This set is in
fact G(f1;f2;:::;fn), so G(f1;f2;:::;fn) has been shown to be connected and the proof is
complete. ?
Theorem 4.4.4. Suppose that Xi is a Hausdorfi continuum for each i and fi(x) is
connected for each x 2 Xi+1. Then Gn is connected for each positive integer n.
Proof: We note that Gn = G(f1;f2;:::;fn) ? Q1i=n+2 Xi. Since G(f1;f2;:::;fn) is
connected (by Theorem 4.4.3) and Q1i=n+2 Xi is connected as well, Gn is connected. ?
Theorem 4.4.5. Suppose X1;X2;:::;Xn+1 is a flnite collection of Hausdorfi continua
and f1;f2;:::;fn is a flnite collection of u.s.c. functions such that fi : Xi+1 ! 2Xi for
1 ? i ? n. If for each i, 1 ? i ? n and each y 2 Xi, fx 2 Xi+1jy 2 fi(x)g is a non-empty,
connected set, then G(f1;f2;:::;fn) is connected.
Proof: To get this result, we shall adjust Mahavier?s proof of Theorem 4.4.3. By
Theorem 4.4.2, the theorem is true for only one bonding function f1.
37
Assume the inductive hypothesis. That is, assume that if for all i and for all y 2 Xi,
fx 2 Xi+1 j y 2 fi(x)g is a non-empty, connected set, then the graph on < n + 1 u.s.c.
functions is connected.
Need: G(f1;f2;:::;fn;fn+1) is connected.
By hypothesis, G(f1;f2;:::;fn) is connected. Assume by way of contradiction that H
and K are mutually separated non-empty point sets with H [K = G(f1;f2;:::;fn;fn+1).
Since the graph is closed, we know that H and K are in fact disjoint closed sets.
Let h : G(f1;f2;:::;fn;fn+1) ! G(f1;f2;:::;fn) be the continuous map deflned by
h(x1;x2;:::;xn;xn+1) = (x1;x2;:::;xn). Since h(G(f1;f2;:::;fn;fn+1)) = h(H [ K) =
G(f1;f2;:::;fn) is connected, there is a point p = (p1;p2;:::;pn;pn+1) belonging to h(H)
and h(K). Note that both h(H) and h(K) are compact and hence, closed.
Thus, f(x1;x2;:::;xn;xn+1;xn+2) 2 G(f1;f2;:::;fn;fn+1)j xi = pi for 1 ? i ? n +
1; xn+2 2 fz 2 Xn+2 j pn+1 2 fn+2(z)gg is a connected set, because it is a product of
connected sets. But this set intersects both H and K, so H and K could not have been
mutually separated. So, we have a contradiction and the proof is complete. ?
Theorem 4.4.6. Let Xi be a Hausdorfi continuum for each positive integer i. Suppose
fi : Xi+1 ! 2Xi is u.s.c. and for each xi 2 Xi, fy 2 Xi+1 j xi 2 fi(y)g is a non-empty,
connected set. Then for each positive integer n, Gn is connected.
38
Proof: By Theorem 4.4.5, G(f1;f2;:::;fn+1) is connected. We note that Gn =
G(f1;f2;:::;fn+1) ? Q1i=n+2 Xi. Since Xi is a continuum for each integer i ? n + 2,
we have that Q1i=n+2 Xi is a continuum. Thus, Gn is a product of two connected sets, and
hence, is connected. ?
Theorem 4.4.7. Suppose that for each positive integer i, Xi is a Hausdorfi 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 Hausdorfi continuum.
Proof: By Theorems 4.3.1 and 4.4.4, for each positive integer n, Gn is a non-empty,
compact, connected set; that is, each Gn is a (Hausdorfi) continuum. Moreover, since
Gn+1 Gn for each n, fGng1n=1 is a monotonic collection of Hausdorfi continua. That
means T1n=1 Gn is a Hausdorfi continuum. But lim??f = T1n=1 Gn, so the result is proven. ?
Theorem 4.4.8. Suppose that for each positive integer i, Xi is a Hausdorfi continuum,
fi : Xi+1 ! 2Xi is an upper semi-continuous function, and for each x 2 Xi, fy 2 Xi+1 j x 2
fi(y)g is a non-empty, connected set. Then lim??f is a Hausdorfi continuum.
Proof: By Theorems 4.3.1 and 4.4.6, for each positive integer n, Gn is a Hausdorfi
continuum. The rest of the proof is the same as the proof of Theorem 4.4.7. ?
39
Next, Mahavier and Ingram give a generalized version of the \space-skipping" theo-
rem seen in Chapter 3. However, we must flrst deflne the notion of composition of u.s.c.
functions. Let X, Y, and Z be compact Hausdorfi spaces, and suppose f : X ! 2Y and
g : Y ! 2Z are u.s.c. functions. Then g ?f : X ! 2Z is deflned by (g ?f)(x) = fz 2 Zj
there exists y 2 Y such that y 2 f(x) and z 2 g(y)g.
Theorem 4.5.1. Suppose X1;X2;:::; is a sequence of compact Hausdorfi spaces and
fi : Xi+1 ! 2Xi is u.s.c. for each positive integer i. If n1;n2;:::; is an increasing sequence
of positive integers, let g1;g2;::: be the sequence of functions with the property that gi =
fni?fni+1?????fni+1?1 for each i. If F : Qi>0 Xi !Qi>0 Xni is given by F(x1;x2;x3;:::) =
(xn1;xn2;xn3;:::), then Fj lim??f is a continuous transformation from lim??f onto lim??g.
Proof: Let O = (Q1i=1 Oni)\lim??g be basic open in lim??g (where Oni is open in Xni, and
for some positive integer k, Oni = Xni for i ? k). Then (Fjlim??f)?1(O) = (Q1j=1 Oj)\lim??f,
where if j = ni for some i, Oj = Oni, and for all other j, Oj = Xj. Since (Fjlim??f)?1(O) is
open in lim??f, Fjlim??f is continuous. The fact that Fjlim??f maps onto lim??g is clear. ?
Theorem 4.5.2. Let X1;X2;::: and Y1;Y2;::: be sequences of compact Hausdorfi
spaces and, for each positive integer i, let fi : Xi+1 ! 2Xi and gi : Yi+1 ! 2Yi be
u.s.c. functions. Suppose further that, for each positive integer i, ?i : Xi ! Yi is a
mapping such that ?i ? fi = gi ? ?i+1. Then the function ? : lim??f ! lim??g given by
?(x) = (?1(x1);?2(x2);?3(x3);:::) is continuous. Moreover, ? is 1-1 (and surjective) if
each ?i is 1-1 (and surjective).
40
Proof: First, we must show that ? maps into lim??g. Let p = (p1;p2;:::) 2 lim??f;
we need to show that ?(p) 2 lim??g. That is, we need to show that for any i, ?i(pi) 2
gi(?i+1(pi+1)). By hypothesis, gi(?i+1(pi+1)) = ?i(fi(pi+1)); however, pi 2 fi(pi+1), so
?i(pi) 2 ?i(fi(pi+1)) = gi(?i+1(pi+1)). Thus, ? does map into lim??g; it remains to show
that ? is continuous.
Let x = (x1;x2;:::) 2 lim??f, and let O = (Q1i=1 Oi)\lim??g be basic open in lim??g, with
O containing ?(x) = (?1(x1);?2(x2);:::). We need to show there exists an open set U in
lim??f containing x so that ?(U) O.
We note that, since Q1i=1 Oi is basic open in Q1i=1 Yi, Oi is open in Yi for each i; also,
for some positive integer k, if i > k, Oi = Yi. Now, since each ?i is continuous, for all i the
set Ui = ??1i (Oi) is open in Xi and contains xi. Hence, the open set U = Tki=1??Ui contains
(x1;x2;:::). To show that ?(U) O, let us assume p = (p1;p2;:::) 2 U and show that
?(p) 2 O. For i ? k, pi 2 Ui = ??1i (Oi), so we have that ?i(pi) 2 Oi. For i > k, since
Oi = Yi, ?i(pi) 2 Oi automatically. So, since p 2 lim??f, ?(p) 2 lim??g and ?(p) 2 O. Thus,
?(U) O, and we have shown that ? is continuous.
Finally, wewill show that if each?i is 1-1 (and surjective), then?is 1-1 (and surjective).
First, suppose ?(x) = ?(y), so that ?i(xi) = ?i(yi) for each i. Since each ?i is 1-1, xi = yi
for each i. Thus, x = y and ? is 1-1. Now suppose that each ?i is also surjective, and let
y = (y1;y2;y3;:::) 2 lim??g. Again, because ?i is surjective for each i, it follows that (for
41
each i) there exists some xi 2 Xi with ?i(xi) = yi. Then ?(x1;x2;x3;:::) = (y1;y2;y3;:::),
but we must verify that (x1;x2;x3;:::) 2 lim??f; i.e., we must show that xi 2 fi(xi+1) for
each i.
We note that yi 2 gi(yi+1) = gi(?i+1(xi+1)) = ?i(fi(xi+1)); thus, ??1i (yi) 2 fi(xi+1).
However, since ?i was 1-1, ??1i (yi) = xi, so xi 2 fi(xi+1), as desired. Thus, x 2 lim??f and
? is surjective. ?
Given that X is a compact Hausdorfi space and f : X ! 2X and g : X ! 2X are u.s.c.
functions, we say f and g are topologically conjugate if there exists a homeomorphism h
with h(X) = X and h?f = g ?h.
Theorem 4.5.3. Suppose X is a compact Hausdorfi space. If f : X ! 2X and
g : X ! 2X are topologically conjugate u.s.c. functions, then lim??f is homeomorphic to
lim??g.
Proof: Since f and g are topologically conjugate, there is a homeomorphism h with
h(X) = X and h ? f = g ? h. Let ? : lim??f ! lim??g be deflned by ?(x1;x2;:::) =
(?1(x1);?2(x2);:::), where ?i = h for all i. Because h ? f = g ? h, each ?i satisfles the
hypothesis of Theorem 4.5.2; thus, ? is continuous. Moreover, since h is 1-1 and surjective,
? is 1-1 and surjective. Therefore, lim??f is homeomorphic to lim??g. ?
42
Examples and Counterexamples
According to Theorem 4.4.7, if (1) each Xi is a Hausdorfi continuum, (2) fi : Xi+1 !
2Xi is an upper semi-continuous function, and (3) for each x in Xi+1, fi(x) is connected,
then lim??f is a Hausdorfi continuum. However, the following example shows that if condition
(3) is omitted, lim??f need not be connected.
Example 1: For each positive integer i, let Xi = [0;1] and let fi : [0;1] ! 2[0;1] be
deflned by the graph consisting of straight line segments connecting the points (0;0) to
(14; 14), (0;0) to (1;0), (1;0) to (1;1), and (34; 14) to (1;1).
Then lim??f is not connected because the space contains an isolated point, namely,
p = (14; 14; 34;1;1;1;:::). To see that p is isolated, we will flnd an open set containing p and
no other point in lim??f. Let O1 ? X1 be (14 ??; 14 + ?), let O2 ? X2 be (14 ??; 14 + ?), let
O3 ? X3 be (34 ??; 34 +?), and let O4 ? X4 be (1??;1], where ? is chosen small enough so
that 0 =2 O1 or O2, 34 =2 O1 or O2, 14 =2 O3, 1112 =2 O3, and 1112 =2 O4. Then p = ??O1\??O2\??O3\??O4,
which is open. ?
Next, it is worth noting that the conclusion of Theorem 4.5.2 is only that the function
Fjlim??f be a continuous transformation, rather than a full- edged homeomorphism. Indeed,
the following example shows that even if the hypotheses of Theorem 4.5.1 apply, Fjlim??f
need not be a homeomorphism.
43
Example 2: For each i, let Xi = [0;1] and let fi : [0;1] ! 2[0;1] be deflned by the graph
consisting of the straight line segments joining the points (0;1) to (12; 12), (0; 12) to (1; 12),
and (1;0) to (1; 12). Then it follows that f ?f : [0;1] ! 2[0;1] is the graph consisting of the
straight line segments joining (0;0) to (0; 12), (0; 12) to (1; 12), and (1; 12) to (1;1). (We will
abbreviate f ?f by f2.)
lim??f is not homeomorphic to lim??f2 because one space contains a triod while the other
space is an arc.
Justiflcation: lim??f contains a triod. For, let A1 be the subset of lim??f consisting of
all points of form (x;1 ? x;1;0;1;0;:::), where x 2 (12;1]. Let A2 be the subset of lim??f
consisting of all points of form (12; 12;x;1?x;1;0;1;:::), where x 2 (12;1]. Finally, let A3
be the subset of lim??f consisting of all points of form (12;x;1;0;1;0;:::), where x 2 [0; 12).
Because ?A1, ?A2 and ?A3 are all arcs with exactly one point, (12; 12;1;0;1;0;:::), in common,
?A1 [ ?A2 [ ?A3 is a triod.
However, lim??f2 is an arc. For, by Theorem 4.4.7, lim??f2 is a Hausdorfi continuum, and
it is easily seen that every point in this space is a cut point except for (0;0;0;:::) and
(1;1;1;:::). So, this continuum has exactly two cut points and is therefore an arc. ?
Ingram and Mahavier give the following example to show \the variety of continua that
can be produced" using inverse limits with u.s.c. bonding functions.
44
Example 3: 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).
Then lim??f is a fan.
Justiflcation: For each positive integer n, let Kn be the set of all points of form
(0;0;:::;0;x;x;x;:::), where the flrst n?1 entries are 0, and x 2 [0;1]. Then each Kn is
an arc, S1n=1 Kn = lim??f, and T1n=1 Kn = (0;0;0;:::), a single point. So lim??f is indeed a
fan. ?
Finally, in the case where Xi = [0;1] and fi : [0;1] ! 2[0;1] is the same u.s.c. bond-
ing function for all positive integers i, lim??f may be not only 1-dimensional or inflnite-
dimensional, but n-dimensional for any positive integer n. Mahavier and Ingram give a
two-dimensional example that is easily generalized:
Example 4: Again, 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 (0; 12), (0; 12) to (12; 12),
(12; 12) to (12;1), and (12;1) to (1;1). Then lim??f consists precisely of all points of form
i) (1;:::;1;1;x; 12; 12;:::; 12;y;0;0;:::), where x 2 [12;1] and y 2 [0; 12],
ii) (1;:::;1;1;x; 12; 12;:::), where x 2 [12;1],
iii) (12;:::; 12; 12;y;0;0;:::), where y 2 [0; 12],
iv) (1;:::;1;1;1;:::),
v) (12;:::; 12; 12; 12;:::),
vi) (0;:::;0;0;0;:::).
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Thus, lim??f is the union of countably many 2-cells, 1-cells, and single points. It follows
that lim??f is 2-dimensional.
The bonding function with two \stair-steps" gives rise to a two-dimensional inverse
limit; an argument similar to the one given in Example 4 shows that a bonding function
with n \stair-steps" gives rise to an n-dimensional inverse limit. 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 (jn; jn) to (jn; j+1n ) and joining (jn; j+1n ) to (j+1n ; j+1n ) for 0 ? j ? n?1. Then lim??f
contains all points of form
(1;:::;1;xn; n?1n ;:::; n?1n ;:::;xn?1; n?2n ;::::::;
i
n;xi;
i?1
n ;:::;xi?1;
i?2
n ;::::::;
1
n;x1;0;:::)
where xi 2 [i?1n ; in] for 1 ? i ? n. Thus, lim??f contains countably many n-cells. Since lim??f in
fact consists of these countably many n-cells and also countably many j-cells where j < n,
it follows that lim??f is n-dimensional.
46
Chapter 5
An Extension of the Inverse Limit with
u.s.c. Bonding Functions
We now expand on the results of Ingram and Mahavier by introducing yet another
generalization of an inverse limit. Suppose that for each integer i, Xi is a compact Hausdorfi
space and fi : Xi+1 ! 2Xi is u.s.c. Then we deflne lim??fXi;figi2Z to be the inverse limit
space consisting of all points of form (:::;x?2;x?1;x0;x1;x2;:::;xk;xk+1;:::), where xi 2
fi(xi+1) for each integer i, and a basis for the topology on the space is
fO \ lim??fXi;figi2Zj O is basic open in Qi2ZXig.
We will often call this space a \two-sided" inverse limit.
If each fi is a continuous function, then the two-sided inverse limit is clearly home-
omorphic to the standard one. However, if each fi is u.s.c., the two-sided inverse limit,
lim??fXi;figi2Z, may be difierent from lim??fXi;figi>0.?? We will provide some examples be-
low, but flrst we prove some basic theorems analogous to the theorems seen in Chapter
4.
Theorem d4:3:2. Suppose that, for each integer i, Xi is a compact Hausdorfi space and
fi : Xi+1 ! 2Xi is u.s.c. Then lim??fXi;figi2Z is non-empty and compact.
47
Proof: For each integer z = 0;?1;?2;:::, the space lim??fXi;figi>z is non-empty and
compact, by Theorem 4.3.2. Thus, for each such integer z, the set ???? Xz?2 ? Xz?1 ?
Xz ? lim??fXi;figi>z is a compact subset of Qi2ZXi. (For convenience, let Kz = ??? ?
Xz?2?Xz?1?Xz?lim??fXi;figi>z.) We note that if w and z are both integers with w < z,
Kw Kz. That means that fKzgz?0 is a monotonic collection of compact (hence, closed)
subsets of Qi2ZXi, a compact space. Thus, Tz?0 Kz is non-empty and compact. But
T
z?0 Kz = lim??fXi;figi2Z, so the proof is complete. ?
Theorem d4:4:7. Suppose that for each integer i, Xi is a Hausdorfi continuum, fi :
Xi+1 ! 2Xi is an upper semi-continuous function, and for each x in Xi+1, fi(x) is con-
nected. Then lim??fXi;figi2Z is a Hausdorfi continuum.
Proof: For each integer z = 0;?1;?2;:::, the space lim??fXi;figi>z is a Hausdorfi
continuum, byTheorem4.4.7. Again, wedeflneKz = ????Xz?2?Xz?1?Xz?lim??fXi;figi>z.
Since each Xi is a Hausdorfi continuum, as is lim??fXi;figi>z for each z, it follows that Kz
is a Hausdorfi continuum for z = 0;?1;?2;::: As before, if w < z, then Kw Kz, so
that fKzgz?0 is a monotonic collection of Hausdorfi continua. It follows that Tz?0 Kz is a
Hausdorfi continuum. Since Tz?0 Kz = lim??fXi;figi2Z, the proof is complete. ?
Wenowpresentanexampletodemonstratehowthetwo-sidedinverselimit, lim??fXi;figi2Z,
may be difierent from the standard inverse limit, lim??fXi;figi>0.
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For each 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). (This bonding function
is the same as in Example 3 in Chapter 4.) If we let Az be the set of all points of form
(:::;0;0;x;x;:::), with 0?s up to the (z ? 1)th slot and x 2 [0;1], then Az is an arc for
each integer z. Let A = f(:::;x;x;x;:::)j x 2 [0;1]g, so that A is also an arc. Thus,
(Sz2ZAz) [ (A) = lim??fXi;figi2Z, and (Tz2ZAz) \ (A) = (:::;0;0;0;:::), a single point.
Thus, lim??fXi;figi2Z is a fan.
However, this fan is not homeomorphic to the fan given by lim??fXi;figi>0 in Example
3. For, as we will show, lim??fXi;figi2Z contains a limit arc while lim??fXi;figi>0 does not.
Consider the arc A in lim??fXi;figi2Z given by f(:::;x;x;x;:::)j x 2 [0;1]g. We will
prove that A consists entirely of limit points of (lim??fXi;figi2Z) n A. To that end, let
O = (Qi2ZOi)\lim??fXi;figi2Z be a basic open set containing some point (:::;x;x;x;:::),
where x 2 [0;1]. If for each i, Oi = Xi, then clearly O contains points not in A. So suppose
O is a proper subset of the space. Since O is open, there must be some least integer i for
which Oi (Xi, and some greatest integer j for which Oj (Xj. If x 6= 0, and ?x lies in the
ith slot, clearly (:::;0;0;:::;0;?x;x;x;:::) 2 O. If x = 0, and ?0 lies in the jth slot, then
(:::;0;0;:::;?0;1;1;:::) 2 O. Either way, O must contain a point in (lim??fXi;figi2Z)nA,
and thus, A is a limit arc.
On the other hand, the space lim??fXi;figi>0 has no limit arc. To see this, consider a
general point (0;0;:::;0;?x;x;x;:::) lying in an arc bA = f(0;0;:::;0;?x;x;x;:::)j x 2 [0;1]g,
49
where ?x lies in the ith slot. If x 6= 0, the point (0;0;:::;0;?x;x;x;:::) cannot be a limit
point of (lim??fXi;figi>0)n bA for the following reason:
Let O1 = [0; x2) ? X1, O2 = [0; x2) ? X2, ???, Oi?1 = [0; x2) ? Xi?1, and Oi =
(x2;1] ? Xi. Then ??O1 \ ??O2 \ ??? \ ???Oi?1 \ ??Oi is open in lim??fXi;figi>0, contains
(0;0;:::;0;?x;x;x;:::), but misses (lim??fXi;figi>0)n bA entirely. ?
50
Chapter 6
An Indecomposable Continuum Produced
by an Inverse Limit on u.s.c. Functions
The Knaster continuum described in example 2 at the end of Chapter 3 is a famous
example of an indecomposable continuum, i.e., a continuum that is not the union of two
proper subcontinua. We conclude this paper with an original example of an inverse limit
on u.s.c. bonding functions that turns out to be an indecomposable continuum.
Example: For each positive integer i, let Xi = [0;1] and let fi : Xi+1 ! 2Xi be deflned
by the graph consisting of the following straight line segments:
1. For each even integer n ? 0, the segment joining the points ( 12n+1; 12n+1) and ( 12n;1).
2. For each odd integer n ? 1, the segment joining the points ( 12n; 12n) and ( 12n+1;1).
3. The vertical line segment joining the points (0;0) and (0;1).
Then lim??f is an indecomposable continuum.
Proof: By Theorem 4.4.7, lim??f is a continuum. It remains to show that lim??f is inde-
composable.
Claim: If H is a proper subcontinuum of lim??f, then there exists some positive integer
N so that if n ? N, ?n(H) 6= Xn.
51
Justiflcation: Suppose not, i.e., suppose H is proper but for every positive integer
N, there exists some n > N such that ?n(H) = Xn. By the way the graph of f is
deflned, it is clear that if ?i(H) = Xi, then ?i?1(H) = Xi?1; thus, we may as well assume
that ?n(H) = Xn for each positive integer n. Since H is proper, there is some point
p = (pi)1i=1 = (p1;p2;p3;:::) such that p 2 lim??fnH. We will show that p is in fact a point
of H.
Case 1: Suppose pi 6= 0 for all positive integers i. Since, for a given positive integer k,
pk 2 ?k(H), there exists some point in H of form
(x1;x2;:::;xk?1;pk;:::).
However, since pk 6= 0, by the way the graph of f is deflned, fk?1(pk) is a unique
non-zero number in Xk?1. That means fk?1(pk) = pk?1. In a similar way, each of
x1;:::;xk?1 is uniquely determined, and that forces xi = pi for 1 ? i ? k. Thus, a
point of form (p1;p2;:::;pk;:::) is in H. Indeed, for each positive integer j, a point of form
(p1;p2;:::;pj;:::) is in H. The point (pi)1i=1 is therefore a limit point of the sequence of
points in H that we just described; hence, because H is closed, (pi)1i=1 2 H.
Case 2: Suppose for some least integer i, pi = 0. Then, since the only possible preimage
of 0 via fi is 0, pn = 0 for each integer n ? i.
Supposep1 = 0. By the above argument, sincef?11 (p1) = p2 = 0, andf?1i (pi) = pi+1 =
0 for i ? 1, the only way that p1 = 0 2 ?1(H) is possible is if (0;0;0;:::) = (pi)1i=1 2 H.
That would be a contradiction.
52
So, suppose instead that pi = 0 for some least integer i > 1. Because H is compact,
the projection of H onto the graph of fi?1 is closed. Thus, since ?i(H) = Xi, by the way
the graph is deflned, each ordered pair (0;x) in G(fi?1) is a limit point of the projection of
H onto G(fi?1). Thus, the ordered pair (0;pi?1) is in that projection. Since pi?1 6= 0 by
assumption, the image of pi?1 via fi?2 is the unique non-zero number pi?2; the image of
pi?2 via fi?3 is the unique non-zero number pi?3, and so forth. Now because ?i(H) = Xi,
we know pi = 0 2 ?i(H); since the only possible preimage of 0 is 0, H must therefore
contain a point of form (h1;h2;:::;hi?1;0;0;:::). However, as we noted, the projection
of H onto G(fi?1) contains the ordered pair (0;pi?1). That is, H contains some point
(h1;h2;:::;hi?1;0;0;:::) where hi?1 = pi?1. But, as previously argued, hi?1 = pi?1 would
force hk = pk for k ? i?1. That is, (p1;p2;:::;pi?1;0;0;:::) = (pi)1i=1 2 H.
In either case, a contradiction has been reached. Thus, if H is a proper subcontinuum,
there exists some positive integer N so that if n ? N, ?n(H) 6= Xn.
Now, suppose by way of contradiction that lim??f = H [ K, a union of two proper
subcontinua. By the above argument, there exists some least positive integer N so that for
all n ? N, ?n(H) 6= Xn and ?n(K) 6= Xn. So, we may assume without loss of generality
that 0 2 ?N(H) and 0 =2 ?N(K). Since the unique preimage of 0 via fN is 0, ?N+1(H) must
contain 0. Thus, because ?N+1(H) is a proper subcontinuum of [0;1] containing 0, but
?N+1(K) 6= [0;1], it follows that ?N+1(H) is some interval of form [0;a] where 0 < a < 1.
However, by the way the graph of fN is deflned, there is some x 2 (0;a] (indeed, inflnitely
many such x) with fN(x) = 1. That means 1 2 ?N(H). Since ?N(H) is a subcontinuum
53
of [0;1] that contains both 0 and 1, it follows that ?N(H) = [0;1] = XN. This is a
contradiction, for we assumed ?N(H) 6= XN. So lim??f is indecomposable. ?
We note that the continuum in this example (lim??f) contains the proper subcontinuum
f(x;y;0;0;:::)j y 2 [0;1];x 2 f1(y)g, a copy of the topologist?s sine curve. Therefore, lim??f
is clearly not homeomorphic to the Knaster continuum (whose proper subcontinua are all
arcs). However, if lim??f is homeomorphic to any known space, it remains an open question
what that space is.
Another open question is the following: if for each positive integer i, Xi = [0;1] and
f = fi : Xi+1 ! 2Xi is u.s.c., are there some necessary conditions the graph of f must
satisfy in order for lim??f to be indecomposable? Are there su?cient conditions?
54
Bibliography
[1] W.T. Ingram, William S. Mahavier, \Inverse Limits of Upper Semi-Continuous Set
Valued Functions," Houston Journal of Mathematics, vol. 32 (1) (2006) 119-130.
[2] K. Kuratowski, Topology, vol. II, Academic Press, New York, 1968.
[3] James R. Munkres, Topology: a flrst course, Prentice-Hall, Englewood Clifis, N.J.,
1975.
[4] R.L. Moore, \Foundations of Point Set Theory," Amer. Math. Soc. Colloq. Publ., vol.
13, Amer. Math Soc., Providence, RI, 1962.
[5] Michel Smith, Notes on Topology, http://www.auburn.edu/?smith01/math7500/ and
http://www.auburn.edu/?smith01/math7510/.
55
Appendix
* In general, for an inverse limit space lim??f with upper semi-continuous bonding functions,
the collection B = f??Oj O is open in some Xig is not a basis for lim??f. Consider the case
where for all positive integers i, Xi = [0;1] and fi : Xi+1 ! 2Xi is given by the graph in
[0;1]?[0;1] consisting of the line segments joining (0;0) to (1;0) and joining (0;1) to (1;1).
Then the open set G = (12;1]?[0; 12)?[0;1]?[0;1]?::: does not contain a member of B
containing (1;0;0;0;:::). For, any such member b of B would have to contain an open set
of form (i) ????[0;a);a < 1 or (ii) ????(a;1];a > 0. In case (i), b would contain (0;0;:::); in case
(ii), b would contain (1;1;:::). In either case, b would fail to be a subset of G, so B cannot
be a basis.
** Thanks to Dr. Stewart Baldwin for suggesting this possibility and encouraging me to
explore it.
56