Indecomposable and Chainable Continua
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 classified information.
Frank Sturm
Certificate of Approval:
Gary Gruenhage
Professor
Mathematics & Statistics
Michel Smith, Chair
Professor
Mathematics & Statistics
Krystyna Kuperberg
Professor
Mathematics & Statistics
Geraldo De Souza
Professor
Mathematics & Statistics
George T. Flowers
Dean
Graduate School
Indecomposable and Chainable Continua
Frank Sturm
A Thesis
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulfillment of the
Requirements for the
Degree of
Master of Science
Auburn, Alabama
August 10, 2009
Indecomposable and Chainable Continua
Frank Sturm
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
Frank Herman Sturm was born October 8, 1983 to Alvin and Rachel Sturm in Pasadena,
TX. Growing up in Pasadena, he graduated from Sam Rayburn High School in 2002 and
then attended San Jacinto Community College for a year before enrolling in the University
of Houston.
In the summer of 2006, Frank graduated from the University of Houston with a BSc.
in Mathematics. After spending a year as a tennis instructor in the Houston area, he left
the big and beautiful Lone Star State for the smaller (yet still beautiful) state of Alabama
to enroll as a graduate student in Auburn University, where he continues his studies in
mathematics.
iv
Thesis Abstract
Indecomposable and Chainable Continua
Frank Sturm
Master of Science, August 10, 2009
(B.Sc., University of Houston, 2006)
67 Typed Pages
Directed by Michel Smith
This thesis covers topics in continuum theory related to indecomposable continua and
chainable continua. Theorems are presented to characterize indecomposable continua and
then chainability is explored in its connection to inverse limit spaces. The end result is to
provide an example of an hereditarily indecomposable continuum.
v
Acknowledgments
I would like to thank my advisor, Dr. Michel Smith, who is always a beacon of en-
couragement and a bag of good questions. Of my commitee, I would like to thank Dr.
Gary Greunhage and Dr. Geraldo DeSouza who have always worked to help me clarify
and simplify my thoughts and reasoning; laborious explanations in this thesis are the sole
responsibility of the author and no fault of these two fine instructors; and to Dr. Krystyna
Kuperberg for gluing mathematics to the fascinating personalities behind the concepts. I
would also like to acknowledge Dr. Gordon G. Johnson, who kept my mathematical curios-
ity from starving, as well as my fellow graduate students who are willing to lend an ear or
patiently explain the obvious.
I also want to thank my mother and father for the sacrifices they didn?t hesitate to
make for my wellbeing and for their unending support. I am also indebted to the ever
beautiful and loving Rosa, who brings me more joy than she might know.
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 (specifically LATEX)
together with the departmental style-file aums.sty.
vii
Table of Contents
0 Background Definitions and Theorems 1
1 Indecomposable Continua 10
2 Forming Chains 21
3 Chainable Continua 26
4 Inverse Limit Spaces and Chainable Continua 32
5 An Hereditarily Indecomposable Continuum 48
Bibliography 59
viii
Chapter 0
Background Definitions
and Theorems
The following definitions and theorems have been compiled from my first graduate
topology course with Dr. Gruenhage [5], the notes Dr. Michel Smith uses in his topology
course [4], as well as the master?s thesis of Scott Varagona [6]. Theorems are stated without
proof, however, a reader unfamiliar with these theorems should be able to find proofs of
equivalent theorems in [1], [2], and [3]
Definition 0.1. Suppose X is a set andT is a collection of subsets of X such that
1. X?T;
2. ??T;
3. IfU?T, then?U?T;
4. IfU?T andU is finite, then?U?T;
then the pair (X,T) is called a topological space with topologyT. Such a topologi-
cal space will often be referred to simply as X when the associated topologyT is understood.
The members ofT are called open sets. If K?X and X\K is open, then K is called a
closed subset of X.
Unless otherwise stated, in this chapter (X,T) is presumed to be a topological space.
Definition 0.2. Suppose M ?X. The closure of M (denoted M) is the intersection of
all closed subsets of X that contain M.
Definition 0.3. Suppose M is a subset of a topological space X. A point p?X is a limit
point of M if every open set containing p contains a point in M different from p.
1
Theorem 0.4. If M ?X, then M = M?{p : p is a limit point of M}.
Definition 0.5. Suppose D ? X. D is dense in X, means that each nonempty open
subset of X contains a point of D. To say that D is somewhere dense in X means that
there is U, a nonempty open subset of X such that D?U is dense in the subspace U.
Lastly, to say that D is nowhere dense in X, means that D is not somewhere dense in
X.
Theorem 0.6. If D?X and D is dense in X, then D = X.
Theorem 0.7. Suppose that S?X and Snegationslash=?. IfTS ={S?O : O?T}, thenTS forms a
topology on the set S.
Definition 0.8. With regards to the topological space (S,TS) described in the previous
theorem, the topologyTS is called the subspace topology on S with respect X and S
is refered to as a subspace of X.
Definition 0.9. Suppose B is a collection of open subsets of X with the property that if
x?X and O is an open subset of X containing x, then there is B?Bsuch that x?B?O.
B is called a basis for the topology on X and an element ofB is called a basic open set
in X.
Theorem 0.10. Suppose B is a collection of subsets of X such that
1. If x?X, there exists some B?B with x?B.
2. If B1 and B2 are in B and x?X such that x?B1?B2, then there is B3 ?B such
that x?B3?(B1?B2).
Then Tprime ={uniontextBprime|Bprime?B} is a topology on X and B is a basis for Tprime.
In the above theorem, a topology such asTprime is said to be generated by the basisB.
Definition 0.11. X is Hausdorff means that if p,q ?X are distinct points in X, then
there are disjoint open sets Op and Oq containing p and q, respectively.
2
X is regular means that if H ? X and p ? X such that p /? H, then there exist
disjoint open sets OH and Op such that H?OH and p?Op.
X is normal, means that if H and K are disjoint closed subsets of X, then there are
disjoint open sets OH and OK containing H and K, respectively.
Definition 0.12. Suppose A and B are sets and f is a function from A to B (denoted
f : A?B). If C ?A, we define f(C) ={f(c)| c?C} and call f(C) the image of C
under f.
If b?B, then the preimage of b (written as f?1(b)) is the collection{a?A|f(a) = b}.
Similarly, if D?B, then f?1(D) ={a?A : f(a)?D}=?{f?1(d) : d?D}and f?1(D)
is called the preimage of D.
Definition 0.13. Suppose A and B are sets and f : A?B.
1. f is onto means that if b?B, then there is a?A such that f(a) = b.
2. f is one-to-one means that if a and aprime are two distinct points in A, then f(a)negationslash= f(aprime).
Often, the inverse of f is denoted as f?1 and explicity stated to be a function to avoid
confusion with the preimage of a point or set.
Theorem 0.14. Suppose A and B are sets and f : A?B is a function that is one-to-one
and onto. Then g : B?A defined as g(f(a)) = a is well defined. The function g is called
the inverse of f.
Definition 0.15. Suppose each of X and Y is a topological space, f : X?Y is a function,
and x?X). To say that f continuous at x means that if V is an open set in Y containing
f(x), there is U an open subset of X such that x?U and f(U)?V. If f is continuous at
each point in X, then f is said to be continuous.
Theorem 0.16. Suppose each of X and Y is a topological space and f : X ? Y . The
following are equivalent:
1. f is continuous.
3
2. If V is an open subset of Y , then the preimage of V , f?1(V) is open in X.
3. If K is a closed subset of Y , then f?1(K) is a closed subset of X.
Definition 0.17. Suppose each of X and Y is a topological space and f : X?Y. f is an
open function means that if U is open in X, then the image of U, f(U), is open in Y. It
is often simply stated that the function f is open.
Theorem 0.18. Suppose that each of X, Y , and Z is a topological space and each of f,g,
and h is a function such that f : X ? Y , g : Y ? Z, h : X ? Z, and h = g?f. For
each of the properties listed below, if both f and g have the given property, then h has this
property.
1. continuous
2. open
3. one-to-one
4. onto
Definition 0.19. If X and Y are topological spaces and f : X ?Y is a function that is
one-to-one, onto, continous, and open, then f is called a homeomorphism and the spaces
X and Y are said to be homeomorphic.
Definition 0.20. Suppose A is a set.
? A is countable, means there is a function f : A?a78 that is one-to-one;
? To say that A is finite, means that there is a positive integer n and a function
f : A?{1,2,...,3}such that f is one-to-one;
? If A is not finite, then A is said to be infinite;
? If A is not countable, then A is said to be uncountable
4
Definition 0.21. Suppose X is a topological space and x?X. IfBx is a collection of open
subsets of X, thenBx is called a local basis at x if
1. for each B?B, x?B;
2. if O is an open set in X and x?O, then there is of B?Bx such that x?B?O.
Definition 0.22. The space X is called first countable if for each x?X, there exists a
countable local basis at x. A space X is called second countable if X has a basis that is
countable.
Definition 0.23. Suppose M ?X. IfU is a collection of subsets of X such that M ??U,
thenU is said to be a cover of M; if each element ofU is an open subset of X, thenU is
said to be an open cover of M. Lastly, ifU is an open cover of M andF?U such that
F covers M, thenF is called a subcover of M from U; ifF is finite, thenF is called a
finite subcover of M from U.
Definition 0.24. The space X is compact means that if U is an open cover of X, then
there is a finite subcover of X fromU.
Theorem 0.25. The interval [0,1], as a subspace of a82 is compact.
Theorem 0.26. Suppose that each of X and Y is a topological space and f : X ? Y is
continuous. If X is compact, then f(X) is a compact subspace of Y .
Definition 0.27. Suppose that S is a collection of subsets of X. S has the finite inter-
section propert (f.i.p.) in X means that if F is a finite nonempty subset of S, then
?Fnegationslash=?.
Theorem 0.28. Suppose that X is a compact Hausdorff space.
1. X is normal.
2. If K is a closed subset of X, then the subspace K is compact.
5
3. If K is a collection of closed subsets of X with the finite intersection property, then
?Knegationslash=?.
Theorem 0.29. Suppose X is compact and Hausdorff. If U is a countable collection of
dense open subsets of X, then ?U is dense in X.
Corollary 0.30. Suppose X is compact and Hausdorff. If C is a countable collection of
nowhere dense closed subsets of X, then ?Cnegationslash= X
Definition 0.31. If I is a nonempty set and for each i ?I, Ai is a nonempty set, then
the product of {Ai : i?I}, denoted producttexti?I Ai, is the collection of functions from I into
?i?IAi, to which the function ? belongs if and only if ?(i)?Ai for each i?I (Note: IfI
is infinite, the existence of producttexti?I Ai depends on the Axiom of Choice.)
If I is finite (respectively countable), it is usually assumed that I ={1,2,...,|I|}, where
|I|is the cardinality ofI (respectively,I= a78). In such a case producttexti?I Ai may be denoted
A1?????An, with n =|I|,
and the elements of this set are considered as ordered n?tuples. In the case that I is
countable, the elements producttexti?I Ai can be thought of as infinite sequences whose ith term is
in Ai.
Theorem 0.32. Suppose n is an integer greater than 1 and for each integer i, 1?i?n,
suppose Xi is a topological space. Let
X =
nproductdisplay
i=1
Xi,
and let B denote the set
{
nproductdisplay
i=1
Oi : Oinegationslash=?,and Oi is open in Xi}
6
B forms a basis for a topology on X.
Definition 0.33. A space such as X, as described in the previous theorem, together with
the topology formed by the basisB (also described previously) is called a finite product
space and its topology is called the finite product topology
Theorem 0.34. Suppose that I is a nonempty set and for each i?I, Xi is a topological
space. Let X = producttexti?I Xi and let B be the set to which producttexti?I Oi belongs if and only if
1. for each i?I, Oinegationslash=? and Oi is an open subset of Xi;
2. there is F a finite subset of I such that Oi = Xi unless i?F;
B forms the basis for a topology on X.
Definition 0.35. X in the above theorem is called a product space and the topology
formed from the basis B is called the product topology on X. In general, if I is a
nonempty set, a product space with index I refers to a space formed in the manner of
X previously described.
Definition 0.36. Suppose X is a product space with indexI. If i?I, then pii : X?Xi
such that if x ? X, then pii(x) = x(i) (recall that x is a function with domain I). The
function pii is called the projection of X onto Xi.
Theorem 0.37. If X is a product space with index I and i ?I, then the function pii is
continuous and onto.
Definition 0.38. Suppose X is a topological space and d : X?X?a82 is a function such
that if each of x,y, and z is in X, then
1. d(x,y)?0, and d(x,y) = 0 if and only if x = y,
2. d(x,y) = d(y,x), and
3. d(x,z)?d(x,y) +d(y,z).
7
The function d is said to be a metric on X, and the ordered pair X,d is called a metric
space. If p?X and epsilon1 > 0, B(p,epsilon1) denotes the set{x?X : d(x,p) < epsilon1}and is called the
open ball of radius epsilon1 centered at p.
Theorem 0.39. Suppose (X,d) is a metric space, then the collectionBdefined as{B(p,epsilon1) :
p?X and epsilon1 > 0} forms a basis for a topology on X.
Definition 0.40. If (X,d) is a metric space, then the topology formed from the basis of
open balls centered at points of X is called the metric topology generated by d. If
(X,T) is a topology and there is a metric d such that the metric topology generated by d
is the same asT, thenT is said to be metrizable.
Theorem 0.41. If (X,d) is a metric space then X is a normal space
Definition 0.42. Suppose (X,d) is a metric space, x?X, and A,B?X. The minimum
distance between x and A is equal to inf({d(x,a) : a ? A}), and the minimum
distance between A and B is equal to inf({d(x,B) : x?A}).
Definition 0.43. A topological space is connected if it is not the union of two nonempty
disjoint open sets. If C?X and C as a subspace of X is connected, then C is also refered
to as a connected subset of X.
Theorem 0.44. The interval [0,1] as a subspace of a82, is connected.
Theorem 0.45. Suppose C is a connected subset of X. If A?X and C?A?C, then A
is also a connected subset of X.
Definition 0.46. Suppose each of H and K is a subset of the space X. H and K are called
mutually separated if H?K = H?K =?.
Theorem 0.47. Suppose M ?X. M is not a connected subset of X if and only if M is
the union of two nonempty mutually separated subsets of X.
Theorem 0.48. If M ?X and M is connected in X, then M is also connectd in X.
8
Theorem 0.49. Suppose C is a collection of connected subsets of X and K is a connected
subset of X such that if C?C, then K and C are not mutually separated. The set K?(?C)
is a connected subset of X.
Theorem 0.50. Suppose each of X and Y is topological space and f : X ? Y is a
continuous function. If X is connected, then f(X), the image of X under f, is a connected
subset of Y .
Definition 0.51. If p?X, then the component of X containing p is the union of all
connected sets in X that contain p. This set is sometimes denoted Cp.
Definition 0.52. A subset of X that is both closed and open in X is called a clopen
subset of X. Notice that the empty set and X are each clopen subsets of X.
Theorem 0.53. The space X contains a proper subset that is clopen if and only if X is
not connected.
Corollary 0.54. Suppose U and V are nonempty disjoint subsets of X such that U?V = X.
If both U and V are open, or if both U and V are closed, then X is not connected.
Definition 0.55. If p ? X, the quasicomponent of X is the intersection of all clopen
subsets of X that contain p. This set is sometimes refered to as Qp.
Definition 0.56. A topological space X is called a continuum if it is non-empty, Haus-
dorff, compact, and connected. If X is a continuum and the topology of X can be generated
by a metric, then X is called a metric continuum.
Corollary 0.57. The interval [0,1], as a subspace of a82, is a continuum.
Definition 0.58. If X is a continuum, A?X, and the subspace A is a continuum, then
A is called a subcontinuum of X. If A is a proper subset of X, then A is a proper
subcontinuum.
9
Chapter 1
Indecomposable Continua
In this chapter some characterizations of indecomposable continua will be developed.
This developement will begin with a useful theorem called the Boundary Bouncing Theorem.
Unless otherwise stated, in this chapter (X,T) is assumed to be a topological space
such that X is a continuum.
Lemma 1.1. If x?X, Cx is the component of X that contains x, and Qx is the quasi-
component of X that contains x, then Cx?Qx
Proof. If K is a clopen subuset of X and x?K, then Cx?K, else the subspace Cx is the
union of two nonempty disjoint clopen subsets of Cx, namely Cx?K and Cx?(X\K).
Thus,
Cx??{K : K is a clopen subset of X and x?K}= Qx.
Lemma 1.2. Suppose X is a compact Hausdorff space andKis a collection of closed subsets
of X. If C =?Kand U is an open set containing C, then there is{F1,F2,...,Fn}, a finite
subset of K, such that ?ni=1Fi?U.
Proof. LetKand C be as defined in the Lemma. Choose an open set O containing C. X\O
is a closed subset of X and therefore it is compact. For each F ?K, X\F is open. Because
?K= C,?{(X\F : F ?K) = X\?K= X\C. C?O implies that X\O?X\C, and
so it follows that the collection{X\F}: F ?K}is an open cover of X\O. Because X\O
is compact, there is a finite subcovering of X\O, {X\F1,X\F2,...,X\Fn}. Because
X\O??n1 (X\Fi) it follows that?n1Fi?O.
10
Lemma 1.3. If X is compact and Hausdorff and x?X, then Qx = Cx.
Proof. By Lemma 1.1, Cx ?Qx. In order to prove the converse relationship it will suffice
to show that Qx is connected.
Suppose that A and B are mutually separated sets such that A?B = Qx; it will
be shown that either A or B must be empty. Without loss of generality, assume x ? A.
Because Qx = A?B and Qx is closed, A ? A?B and B ? A?B. Since A and B are
mutually separated it must be that A?A and B?B, hence each of A and B is closed.
X is compact and Hausdorff, therefore it is normal. Because A and B are mutually
separated, they are disjoint; thus U and V may be chosen to be disjoint open subsets of X
such that A?U and B?V.
Qx is the intersection of clopen sets, andso by Lemma 1.2 the finite collection{F1,F2,...,Fn}
may be chosen so that?ni=1Fi?U?V. If we define F as?ni=1Fi, then F is the interesection
of clopen sets and therefore F is clopen. Define Uprime = F?U. A?U?F and x?A, therefore
x?Uprime. U and F are open, so Uprime must be open, and because V is open and F is closed,
F\V = U is closed as well. It follows that Uprime is clopen and contains x, hence Qx ?Uprime.
This means V ?Qx =?. A?V ?Qx, hence, A =?.
From the preceeding argument it follows that Qx is not the union of two nonempty
mutually separated subsets, hence Qx is connected. Since x?Qx, Qx ?Cx. Combining
this with the result of Lemma 1.1 yields Qx = Cx.
Theorem 1.4 (Boundary Bouncing Theorem). Suppose X is a continuum, a?X, and O
is a nonempty open subset of X such that a /?O. There is C, a connected subset of X, such
that a?C, C?O =?, and C?Onegationslash=?.
Proof. Suppose a ? X and O is a nonempty open subset of X such that a /? O. Let
Y = X\O. Because X is connected Y ?Onegationslash=?, else O would be open meaning O = O and
it follows that Y and O are two disjoint clopen sets whose union is X, which would mean
X is not connected and not a continuum. As a subspace Y is a closed subset of X, thus,
as a subspace Y is compact and Hausdorff. Because a /?O, a?Y. Now suppose U is a
11
clopen subset of Y. It will be shown that U?Onegationslash=?.
Case 1: If U = Y, then U?O = Y ?Onegationslash=?.
Case 2: Suppose U negationslash= Y. Let V = Y \U; it follows that V is a clopen set such that
U?V negationslash=? and U?V = Y. Thus, X = O?Y = O?U?V. Since Y is closed in X and
each of U and V is closed in Y, each of U and V is closed in X. Thus U and V ?O are
closed subsets of X whose union is X. Because X is a continuum, X is connected, which
means U?(V ?O)negationslash=?. Since U?V =?it follows that U?Onegationslash=?. Continuing with the
main proof, letQa ={Q : a?Q and Q is clopen in Y}and letK={Q?O : Q?Qa (note
that ?Qa is the quasicomponent of Y that contains a). It will now be shown that K has
the f.i.p.
IfF is a finite subset ofK, then there isFprime a finite subset ofQa such that (?Fprime)?O =
?F; ?Fprime is clopen and contains a, therefore (?Fprime)?Onegationslash=?and so?F negationslash=?. The result of
the above facet is that?Knegationslash=?, since Y is compact and Hausdorff. Since?K= (?Qa)?O,
it means that the quasicomponent of Y that contains a intersects O. Again the fact that
Y is compact and Hausdorff means that the quasicomponent containing a is the connected
component of y that contains a, thus this component is a connected subset of Y (and thus
a connected subset of X) that contains a and intersects O.
Definition 1.5. Suppose the space X is a continuum. X is indecomposable means that
if H and K are proper subcontinua of X, then H?Knegationslash= X.
Many familiar continua are not indecomposable, which is one reason indecomposable
continua are interesting. For instance, the closed interval [0,1] with the subspace topology
derived from a82 is not indecomposable because [0, 12] and [12,1] are both proper subcon-
tinua and [0, 12]?[12,1] = [0,1]. The following theorem is a further characterization of an
indecomposable continuum.
Theorem 1.6. X is an indecomposable continuum if and only if each proper subcontinuum
of X is nowhere dense in X.
12
Proof. (?) By way of contrapositive it will be shown that if X is a continuum and C is a
proper subcontinuum of X that is somwhere dense in X (ie not nowhere dense), then X is
not indecomposable.
Let U be a nonempty open set in X such that every nonempty open subset of U
intersects C. This means that C is dense in U. Because C is closed, U ?C.
Let S be the subspace X\U. S is a closed subset of C, thus S is compact. Let a?S\C
and let Ca be the component of S containing a. Because S is closed, Ca is closed in X, and
thus Ca is a p.s.c. of X.
If (X\C)?Ca, then X = Ca?C and thus X is not indecomposable.
If X\C is not a subset of Ca, let b?X that is not in C?Ca. S\{a}is open in S, so
by the Boundary Bouncing Theorem, we may choose Q to be a subset of S\{a}such that
Ca?Q and Q is clopen in S.
Let C1 = Q?C and C2 = (S\Q)?C. Both C1 and C2 are connected for if x?Q
(respectively S\Q ), then Cx, the component of x in S, is a subset of Q (resp. S\Q).
Because Cx?U negationslash=?, Cx?Cnegationslash=?; thus C1 (resp C2 ) is the union of a collection of connected
sets, whose intersection is nonempty. It follows that both C1 and C2 are connected subsets
of X.
Because Q and S\Q are closed subsets of a subspace that is closed in X, Q and S\Q
are closed in X, thus C1 and C2 are closed in X; from this it follows that each of C1 and
C2 is a subcontinua of X.
The point a?C1 is not in (S\Q)?C = C1, and the point b?C2 is not in Q?C = C2,
which means C1 and C2 are proper subcontinua of X. Lastly,
C1?C2 = (Q?C)?((S\Q)?C) = (S?C) = X
and so we have that X is not indecomposable.
(?) To prove the converse, suppose that every proper subcontinuum of X is nowhere dense
and that C1 and C2 are proper subcontinuum of X. C1 is nowhere dense, which means U,
13
a nonempty open subset of U may be chosen so that U ?C1 = ?. C2 is nowhere dense,
which means there is a nonempty open subset of U that does not intersect C2. Call such a
set Uprime. Uprime?(C1?C2) =?, thus C1?C2negationslash= X. Because C1 and C2 are chosen arbitrarily, if
follows that X is indecomposable.
Definition 1.7. If the space X is a continuum and p ? X, then the compossant of X
that contains p is the union of all proper subcontinuum of X that contain p.
Theorem 1.8. If X is an indecomposable continuum and each of C and D is a compossant
of X, then C = D or C?D =?.
Proof. Suppose that X is an indecomposable continuum, p1 and p2 are points in X, K1 is
the compossant of X at p1, and K2 is the compossant of X at p2. Let Q1 be the collection
to which C belongs if and only if C is a proper subcontinuum of X containing p1; define
Q2 similarly with respect to p2.
If C1?Q1, C2?Q2, and C1?C2negationslash=?, then C1?C2 is a proper subcontinuum containing
p1 and p2; hence, K1 = K2, for if x?Ki, where i is 1 or 2, and B is a proper subcontinuum
containing x and pi, then let Cprime = B?(C1?C2). Cprime is a proper subcontinuum because it
is connected and is the finite union of proper subcontinua, thus x?K1?K2.
From the above paragraph, if K1negationslash= K2, then no element of Q1 intersects an element of
Q2. Therefore,?Q1intersectiontextC =?for every C ?Q2; thus,?Q1intersectiontext?Q2 =?. Because K1 =?Q1
and K2 =?Q2, K1?K2 =?.
Definition 1.9. The continuum X has the Countable Compossant Property (CCP)
if each compossant of X is the union of countably many subcontinua of X
Theorem 1.10. If X is an indecomposable continuum and every compossant of X can be
written as a countable collection of proper subcontinua, then X has an uncountable number
of compossants.
14
Proof. It will be shown by way of contrapositive, that if X has at most countable many
compossants and each compossant of X is a union of countably many subcontinua of X,
then X is not indecomposable.
Suppose that X is an indecomposable continuum and K1,K2,... are distinct compos-
sants of X. If n?a78, define Qn ={Kn,1,Kn,2,...}, where Kn,i is a proper subcontinuum,
such that?Qn = Kn. Because Qn is countable for each n?a78 it follows that Q = uniontext?i=1 Qn
is countable. Let {C1,C2,...} be an enumeration of Q. Because X is indecomposable, if
n?a78, then Ci is nowhere dense (this follows from Theorem 1.6).
Define Un = X \Ci for each n ? a78. It follows that Un is open and dense in X.
By Baire?s theorem, the intersection of a countable collection of dense open subsets of a
compact Hausdorff space is dense; thus??i=1Uinegationslash=?and I can conclude that
?uniondisplay
i=1
Ci = X\
?intersectiondisplay
i=1
Uinegationslash= X.
Because uniontext?i=1 Ki = uniontext?i=1 (?Qi) = uniontext?i=1 Ci, there must be a point of X not contained in any
of the listed compossants. Thus the number of compossants cannot be countable.
Theorem 1.11. Suppose X is a metric continuum, with metric d. If X is nondegenerate
and indecomposable, then each compossant of X is the union of a countable collection of
subcontinua of X.
Proof. Let p ? X and let A be a dense countable subset of X. If a ? A and i ? a78, let
Pi(a) be the set to which C belongs, if and only C is a subcontinuum of X, p ? C, and
C?B(a, 1i) =?. Now define Pi(a) as the set?Pi(a).
Pi(a) is a union of connected sets containing p, hence Pi(a) is connected and thus
Pi(a) is connected. Pi(a) ? X \B(a, 1i), thus Pi(a) ? X \B(a, 1i), which means
Pi(a)?Pi(a), thus Pi(a) is a proper subcontinuum of X.
Let Cp = { Pi(a) : i ? a78, a ? A}. It will be shown that ?Cp is the compossant of
X that contains p. First note that ?Cp is the union of proper subcontinua of X, each of
which contains p (ie ?Cp is a subset of the compossant of X containing p). Now suppose
15
C is a proper subcontinuum of X and p ? C. C is closed and C negationslash= X, hence X\C is a
nonempty open set. A is dense in X so a?A may be chosen so that a?X\C (ie a /?C).
Because a /?C, and C is closed d(a,C) > 0. Choose i?a78 so that 1i < d(a,C). It follows
that C?B(a, 1i) = ?, hence C ?Pi(a), meaning C ? Pi(a) ??Cp. Thus, ?Cp is the
compossant of X that contains p.
Corollary 1.12. If X is a metric continua, then X has an uncountable number of com-
possants.
Theorem 1.13. If p?X, then the compossant of X containing p is dense in X
Proof. Let p?X and suppose U is a nonempty open subset of X.
Case 1: If p?U, then{p}?U?Kp
Case 2: Suppose p /?U, and let q?U. X is regular, so Uprime may be chosen such that q?Uprime
and Uprime ?U. By Theorem 1.4, C may be chosen to be a connected subset of X such that
p?C, C?Uprime =?and C?Uprimenegationslash=?. It follows that C is connected and q /?C; hence, C is a
proper subcontinuum of X that contains p. C?Uprimenegationslash=?and Uprime?U, so it also follows that
C?U negationslash=?.
Definition 1.14. Suppose X is a continuum and a and b are two points in X. X is
irreducible between a and b means that no proper subcontinuum of X contains both a
and b.
Theorem 1.15. Suppose X has the CCP. X is indecomposable if and only if there are three
points a, b, and c in X such that X is irreducible between each pair in {a,b,c}.
Proof. (?) Suppose {a,b,c}? X and X is irreducible between any pair in {a,b,c}, and
suppose A and B are proper subcontinuums of X and A?B = X.
This means a?A?B; let?s assume that a?A. b?A?B and b /?A because A is a
p.s.c containing a; let?s assume b?B. Of course this means that c is not contained in A
16
or B because each is a p.s.c. one contains a and the other contains b. Thus c /?A?B and
A?Bnegationslash= X.
(?) Supposing now that X is indecomposable. By the previous theorem, X has an
uncountable number of disjoint compossants and so I may choose three nonempty disjoint
compossants A,B,C and choose a?A, b?B, c?C and X is irreducible between each
pair in{a,b,c}.
Definition 1.16. Suppose X is a continuum and p ? X. End(p,X) is defined as the
collection
End(p,X) ={q?X : X is irreducible between p and q}.
Theorem 1.17. Suppose X is a nondegenerate indecomposable continuum and p?X. If
End(p,X)negationslash=?, then End(p,X) is dense in X.
Proof. Let q ?End(p,X). Let Kp and Kq denote the compossant of X containing p and
the compossant of X containing q, respectively. Because X is irreducible between p and q,
q /?Kp; thus Kpnegationslash= Kq, so by Theorem 1.8 Kp?Kq =?.
Theorem 1.18. If a space X has a connected dense subset, then X is connected.
Proof. Let X be a space and let K be a connected dense subset of X. K is connected
therefore K is connected, by 0.45. K is dense, therefore K = X (by 0.6); hence, X is
connected.
Theorem 1.19. Suppose X is a nondegenerate indecomposable continuum and X has the
CCP. If K is the union of countably many proper subcontinua of X, then X\K is connected.
Proof. From Theorem 1.10, X has an uncountable number of compossants, hence, a com-
possant C may be chosen such that C?K =?, where K is as described in the theorem.
A compossant such as C is connected and dense, hence it follows that any subspace of X
that contains C is connected. Thus, X\K is connected.
17
Theorem 1.20. Suppose C a collection of subcontinua of X such that if F is a finite
nonempty subset of C, then there is C ?F, such that C ??F. ?C is a nonempty subcon-
tinuum of X
Proof. Notice that C is a collection of closed subsets of a compact space such that the
intersection of a finite (and nonempty) subset ofCis nonempty; thus,?Cnegationslash=?. Let V =?C.
To show V is connected, suppose V is not connected. If V is a closed set that is not
connected, then V is the union of two disjoint closed sets (proved previously). Call two such
sets A and B. Because A and B are each disjoint closed subsets of the compact Hausdorff
space X, there are disjoint open setsO1 andO2 such that A?O1 and B?O2. If C?C,
define Cprime = C\(O1?O2); Cprimenegationslash=?, because C is a nonempty connected set that intersects
A and B. Let Cprime = {Cprime : C ?C}; Cprime will also have the finite intersection property, for if
Fprime?Cprime is finite and F ?C corresponds to Fprime, then there is C?C such that C??F, thus
Cprime ?Fprime. Let V prime =?Cprime. V prime is nonempty and V prime ?V, thus V = A?B is not a subset of
O1?O2, which contradicts an implication of the assumption.
Theorem 1.21. Suppose that X is a continuum, p,q?X, X is irreducible between p and
q, and each nondegenerate subcontinuum containing q is not indecomposable. End(p,X) is
a continuum.
If K is a proper subcontinuum of X containing p, defineOK to be X\K.
(i) OK is connected.
Proof. By way of contrapositive, suppose thatOK is not connected. Let U and V be
nonempty disjoint sets open inOK such that U?V =OK; notice that U and V are
open in X as well becauseOK is open in X. WLOG, assume q?U and let C be the
component of X\V containing q. By the boundary bouncing theorem, C intersects
Bd(V). Because Bd(V)?K, C?Knegationslash=??C?K is a subcontinuum of X. Because
18
V is nonempty, it must be that C?K is a proper subcontinuum of X containing p
and q, which is against the original hypothesis.
From the above argument, it follows that if K is a proper subcontinuum of X containing
p, thenOK is subcontinuum of X that intersects K at its boundary. DefineC as
C={OK : K is a proper subcontinuum of X containing p}.
(ii) If F is a nonempty finite subset of C, then there is C?C such that C??Fnegationslash=?.
Proof. Let Fprime be a finite nonempty subset ofC; call the elements of Fprime 1prime,2prime,3prime,...,nprime
(ie Fprime ={1prime,2prime,3prime,...,nprime}). If iprime?Fprime let i be a proper subcontinuum of X containing
p such thatOi = iprime; define F ={i : iprime?Fprime}. If i?F, then i is a proper subcontinuum
containing p and not q; thus?F is a proper subcontinuum of X. Let K =?F.
If i?F, then i?K ?OK ?Oi; thus
OK ??{Oi : i?F}??{Oi : i?F}=?Fprime.
With (ii), it follows from Theorem 1.20 that?C is a subcontinuum of X. Let V =?C.
Notice that if K is a proper subcontinuum containing p, then End(p,X)?OK; thus
End(p,X)?V.
If we are fortunate enough that End(p,X) = V then the theorem is proved. But
suppose End(p,X) negationslash= V. To begin, we know that V is not indecomposable because it
contains q. Let A and B be proper subcontinuum of V such that A?B = V; assume that
q?A.
(iii) A?End(p,X)
19
Proof. By way of contradiction, suppose A is not a subset of End(p,X) and that t
is an element of A such that t /?End(p,X). Let K be a proper subcontinuum of X
containing p and t.
If there is b?OK such that b /?A, then it follows that A?K is a proper subcontinuum
of X that contains p and q and so X is not irreducible between the two points; thus,
we may assume OK ?A. A is closed ?OK ?A, and because V ?OK it must be
that V ?A; however, this means V = A, which conflicts with our assumption that A
is a proper subcontinuum of V.
(iv) End(p,X)?A
Proof. By way of contradiction, suppose b ? V and b /? A. This necessitates that
b?B.
Because we are under the assumption that End(p,X) negationslash= V we will let t ? V such
that t /?End(p,X), and we will let K be a proper subcontinuum of X that contains t
and p. The set K?B is the union to two intersecting subcontinua of X, thus K?B
is subcontinuum of X. Recall that B is a proper subcontinuum of V, so we may let
a?A such that a /?B. From (iii) above, a is necessarly in End(p,X) and so a /?K;
hence, K?B is a proper subcontinuum of X containg p and b?b /?End(p,X).
From (iii) and (iv) above we have that End(p,X) = A. Because A is a subcontinuum
of a subcontinuum of X, End(p,X) is a subcontinuum of X.
20
Chapter 2
Forming Chains
Unless otherwise stated, in the chapter it is assumed that (X,T) is a topological space.
Definition 2.1. Suppose n?a78, andC is a collection formed from n subsets of X. To say
thatC forms a chain, means thatC can be enumerated by the integers 1,2,...,n (that is
C ={C1,C2,...Cn}) so that if i,j?n, then Ci?Cj negationslash=?if and only if|i?j|?1. If the
enumeration ofC is defined (or understood), then it is said thatC is a chain in X. The
length ofCis the number of elements inCand is denoted by|C|. The elements of the chain
C are called links ofC, where the first and last links ofC are C1 and C2, respectively. An
interior link ofC is a link that is not a first or last link ofC. If each of C and D is a link
inC and C?Dnegationslash=?, then C and D are called adjacent links in C.
Note: Although the definition of a chain is general enough to allow for links of a chain
to be any nonempty subset of X, it will be more convenient for the purposes of this paper
to assume (unless stated otherwise) that links of a chain in X are open subsets of X.
The following is a list a of conventions used when speaking of chains:
1. Chains are denoted with capital script letters.
2. Links of a chain are denoted with plain capital letters (usually the same letter used
to denote the chain).
3. Unless stated specifically, if it is said thatC is a chain of length n, then it is assumed
that{C1,C2,...Cn}is the enumeration ofC.
4. A sequence of chains is indexed with superscripts (ie{Ci}?i=1).
The next theorem will be given with out proof.
21
Theorem 2.2. If C is a chain in X of length n, and D is the enumerated collection
{D1,D2,...,Dn}, where Di = Cn?i+1, then D is a chain in X.
Definition 2.3. Suppose the chainsC andD are as described in the previous theorem. D
is calledthe reverse of C, and will be denoted as?C={?C1,?C2,...,?Cn}.
Theorem 2.4. IfC is a chain in X andC1 andC2 are disjoint nonempty subsets ofC such
that C1?C2 =C, then (?C1)?(?C2)negationslash=?.
Proof. Suppose C ={C1,...,Cn} is a chain and each of C1 and C2 are nonempty disjoint
subsets ofCsuch thatC1?C2 =C. Without loss of generality, assume that C1, the first link
ofC, is inC1. Let l be the least index fromC such that Cl?C2. l > 1, which means l?1 is
an index of a link inC; furthermore Cl?1 ?C1 since l is the least index such that Cl ?C2.
Cl?1 and Cl are adjacent which means Cl?1?Clnegationslash=?, hence (?C1)?(?C2)negationslash=?.
Definition 2.5. If X is a topological space and each of V and U is a collection of open
sets, then to say that V refines (respectively properly refines) U, means that if V ?V
then there is U ?U such that V ?U (respectively V ?U).
Lemma 2.6. Suppose thatBis a base for X, M ?X, andU is an open cover of M. There
is V an open cover of X that refines U such that V?B.
Proof. For each x?M, let Ux?Usuch that x?Ux, and let Bx?Bsuch that x?Bx?Ux.
LetV={Bx : x?M}.
For each x?X, Bx is defined; thus?V= X, soV covers X.
If V ? V, there is x ? X such that V = Bx. Thus there is Ux ? U such that
Bx?Ux?U. Hence,V refinesU.
Definition 2.7. Suppose that n?a78 andC is a chain of length n in X. If l,m?a78 (with
l,m?n), then the segment of C from l to m is the collection{Ci : i?a78, min(l,m)?
i?max(l,m)}, and is denotedC(l,m).
The following lemma is also intuitive and will be given without proof.
22
Lemma 2.8. If C is a chain in the space X, then each segment of C forms a chain in X.
Definition 2.9. Suppose now thatC(h,j) is a segment from the chainC andD(k,m) is a
segment from the chainD. To say thatD(k,m) is anchored inC(h,j) means that Dk ?Ch
and Dm?Cj.
To say thatDis anchored inC means that the first link ofDis a subset of the first
link ofC and the last link ofD is a subset of the last link ofC.
Theorem 2.10. Suppose that C is a chain in X and D is a chain that refines C. Suppose
also that C(h,j) is a segment from C and D(k,m) is a segment from D such that D(k,m)
is anchored in C(h,j). If Ci is an interior link in C(h,j), then there is Dl ?D(k,m) such
that Ci is the only link in C such that Dl?Ci.
Proof. Let D1 = {D ? D(k,m) : D ? ?C(1,i?1) and D2 = {D ? D(k,m) : D ?
?C(i + 1,|C|)}. D1 and D2 are nonempty Dk ?Ch and Dm ?Cj and i is between h and
j. D1 andD2 are disjoint, for if D?D1 and Ca ?C(1,i?1) such that D?Ca, if Cb ?C
such that D?Cb then |a?b|?1, which means b?a + 1?i, and so Cb /?C(i + 1,|C|),
and therefore D /?D2.
It follows that D1 and D2 are nonempty disjoint subsets of D(k,m), which means
D1?D2 negationslash= D(k,m), since D(k,m) forms a chain. Thus, there is Dl ?D(k,m) such that
Dl /?D1?D2. BecauseDrefinesC, there is a link inCthat contains Dl, thus Ci is the only
link inC such that Dl?Ci.
Theorem 2.11. Suppose C is a chain in the the space X and D is a chain that refines
C. Suppose also that D(k,m) is a segment from D. The collection Cprime = {C ?C : C?
(?D(k,m))} is a segment in C.
Proof. Let Ch,Cj ?Cprime such that h < j and if Ci ?Cprime, then h < i < j; thus Cprime ?C(h,j).
It will now be shown that if Ci ?C(h,j), then Ci ?Cprime. Let Da,Db ? D(k,m) such that
Da?Ch negationslash=? and Db?Cj negationslash=?. Let Chprime ?Cprime such that Da ?Chprime, similarly, choose Cjprime so
that Db?Cjprime. Because Chprime contains Da and Da?Chnegationslash=?, Chprime ?Chnegationslash=?, thus|h?hprime|?1,
23
which means hprime = h or hprime = h+1. In a similar manner, it may be concluded that jprime = j or
jprime = j?1.
Thus, D(a,b) is a segment in D anchored in C(hprime,jprime). Because Ch,Cprimeh,Cj, and Cprimej
each intersect Da or Db, these links are necessarily in Cprime. Suppose that Ci ?C(h,j) and
hprime < i < jprime. Ci is therefore an interior link ofC(hprime,jprime), so by 2.10 there is Dl?D(a,b) such
that Dl ?Ci. Because Da,Db ?D(k,m),D(a,b)?D(k,m), thus Dl ?D(k,m) such that
Dl?Cinegationslash=?; hence, Ci?Cprime.
It follows thatC(h,j)?Cprime, thusCprime is the segmentC(h,j).
Theorem 2.12. Suppose n?a78 and C is a chain of length n in X. Also suppose that U is
an open subset of X. If i?a78 (with i?n) and for each C ?C, U?C negationslash=? if and only if
Ci?Cnegationslash=?, then D={D1,D2,...,Dn}, where Di = U and Dj = Cj (if jnegationslash= i), is a chain
in X.
Proof. Suppose j ?a78 (with j ?n and j negationslash= i). Dj ?Di = Cj ?U and Cj ?U negationslash=? if and
only if Cj?Ci negationslash=?if and only if|j?i|?1; thus, Dj?Di negationslash=?if and only if if and only if
|j?i|?1. If k?a78 (with k?n and knegationslash= i), then Dj?Dk = Cj?Ck; thus Cj?Dk negationslash=?if
and only if|j?k|?1.
Definition 2.13. In the previous theorem, the chainDis called the chain formed from
C by replacing link Ci with U, and is denotedC(Ci,U).
Definition 2.14. SupposeC is a chain in X. To say thatC is a spaced chain means that if
C,D?C, then C?D =?if and only if C?D =?. In other words, the closure of two links
inC intersect if and only if the two links are adjacent. If X is a metric space with metric
d, then spacing of C is defined as
S(C) = min({d(C,D) : C,D?C and C?D =?}.
For the following chapter it will be useful to note that when the metric space X is
compact or each link in the chainC is bounded, that S(C) > 0.
24
Theorem 2.15. Suppose X is normal, A is a closed subset of X, and C={C1???C|Cprime|} is
a chain in X that covers A. There is D, a spaced chain of length n, such that D properly
refines C and if Di?D, then Di?Ci.
Proof. Begin by choosing b2,b3,...,b|C| such that, if i ? a78 (with 2 ? i ?|C|), then bi ?
Ci?1?Ci (this is possible since|(i?1)?i|?1).
If j?a78, and j?|C|, letUj =C\{Cj}, and define Bj as Bj ={bi : Ci?Cj negationslash=?. Thus
Bj intersects each link inC that is adjacent to Cj; finally, define Aj as Aj = Bj?(A\?U).
Thus, for each j?a78 (with j?|C|)?Uj is open and Bj is finite and is threfore closed,
which means Aj is the union of two closed subsets of Cj; hence, Aj is a nonempty closed
subset of Cj. Because X is normal, Dj may be chosen to be an open set such that Aj ?Dj
and Dj ?Cj.
LetD={D1,D2,...,D|C|}. By the selection of each element ofD, it is hopefully clear
that if Di ?D, then Di ?Ci, which will also yield thatD is a proper refinement ofC. To
show thatD is a chain, suppose that Dk,Dl ?D. Without loss of generality, assume that
k?l.
If |k?l|?1, then l = k or l?1 = k. If k = l, then Dk?Dl negationslash=?, since Dk negationslash=?. If
k = l?1, then bl?Ck?Cl. Ck and Cl are each adjacent to Cl, thus
bl?Bk?Bl?Ak?Al?Dk?Dl.
If|k?l|> 1, then Ck?Cl =?. Because Dk ?Ck and Dl?Cl, Dk?Dl =?as well.
Thus,D is a chain.
25
Chapter 3
Chainable Continua
With the notions developed in the previous chapter regarding chains, the relationship
between chains and continuum will follow in this chapter. First, it must be stated what is
meant for a subset of a topological space to be chainable.
Definition 3.1. Suppose that X is a topological space and that M ?X. To say that M
is chainable, means that ifU is an open cover ofM, then there is a chainC that refinesU
and covers M.
Before showing that the interval [0,1] is chianable a useful lemma will be proven.
Lemma 3.2. Suppose that (X,d) is a metric space. If C is a compact subset of X and U
is a open cover of C, then there is epsilon1 > 0 such that if p?C, then there is U ?U such that
B(p,epsilon1)?U.
Proof. Suppose C is a compact subset of X and U is an open cover of C. If ? > 0 let C?
denote the subset of C to which p belongs if and only if no element ofU contains B(p,?).
Notice that if 0 < a < b then Cb?Ca
Thus, {C1
n
: n?a78} is a nested collection of closed subsets of C. ?{Cc : c > 0}=?
because for each p?C there is c > 0 such that B(p,c) is contained in some U ?U; thus,
there is n?a78, such that C1
n
=?since C is compact. It follows that if p?C, then B(p, 1n)
is contained in some element ofU.
Theorem 3.3. The interval [0,1], with the subspace topology from a82 is chainable.
Proof. It is assumed that the metric on [0,1] is the conventional abosolute difference metric
(ie d(x,y) = max(x?y,y?x)).
26
SupposeU is an open cover of [0,1]. [0,1] is compact, so by Theorem 3.2, epsilon1 > 0 may be
chose so that if x?[0,1], then then B(x,epsilon1) is contained in an element ofU. Let n?a78 such
that 1n < epsilon1. If i?a78, and 1?i?n + 1, let Fi = B(i?1n , 1n). Thus,F ={F1,F2,...,Fn+1}
is a refinement ofU.
As intervals, F1,F2,...,Fn+1 may be writtens as follows:
F1 = [0, 1n), Fn+1 = (n?1n ,1], and if i?a78 and 1 < i < n,
Fi = (i?1n ? 1n, i?1n + 1n) = (i?2n , in).
Note that if i?a78 and i?n, then the right endpoint of Fi is in and the left endpoint of
Fi+1 is i?1n .
To showF is a chain, choose j,k?a78 such that j?k?n+ 1.
If|j?k|?1, then it follows that j = k or j + 1 = k. If j = k, thenFj?Fk = Fj negationslash=?
and if j + 1 = k, then
Fj?Fk ?
bracketleftbiggi?1
n ,
i
n
parenrightbigg
?
parenleftbiggi?1
n ,
i
n
bracketrightbigg
=
parenleftbiggi?1
n ,
i
n
parenrightbigg
negationslash=?.
If|j?k|> 1, then j + 2?k, which means
Fj?Fk ?
bracketleftbigg
0, jn
parenrightbigg
?
parenleftbigg(j + 2)?2
n ,1
bracketrightbigg
=
bracketleftbigg
0, jn
parenrightbigg
?
parenleftbiggj
n,1
bracketrightbigg
=?.
Thus, Fj?Fk negationslash=?if and only if|j?k|?1.
Corollary 3.4. If X is a nonempty subcontinuum of [0,1], then X is chainable.
Proof. Suppose X is a subcontinuum of [0,1]. X is either a singleton or X is homeomorphic
to [0,1]. If X is a singleton, then X is chainable, for if U is an open cover of X, you can
pick U ?U such that X ?U, and{U}will form a chain of length one that covers X and
refinesU.
27
If X is homeomorphic to [0,1], pick h : [0,1] ?X such that h is a homeomorphism.
SupposeU is an open cover of X. LetUprime ={h?1(U) : U ?U}. h is continuous, hence each
element ofUprime is open in [0,1]; X is range of X and?U = X, hence, [0,1] =?Uprime. It follows
thatUprime is an open cover of [0,1].
[0,1] is chainable, so Cprime, a chain covering [0,1] and refining Uprime may be chosen. Let
C = {h(Cprimei) : Cprimei ?Cprime}. Since h is open, C is an open collection. Since h is onto and
?Uprime = [0,1], X = ?U. Lastly, to show C to show that C is a chain, if Cprimei ? Cprime, let
Ci = h(Cprimei). This provides an enumeration ofC. If Ci,Cj ?C, then because h is one-to-one
and onto, Ci?Cj negationslash=?if and only if h?1(Ci)?h?1(CJ)negationslash=?if and only if|i?j|?1; thusC
is a chain covering X.
C will refineU, for if Ci?C, then h?1(Ci)?Cprime. There is Uprime?Uprime such that h?1(Ci)?
Uprime, thus h(Uprime)?U such that Ci?h(Uprime).
Definition 3.5. If (X,d) is a metric space and C is a chain in X, then the mesh of C is
defined as mesh(C) = min({diam(C) : C?C}), where diam(C) = sup({d(x,y) : x,y?C})
for each C ?X. If each link inC is bounded and s > 0 such that s?mesh(C), thenC is
called an s?chain.
Theorem 3.6. Suppose that X is a compact metric space with metric d, and {Ci}?i=1 is a
sequence of chains of open sets such that
1. For each i the chain Ci+1 is a proper refinement of Ci, and
2. For each i, Ci is an 1i ?chain.
The set ??i=1(?Ci) is chainable and a subcontinuum of X.
Proof. For ease, let M =??i=1?Ci. We may assume without of generality, that if n?a78
and C is a first or last link in the chainCn, then C?M negationslash=?, and C?M is not a subset of
another link in Cn (else, Cn\{C}can be enumerated to form a chain that covers M).
28
Note that for each positive integer i, if C ?Ci+1, then C ??Ci, since Ci+1 properly
refinesCi ; thus
?Ci+1 =
uniondisplay
C?Ci+1
C ??Ci,
and therefore M =??i=1?Ci is a nonempty closed set. Hence M is compact.
Secondly, M is connected. If M were not connected, then because M is compact, there
would be disjoint, nonempty, closed sets H and K, such that M = H?K. X is normal,
thus we can choose disjoint open sets U and V, such that H ?U and K ?V. Without
losing generality, we may assume U?V =?. Let epsilon1 = min(D(H,U),d(K,V),(U,V)).
Notice that if n?a78 such that 1n < epsilon1, then a link inCn cannot intersect U and V, for
if x?U and y?V, then d(x,y) > e(U,V)?epsilon1 > 1n. This means there is a link inCn that
is not a subset of U?V; thus?Cn\(U?V) is a nonempty closed set. Because
?Cn+1??Cn,
it follows that parenleftBig
?n>1
epsilon1
?Cn
parenrightBig
\(U?V)negationslash=?.
Of course the above set is also a subset of M and is therefore a subset of U and V (contra-
diction).
Before showing chainability notice that each link in each chain must intersect M, for
ifCn ={C1,...,CN}and j ?a78 (with j ?N) such that Cj?M =?, then 1 < j < n (by
one of the initial assumptions) and U =?j?1i=1Ci and V =?Ni=j+1Ci are disjoint open sets,
each intersecting M, whose union contains M, hence M is not connected.
To show chainability, suppose U is an open cover of C. Since X is a compact metric
space and M is a closed subset of X, Theorem 3.7 states that epsilon1 > 0 may be chosen such
that if p ? M, then {B(p,epsilon1)}? U for some U ?U. Let n ? a78 such that 1n < epsilon12. If C
is a link in Cn, then by the previous argument, C?M negationslash= ?. Let pprime ? C?M. Because
mesh(Cn) < 1n < epsilon12, diam(C) < epsilon12; thus d(pprime,c) < 2epsilon12 = epsilon1 for each c ? C, which means
29
C?B(pprime,epsilon1). By our choice of epsilon1 there is U ?U such that B(p,epsilon1)?U; hence, there is U ?U
such that C?U. ThereforeCn refinesU and M is chainable.
Lemma 3.7. Suppose (X,d) is a metric space, M ? X, U is an open cover of M, and
epsilon1 > 0. Let B(epsilon1) = {B(x,?) : x ? M and ? ? epsilon1}. There is V, an open cover of M that
refines U such that V?B.
Proof. The proof is similar to that of 2.6. It will be shown that each element of x is
contained in an element ofB(epsilon1), which is contained in an element ofU. Choose x?M and
U ?U, such that x ? U. There is ? > 0 such that B(x,?) ? U, since U is open. Thus
B(x,min(?,epsilon1))?B(epsilon1) and B(x,min(?,epsilon1))?B(x,?)?U.
Theorem 3.8. Suppose M is a chainable continuum lying in the metric space (X,d) and
{ai}?i=1 is a decreasing sequence converging to zero. There exists a sequence of spaced chains
{Cn}?n=1, each of which covers M, such that if i?a78, then
1. Ci+1 properly refines Ci,
2. mesh(Ci)?ai ,
3. M =??n=1(?Cn).
Proof. To begin, let U1 = {B(x,1) : x ? M}. U1 is an open cover of M, so D1 may be
chosen to be a chain that refines U1 and covers M. Since metric spaces are normal and M
is chainable and thus closed, by 2.15, C1 may be a spaced chain that properly refinesD1.
The remaining chains will be defined inductively. SupposeCi is defined as spaced chain
covering M. By 3.7,Ui+1 may be chosen to be a refinement ofCi, such thatUi+1?B(ai+1),
whereB(ai+1 ={B(x,ai+1) : x?M}. LetDi+1 be a chain refiningUi+1 and covering M.
By 2.15,Ci+1 may be chosen to be a spaced chain that properly refinesDi+1.
The sequence of spaced chains{Ci}?i=1 is now defined and each chain in the sequence
does cover M. It remains to show that properties 1, 2, and 3 are satisfied by this sequence.
Suppose i ? a78. Di+1 refines Ci and Ci+1 properly refines Di+1, thus Ci+1 properly
refinesCi; hence, 1 is satisfied.
30
BecauseUi ?B(ai), if D?Di, diam(D)?ai; thus mesh(Di)?ai, meaningDi is an
ai-chain (satisfying 2).
Lastly, each chain covers M, so M ? ??i=1(?Ci). To prove the converse, suppose
x???i=1. It will be shown that x is a limit point of M.
Suppose O is open and x ? O. epsilon1 > 0 may be chosen so that B(x,epsilon1) ? O ( since
{B(p,?) : p?X, and ? > 0} is a base for X). Because {ai}?i=1 converges to zero, n?a78
may be chosen such that an < epsilon1. Let Cnk ?Cn such that x?Cnk. Cnk is contain is contained
in some element in Dn, which is contained in some element in Un; so B(p,epsilon1prime) ?Un may
be chosen such that Cnk ?B(p,epsilon1prime). By construction of Un, p?M and epsilon1prime < an < epsilon1. Thus
p?B(x,epsilon1). We have that x is a limit point of M. M is a continuum and thus closed in X.
This means it contains all of its limit points, yeilding that??i=1(?Ci)?M.
31
Chapter 4
Inverse Limit Spaces and Chainable Continua
This chapter will offer a further characterization of chainable subsets of metric spaces.
It is shown in the previous chapter that a chainable subset of a metric space is a continuum.
In this chapter it will be shown that a subset of a metric space is chainable if and only if it
is homemorphic a specific type of inverse limit space. The notion of an inverse limit space
will now be developed.
Suppose that if i ? a78, then Xi is a topological space. Suppose also that fi is a contin-
uous function from Xi+1 to Xi. Let lim??{Xi,fi}?i=1 denote the subset of producttext?i=1 Xi, to which
the sequence{xi}?i=1 belongs if and only if xi = fi(xi+1), for each i?a78.
If i ? a78 and Oi ? Xi, let ??Oi denote the collection {x ? X| xi ? Oi}; thus, ??Oi ?
lim??{Xi,fi}?i=1.
Theorem 4.1. The collection B, defined as
B={??Oi : i?a78 and Oi is an open subset of Xi},
is a basis for a topology on lim??{Xi,fi}?i=1.
The proof to the above theorem can be found in [6].
Definition 4.2. The space X = lim??{Xi,fi}?i=1 with topology generated by B (as defined
in 4.1) is called an inverse limit space. If i?a78, Xi is called the ith factor spaces and
fi is called the ith bonding map. Furthermore, if Oi ?Xi and x???Oi, then x is said to
pass through the set Oi in Xi.
32
Inverse limit spaces are a valuable commodity in topology and are dealt with thouroughly
in [6]. For the purposes of this chapter, the only inverse limit spaces that will be considered
are those whose factor spaces are [0,1]. For this reason, necessary theorems, whose proofs
can be found in [6], will be stated (without proof), and afterwards, it will be shown (with
proof) that each chainable continua is in fact homeomorphic to an inverse limit space whose
factor spaces are each [0,1].
Theorem 4.3. Suppose that if i ? a78, Xi is a topological space and fi : Xi+1 ? Xi is a
continuous function.
1. lim??{Xi,fi}?i=1 is Hausdorff if Xi is Hausdorff for each i?a78.
2. lim??{Xi,fi}?i=1 is compact if Xi is compact for each i?a78.
The proof of 1 and 2 may be found respectively in 3.1 and 3.4 in [6].
When considering an inverse limit space lim??{Xi,fi}?i=1, it is useful to look at compo-
sitions of the bonding maps. In such cases, the following convention will be used:
If i,j?a78 and i < j, fji : Xj ?Xi, such that
fji = fi?fi+1?????fj?2?fj?1.
Notice that with this notation, fi+1i = fi. Furthermore, fji is a composition of contin-
uous functions, hence fji is continuous by an extension of 0.18.
For the following lemmas and theorems, suppose that if i?a78, then Xi = [0,1] and if
fi : Xi+1?Xi is a continuous function, let X denote the space lim??{Xi,fi}?i=1.
Lemma 4.4. Suppose n ? a78, On is an open subset of Xn. If i ? a78 and On+i =
(fn+in )?1(On) (ie On+i is the preimage of On under fn+in ), then ??On =????On+i.
Proof. Because On+i = (fn+in )?1(On), it follows that if x?X, then xn ?On if and only if
xn+i?On+i. Hence x??On if and only if x?????On+i, meaning??On =????On+i.
33
The next lemma could be stated as a corollary to theorems in [6], however, due to the
general nature of [6], it is felt that proving the following theorem for this specific instance
is beneficial.
Lemma 4.5. If n?a78, then pin(X) is a subcontinuum of [0,1].
Proof. Suppose n ? a78. Let K = ??i=1fn+in ([0,1]). Because fn+jn is continuous for each
j ?a78, and Xn+j = [0,1] is a continuum, it follows that fn+jn (Xn+j) is a subcontinuum of
Xn. Since fn+j+1n+j (Xn+j+1])?Xn+j, for each j?a78,
[0,1]?fn+1n (Xn+1)?fn+2n (Xn+2)????,
and since fn+jn (Xn+j)negationslash=?for each j ?a78, it follows that K is in fact the intersection of a
nested collection of nonempty subcontinua of Xn. By 1.20, K must be a subcontinuum of
Xn = [0,1].
It will now be shown that pin(X) = K. First note that if x?X (ie xn?pin(X)), then
fn+in (xn+i)?K for each i?a78; thus xn?K. It follows that pin(X)?K.
To show K?pin(X), a less straightforward proof is recquired. If i?a78, define Ki as
Ki =??j=1fn+i+jn+i (Xn+i+j).
For the same reasons that K is a nonempty subcontinuum of Xn, Ki is a nonempty sub-
continuum of Xn+i. Now suppose that yn ?K. Since yn is in the image of fn+in for each
i ? a78, and fn+in = fn?fn+in+i, it must be that (fn)?1(yn)?fn+i+1n+1 (Xn+i+1) negationslash= ? for each
i?a78. Because
(fn)?1(yn)?K1 =??j=1(fn)?1(yn)?(fn+j+1n+1 (Xn+j+1)),
(fn)?1(yn)?K1 is the intersection of a decreasing collection of nonempty compact sets, thus
(fn)?1(yn)?K1negationslash=?. Choose yn+1?(fn)?1(yn)?K1.
34
For eachi?a78, ifyn+i is chosen, then for reasons similar to those above, (fn+i)?1(yn+i)?
Ki+1 is also the intersection of a decreasing sequence of nonempty compact subsets of
Xn+i+1, meaning (fn+i)?1(yn+i)?Ki+1negationslash=?. Choose yn+i+1?(fn+i)?1(yn+i)?Ki+1.
Consider the sequence{xi}?i=1, where
(a) xi = fni (yn), if i < n,
(b) xi = yi, if i?n (where yi is as chosen previously)
To show the sequence is in X, pick i?a78
(a) If i < n, then xi = fni (yn) = fi(fni+1(yn)) = fi(xi+1).
(b) If i?n, then xi = yi. yi+1?(fi)?1yi, thus xi = yi = fi(yi+1) = fi(xi+1).
Because xn = yn, it follows that K?pin(X). Thus pin(X) = K, which means pin(X) is
a continuum.
Theorem 4.6. X is chainable
Proof. Suppose U is an open cover of X; by Lemma 2.6, V, an open cover of X may be
chosen so thatV refinesU, andV?B, whereB is as defined in 4.1.
By Theorem 4.3 part 2, X is compact, thusF may be chosen as a finite subcover of X
fromV. If F ?F, then F is a basic open subset of X, therefore a positive integer nF and
Onf, an open subset of [0,1], may be chosen so that F =???OnF .
Let N = max(nF : F ?F). If F ?F, let OF = (fNnF )?1(Onf)); by Lemma 4.4,
F = ???Onf = ??OFN. Let XN = piN(X); XN is a subcontinuum of [0,1] by 4.5. Let FN =
{OFN?[0,1] : F ?F}. SinceF covers X,FN covers XN. Because projection mappings are
open, and the image of piN (restricted to X) is XN, it follows that piN(F) is an open subset
of XN for each F ?F, thus FN is an open cover of XN. The corollary to 3.3 yields that
XN is chainable, and thatCN may be chosen to be a chain coverring XN that refinesFN.
If CNi is a link in CN, let Ci =??CNi . It will first be shown that if CNi ?CN, then Ci
is open in X. Suppose CNi ?CN. CNi is open in XN and XN is a subspace of [0,1], thus
35
there is ON open in [0,1] such that ON ?XN = CNi . CNi ?O, hence Ci =??CNi ???ON. If
x???ON, then xn?On?piNX, which means xn?CNi , thus??ON ???CNi = Ci. It follows that
C is a collection of basic open subsets of X.
LetC denote the collection{Ci : CNi ?CN}. It will now be shown thatC is a chain,C
covers X, andC refinesU.
Suppose Ci,Cj ?C and Ci?Cj negationslash=?. x?Ci?Cj if and only if xn ?CNi ?CNj if and
only if|i?j|?1; thusC is a chain.
If x ? X, then xN ? piN(X) = XN, which means CNi ?CN may be chosen so that
xN ?CNi . It follows that x???CNi = Ci. Thus,?C= X and thereforeC covers X.
Lastly, C refinesU. Suppose Ci ?C and let CNi be the corresponding link inCN. CN
refinesFN ={OFN?[0,1] : F ?F}, so Fprime?F may be chosen so that CNi ?OFprimeN ; this means
that
Ci =
??
CNi ?
???
OFprimeN = F.
F?V, so V prime ?V may be chosen so that Fprime ?V prime. V refines U, so Uprime ?U may be chosen
so that V prime?Uprime. Thus, Fprime?Uprime?U. It may be concluded thatC refines the open coverU
picked originally.
Hence, ifU is an open cover of X, there is a chainC that covers X and refinesU.
It has now been shown that that an inverse limit space whose factor spaces are each
[0,1] is a chainable continuum. The rest of the chapter is devoted to showing that each
chainable subset of a metric space is homeomorphic to an inverse limit whose factor spaces
are each [0,1].
Definition 4.7. IfCis a chain, the indexing set forCis the collectionI={1,2,...,|C|}.
SupposeC is a chain andI is the indexing set forC. If A?I, and j,k?A (with j < k),
then to say that j and k are consecutive elements in A, means that if l?A, then l?j or
l?k; Cj and Ck may be referred to as consecutive links in terms of A.
36
Definition 4.8. If n?a78 andK={[i?1n , in] : i?a78, 1?i?n}, thenKis called the rusted
chain of length n and if j?a78 and j?n then Kj = [j?1n , jn] is the jth rusty-link of K.
K is a chain in the general sense, however, it is unlike the chains used previously,
because each link in K is not an open subset of [0,1]. The term ?rusted ? came to mind
defining such chains, because unlike chains in normal spaces whose links are open subsets,
a link in a rusted chain cannot be replaced by a proper subset of the link and still cover
[0,1]; thus, there is less flexibility.
The following construction will help describe how to ?refine? a rusted chain with an-
other rusted chain.
SupposeCis a chain andDis a chain that refinesCsuch that the union of two adjacent
links inDdoes not intersect more than two links inC. LetI be the indexing set ofC, and
letJ be the indexing set ofD. Define T as
T ={j?J : Dj intersects two links inC},
and define U as
U ={j?J : j + 1?T}.
Let F ={jn : j ?T ?U}?{0,1}. For each j ?J, let i(j) be the least element ofI such
that Dj?Ci(j)negationslash=?. Define f : F ?[0,1] defined as follows:
if j?T, then f(jn) = i(j)m , where m =|C|;
if j?U\T, then f(jn) = f(j+1n ).
If 0 /?T?U, let j0 = min(T) and let f(0) = f(j0n ).
If 1 /?T?U, let j1 = max(T?U) and let f(1) = f(j1n ).
Let f be the piecewise linear expansion of f.
Definition 4.9. In the previous construction,f is called the bending function relative to
D inC. The collection F used to define f is called the defining set for f.
37
Let K be the rusted chain of length n =|D|. Because F is a subset of the endpoints
of links in K it follows that if j?J, f is linear over Kj.
By the initial condition, that the union of two adjacent links inDintersect at most two
links inC, if j?T?U, then Dj and Dj+1 intersect the same two links inC(else Dj?Dj+1
intersects more than two links inC); thus f(jn) = i(j)m for every j?T. Notice that this also
implies that if j+1?T, then then f(jn) = f(j+1n ) regardless of whether j?T or j?U\T.
Lemma 4.10. Suppose (X,d) is a metric space,C is a spaced chain in X, andDis a chain
that refinesC. Suppose also thatD(k,l) is a segment inD. If mesh(D) < S(C)4 , where S(C)
is defined in 2.14, and |k?l|?2, then ?D(k,l) intersects at most two links in C.
Proof. If |k?l|?2, then there are at most three links in D(k,l). If D(k,l) has no more
than two links, then because Dk?Dlnegationslash=?,
diam(?D(k,l)?diam(Dk) +diam(Dl)? S(C)2 ) < S(C).
If D(k,l) contains three links, let Dj denote the link that is not Dk or Dl. Thus, Dj
intersects both Dk and Dl, and
diam(?D(k,l))?diam(Dk?Dj) +diam(Dl)? S(C)2 + S(C)4 < S(C).
If Ci,Cj ?C and each intersects?D(k,l), then d(Ci,Cj) < S(C). If d(Ci,Cj) < S(C),
then Ci and Cj must be adjacent.
Lemma 4.11. Suppose (X,d) is a metric space, C is a spaced chain, D is a chain that
refines C, and mesh(D) < S(C)4 . If C(h,j) is a segment of C, D(k,m) is a segment of D
that is anchored inC(h,j), and Ci is an interior link ofC(h,j), then there is Dl?D(k,m)
such that Dl only intersects Ci.
Proof. Let D1 = {D ?D(k,m) : D?(?C(1,i?1)) negationslash= ?}, and let D2 = {D ?D(k,m) :
D?(?C(i + 1,|C|))negationslash=?}. It will be shown thatD1 andD2 are disjoint. Suppose D?D1,
38
and Ca ?C(1,i?1) such that D?Ca negationslash= ?. From 4.10, D cannot intersect a link in C
that is not adjacent to Ca; hence, if Cb ?C and D?Cb negationslash=?, then b?a + 1?i, meaning
Cb /?C(i+ 1,|C|).
By 2.11,D1?D2 cannot contain every link inD(k,m). Pick Dl ?D(k,m), such that
Dl /?D1?D2. It follows that Dl cannot intersect a link inC(1,i?1)?C(i+1,|C|), thus Dl
only intersects Ci.
For Theorem 10 through Theorem 13, the following are assumed:
(1) m?a78 andC is a spaced chain of length m
(2) n?a78 andDis a spaced chain of length n that properly refinesCsuch that mesh(D) <
S(C)
4 .
(3) I andJ are the respective indexing sets ofC and D.
(4) T ?J to which j belongs if and only if Dj intersects two links inC.
(5) U ?J defined as U ={j : j + 1?T}.
(6) F = T?U?{0,1}
(7) f is the bending function relative toD laying inC.
(8) KC is the rusted chain of length m and KD is a rusted chain of length n.
Notice that by assumption (2) and Theorem 4.10, a link in D cannot intersect more
than two links inC, so the function f in assumption (7) is definable.
Theorem 4.12. If j,k?F (with j < k )are consecutive indices of F, then|f(jn)?f(kn)|?
1
m; furthermore, if l ? J (with j < l < k) and i ? I such that Dl ?Ci negationslash= ?, then
Dl+1?Cinegationslash=?.
Proof. First note that the hypothesis is true when f(jn) = f(kn) and that this occurs if
(a) k?T and j + 1 = k, or
39
(b) j = 0 and 0 /?T?U, or
(c) k = n and n /?T?U.
The final case to consider is when the negation of (a), (b), and (c) occur. Not (a)
implies that k?U\T (else, j < k?1 < k and k?1?U ?F is between j and k). Not
(b) implies j > 0, and therefore j?T?U. j /?U (else, j +1?T ?F and j < j +1 < k),
which means j?T. Let h be the least index inIsuch that Ch?Dj negationslash=?. Because k?U\T,
knegationslash= n, and therefore k + 1?T; let i be the least index ofI such that Ci?Dk+1negationslash=?.
By induction, it will be shown that Dj+1,Dj+2,...Dk are each contained in the same
link in C. Since j + 1 /? F, Dj+1 intersects exactly one link in C; let g be the index of
the link in C that intersects Dj+1. Let l = k?j. If p ? a78, 1 ? p < l, and Dj+p ? Cg,
then j < j + p + 1?k implies that j + p + 1 /?T, and so Dj+p+1 intersects only one link
in C; Dj+p+1 intersects Dj+p and Dj+p ? Cg, hence, Dj+p+1 ? Cg. Thus, Dj+1,...,Dk
(equivalently Dj+1,...,Dj+l = Dk) are each a subset of Cg.
Dk+1?Dk negationslash=?andDk ?Cg, henceDk+1?Cg negationslash=?. Ifh = i, thenf(jn) = hm = im = f(kn)
and the hypothesis of the theorem is true.
Suppose now that h < i; this means h < h + 1 ? i, and by Theorem 2.10 , there is
q?J such that j < q < k +1 and Dq ?Ch+1. Because Dj+1,...Dk are each contained in
Cg, g = h+1, and therefore h+1 is the least index inI such that Dk+1?Ch+1negationslash=?; thus,
f(kn) = f(k+1n ) = h+1m and|f(jn)?f(kn)|=|hm?h+1m |= 1m.
Lastly, suppose that i < h. Dj ?Ch+1 = ?, or else Dk+1 intersects Ch+1 and Ci,
which is not possible by 4.10 because Ci and Ch+1 are disjoint (nonadjacent). Therefore,
Dj ?Ch negationslash= ? (else Dj would intersect three links in C), meaning g = h. It follows that
i = h?1 and|f(jn)?f(kn)|=|hm?h?1m |= 1m.
Corollary 4.13. diam(f(KDj ))? 12m for each j?J.
Proof. Suppose l ? J. Let j and k be two elements of F such that j ? l ? k and
if q ? F, q ? j or q ? k; f is defined to be linear between consecutive points in F,
40
thus diam(f([jn, kn])) = |f(jn)?f(kn)|. If f(jn) = f(kn), then the diameter of f(Knl ) is 0
because KDl ? [jn, kn]. If f(jn) negationslash= f(kn), then let h ?I such that Ch is the least link in
C intersecting Dj. As in the proof of the previous theorem we can conclude that j ? T
and k ? U \T; let i ?I such that Ci is the least link in C intersecting Dk+1. Because
the mesh ofD is less than S(C)4 , there are at least three links inD between Dj and Dk+1,
thus there is at least two links of D between Dj and Dk. Because f is defined to be
linear between jn and kn, if x,y ? [jn, kn], then |f(x)?f(y)|? 1m |x?y||j
n?
k
n|
? n|x?y|2m . Thus,
diam(f(KDl )) =|f( ln)?f(l?1n )|? n2m 1n = 12m.
Theorem 4.14. If j?J, i?I, and Dj?Cinegationslash=?, then f(Knj )?Kmi .
Proof. First suppose that j ? T. Let h be the least index of I such that Ch?Dj negationslash= ?.
f(jn) = hm and because j ?1 ? U, f(j?1n ) = f(jn) = hm. Because f is defined to be
linear between j?1n and jn, f must be constant over Knj = [j?1n , jn]; thus f(Knj ) ?{hm}.
j is assumed to be in T, meaning Dj intersects two link in C; furthermore, these two
links must be adjacent. Ch is the least such link, meaning Dj must also intersect Ch+1.
h
m ?K
m
h ?K
m
h+1, therefore f(K
nj )?Km
h ?K
m
h+1.
The remainder of the proof will follow by induction. Beginning by showing the hypoth-
esis holds if j = 1.
If 1?T, then the hypothesis holds by the previous argument. If 1 /?T, let l be the
least element of T and let i?Isuch that Dl intersects Ci and Ci+1. D1?Ci or D1?Ci+1,
for if not, D1?Ci?1 (or D1?Ci+2), which means there is an index lprime, with 1?lprime < l, such
that Dlprime intersects Ci?1 and Ci ( or Dlprime intersects Ci+1 and Ci+2). Thus, lprime?T and lprimenegationslash= l
because no link inDintersects more than one link inC, which means lprime < l and therefore l
is not the least element of T.
By definition, f(0) = f( ln) = f(l?1n ); thus, f is constant over [0, l?1n ] (note: l?1?U
and is therefore also defined as f( ln)). It follows that f(Kn1 )?Kmi ?Kmi+1?Kmi .
Suppose now that j?J, and for each k?J (with k < j ), if i?I and Dk ?Cinegationslash=?,
then f(Knk)?KCi .
41
If j ? T, then the hypothesis of the theorem holds by the initial argument of the
theorem. If j /? T, let k and l be consecutive links in F such that k ? j + 1 ? l. If
k ? j + 1 ? l, then h ?I may be chosen so that Dj+1?Ch negationslash= ?, and let l be the least
index in F that is greater than j + 1. Because Dj+1 only intersects the ith link in C, Dl
must intersect a link adjacent to Ch; thus, Dl intersects two links inCand Ch is one, which
means f( ln) ? f(Knl ) ? Kmh . Dj ?Ch negationslash= ? and thus, f(jn) ? f(Knj ) ? Kmh . It follows
that the left most and right most point of Knj+1 are each in Kmh , therefor f(Knj+1)?Kmh ,
because f is linearly defined over Knj+1.
If j /?T, let k and l be two consecutive indices in F such that k ?j ?l. Let i?I
such that Dj?Cinegationslash=?. By Theorem 4.10, Dk?Cinegationslash=?and Dl?Ci; thus, by the inductive
hypothesis f(Knk) ? Kmi . If l ? U, l + 1 ? T and Dk+1 intersects Ci, which means that
f( ln) = f(l+1n ) ? Kmi ; if l /? U, then l = n and therefore f( ln) = f(1) = f(kn) ? Kmi .
Regardless of the case, both f(kn) and f( ln) are in Kmi , which means that f(Knj ) ? Kmi
because Kmi ?[kn, ln] and f is linear over this interval.
For Theorem 4.15 through Corollary 4.19, suppose the following
(1) X is a metric space and M is a chainable subset of X;
(2) C1,C2,...is a sequence of spacedchains inX, with indexing sets respectiveI(1),I(2),...,
and respective lengths n(1),n(2),... such that
(a) Ci+1 properly refinesCi;
(b) mesh(Ci+1) < min( 1i+1, S(Ci)4 ),
(c) ??i=1(?Ci) = M;
(3) Ki is a rusted chain of length n(i)
(4) fi : I?I is the bending function relative toCi+1 insideCi;
(5) Fi is the defining set of fi;
42
(6) If g?M, i(g) is the least index inI(i) such that the i(g)th link inCi contains g. For
ease, the i(g)th link in Ci will be referred to as C(g,i) (ie C(g,i) is the first link inCi
that contains g), and the i(g)th link in Ki, (Kii(g)), will be referred to as K(g,j).
For each g?M, let hi(g) =?j>ifji (K(g,i)).
Theorem 4.15. If g?M, and i?a78, hi(g) is a singleton.
Proof. For each i?a78,Ci+1 properly refinesCi, so by Theorem 4.10 fi(K(g,i+1))?K(g,i)
because C(g,i+ 1)?C(g,i)?{g}negationslash=?. Hence,
K(g,i)?fi+1i (K(g,i))?fi+2i (K(g,i+ 2))????,
which implies that hi(g) is the intersection of a nested collection of nonempty closed subsets
of [0,1], meaning hi(g)negationslash=?.
To show that hi(g) is in fact a singleton, for each i?a78, the case for h1(g) will be made
and then generalized. First note that f1 = f21 is a function that is linear over K(g,2) and
that diam(f1(K(g,2)))? 12n(1) = (12)2?1? 1n(1).
If j?a78 (with j?2), and fj1 is linear over K(g,j) such that
diam(fj1(K(g,j)))?
parenleftbigg1
2
parenrightbiggj?1
? 1n(1),
then fj+11 = fj1 ?fj is linear over K(g,j + 1) because fj is linear over K(g,j + 1) and
f(K(g,j + 1))?K(g,j), and
diam(fj+11 (K(g,j + 1))) ? diam(fj(K(g,j+1)))diam(K(g,j)) ?diam(fj1(K(g,j)))
? 12?parenleftbig12parenrightbigj?1? 1n(1)
?parenleftbig12parenrightbigj? 1n(1)
43
because diam(fj(K(g,j + 1)))? 12n(j) = 12?diam(K(g,j)). It follows that
limj??diam(fj1(K(g,j))) = limj??
parenleftbigg1
2
parenrightbiggj?1parenleftbigg 1
n(1)
parenrightbigg
= 0,
meaning h1(g) is a singleton.
In general, note that fi = fi+1i is a function that is linear over K(g,i + 1) and that
diam(fi(K(g,i+ 1)))? 12? 1n(i) = (12)1 1n(i).
If j?a78 (with j?i+ 1), and fji is linear over K(g,j) such that
diam(fji (K(g,j)))?
parenleftbigg1
2
parenrightbiggj?i
? 1n(i),
then fj+1i = fji ?fj is linear over K(g,j + 1) because fj is linear over K(g,j + 1) and
fj(K(g,j + 1))?K(g,j), and
diam(fj+1i (K(g,j + 1))) ? diam(fj(K(g,j+1)))diam(K(g,j)) ?diam(fji (K(g,j + 1)))
? 12?parenleftbig12parenrightbigj?i? 1n(i)
?parenleftbig12parenrightbigj? 1n(i)
because diam(fj(K(g,j + 1)))? 12n(j) = 12?diam(K(g,j)). It follows that
limj??diam(fji (K(g,j))) = limj??
parenleftbigg1
2
parenrightbiggj?iparenleftbigg 1
n(i)
parenrightbigg
= 0,
meaning hi(g) is a singleton.
Theorem 4.16. Suppose g?M. For each i?a78, let ai ?hi(g). {aj}?j=1 is in the inverse
limit space lim??{Xi,fi}?i=1, where Xi = [0,1] and fi is the bending function described prior
to 4.15.
44
Proof. Let g?M and{aj}?j=1 be as suggested in the theorem. If i?a78, then for each k?a78
(with k?i+ 1),
fi(ai+1) = fi(hi+1(g))?fi(fji+1(K(g,j)))?fji (K(g,j)),
thus fi(ai+1)???j=1fi+ji (K(g,i+j)) = hi(g), so by definition fi(ai+1) = ai.
By the previous two theorems, it follows that for each i?a78, hi can be thought of as
a function from M into I. and h : M ? lim??{Xi,fi}?i=1, defined as h(g) = {hi}?i=1, is a
function from M into lim??{Xi,fi}?i=1.
Theorem 4.17. The function h as defined above is continuous and one-to-one.
Proof. h is one-to-one, for if p,q?M and pnegationslash= q, then there is i?a78 such that 1i < d(p,q)2 ,
which means if C,D?Ci, p?C and q?D, then there is one link ofCi between C and D,
meaning K(p,i)?K(q,i) =?and hi(p)negationslash= hi(q).
To show continuity, suppose p?M, and U is open in lim??{Xi,fi}?i=1 such that h(p)?U.
Without losing generality, it may be assumed that U is a basic open set. Let i?a78, and let
Ui be an open subset of Xi such that U =??Ui.
By the topological nature of [0,1], epsilon1 > 0 may be chosen so thatXi?(hi(p)?epsilon1,hi(p)+epsilon1)?
Ui, and j ?a78 may be chosen so that parenleftbig12parenrightbigj?1 < epsilon1. C(p,i + j)?M is an open subset of M
and if x?C(p,i+j), then C(x,i+1)?C(p,i+j)negationslash=?; thus, K(x,i+j)?K(p,i+j)negationslash=?
and fi+ji (K(x,i + j)?fj+1i (K(p,i + j))negationslash=?. The diameter of the fi+ji - image of a rusty
link in Kn(i+j) is not greater than parenleftbig12parenrightbigj, meaning
|hi(x)?hi(p)|?
parenleftbigg1
2
parenrightbiggj
+
parenleftbigg1
2
parenrightbiggj
=
parenleftbigg1
2
parenrightbiggj?1
< epsilon1.
Theorem 4.18. If {aj}?j=1?lim??{Xi,fi}?i=1, there is g?M such that hj(g) = aj for each
j?a78.
45
Proof. Let a ={aj}?j=1?lim??{Xi,fi}?i=1. For each j?a78, let K(a,j) be the lowest indexed
rusted link in Kj that contains aj; let C(a,j), be the link inCj corresponding to K(a,j).
Let V(j) ={Kji : Kji is adjacent to K(a,j)}.
It will now be shown that for each i?a78, fi(?V(i + 1))??V(i) and each rusty link
in Ki+1 is only contained in a link in V(i). Let K be the union of K(a,i+1) and another
rusted link in V(i + 1). Thus K is connected and contains ai+1. By Corollary 3.5, the
diameter of the image of a link in Ki+1 under fi does not exceed 12(n(i)); because K is
connected and fi is continuous, fi(K) is connected, and
diam(fi(K))? 12n(i) + 12n(i) = 1n(i).
Because f(K)?K(a,i) contains fi(ai+1) = ai, and the diameter of fi(K) does not exceed
1
n(i), each point in fi(K) must be in K(a,i) or a rusted link in K
i that is adjacent to K(a,i);
hence fi(K)??V(i). Because the choice of the adjacent link used to form K is arbitrary,
it follows that fi(V(i+1))?Vi. Furthermore, because ai?K(a,i), ai is at least 1n(i) from
the boundary of?V(i) we know that if L?Ki+1, fi(L) is not a subset of Bd(?V(i)) and
therefore fi(L) is not a subset of any link in Ki that is not in V(i).
Analogous to V(j) above, if j ? a78, let W(j) = {Cji : Cji is adjacent to C(a,j)}; in
other words, each link in W(j) corresponds to a link in V(j). We now want to show that
?W(i+ 1)??W(i) for each i?a78.
From Theorem 3.6, if D?Ci+1 and C?Ci such that D?Cnegationslash=?, then if KD is the link in
Ki+1 corresponding to D and KC is the link in Ki corresponding to C, then fi(KD)?KC.
By the previous argument, each link of V(i + 1) is only a subset of a link in V(i), thus a
link inCi+1 can only intersect a link (or links) in W(i); hence,?W(i+1)??W(i) for each
i?a78.
From the above argument,{?W(i) : i?a78}is a decreasing collection of nonempty sets
and??i=1(?W(i))negationslash=?and we can choose g???i=1(?W(i)). Note that W(i) is a segment in
46
Ci and diam(?W(i))? 3n(i); this means
limi??diam(?W(i)) = 0
and g is the only element in the intersection.
The current claim is that h(g) ={aj}?j=1. We know that g??W(i), however C(g,i)
may not be a link in W(i); if i?a78, let Wprime(i) ={C(g,i)}?W(i), and let V prime(i) ={K(g,i)}?
V(i). Because each link in Wprime(i) corresponds to a link in V prime(i) and Wprime(i) is segment in Ci
with length at most four, it follows that V prime(i) is a segment of rusty links in Ki with length
at most four. Furthermore, fi(?V(i+1))?V(i) and fi(V(g,i+1))?V(g,i), which means
that fi(?V prime(i+ 1))?V prime(i).
We now have that for each i ?a78, hi(x) ??V prime(i). By Corollary 3.5, for each j ?a78
(with j > i),
diam(fji (?V prime(j)))?4?(12)j? 1n(i),
thus, {fji (?V prime(j)) : j ?a78, j > i} is a decreasing sequence of nonempty closed connected
sets whose diameters converge to 0 and?j>ifji (?V prime(j)) contains exactly one point. By our
construction hi(g) must be in this intersection because K(g,j)??V prime(j) for each j > i, and
ai is in this intersection because K(a,j)?V(j)?V prime(j) for each j > i; thus hi(g) = ai.
Corollary 4.19. h : M ?lim??{Xi,fi}?i=1, defined previously is a homeomorphism from M
onto lim??{Xi,fi}?i=1.
47
Chapter 5
An Hereditarily Indecomposable Continuum
Definition 5.1. A continuum X is said to be hereditarily indecomposable, if each subcon-
tinuum of X is indecomposable.
In this section an hereditarily indecomposable continuum will be constructed using the
notions of chainability.
Lemma 5.2. If C is a chain that covers the continuum K, and Cprime ?C containing exactly
those links in C that intersect K, then Cprime is a segment in C.
Proof. Let K,C andCprime, be as described. Let m denote the minimum index for a link inCprime
and let M denote the maximum index for a link in Cprime. If Cprime is not a segment, then there
is j ?a78, (m < j < M) such that Cj /?C\Cprime; it would then follow, that if U = ?j?1i=mCi
and V = ?Mi=j+1Ci, then U and V are disjoint open sets, such that each intersect K and
K??Cprime?U?V. Hence, K could not be connected.
Lemma 5.3. If M is chainable, then each subcontinuum of M is chainable.
The above lemma is given without proof, however, the following reasoning is provided.
Suppose {Cn}?n=1 is a sequence of chains covering M, as described in 3.8, and K is a
subcontinuum of M. If ?Cn is described as the collection of links in Cn that intersect K,
then by 5.2, ?Cn is a segment from Cn, and can therefore be thought of as a chain as well.
The sequence {?Cn}?n=1, will have all the properties necessary in 3.6 to ensure that K is
chainable.
Definition 5.4. SupposeC is a spaced chain with|C|?6 andD is a chain that refinesC.
To say that D is doubly coiled in the interior of C means that if Cg and Ch are links of
48
C, 3?g,h?|Cg|?2, and|g?h|?2, then there are links Di,Dj, and Dk inD such that
i < j < k, (Di?Dk)?Cg and Dj ?Ch.
Lemma 5.5. Suppose C is a spaced chain with m =|C|?6 and D is chain refining C. D
is doubly coiled in the interior of C if and only if there are links Dt,Du,Dv, and Dw in D
such that t < u < v < w and one of the following holds:
(a) Dt?Dv ?C3 and Du?Dw ?Cm?2, where m =|C|, or
(b) Dt?Dv ?Cm?2 and Du?Dw ?C3.
Proof. Suppose that C is a spaced chain (with |C|? 6), D is a chain that refines C; let
m =|C|.
(?) IfD is doubly coiled in the interior ofC, then because C3 and Cm?2 (the second and
second to last links ofC respectively) are interior links ofC, and|(m?2)?2|= m?4?2,
there are links Di,Dj, and Dk inD such that i < j < k, Di?Dk ?C3, and Dj ?Cm?2.
Similarly there are links Diprime,Djprime, and Dkprime in D such that iprime < jprime < kprime, Diprime ?Dkprime ?Cm?2
and Djprime ?C3. Let t = min(i,iprime).
If t = i, then let u = min(iprime,j), let v = min(jprime,k) and let w = max(kprime,j). i < j < k
and i < iprime < jprime < kprime, so it follows that t = i < u < v. If v = jprime, then v < kprime ? w, and
similarly if v = k, then v?jprime < kprime?w; hence, t < u < v < w. Because Diprime ?Dj ?Cm?1,
Djprime ?Dk ?C3 and Dkprime ?Dj ?Cm?2, it follows that Dt?Dv ?C3 and Du?Dw ?Cm?2.
If t = iprime, a similar argument may be used by letting u = min(i,jprime), v = min(j,kprime),
and w = max(k,jprime), and showing that t < u < v < w and that Dt?Dv ? Cm?2 and
Du?Dw ?C3.
(?) Suppose that there are links Dt,Du,Dv and Dw in D such that t < u < v < w,
Dt?Dv ?C3 and Du?Dw ?Cm?2 (as in part (a) of the theorem). Suppose Cg and Ch are
links inC such that 2?g,h?m?2 and|g?h|?2. Without loss of generality, suppose
that g < h; notice that this means 3 ? g < h ? m?2. By Theorem 2.10, because C is
spaced, 3?g?m?2, t < u, Dt?C3, and Du?Cm?2, there is a link Di inD such that
49
Di?Cg and t?i?u. Because|g?h|?2,|g?(m?2)|?2; thus, Cg and Cm?2 are not
adjacent, meaning inegationslash= u and therefore i < u.
A similar argument may be used to show that, there is Dj and Dk in D such that
i?j ?u, u?k?v, Dj ?Ch, and Dk ?Cg. It is now shown that there is Di,Dj, and
Dk in D such that Di?Dk ?Cg and Dj ?Ch. To conclude the argument for this case,
now pick Dl such that v?l?w and Dl ?Ch; it follows that Dj,Dk, and Dl are links in
D such that j < k < l, Dj?Dl?Ch and Dk ?Cg.
In the case that there are links Dt,Du,Dv, and Dw in D such that t < u < v < w,
Dt?Dv ?Cm?1 and Du?Dw ?C2 (as in case (b) of the theorem) an argument similar to
the previous one will show thatD is doubly coiled in the interior ofC.
Theorem 5.6. Suppose {Cn}?n=1 is a sequence of chains such that if n?a78, then
(i) Cn+1 properly refines Cn,
(ii) Cn is a 1n?chain, and
(iii) Cn+1 is doubly coiled in the interior of Cn.
If M =??n=1(?Cn), then M is indecomposable.
Proof. Suppose K is a proper subcontinuum of M. Let ?Ci ?Ci containing exactly those
links in Ci that intersect K. It will be shown that no epsilon1?ball centered at a point in K is
contained in K; hence, K has no interior.
Let p ? K and epsilon1 > 0. If n ?a78 such that 1n < epsilon13, then if Cna, Cnb , and Cnc are three
consecutive links inCn and one contains p, then Cna ?Cnb ?Cnc ?B(p,epsilon1) (the open ball of
radius epsilon1 centered at p). Let q ? M \K and choose N to be a positive integer such that
1
N < min(
epsilon1
3,
d(q,K)
4 ). Because mesh(CN) <
1
N <
d(p,q)
7 , the union of six adjacent links in
CN+1 cannot contain both p and q, and a link inCN containing q will not intersect K. Let
w be the index of a link inCN that contains p and let z be the index of a link inCN that
50
contains q. It is possible that w or z is not in {3,...,|CN|?2}, so choose x and y to be
indices of links inCN such that|w?x|?2,|z?y|?2, and w,z?{3,...,(|CN|?2)}.
Because no six adjacent links inCN cover both p and q,|w?z|?6. Since|x?w|?2
and |y?z|? 2, |x?y| > |w?z|?|x?w|?|y?z|? 6?4 = 2 and so by the initial
assumptions r,s,t?a78 (with r < s < t) may be chosen so that each is an index of a link in
CN+1, CN+1r ?Cn+1t ?CNx , and CN+1s ?CNy . Recall that the union of four adjacent links
in CN cannot contain q and cover K; because |y?z|? 2 and Cy contains q, CNy cannot
intersect K, and because CN+1s ?CNy , CN+1s ?K =?as well. It follows that CN+1s /?CprimeN+1
(whereCprimeN+1 is the segment fromCN+1 containing exactly those links intersecting K); this
means that CN+1r or CN+1t is not an element of CprimeN+1 since CprimeN+1 is a segment. Because
|w?x|?2 and the diameters of each link inCN is less than epsilon13, the diameter of CNw ?CNx
is less than epsilon1. p?CNw so CNw ?CNx ?B(p,epsilon1). Both CN+1r and CN+1t are disjoint subsets
of CNx , and therefore each is a subset of B(p,epsilon1). Because CN+1r or CN+1t is not inCprimeN+1, it
follows that B(p,epsilon1) contains an open set that is not a subset of K.
Hence, a proper subcontinuum of M must be nowhere dense in M, meaning M is
indecomposable.
Corollary 5.7. Suppose {Cn}?n=1 is a sequence of spaced chains such that for each n?a78,
(i) Cn is a 1n?chain,
(ii) Cn+1 properly refines Cn, and
(iii) if m = |Cn|, then there are integers t,u,v,w such that 1 < t < u < v < w <|Cn+1|
such that Cn+1t ?Cn+1v ?Cn2 and Cn+1u ?Cn+1w ?Cnm?1, or Cn+1t ?Cn+1v ?Cnm?1 and
Cn+1u ?Cn+1w ?Cn2 ;
then M =??i=1(?Cn) is an indecomposable continuum.
Proof. By Lemma 4.6, property (iii) in the Corollary is equivalent to property (3) in The-
orem 4.7, thus M is indecomposable.
51
Definition 5.8. SupposeC is a chain with length greater than five, andD is a chain that
refinesC. To say thatDis very crooked inC means that if Cr,Cs?C (with|r?s|?5),
and t and w are each indices from D, such that Dt ? Cr and Dw ? Cs, then there are
indices u and v such that t < u < v < w, and
1. if r < s, then Du?Cs?1 and Dv ?Cr+1;
2. if r > s, then Du?Cs+1 and Dv ?Cr?1.
Theorem 5.9. Suppose that{Cn}?n=1, is a sequence of chains with respective lengths{ln}?n=1,
such that l1 =|C1|?6 and for each n?a78,
(a) Cn is a 1n?chain,
(b) Cn+1 properly refines Cn,
(c) Cn+1 is very crooked in Cn, and
(d) Cn+11 ?Cn1 and Cn+1ln+1 ?Cnln.
If M =??i=1(?Ci) and K is a subcontinuum of M, then K is indecomposable.
Proof. Let M = ??i=1(?Ci) and suppose K is a proper subcontinuum of M. If K is a
singleton, then there do not exist two nonempty proper subcontinuums of K; hence K is
indecomposable. Suppose then, that K is not a singleton.
For each i?a78, let ?Ci denote the collection of links in Ci that intersect K. It follows
from 5.2 that ?Ci is a segment inCi.
Let Ki denote the chain formed by reenumerating the links in ?Ci. It follows that
??i=1(?Ki) = K, for if x is in the intersection, then d(x,K) < 1i for each i?a78, thus x is a
limit point of K?x?K. Kn is spaced becauseCn is spaced. Lastly,Kn+1 is a refinement
of Kn, for if Kn+1i is a link in Kn+1 and Cn+1iprime is the link in Cn+1 corresponding to Kn+1i ,
then there is a link Cnjprime ?Cn that contains Cn+1iprime , but this means that Cnjprime intersects K, and
so there is a link Knj ?Kn that contains Cn+1iprime = Kn+1i , therefore,Kn+1 is a refinement of
Kn.
52
Let N ?a78 such that 1N ? diam(K)6 ; thus, for each integer n?N, mesh(Kn)? diam(K)6 ,
meaning|Kn|?6 in order forKn to cover K.
Now suppose n is an integer and n ? N, and let m = |Kn|. Kn2 and Knm?1 correspond
to links in in ?Cn (and thusCn as well), call these corresponding links Cngprime and Cnhprime, respec-
tively. It follows that Kn1 = Cngprime?1 and Knm = Cnhprime+1, since Kn corresponds to a segment
from Cn; further, it is established that gprime < hprime. Because Kn is a spaced chain that covers
the continuum K and because Kn2 and Knm?1 are interior links of Kn, points p and q in
K can be chosen, such that Kn2 is the only link inKn that covers p and Knm?1 is the only
link inKn that covers q. p and q are each covered by links inKn+1, so letKn+1s and Kn+1x
be links in Kn+1 containing p and q respectively. Since p?Kn+1s , Kn+1s contains a point
contained in exactly one link inKn (ie Kn2 ); for this reason and the fact thatKn+1 refines
Kn, it follows that Kn+1s ?Kn2 . By a similar argument, Kn+1x ?Knm?1 because q?Kn+1x .
Letting Cn+1sprime and Cn+1xprime each be the links in Cn+1 corresponding to Kn+1s and Kn+1x , it
follows that Cn+1sprime ?Cng and Cn+1xprime ?Cnh.
Case 1: Suppose sprime < xprime...., then by the initial assumptions, there is Cn+1uprime and Cn+1vprime inCn+1
such that sprime < uprime < vprime < xprime, Cn+1uprime ?Cnhprime?1, and Cn+1vprime ?Cngprime+1 (remember that gprime < hprime). Let
Kn+1u and Kn+1v be links inKn+1 corresponding to Cn+1uprime and Cn+1vprime . Thus, s < u < v < x,
Kn+1u ? Knm?2 and Kn+1v ? Kn3 . Since s < u, Kn+1s ? Kn2 and Kn+1u ? Knm?2, there is
Kn+1t such that s < t < u, Kn+1t ?Kn3 ; similarly, there is Kn+1w such that v < w < x and
Kn+1w ?Knm?2. It follows that Kn+1t ,Kn+1u ,Kn+1v and Kn+1w are links in Kn+1 such that
t < u < v < w, Kn+1t ?Kn+1v ?Kn3 and Kn+1u ?Kn+1w ?Knm?2.
Case 2: Ifxprime < sprime, an argument similar to that of Case 1, will choose linksKn+1w , Kn+1v , Kn+1u
and Kn+1t inKn+1 such that w < v < u < t, Kn+1w ?Kn+1u ?Kn2 and Kn+1v ?Kn+1t ?Knm?2.
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Because the above argument holds for each integer n such that n ? N, it follows that
{Kn}?n=N is a collection of spaced chains satisfying conditions (i), (ii), and (iii) in Corollary
4.8; thus, M =??n=N(?Kn) is indecomposable.
From the above theorem, if a chainable continuum is formed from a sequence of chains
as described in the theorem, then such a continuum is hereditarily indecomposable. The
final step is to show that such a continuum exists.
Some modifications to previous terms will come in handy.
Definition 5.10. SupposeCis a chain in a822. To say thatCis a rectangular chain, means
that if C ?C, then there are real numbers a,b,c, and d such that C = (a,b)?(c,d). To
say that the rectangular chain C is a straight rectangular chain means that there are
numbers c,d and if Ci?C and Ci = (ai,bi)?(ci,di), then ci = c and di = d.
Definition 5.11. SupposeC is a chain andDis a chain that refinesC. To say thatDis a
snug refinement ofC means that each link ofD is a subset of exactly one link inC.
Definition 5.12. If C is a chain and D is a chain that is anchored in C, then to say that
Dis securely anchored in C, means that the first link ofDis only contained in the first
link ofC, the last link ofDis only contained in the last link inC, and an interior link ofD,
is only contained in an interior link inC.
Definition 5.13. SupposeCis a spaced rectangular chain. Cis straight, means that there
is c,d?a82 (with c < d) so that if Ci?C and Ci = (ai,bi)?(ci,di), then ci = c and di = d.
Theorem 5.14. IfCis a spaced rectangular chain that is straight, then there is a rectangular
chain D, such that D is securely anchored in C, and D is very crooked in C.
Proof. First note, that for a very crooked chain inC to have any novel qualities, that there
must be two links inC whose indices differ by five. If|C|?5, then the fact thatC is very
crooked inC is vacuously true. Thus, if n?5 andC is a spaced rectangular chain that is
straight, then C is refined by a rectangular chain that is securely anchored in C and very
crooked inC.
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The remainder of the proof will be done inductively. Suppose that n?a78 (with n > 5)
and for each m ? a78, m < n, it is known that if Cprime is a spaced rectangular chain that is
straight and |Cprime| = m, then there is a rectangular chain that refines Cprime that is securley
anchored inCprime and very crooked inCprime.
It will know be shown that if C is a spaced, rectangular chain that is straight and
|C|= n, then there is a rectangular chain that is securely anchored inC and very crooked
inC).
With a brief slight of hand, the author now focuses the audience?s attention to the
specific case of the chainC={C1,C2,...,Cn}, where
Ci = (i?23,i+ 23)?(0,3).
It is hoped that the reader will accept a validation of this specific case to cary over to all
other spaced rectangular chains that are straight.
Define the collectionsCa,Cb, andCc as follows:
(a) Ca ={Cai = Ci?(a82?(2,3)) : 1?i?n?1}
(b) Cb ={Cbi = Ci+1?(a82?(1,2)) : 1?i?n?2}
(c) Cc ={Cci = Ci+1?(a82?(0,1)) : 1?i?n?1}.
It will be taken for granted that each ofCa,Cb, andCc forms a chain that is spaced, rectan-
gular, straight and with length less than n. By the induction hypothesis,Da,Db, andDc,
may be chosen to be rectangular chains that are very crooked and securely anchored in the
respective chainsCa,Cb, andCc.
Let x =|Da|,y =|Db|, and z =|Dc|, and construct the chainD as follows:
(a) if 1?i?x, let Di =Dai ;
(b) let Dx+1 = (n?1?13,n?1 + 13)?(1,3);
(c) if x+ 2?i?x+y + 1, let Di =?Dbi?(x+1);
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(d) let Dx+y+2 = (2?13,2 + 13)?(0,2);
(e) if x+y + 3?i?x+y +z + 2, let Di = Dbi?(x+y+2).
Notice that Dx+1 ?Cn?2 negationslash=?, and that Dx+y+2?C3 negationslash=?; thus Dx+1 and Dx+y+2 cannot
be subsets of C1 or C2. BecauseDa andDb are securely anchored inCa andCb, respectively,
and because Dx+1 only intersects Cn??(2,3)?a82, Dx+1 only intersects the last links ofDa
andDb. Similarly, Dx+y+2 only intersects the first ofDb and the first link ofDc.
D is securely anchored in C. To show this, first note that Dx+1 and Dx+y+2 only
intersect Cn?2 and C3, respectively, so neither can be a subset of D1 or Dx+y+z+2. Da is
the only defining chain for D that intersects C1, thus a link of D that is contained in C1
must be from Da. Da is securely anchored in Ca, so Da1 is the only link contained in Ca1.
By the construction of Ca1, if a link inDa does not lie inside Ca1, then it will not lie inside
C1; hence D1 = Da1 ?Ca1 ?C1 and D1 is the only link inD that is a subset of C1.
In a similar fashion it can be shown that Dx+y+z+2 ? Cn and Dx+y+z+2 is the only
link ofD that is a subset of Cn.
The final step is to prove that D is very crooked. Suppose Cr.Cs ? a78 such that
|r?s|?5, and that Dt,Dw ?Dsuch that Dt?Cr and Dw ?Cs. It will be shown that if
r < s), then there is Du,Dv ?D, such that t < u < v < w and Du?Cs?1 and Dv ?Cr+1;
the case when Dt?Cs and Dw ?Cr can be proven in a similar manner.
For the moment, suppose that t /?{x+ 1,x+y + 2}and let q?{a,b,c}such that Dt
is chosen from the defining chainDq.
If Dw is also defined from a link inDq, then it follows that appropriate links Du and
Dv exist, sinceDq is very crooked inCq, andCq refinesC.
If Cs contains a link in Dq, then Dwprime may be chosen to be a link defined from a link
in Dq such that Dwprime ?Cs; furthermore, because the last link of Dq is contained in a link
in C with index greater than or equal to s, it may be assumed that t < wprime. From the
argument in the previous paragraph, links Du and Dv may be chosen, so that Du ?Cs?1
and Dv ?Cr+1.
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Lastly, if Cs does not contain a link inDq, then it follows that s = n and qnegationslash= c. Only
the last link of D is contained in Cn, thus w = x + y + z + 2. If t = 1, then r = 1, and
Dx+1 and Dx+y+2 are links inD such that t < x + 1 < x + y + 2 < w, Dx+1 ?Cn?1 and
Dx+y+2 ?C2. If t > 1, then r > 1 and Cr contains a link inDc. Let Dczprime be the first link
ofDc that is contained in Cr, and define tprime as tprime = x+y + 2 +zprime. Because Dtprime and Dw are
both defined from links inDc andDc is very crooked inC, there are links Du and Dv such
that Du?Cn?1 and Dv ?Dr+1.
Earlier, t was excused from being equal to x + 1 or x + y + 2; these cases shall now
be unexcused. t negationslash= x + 1, since this would mean r ? n?3 and thus s would have to be
greater than r + 5 = n+ 2, meaning s > n. If t = x+y + 2, then let tprime = x+y + 3. Dtprime is
defined by the first link inDc, which is very crooked inCc; thus, there is Du and Dv such
that tprime < u < v < w, Du ?Cn?1 and Dv ?C3. Since t < tprime, and Dt only intersects C2, it
follows that Du and Dv are appropriate choices for t as well.
It is now concluded that if n?a78 andC is a spaced rectangular chain of length n that
is straight, then there is a rectangular chain that is very crooked inC.
The following theorem is not so much a corollary, as it is a theorem that would have
been preferable to prove using a technique similar to the previous proof. A sketch of an
argument will be given, but a solid proof requires further development of the properties of
a822.
Corollary 5.15. If C is a chain of convex open subsets of a822, then there is D, a chain of
convex open subsets of a822, such that D is anchored in C and D is very crooked in C.
Sketch: Because each link in C is convex ?C is path connected and there is an arc con-
tained in ?C that begins in the first link in C and ends in the last link in C. This arc is
?thickened? so that it remains inside ofC. Let Ai denote the intersection of this thickened
arc with Ci ?C, and let A={Ai : Ci ?C}. A is a chain and there is a homeomorphism
h :?R|A| ??A, where R|A| is a spaced rectangular chain of length |A|. From the prior
theorem, there is n?a78 such that R|A| can be refined by a rectangular chain of length n
57
that is very crooked in R|D|; denote such a chain as Rn. For each Rni ?Rn, define Bi as
Bi = h?1(Ri) and letB={Bi : Rni ?Rn}. B is a chain that refinesC and is very crooked
inC. Although the links inB may not be convex, ?B is path connected and there is a an
arc contained in ?B that begins in the first link of B and ends in the last link of B. This
arc may be covered byDa chain of convex open subsets of a822 such thatDrefinesB; hence,
D is very crooked inC.
With the above ?corollary? in mind, the following sequence of chains may be defined.
Let C1 be a chain of length six, whose links are open balls with radius 12. For each
i?a78, ifCi is defined as a chain whose links are open convex subsets of a822, letDi be a chain
properly refinesCi such that each link inDi is convex and open, and mesh(Di) < 1i+1. Let
Ci+1 be a chain of convex open subsets of a822, such thatCi+1 is very crooked inDi.
For each i?a78,
1. Di properly refinesCi andCi+1 refinesDi, thereforeCi+1 properly refinesCi;
2. mesh(C1) = 1 and mesh(Ci+1)?mesh(Di) < 1i+1;
3. Di refinesCi andCi+1 is very crooked inDi, thereforeCi+1 is very crooked inCi.
Thus, by 5.9,??i=1(?Ci) is an hereditarily indecomposable continuum.
58
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[1] R.L. Moore, ?Foundations of Point Set Theory,? Amer. Math. Soc. Colloq. Publ., vol.
13, Amer. Math Soc., Providence, RI, 1962.
[2] K. Kuratowski, Topology, vol. II, Academic Press, New York, 1968.
[3] James R. Munkres, Topology: a first course, Prentice-Hall, Englewood Cliffs, N.J.,
1975.
[4] Michel Smith, Notes on Topology, http://www.auburn.edu/?smith01/math7500/ and
http://www.auburn.edu/?smith01/math7510/.
[5] Gary Gruenhage, Topology Notes, http://www.auburn.edu/?gruengf/fall08.html.
[6] Scott Varagona, Inverse Limit Spaces, http://hdl.handle.net/10415/1486.
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