c 2012, nadeeka de silva - tdl
TRANSCRIPT
Span of Subcontinua
by
Nadeeka de Silva, M.Sc.
A Dissertation
In
Mathematics and Statistics
Submitted to the Graduate Facultyof Texas Tech University in
Partial Fulfillment ofthe Requirements for the Degree of
Doctor of Philosophy
Approved
Dr. Wayne Lewis
Dr. Razvan Gelca
Dr. Magdalena Toda
Dr. Robert Byerly
Dr. Peggy Gordon MillerDean of the Graduate School
August 2012
c©2012, Nadeeka de Silva
Texas Tech University, Nadeeka de Silva, August 2012
ACKNOWLEDGEMENTS
First I would like to express my special appreciation and thanks to my adviser Dr.
Wayne Lewis for his guidance in Mathematics and other aspects, for being patient with
me and being there whenever I needed advice or help. I would also like to thank Dr.
Magdalena Toda for being in the committee and for the help and encouragement.
Without their help this thesis would not have been possible.
I would like to thank other committee members Dr. Robert Byerly and Dr. Razvan
Gelca for their advice and help. I also take this opportunity to thank all of my teachers
for their aid in my development as a mathematician. Thanks also go out to my friends
Matthew Buyum, Mervyn Ekanayake and Indika Wijayasinghe for their help in proof
reading.
Finally I would like to thank my family for being encouraging to my dreams and for
being there for me.
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TABLE OF CONTENTS
ACKNOWLEDGEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi
1. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Continua . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.2 Span of Continua . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Span, semispan, surjective span and surjective semispan of a con-
tinuum consisting of a ray limiting to a continuum . . . . . . . 5
1.2.2 Span of Subcontinua . . . . . . . . . . . . . . . . . . . . . . . . 6
2. SPAN, SEMISPAN, SURJECTIVE SPAN AND SURJECTIVE SEMISPAN
OF A CONTINUUM CONSISTING OF A RAY LIMITING TO CON-
TINUUM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.1 Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Other versions of span . . . . . . . . . . . . . . . . . . . . . . . . . 10
3. SPAN OF SUBCONTINUA . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.1 Construction of a continuum Y such that the set given by the span
of subcontinua of Y is equal to a given special closed set . . . . . . . 12
3.1.1 Construction of a continuum YF such that the set given by the
span of subcontinua of YF is equal to a given finite set F . . . . 12
3.1.2 Construction of a continuum YT such that the set given by the
span of subcontinua of YT is equal to a set which is the union of
{0} and a given closed interval that does not contain 0 . . . . . 16
3.1.3 Construction of a continuum YK such that the set given by the
span of subcontinua of YK is equal to the Cantor set . . . . . . . 19
3.2 Construction of a continuum YM such that the set given by the span
of subcontinua of YM is equal to a given closed set . . . . . . . . . . 23
4. CONCLUSION AND FURTHER QUESTIONS . . . . . . . . . . . . . . . . . 26
4.1 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.1.1 Ray limiting to a continuum . . . . . . . . . . . . . . . . . . . . 26
4.1.2 Span of subcontinua . . . . . . . . . . . . . . . . . . . . . . . . 26
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4.2 Further Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
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ABSTRACT
Let Y be continuum consisting of a ray limiting to continuum X. We prove that
σ(Y ) ≤ max{σ(X), σ∗0(X)}. When σ(X) = 0 or when X is a simple closed curve, we
have that σ(Y ) = σ(X). Using this, we construct for each closed subset G of [0, 1] with
0 ∈ G a one-dimensional continuum YG such that the set of values of span of subcontinua
of YG is the set G. Some other results related to this are also presented.
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LIST OF FIGURES
3.1 Continuum YF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.2 Continuum YT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.3 Continuum YK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
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CHAPTER 1
INTRODUCTION
1.1 Preliminaries
1.1.1 Continua
Below are listed some basic definitions in continuum theory.
Definition 1 (Continuum). A continuum is a nonempty, compact, connected metric
space. A subcontinuum is a continuum which is a subset of a continuum.
Definition 2 (Arc). An arc is any continuum which is homeomorphic to the closed
interval [0, 1].
Definition 3 (Simple closed curve). A simple closed curve is any continuum which is
homeomorphic to a circle.
Definition 4 (Triod). A continuum X is a triod provided that there is a subcontinuum
Z of X such that X − Z has at least 3 components.
Definition 5 (Simple triod). A simple triod is a continuum which is the union of three
arcs which have a common endpoint and are otherwise pairwise disjoint. The three arcs
are called the legs of the triod and the common point is called the branch point of the
triod.
Definition 6 (Ray). A ray is a space homeomorphic to (0, 1].
Definition 7 (Atriodic). A continuum is atriodic if it contains no triod.
Definition 8 (Coincidence Point). If f and g are continuous functions from X to Y ,
then f and g have a coincidence point if there is a p ∈ X with f(p) = g(p).
In addition, below are listed some known theorems. They will be employed later.
Theorem 1.1.1. (Coincidence Point Theorem) Let X be a continuum and f and g be
continuous functions from X into [0, 1]. If f(X) ⊆ g(X) then f and g have a coincidence
point.
Theorem 1.1.2. (Boundary Bumper Theorem) If U is a nonempty proper open subset of
the compact and connected Hausdorff space X, then every component of U has a limit
point on the boundary of U .
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The following definition is crucial to this thesis and will be used frequently.
Definition 9 (ray limiting to X). Let X be a continuum and R be a ray. R is a ray
limiting to X if
1. X ∩R = ∅,
2. X ∪R = R and
3. R is compact.
1.1.2 Span of Continua
The notion of span of a metric space was introduced by A Lelek [4]. He proved that
chainable continua have span zero, and in 1971 he asked whether the converse holds. It
became a topic of interest in continuum theory, in part because there are few means
presently available to decide whether a given continuum is chainable. It remained an
open question for a long time and there were many attempts to solve it in both
directions. Many positive partial results were obtained [2]. In 2010 L.C. Hoehn gave an
example showing that in general span zero does not imply chainability [2]. Though this
is a disappointing end to the attempt of characterizing chainability using span, there are
many other reasons to be interested in span. A number of properties of chainable
continua have been established for span zero continua. After Hoehn’s example there is
more inspiration to continue studies in this direction. Also, there are many other
interesting problems dealing with span.
The interactions between span and other concepts of metric topology, particularly in
the theory of continua, have been investigated. Relationships between the span and the
dimension of the space, and between span and essential maps, have been investigated [6].
There are various modified versions of span. Some open problems related to span can be
solved for these versions of span. Studying the relationships between different versions of
span is also a topic of interest. We are interested in studying the set given by the spans
of subcontinua of a continuum.
Definition 10 (S(A)). Let X be a metric space with distance d. Let A be a nonempty
subset of X ×X. S(A) is defined by:
S(A) = inf{d(x, y) : (x, y) ∈ A}.
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Definition 11 (Span). Let X be a continuum with distance d. For each nonempty subset
A of X ×X, let S(A) be as in the above definition. The span of continuum X is
defined by:
span(X) = σ(X) = sup{S(A) : A ⊂ X ×X, A is connected and π1(A) = π2(A)}
where π1, π2 are the projection maps.
The following facts are straightforward from the definition.
• If H ⊂ X then σ(H) ≤ σ(X).
• The set A in this definition can be considered to be closed.
• The span of a continuum depends on the metric, but having span zero is a
topological property.
The span of a continuum X can be visualized as the maximal number d such that two
persons can pass over the same part of X always staying at least a distance d apart.
In [5], Lelek defined three other versions of span called surjective span, semispan, and
surjective semispan. Their defintions are listed below. Note that in each case S(A) is as
in Definition 10.
Definition 12 (Semispan). Semispan of a continuum X is denoted by σ0(X) and
defined by:
σ0(X) = sup{S(A) : A ⊂ X ×X,A is connected and π1(A) ⊂ π2(A)}.
Definition 13 (Surjective span). Surjective span of a continuum X is denoted by
σ∗(X) and defined by:
σ∗(X) = sup{S(A) : A ⊂ X ×X,A is connected and π1(A) = π2(A) = X}.
Definition 14 (Surjective semispan). Surjective semispan of a continuum X is
denoted by σ∗0(X) and defined by:
σ∗0(X) = sup{S(A) : A ⊂ X ×X,A is connected and π1(A) ⊂ π2(A) = X}.
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It follows directly from the definitions that the following inequalities hold:
0 ≤ σ∗(X) ≤ σ(X) ≤ σ0(X) ≤ diam(X) (1.1)
0 ≤ σ∗(X) ≤ σ∗0(X) ≤ σ0(X) ≤ diam(X) (1.2)
σ(H) ≤ σ(X), σ0(H) ≤ σ0(X) for H ⊂ X. (1.3)
For each arc, as well as for each arc-like continuum, all of these four quantities are
equal to zero (cf. [5], Propositions 1.3 and 2.1). Nevertheless, they are quite useful in the
theory of tree-like continua. There is an easy example of a continuum X such that
σ(X) = σ0(X) = 1 and σ∗(X) = σ∗0(X) =
1
2(see [5], Example 1.4). Its existence shows,
among other things, that the analogues of inequalities (1.3) for surjective span and
surjective semispan instead of span and semispan, respectively, do not necessarily hold.
Theorem 1.1.3. (J. F. Davis [1]). For each continuum X, σ0(X) = 0 if and only if
σ(X) = 0.
Thus for each continuum X, σ(X) = 0 implies that all versions for span of X are zero.
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1.2 Results
1.2.1 Span, semispan, surjective span and surjective semispan of a continuum consisting
of a ray limiting to a continuum
For arcs and arc-like continua, all versions of span are zero. Consider a continuum Y
consisting of a ray limiting to continuum X. Does σ(X) = σ(Y )? We obtain a partial
result in answering this question.
Theorem 1.2.1. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
Then σ(Y ) ≤ max(σ(X), σ∗0(X)).
Corollary 1.2.2. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
If σ(X) = σ∗0(X) then σ(Y ) = σ(X).
Corollary 1.2.3. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
If σ(X) = 0 then σ(Y ) = σ(X).
Tina Sovic has recently given an independent proof of the result in Corollary 1.2.3.
Corollary 1.2.4. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
If X is a simple closed curve then σ(Y ) = σ(X).
It is also worth mentioning the following observation.
Theorem 1.2.5. Let Y be a continuum consisting of a ray limiting to continuum X. Let
R be a ray limiting to Y . Let Z = Y ∪R. Then σ(Z) ≤ max(σ(X), σ∗0(X)).
We also explore how the result in Theorem 1.2.1 can be applied to other versions of
span.
Theorem 1.2.6. Let X be a continuum and R be a ray limiting to it. Let Y = X ∪R.
Then σ0(Y ) = σ0(X).
Theorem 1.2.7. Let X be a continuum and R be a ray limiting to it. Let Y = X ∪R.
Then σ∗(Y ) ≤ σ∗0(X).
Theorem 1.2.8. Let X be a continuum and R be a ray limiting to it. Let Y = X ∪R.
Then σ∗0(Y ) ≤ σ∗
0(X).
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1.2.2 Span of Subcontinua
For a given continuum X our focus is directed on the set of values of span of
subcontinua of X.
Definition 15 (B(X)). Let X be a continuum. B(X) is defined by:
B(X) = {σ(H) : H is a subcontinuum of X } .
In most of the cases I will assume that the span of X is 1. The following facts are
straightforward.
• If X is a simple closed curve with σ(X) = 1, then B(X) = {0, 1}.
• If X is a simple triod with σ(X) = 1, then B(X) = [0, 1].
• 0 ∈ B(X).
Due to the above observations, we study the possibilities of B(X) between the two
extreme cases of {0, 1} and [0, 1]. This lead us to the following question.
Question 1. Given that M ⊂ [0, 1] with 0, 1 ∈ M , does there exist a continuum XM such
that B(XM) = M?
In answering this question we obtain the following results.
• For each closed subset M of [0, 1] with 0, 1 ∈ M , there exits a one-dimensional
continuum YM such that the set of values of span of subcontinua of YM is the set
M . There are uncountably many incomparable continua for each case.
• Except for the case where M has an infinite number of nondegenerate components,
the examples constructed are planar.
Question 2. What conditions on a set G ∈ [0, 1] will guarantee that there will be a
continuum XG such that B(XG) = G?
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CHAPTER 2
SPAN, SEMISPAN, SURJECTIVE SPAN AND SURJECTIVE SEMISPAN OF A
CONTINUUM CONSISTING OF A RAY LIMITING TO CONTINUUM
In this chapter we obtain bounds on the span, semispan, surjective span and surjective
semispan of continuum Y consisting of a ray limiting to continuum X.
2.1 Span
First we list the following propositions which will be used later.
Proposition 2.1.1. Let X be a continuum and R be a ray limiting to X. Let
Y = X ∪R. There exists a continuous surjection f : Y → [0, 1] such that
1. y ∈ X if and only if f(y) = 0 and
2. f is one-to-one on R.
Proposition 2.1.2. Let X be a continuum and R be a ray limiting to X. Let
Y = X ∪R and let A be a subcontinuum of Y × Y such that S(A) > 0 and
π1(A) ⊆ π2(A). Then A ∩ (X ×X) 6= ∅.
Proof. Note that πi(A) 6⊂ R, for i ∈ 1, 2. (Every subcontinuum of R has span zero.) Let
f be a continuous function as in Proposition 2.1.1. Define the functions gi : A → [0, 1] by
gi = f ◦ πi, for i = 1, 2. The functions g1 and g2 satisfy the hypotheses of the Coincidence
Point Theorem. Therefore there exists a point (x, y) in A such that g1(x, y) = g2(x, y),
which implies f(x) = f(y). Since f(x) = f(y), (x, y) ∈ (X ×X) ∪ (R×R) . If
(x, y) ∈ R×R, then, since f is one-to-one on R, x = y and S(A) = 0, which is a
contradiction. Therefore (x, y) ∈ X ×X and A ∩ (X ×X) 6= ∅.
Proposition 2.1.3. Let X be a continuum and R be a ray limiting to X. Let
Y = X ∪R and D be a subcontinuum of Y × Y . If πi(D) ∩X 6= ∅ and πi(D) ∩R 6= ∅where i ∈ {1, 2}, then X ⊂ πi(D).
Theorem 2.1.4. Let X be a continuum and R be a ray limiting to it. Let Y = X ∪R.
Then σ(Y ) ≤ max(σ(X), σ∗0(X)).
Proof. If σ(Y ) = 0, then since X ⊂ Y we have that σ(Y ) = σ(X) = 0. Therefore we can
assume that σ(Y ) > 0.
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From the definition of span there exists a subcontinuum A of Y × Y such that
S(A) = σ(Y ) and π1(A) = π2(A). From Proposition 2.1.2, A ∩ (X ×X) 6= ∅.Case 1: A ⊂ X ×X
In this case σ(Y ) = S(A) ≤ σ(X) ≤ σ(Y ), so σ(Y ) = σ(X).
Case 2 : A 6⊂ X ×X
Let C be any component of A ∩ (X ×X). For each ǫ > 0, let
N(C, ǫ) = {t ∈ A : d(t, C) < ǫ}. Let Cǫ be the component of N(C, ǫ) that contains C.
From the Boundary Bumper Theorem Cǫ has a limit point on the boundary of N(C, ǫ).
Therefore Cǫ is a connected subset of A that properly contains C. So Cǫ contains a point
in the complement of X ×X, i.e. π1(Cǫ) ∩R 6= ∅ or π2(Cǫ) ∩R 6= ∅. Consequently using
Proposition 2.1.3 it follows that, for each ǫ > 0, X ⊂ π1(Cǫ) or X ⊂ π2(Cǫ). Without loss
of generality, there exists a sequence {ǫn}∞n=1 converging to zero, such that, for all ǫn,
X ⊂ π1(Cǫn). Hence X ⊂ π1(C). Note that C is a connected subset of X ×X and that
C satisfies the condition π2(C) ⊂ π1(C) = X. Therefore in this case we conclude that
σ(Y ) ≤S(C)≤ σ∗0(X).
In general, we have that σ(Y ) ≤ σ∗0(X) or σ(Y ) ≤ σ(X) and hence
σ(Y ) ≤ max{σ∗0(X), σ(X)}.
In [3] L.C. Hoehn and A. Karasev gave an example of a continuum X such that
σ(X) < σ∗0(X). Thus it remains a open question whether σ(Y ) = σ(X) in general.
However equality can be obtained in the following special cases.
Corollary 2.1.5. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
If σ∗0(X) ≤ σ(X) then σ(Y ) = σ(X).
Proof. From the condition σ∗0(X) ≤ σ(X) and from Theorem 2.1.4 it follows that
σ(Y ) ≤ σ(X), but X ⊂ Y so σ(Y ) = σ(X).
Corollary 2.1.6. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
If X is a simple closed curve, then σ(Y ) = σ(X).
Proof. If X is a simple closed curve, then σ(X) = σ∗0(X). Hence from Corollary 2.1.5,
σ(Y ) = σ(X).
Corollary 2.1.7. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
If σ(X) = 0 then σ(Y ) = σ(X).
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Proof. If σ(X) = 0, then from Theorem 1.1.3 and inequalities (1.1) and (1.2), σ∗0(X) = 0.
Hence from Corollary 2.1.5 it follows that σ(Y ) = σ(X).
Tina Sovic has recently obtained independently and by a different argument the result
in Corollary 2.1.7.
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2.2 Other versions of span
Here, similar results to the one in Theorem 2.1.4 are obtained for the other versions of
span.
Theorem 2.2.1. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
Then σ0(Y ) = σ0(X).
Proof. If σ0(Y ) = 0, then since X ⊂ Y we have that σ0(Y ) = σ0(X) = 0. Therefore we
can assume that σ0(Y ) > 0. From the definition of semispan there exists a subcontinuum
A of Y × Y such that S(A) = σ0(Y ) and π1(A) ⊆ π2(A). From Proposition 2.1.2
A ∩ (X ×X) 6= ∅.Case 1: A ⊂ X ×X
In this case σ0(Y ) = S(A) ≤ σ0(X) ≤ σ0(Y ), so σ0(Y ) = σ0(X).
Case 2: A 6⊂ X ×X
In a similar way as in the proof of Theorem 2.1.4, we obtain that:
σ0(Y ) ≤ σ∗0(X). (2.1)
From inequalities (1.2) and (2.1) it follows that σ0(Y ) ≤ σ0(X). Since X ⊂ Y ,
σ0(X) ≤ σ0(Y ). Thus σ0(Y ) = σ0(X).
Theorem 2.2.2. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
Then σ∗0(Y ) ≤ σ∗
0(X).
Proof. If σ∗0(Y ) = 0, then σ∗
0(Y ) ≤ σ∗0(X). Therefore we can assume that σ0(Y ) > 0.
From the definition of surjective semispan there exists a subcontinuum A of Y × Y such
that S(A) = σ∗0(Y ) and π1(A) ⊆ π2(A). From Proposition 2.1.2, A ∩ (X ×X) 6= ∅.
Case 1: A ⊂ X ×X
In this case σ∗0(Y ) = S(A) ≤ σ∗
0(X).
Case 2: A 6⊂ X ×X
In a similar way as in the proof of Theorem 2.1.4, we obtain that σ∗0(Y ) ≤ σ∗
0(X).
Corollary 2.2.3. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
Then σ∗(Y ) ≤ σ∗0(X).
Proof. From Theorem 2.2.2 and inequality (1.2) it follows that σ∗(Y ) ≤ σ∗0(X).
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Corollary 2.2.4. Let X be a continuum and R be a ray limiting to X. Let Y = X ∪R.
If σ(X) = 0 then all versions of span values of Y are zero.
Proof. If σ(X) = 0, then from Theorem 1.1.3 and inequalities (1.1) and (1.2) all versions
of span value of X are zero. Hence from the above theorems it follows that all versions of
span value of Y are zero.
Theorem 2.2.5. Let Y be continuum consisting of a ray limiting to continuum X. Let
R be a ray limiting to Y . Let Z = Y ∪R. Then σ(Z) ≤ max(σ(X), σ∗0(X)).
Proof. By applying Theorem 2.1.4 to Z:
σ(Z) ≤ max(σ(Y ), σ∗0(Y )). (2.2)
By applying Theorem 2.1.4 to Y:
σ(Y ) ≤ max(σ(X), σ∗0(X)). (2.3)
By applying Theorem 2.2.2 to Y:
σ∗0(Y ) ≤ σ∗
0(X). (2.4)
From inequalities (2.2), (2.3) and (2.4), σ(Z) ≤ max(σ(X), σ∗0(X)).
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CHAPTER 3
SPAN OF SUBCONTINUA
In this chapter, we construct for each closed subset G of [0, 1] with 0 ∈ G a
one-dimensional continuum YG such that the B(YG) = G. Except for the case where G
has an infinite number of nondegenerate components, all of the examples are planar.
First, we will construct the planar examples.
3.1 Construction of a continuum Y such that the set given by the span of subcontinua
of Y is equal to a given special closed set
In this section we construct planar, one-dimensional, atriodic continua such that B(X)
is a:
1. finite set,
2. set which is the union of {0} and a closed interval that does not contain 0, or
3. Cantor set.
3.1.1 Construction of a continuum YF such that the set given by the span of
subcontinua of YF is equal to a given finite set F
Let F be any finite subset of [0, 1] that contains 0 and 1. In this section we construct a
continuum YF such that B(YF ) = F . In constructing this continuum we use an extended
version of Theorem 2.1.4. The following propositions will be helpful.
Proposition 3.1.1. Let Cn, n = 1, 2, 3, . . . ,m, be disjoint continua. Let
Rn, n = 1, 2, 3, . . . ,m− 1, be pairwise disjoint copies of (0, 1) such that the end of Rn
corresponding to 0 limits to Cn and the other end limits to Cn+1 and the Rn’s are disjoint
from the Cn’s. Let Yn = Rn = Cn ∪Rn ∪ Cn+1 and Y =m−1⋃
n=1
Yn. Then Y is a continuum.
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Proposition 3.1.2. Let Y be a continuum as in Proposition 3.1.1. There exists a
continuous surjection f : Y → [0, 1] such that
1. y ∈ Cn if and only if f(y) = rn, for n = 1, 2, 3, . . . m and
2. y ∈ Rn if and only if f(y) = hn(y), for n = 1, 2, 3, . . . ,m− 1,
where 0 = r1 < r2 < r3 < ... < rm = 1 and for each n = 1, 2, 3, 4, . . . m− 1,
hn : Rn → (rn, rn + 1) is a homeomorphism.
Proof. Consider the function defined by:
f(y) =
rn, if y ∈ Cn
hn(y), if y ∈ Rn.
Then f is a continuous function that satisfies the properties 1 and 2.
Proposition 3.1.3. Let Y be a continuum as in Proposition 3.1.1. If A is a
subcontinuum of Y × Y such that S(A) > 0 and π1(A) = π2(A), then
A ∩(
m⋃
n=1
(Cn × Cn)
)
6= ∅.
Proof. Note that πi(A) 6⊂m−1⋃
n=1
Rn. Let f be a continuous function as in Proposition 3.1.2.
Define the functions gi : A → [0, 1] by gi = f ◦ πi for i = 1, 2. The functions g1 and g2
satisfy the hypotheses of the Coincidence Point Theorem (Theorem 1.1.1). Therefore
there exists a point (x, y) in A such that g1(x, y) = g2(x, y), which implies f(x) = f(y).
Since f(x) = f(y), (x, y) ∈(
m⋃
n=1
(Cn × Cn)
)
∪(
m−1⋃
n=1
(Rn ×Rn)
)
. Assume that
(x, y) ∈m−1⋃
n=1
(Rn ×Rn). Then x = y and hence S(A) = 0, which is a contradiction.
Therefore (x, y) ∈m⋃
n=1
(Cn × Cn) and hence A ∩(
m⋃
n=1
(Cn × Cn)
)
6= ∅.
Proposition 3.1.4. Let Y be a continuum as in Proposition 3.1.1. Let D be a
subcontinuum of Y × Y . If there exists an n such that πi(D) ∩ Cn 6= ∅ and πi(D) 6⊂ Cn
for i = 1 or 2, then Cn ⊂ πi(D).
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Texas Tech University, Nadeeka de Silva, August 2012
Theorem 3.1.5. Let Y be a continuum as in Proposition 3.1.1. There exists
n0 ∈ {1, 2, 3, . . . ,m} such that σ(Y ) ≤ max(σ(Cn0), σ∗
0(Cn0)).
Proof. If σ(Y ) = 0, then, since for each n, Cn ⊂ Y , we have that σ(Y ) = σ(Cn) = 0.
Therefore we can assume that σ(Y ) > 0.
From the definition of the span there exists a subcontinuum A of Y × Y such that
S(A) = σ(Y ) and π1(A) = π2(A). Note that from Proposition 3.1.3 there exists n0 such
that A ∩ (Cn0× Cn0
) 6= ∅. Consider the two cases.
Case 1 : A ⊂ Cn0× Cn0
:
In this case σ(Y ) = S(A) ≤ σ(Cn0) ≤ σ(Y ), so σ(Y ) = σ(Cn0
).
Case 2 :A 6⊂ Cn0× Cn0
:
Let C be any component of A ∩ (Cn0× Cn0
). For each ǫ > 0, let
N(C, ǫ) = {t ∈ A : d(t, C) < ǫ}. Let Cǫ be the component of N(C, ǫ) that contains C.
From the Boundary Bumper Theorem (Theorem 1.1.2), Cǫ has a limit point on the
boundary of N(C, ǫ). Therefore Cǫ is a connected subset of A that properly contains C.
So Cǫ contains a point in the complement of Cn0× Cn0
, i.e.
π1(Cǫ) 6⊂ Cn0or π2(Cǫ) 6⊂ Cn0
. Consequently using Proposition 3.1.4 we obtain that, for
each ǫ > 0, Cn0⊂ π1(Cǫ) or Cn0
⊂ π2(Cǫ). Without loss of generality, there exists a
sequence {ǫn}∞n=1 converging to zero, such that, for all ǫn, Cn0⊂ π1(Cǫn). Hence
Cn0⊂ π1(C). Note that C is a connected subset of Cn0
× Cn0and that C satisfies the
condition π2(C) ⊂ π1(C) = Cn0. Therefore in this case we have that
σ(Y ) ≤ S(C) ≤ σ∗0(Cn0
).
In general we have that σ(Y ) ≤ σ∗0(Cn0
) or σ(Y ) ≤ σ(Cn0), and hence
σ(Y ) ≤ max{σ∗0(Cn0
), σ(Cn0)}.
Corollary 3.1.6. For n = 1, 2, 3, . . . ,m, let Cn be disjoint continua such that
σ∗0(Cn) = σ(Cn). Let Y be a continuum as in Proposition 3.1.1. Then there exists
n0 ∈ {1, 2, 3, . . . ,m} such that σ(Y ) = σ(Cn0). Furthermore
σ(Cn0) = max{σ(Cn) : n = 1, 2, 3, . . . ,m}.
Proof. From Theorem 3.1.5 and from the condition σ∗0(Cn) = σ(Cn) for each n, it follows
that σ(Y ) = σ(Cn0) for some n0 ∈ {1, 2, 3, . . . ,m}. But σ(Cn) ≤ σ(Y ) = σ(Cn0
) for all n.
Hence σ(Cn0) = max{σ(Cn) : n = 1, 2, 3, . . . ,m}.
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Texas Tech University, Nadeeka de Silva, August 2012
Theorem 3.1.7. For any finite subset F of [0, 1] that contains 0 and 1, there exists a
continuum YF such that B(YF ) = F .
Proof. First we construct the continuum YF as follows. Let F be {r1, r2, r3 . . . rm} where
0 = r1 < r2 < r3 < . . . < rm = 1. For each n, let Cn be a circle in R2 with center (0, 0)
and diameter rn. For n = 1, 2, 3, . . . ,m− 1, let Rn be disjoint copies of (0, 1) that lie in
R2, such that the end corresponding to 0 limits to Cn and the other end limits to Cn+1.
Let Yn = Rn = Cn ∪Rn ∪ Cn+1 and YF =m−1⋃
n=0
Yn. (See Figure 3.1.)
Figure 3.1. Continuum YF
We claim that B(YF ) = {σ(S) : S is a subcontinuum of YF} = F . The proof is given
below.
Consider any subcontinuum S of YF . Let f : YF → [0, 1] be a function defined as in
Proposition 3.1.2. Then f(S) ⊆ I is compact and connected.
Case 1 : f(S) is a singleton and f(S) 6∈ F .
In this case S is a singleton, thus σ(S) = 0.
Case 2 : f(S) is a singleton and f(S) ∈ F .
In this case, S ⊆ Cr for some r ∈ F . Hence σ(S) = diam{Cr} or σ(S) = 0.
Case 3 : f(S) is a closed interval and f(S) ⊂ FC .
In this case, S is an arc. Hence σ(S) = 0.
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Texas Tech University, Nadeeka de Silva, August 2012
Case 4: f(S) is a closed interval and f(S) ∩ F 6= ∅.In this case, S is a continuum consisting of circles and rays limiting to them. Let
L = f(S). f |S is a continuous function from S onto L. Therefore in a similar manner as
in the proof of Proposition 3.1.3, we obtain the following result:
If A is a subcontinuum of S × S such that S(A) > 0 and π1(A) = π2(A), then
A ∩(
⋃
r∈F∩L
Cr × Cr
)
6= ∅.Using this result, Proposition 3.1.4 and using similar arguments as in the the proof of
Theorem 3.1.5, we conclude that σ(S) = max{diam(Cr) : r ∈ F ∩ L}.In any case σ(S) ∈ F . Hence B(Y ) ⊂ F . For any m ∈ F there exists a circle Cm ⊂ Y
such that diam(Cm) = m. Therefore F ⊂ B(Y ), hence F = B(Y ).
In [7] Z. Waraszkiewicz showed that there is an uncountable family of pairwise
incomparable spirals that limit onto a given circle. So we get the following result.
Corollary 3.1.8. For any finite subset F of [0, 1] that contains 0 and 1, there exists an
uncountable family YF , of pairwise incomparable continua such that, for each member
Y ∈ YF , B(Y ) = F .
3.1.2 Construction of a continuum YT such that the set given by the span of
subcontinua of YT is equal to a set which is the union of {0} and a given closed
interval that does not contain 0
Theorem 3.1.9. Let d(x, y) be the Euclidean distance in R2 and YT = L1 ∪ L2 ∪ L3 ∪ U ,
where
• L1 = {(x, 0) : x ∈ [2, 4]} ∪ {(
(1 + e−θ) cos(θ), (1 + e−θ) sin(θ))
: θ ∈ [0,∞)},
• L2 = {(x, y) : y = −√3x : x ∈
[−2,−1]} ∪{(
(1 + e−θ) cos(θ +2π
3), (1 + e−θ) sin(θ +
2π
3)
)
: θ ∈ [0,∞)
}
,
• L3 = {(x, y) : y =√3x : x ∈
[−2,−1]} ∪{(
(1 + e−θ) cos(θ +4π
3), (1 + e−θ) sin(θ +
4π
3)
)
: θ ∈ [0,∞)
}
and
• U is the unit circle.
Then YT is a continuum and B(YT ) = {0} ∪ [2, σ(YT )].
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Texas Tech University, Nadeeka de Silva, August 2012
Figure 3.2. Continuum YT
Proof. Let l1,l2 and l3 be the endpoints of L1,L2,L3 respectively. Consider the
subcontinuum A of YT × YT described by A =6⋃
n=1
An, where
1. A1 = {l1} × (L2 ∪ U ∪ L3) ,
2. A2 = (L1 ∪ U ∪ L2)× {l3},
3. A3 = {l2} × (L1 ∪ U ∪ L3) ,
4. A4 = (L2 ∪ U ∪ L3)× {l1},
5. A5 = {l3} × (L1 ∪ U ∪ L2) and
6. A6 = (L1 ∪ U ∪ L3)× {l2}.
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Texas Tech University, Nadeeka de Silva, August 2012
It is clear that A is a connected subset and that π1(A) = π2(A) ,
inf{d(x, y) : (x, y) ∈ A1} = min{d(l1, L2 ∪ U ∪ L3)},inf{d(x, y) : (x, y) ∈ A2} = min{d(l3, L1 ∪ U ∪ L2)},inf{d(x, y) : (x, y) ∈ A3} = min{d(l2, L1 ∪ U ∪ L3)},inf{d(x, y) : (x, y) ∈ A4} = min{d(l1, L2 ∪ U ∪ L3)},
inf{d(x, y) : (x, y) ∈ A5} = min{d(l3, L1 ∪ U ∪ L2)} and
inf{d(x, y) : (x, y) ∈ A6} = min{d(l2, L3 ∪ U ∪ L1)}.
From the above equations it follows that:
inf{d(x, y) : (x, y) ∈ A} = min{d(li, Li+1 ∪ U ∪ Li+2 : i = 1, 2, 3}.
Therefore
σ(YT ) ≥ inf{d(x, y) : (x, y) ∈ A} = min{d(li, Li+1 ∪ U ∪ Li+2 : i = 1, 2, 3}.
The arms are extended far enough such that σ(YT ) > 2.
Consider any subcontinuum S of YT .
Case 1 : There exists n in {1, 2, 3} such that Ln ∩ S = ∅.If S is a point or an arc then σ(S) = 0. Otherwise from results in previous sections
σ(S) = 2.
Case 2 : Li ∩ S 6= ∅ for each i ∈ {1, 2, 3}.In this case following similar arguments as in the case of YT , we get that σ(S) ≥ 2.
Hence B(YT ) ⊂ {0} ∪ [2, σ(YT )].
There is a deformation retract of the entire continuum YT which is the identity except
on the shortest arm, which it shortens further. For any subcontinuum A of YT × YT , this
deformation retract induces a homotopy of A. Under this homotopy S(A) changes
continuously. Thus σ(YT ) changes continuously as the arm shrinks. Hence for each value
r in [2, σ(YT )], there exists a subcontinuum S such that σ((S) = r. Therefore
B(YT ) = {0} ∪ [2, σ(YT )].
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Texas Tech University, Nadeeka de Silva, August 2012
3.1.3 Construction of a continuum YK such that the set given by the span of
subcontinua of YK is equal to the Cantor set
Proposition 3.1.10. Let K denote the Cantor set and K = {r/2 : r ∈ K}. For each
r ∈ K, let Cr be the circle in R2 with center (0, 0) and radius r. Consider the disjoint
collection of open intervals G such that⋃
I∈G
I is the complement of K in [0, 1/2]. For each
(a, b) ∈ G, let Rab be a copy of (0, 1), such that the end that corresponds to 0 limits to Ca
and the other end limits to Cb, the Rab’s are pairwise disjoint and the Rab’s are disjoint
from the Cr’s. Let Yab = Rab = Ca ∪Rab ∪ Cb and YK =
(
⋃
(a,b)∈G
Yab
)
∪(
⋃
r∈K
Cr
)
. (See
Figure 3.3). Then YK is a continuum.
Figure 3.3. Continuum YK
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Texas Tech University, Nadeeka de Silva, August 2012
Proposition 3.1.11. Let YK be a continuum as in Proposition 3.1.10. There exists a
continuous surjection f : YK → [0, 1] such that
1. y ∈ Cr if and only if f(y) = r,
2. y ∈ Rab if and only if f(y) = hab(y), where, for each (a, b) ∈ G, hab : Rab → (a, b) is
a homeomorphism and
Proof. For each (a, b) ∈ G, let hab : Rab → (a, b) be a homeomorphism. Consider the
function defined by
f(y) =
r if y ∈ Cr
hab(y) if y ∈ Rn.
It is straightforward that f is a continuous function that satisfies the properties 1 and
2.
Proposition 3.1.12. Let YK be a continuum as in Proposition 3.1.10. If A is a
subcontinuum of YK × YK such that S(A) > 0 and π1(A) = π2(A), then
A ∩(
⋃
r∈K
Cr × Cr
)
6= ∅.
Proof. Note that πi(A) 6⊂⋃
(a,b)∈G
Rab. (Every subcontinuum of YK which is a subset of
⋃
(a,b)∈G
Rab has span zero.) Let f : YK → [0, 1] be a continuous surjection as in Proposition
3.1.11. Consider the functions gi : A → [0, 1] defined by gi = f ◦ πi, where i = 1, 2. The
functions g1 and g2 satisfy the hypotheses of the Coincidence Point Theorem (Theorem
1.1.1). Therefore there exists a point (x, y) in A such that g1(x, y) = g2(x, y), which
implies f(x) = f(y). Since f(x) = f(y), (x, y) ∈(
⋃
r∈K
Cr × Cr
)
∪(
⋃
(a,b)∈G
Rab ×Rab
)
.
Assume that (x, y) ∈ ⋃
(a,b)∈G
Rab ×Rab. Then x = y and hence S(A) = 0, which is a
contradiction. Therefore (x, y) ∈(
⋃
r∈K
Cr × Cr
)
and A ∩(
⋃
r∈K
Cr × Cr
)
6= ∅.
Proposition 3.1.13. Let YK be a continuum as in Proposition 3.1.10. Let D be a
subcontinuum of YK × YK. If there exists r such that πi(D) ∩ Cr 6= ∅ and
πi(D) ∩(
⋃
(a,b)∈G
Rab
)
6= ∅ where i ∈ {1, 2}, then Cr ⊂ πi(D).
Theorem 3.1.14. Let YK be a continuum as in Proposition 3.1.10. Then there exists
r0 ∈ K such that σ(YK) = σ(Cr0) = diam(Cr0) = max{diam(Cr) : r ∈ K} = 1.
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Texas Tech University, Nadeeka de Silva, August 2012
Proof. If σ(YK) = 0, then, since for each r ∈ K, Cr ⊂ YK , we have that
σ(YK) = σ(Cr) = 0. Therefore we can assume that σ(YK) > 0.
From the definition of span there exists a subcontinuum A of YK × YK such that
S(A) = σ(YK) and π1(A) = π2(A). Note that from Proposition 3.1.12 there exists r0 such
that A ∩ (Cr0 × Cr0) 6= ∅. Consider the following cases.
Case 1 : A ⊂ Cr0 × Cr0
In this case σ(YK) = S(A) ≤ σ(Cr0) ≤ σ(YK), so σ(YK) = σ(Cr0) = diam(Cr0).
Case 2 :A 6⊂ Cr0 × Cr0
Let C be any component of A ∩ (Cr0 × Cr0). For each ǫ > 0, let
N(C, ǫ) = {t ∈ A : d(t, C) < ǫ}. Let Cǫ be the component of N(C, ǫ) that contains C.
From the Boundary Bumper Theorem (Theorem 1.1.2), Cǫ has a limit point on the
boundary of N(C, ǫ). Therefore Cǫ is a connected subset of A that properly contains C.
So Cǫ contains a point in the complement of Cr0 × Cr0 , i.e.
π1(Cǫ) ∩(
⋃
(a,b)∈G
Rab
)
6= ∅ or π2(Cǫ) ∩(
⋃
(a,b)∈G
Rab
)
6= ∅. Consequently using Proposition
3.1.13 we obtain that, for each ǫ > 0, Cr0 ⊂ π1(Cǫ) or Cr0 ⊂ π2(Cǫ). Without loss of
generality, there exists a sequence {ǫn}∞n=1 converging to zero, such that, for all ǫn,
Cr0 ⊂ π1(Cǫn). Hence Cr0 ⊂ π1(C). Since ǫn converges to zero and Cr0 ⊂ π1(Cǫn), then
Cr0 ⊂ π1(C). Note that C is a connected subset of Cr0 × Cr0 and that C satisfies the
condition π2(C) ⊂ π1(C) = Cr0 . Therefore in this case we have that
σ(YK) ≤ S(C) ≤ σ∗0(Cr0).
In general we have that σ(YK) ≤ σ∗0(Cr0) or σ(YK) ≤ σ(Cr0) and
σ(YK) ≤ max{σ∗0(Cr0), σ(Cr0)}.
Since Cr0 is a circle it follows that σ(YK) = diam(Cr0). But we know that
diam(Cr) = σ(Cr) ≤ σ(YK) for all r ∈ K. Hence
σ(YK) = diam(Cr0) = max{diam(Cr) : r ∈ K} = 1.
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Texas Tech University, Nadeeka de Silva, August 2012
Theorem 3.1.15. Let K denote the Cantor Set. Let YK be a continuum as in
Proposition 3.1.10. Then B(YK) = {σ(S) : S is a subcontinuum of YK} = K.
Proof. Consider any subcontinuum S of YK . Let f : YK → [0, 1] be a function defined as
in Proposition 3.1.11. Then f(S) ⊆ I is compact and connected.
Case 1: f(S) is a singleton and f(S) 6∈ K.
In this case, S is a singleton, thus σ(S) = 0.
Case 2 : f(S) is a singleton and f(S) ∈ K.
In this case, S ⊆ Cr0 for some r0 ∈ K. Hence σ(S) = diam{Cr0} or σ(S) = 0.
Case 3 : f(S) is a closed interval and f(S) ⊂ KC .
In this case, S is an arc. Hence σ(S) = 0.
Case 4 : f(S) is a closed interval and f(S) ∩K 6= ∅.In this case, S is a continuum consisting of circles and rays limiting to them. Let
L = f(S). f |S is a continuous function from S onto L. Therefore, in a similar manner as
in the proof of Proposition 3.1.12, we obtain the following result.
If A is a subcontinuum of S × S such that S(A) > 0 and π1(A) = π2(A), then
A ∩(
⋃
r∈K∩L
Cr × Cr
)
6= ∅.Using this result, Proposition 3.1.13 and using similar arguments as in the proof of
Theorem 3.1.14, we conclude that σ(S) = max{diam(Cr) : r ∈ K ∩ L}.In any case σ(S) ∈ K. Hence B(YK) = {σ(A) : A is a subcontinuum ofYK} ⊂ K. For
each r ∈ K there exists a circle Cr/2 ⊂ YK such that diam(Cr/2) = r. Therefore
K ⊂ B(Yk), hence K = B(YK).
As before, due to a result by Z. Waraszkiewicz in [7] we get the following result.
Corollary 3.1.16. Let K denote the Cantor Set. There exists an uncountable family
YK, of pairwise incomparable continua, such that, for each member Y ∈ YK, B(Y ) = K.
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Texas Tech University, Nadeeka de Silva, August 2012
3.2 Construction of a continuum YM such that the set given by the span of subcontinua
of YM is equal to a given closed set
Theorem 3.2.1. Let M be a nonempty closed subset of [0, 1] and let K denote the
Cantor set. There exists a continuous surjection g : K → M .
Proposition 3.2.2. Let K denote the Cantor set and M be a nonempty closed subset of
[0, 1] which contains 0 and 1. Let g : K → M be a continuous surjection such that
g(0) = 0. For each r ∈ K, let Cr be the circle parallel to the yz plane in R3 with center
(r, 0, 0) and diameter g(r). Consider the disjoint collection of open intervals G such that⋃
I∈G
I is the complement of K in [0, 1]. For each (a, b) ∈ G, let Rab be a copy of (0, 1),
such that the end that corresponds to 0 limits to Ca and the other end limits to Cb, the
Rab’s are pairwise disjoint and the Rab’s are disjoint from the Cr’s. Let
Yab = Rab = Ca ∪Rab ∪Cb and YM =
(
⋃
(a,b)∈G
Yab
)
∪(
⋃
r∈K
Cr
)
. Then YM is a continuum.
Proposition 3.2.3. Let YM be a continuum as in Proposition 3.2.2. There exists a
continuous surjection f : YM → [0, 1] such that.
1. y ∈ Cr if and only if f(y) = r and
2. y ∈ Rab if and only if f(y) = hab(y), where, for each (a, b) ∈ G, hab : Rab → (a, b) is
a homeomorphism
Proof. For each (a, b) ∈ G, let hab : Rab → (a, b) be a homeomorphism. Consider the
function defined by
f(y) =
r if y ∈ Cr
hab(y), if y ∈ Rn.
Then f is a continuous function that satisfies the properties 1 and 2.
Proposition 3.2.4. Let YM be a continuum as in Proposition 3.2.2. If A is a
subcontinuum of YM × YM such that S(A) > 0 and π1(A) = π2(A), then
A ∩(
⋃
r∈K
Cr × Cr
)
6= ∅.
Proof. Note that πi(A) 6⊂⋃
(a,b)∈G
Rab. (Every subcontinuum of⋃
(a,b)∈G
Rab has span zero.)
Let f : YM → [0, 1] be a continuous surjection as in Proposition 3.2.3. Consider the
functions gi : A → [0, 1] defined by gi = f ◦ πi, where i = 1, 2. The functions g1 and g2
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Texas Tech University, Nadeeka de Silva, August 2012
satisfy the hypotheses of the Coincidence Point Theorem (Theorem 1.1.1). Therefore
there exists a point (x, y) in A such that g1(x, y) = g2(x, y), which implies f(x) = f(y).
This is possible only if (x, y) ∈(
⋃
r∈K
Cr × Cr
)
∪(
⋃
(a,b)∈G
Rab ×Rab
)
. Assume that
(x, y) ∈ ⋃
(a,b)∈G
Rab ×Rab. Then x = y and S(A) = 0, which is a contradiction. Therefore
(x, y) ∈ ⋃
r∈K
Cr × Cr. Hence A ∩(
⋃
r∈K
Cr × Cr
)
6= ∅.
Proposition 3.2.5. Let YM be a continuum as in Proposition 3.2.2. Let D be a
subcontinuum of YM × YM . If there exists r such that πi(D) ∩ Cr 6= ∅ and
π1(D) ∩(
⋃
(a,b)∈G
Rab
)
6= ∅ where i ∈ {1, 2}, then Cr ⊂ πi(D).
Theorem 3.2.6. Let M be a closed subset of [0, 1] containing 1 and 0 and let YM be a
continuum as in Proposition 3.2.2. There exists r0 ∈ K such that
σ(YM) = diam(Cr0) = max{diam(Cr) : r ∈ K} = 1.
Proof. If σ(YM) = 0, then for each r ∈ K, since Cr ⊂ YM , we get that
σ(YM) = σ(Cr) = 0. Therefore we can assume that σ(YM) > 0.
From the definition of span there exists a subcontinuum A of YM × YM such that
S(A) = σ(YM) and π1(A) = π2(A). Note that from Proposition 3.2.4 there exists r0 such
that A ∩ (Cr0 × Cr0) 6= ∅.Case 1 : A ⊂ Cr0 × Cr0
In this case σ(YM) = S(A) ≤ σ(Cr0) ≤ σ(YM), so σ(YM) = σ(Cr0) = diam(Cr0).
Case 2 :A 6⊂ Cr0 × Cr0
Let C be any component of A ∩ (Cr0 × Cr0). For each ǫ > 0, let
N(C, ǫ) = {t ∈ A : d(t, C) < ǫ}. Let Cǫ be the component of N(C, ǫ) that contains C.
From the Boundary Bumper Theorem (Theorem 1.1.2), Cǫ has a limit point on the
boundary of N(C, ǫ). Therefore Cǫ is a connected subset of A that properly contains C.
So Cǫ contains a point in the complement of Cr0 × Cr0 , i.e.
π1(Cǫ) ∩(
⋃
(a,b)∈G
Rab
)
6= ∅ or π2(Cǫ) ∩(
⋃
(a,b)∈G
Rab
)
6= ∅. Consequently using Proposition
3.2.5 we obtain that, for each ǫ > 0, Cr0 ⊂ π1(Cǫ) or Cr0 ⊂ π2(Cǫ). Without loss of
generality, there exists a sequence {ǫn}∞n=1 converging to zero, such that, for all ǫn,
Cr0 ⊂ π1(Cǫn). Hence Cr0 ⊂ π1(C). Since ǫn converges to zero and Cr0 ⊂ π1(Cǫn), then
Cr0 ⊂ π1(C). Note that C is a connected subset of Cr0 × Cr0 and that C satisfies the
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Texas Tech University, Nadeeka de Silva, August 2012
condition π2(C) ⊂ π1(C) = Cr0 . Therefore in this case we have that
σ(YM) ≤ S(C) ≤ σ∗0(Cr0).
In general we have that σ(YM) ≤ σ∗0(Cr0) or σ(YM) ≤ σ(Cr0), i.e.
σ(YM) ≤ max{σ∗0(Cr0), σ(Cr0)}. Since Cr0 is a circle it follows that σ(YM) = diam(Cr0).
But we know that diam(Cr) = σ(Cr) ≤ σ(YM) for each r ∈ K. Hence
σ(YM) = diam(Cr0) = max{diam(Cr) : r ∈ K} = max(M) = 1.
Theorem 3.2.7. Let M be a closed subset of [0, 1] containing 1 and 0 and let YM be a
continuum as in Proposition 3.2.2. Then
B(YM) = {σ(S) : S is a subcontinuum of YM} = M .
Proof. Consider any subcontinuum S of YM . Let f : YM → [0, 1] be a function defined as
in Proposition 3.2.3. Then f(S) ⊆ I is compact and connected.
Case 1 : f(S) is a singleton and f(S) 6∈ K.
In this case, S is a singleton, hence σ(S) = 0.
Case 2 : f(S) is a singleton and f(S) ∈ K.
In this case, S ⊆ Cr0 for some r0 ∈ K. Hence σ(S) = diam{Cr0} or σ(S) = 0.
Case 3 : f(S) is a closed interval and f(S) ⊂ KC . In this case S is an arc. Hence
σ(S) = 0.
Case 4 : f(S) is a closed interval and f(S) ∩K 6= ∅. In this case S, is a continuum
consisting of circles and rays limiting to them. Let L = f(S). f |S is a continuous
function from S onto L. Therefore in a similar manner as in the proof of Proposition
3.2.4 we obtain that:
if A is a subcontinuum of S × S such that S(A) > 0 and π1(A) = π2(A), then
A ∩(
⋃
r∈K∩L
Cr × Cr
)
6= ∅.Using this result, Proposition 3.2.5 and using similar arguments as in the proof of
Theorem 3.2.6, we conclude that σ(S) = max{diam(Cr) : r ∈ K ∩ L}.In any case σ(S) ∈ M . Hence B(YM) = {σ(A) : A is a subcontinuum of YM} ⊂ M . For
any m ∈ M there exists a circle Cr ⊂ YM such that diam(Cr) = g(r) = m. Therefore
M ⊂ B(YM), hence M = B(YM).
As before, due to a result by Z. Waraszkiewicz in [7] we get the following result.
Corollary 3.2.8. Let M be a closed subset of [0, 1] containing 1 and 0. There exists an
uncountable family YM , of pairwise incomparable continua, such that, for each member
Y ∈ YM , B(Y ) = M .
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Texas Tech University, Nadeeka de Silva, August 2012
CHAPTER 4
CONCLUSION AND FURTHER QUESTIONS
4.1 Conclusion
4.1.1 Ray limiting to a continuum
Let Y be a continuum consisting of a ray limiting to continuum X. We proved that:
σ(Y ) ≤ max(σ(X), σ∗0(X)). (4.1)
Also we obtained similar results for other versions of span.
Furthermore we proved that if Y is a continuum consisting of a ray limiting to
continuum X and Z is the continuum formed by a ray limiting to Y then
σ(Z) ≤ max(σ(X), σ∗0(X)). Thus if we start from a continuum and keep adding limiting
rays in this fashion the span of the resulting continuum is bounded by the maximum of
the span and the surjective semispan of the starting continuum. Also we proved that the
inequality (4.1) can be improved to σ(Y ) = σ(X), under any of the following conditions.
• σ(X) = 0.
• X is a simple closed curve.
• σ(X) = σ∗0(X).
We used the above results to construct continua such that the set of values of span of
subcontinua is equal to specific sets. There are other uses of these results. Tina Sovic has
recently given an independent proof of the span zero case and she is using it to construct
more examples of nonchainable continua with span zero.
4.1.2 Span of subcontinua
Our main goal was to study the set:
B(X) = {σ(H) : where H is a subcontinuum of X } .
We proved following results.
For each closed subset M of [0, 1] containing 1 and 0, we constructed an uncountable
family YM , of pairwise incomparable continua such that, for each member Y ∈ YM the
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Texas Tech University, Nadeeka de Silva, August 2012
set of values of span of subcontinua of Y is the set M . Furthermore each member of this
family is atriodic and one-dimensional.
Except for the case where M has infinitely many nondegenerate components, we
constructed planar examples. In the planar examples, when M contains intervals the
constructed examples contain triods.
4.2 Further Questions
Here some interesting questions are stated which would extend the results of this
dissertation in many different directions.
We observed that inequality (4.1) can be improved to σ(Y ) = σ(X), when
σ(X) = σ∗0(X).
Question 3. Let X be a continuum and R be a ray limiting to it. Let Y = X ∪R. What
conditions on X will guarantee that σ(Y ) = σ(X)? σ(X) = σ∗0(X) is such a condition. Is
this condition necessary for σ(Y ) = σ(X)?
All examples we constructed involved closed sets.
Question 4. What conditions on a set G ⊆ [0, 1] will guarantee that there will be a
continuum X such that B(XG) = G? Must G be a closed set?
The examples which we constructed contain rays limiting to simple closed curves and
hence are not arcwise connected.
Question 5. If X is an arcwise connected continuum, what closed sets can B(X) be?
Except for the case of a closed set G with infinitely many nondegenerate components,
we have constructed a planar example of a continuum XG with B(XG) = G.
Question 6. Does there exist a planar example for every closed set G?
The following results are straightforward.
• If X is a simple closed curve B(X) = {0, σ(X)}.
• If X is a simple triod B(X) = [0, σ(X)].
• 0 ∈ B(X).
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Texas Tech University, Nadeeka de Silva, August 2012
The above results are independent of the metric on X. When X is a simple closed
curve or when X is a triod, if Z ≈ X then B(Z) ≈ B(X). For some continua X, the set
B(X) depends on the metric.
Question 7. For which continua X, is B(X) is a topological invariant ofX?
Question 8. What properties of B(X) are topological invariants of X?
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BIBLIOGRAPHY
[1] J. F. Davis,The equivalence of zero span and zero semispan, Proceedings of theAmerican Mathematical Society 90 (1984) 133-138.
[2] L.C. Hoehn, A non-chainable plane continuum with span zero, FundamentaMathematicae 211(2011) 149-174.
[3] L.C. Hoehn and A.Karasev, Equivalent metrics and the spans of graphs, ColloquiumMathematicum 114 (2009) 135-153.
[4] A. Lelek, Disjoint mappings and the span of spaces, Fundamenta Mathematicae 55(1964) 199-214.
[5] A. Lelek, On the surjective span and semispan of connected metric spaces,Colloquium Mathematicum 37 (1977) 35-45.
[6] A. Lelek and T. West, Spans of continua and their applications, Continua with theHouston Problem Book, Lecture Notes in Pure and Applied Mathematics170,Marcel Dekker Inc., New York, Basel, Hong Kong. 127-132
[7] Z. Waraszkiewicz Une famille indenombrable de continus plans dont aucun n’estl’image continue d’un autre, (in French), Fundamenta Mathematicae 18 (1932)118-137.
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