c 2012, nadeeka de silva - tdl

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

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.

ii


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

iii


4.2 Further Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

iv


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.

v


LIST OF FIGURES

3.1 Continuum YF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.2 Continuum YT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.3 Continuum YK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

vi


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 .

1


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}.

2


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}.

3


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.

4


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).


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.


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).


Then σ∗(Y ) ≤ σ∗0(X).


Then σ∗0(Y ) ≤ σ∗

0(X).

5


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?

6


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.


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= ∅.


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).


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.

7


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.


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).


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).


If σ(X) = 0 then σ(Y ) = σ(X).

8


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.

9


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.


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).


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).


Then σ∗(Y ) ≤ σ∗0(X).

Proof. From Theorem 2.2.2 and inequality (1.2) it follows that σ∗(Y ) ≤ σ∗0(X).

10



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)).

11


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.

12


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).

13


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


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}.

14


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.

15


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 )].

16


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}.

17


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 )].

18


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

19


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.

20


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




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.

21


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.

22


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

23


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




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

24


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 .

25


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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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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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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