A Brief Exploration of the Sorgenfrey Line
and the Lexicographic Order
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.
Regina Greiwe
Certificate of Approval:
Edward Slaminka
Associate Professor
Department of Mathematics and
Statistics
Jo Heath, Chair
Professor
Department of Mathematics and
Statistics
Randall Holmes
Associate Professor
Department of Mathematics and
Statistics
Anotida Madzvamuse
Assistant Professor
Department of Mathematics and
Statistics
Stephen L. McFarland
Dean
Graduate School
A Brief Exploration of the Sorgenfrey Line
and the Lexicographic Order
Regina Greiwe
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
May 11, 2006
A Brief Exploration of the Sorgenfrey Line
and the Lexicographic Order
Regina Greiwe
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
Regina Marie Greiwe, daughter of Michael Greiwe and Karen Broome, was born
February 4, 1982. She attended Tri-County High School in Buena Vista, Georgia where
she graduated fourth in her class in June, 1999. She then entered Columbus State
University under an Honors scholarship where she graduated Magna cum Laude in June,
2003. She was awarded with a Bachelor of Science degree in Applied Mathematics.
While at Columbus State University, she was president of the local chapter of the MAA
and was awarded the Mathematics Award in 2003. The following year she entered
Auburn Univerty to pursue a Master of Science degree in Mathematics. While at Auburn
University, she served as president of the local chapter of SIAM and has been involved
in Science Olympiad and mathematics education. She is the mother of Laura Jeanette
Amilia Greiwe.
iv
Thesis Abstract
A Brief Exploration of the Sorgenfrey Line
and the Lexicographic Order
Regina Greiwe
Master of Science, May 11, 2006
(B.S., Columbus State University, 2003)
45 Typed Pages
Directed by Jo Heath
The author will compare properties of the usual topology on the real line with the
properties of the Sorgenfrey line. After that she will look at properties of the Lexi-
cographic order on R ? [0,1]. The main properties of interest will be the separation
properties, along with compact and connected subsets.
v
Acknowledgments
I would like to thank my family for helping me take care of ?life?s little bumps? so
that I could concentrate on my work. I would also like to thank my topology professors
for providing me with inspiration along the way.
I would like to thank my advisor for her patience with every attempt at under-
standing that I presented to her. Because of her unwavering ability to not give away the
answers, I have made an expansive growth in my mathematical maturity. I would also
like to thank you, the reader, for taking the time to look at my thoughts and work.
vi
Style manual or journal used Journal of Approximation Theory (together with the
style known as ?aums?). Bibliograpy 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
1 Introduction 1
2 Fundamental Definitions and Theorems 3
3 The Sorgenfrey Line 12
4 The Lexiographic Order on R?[0,1] 23
Bibliography 37
viii
Chapter 1
Introduction
In each area of topology, there are interesting spaces that many people are familiar.
In this paper, I will be presenting two such spaces that I took a personal interest in.
When I began my research, I was very new to the study of topology in general.
After a semester in my first topology course, I began my work by investigating some
basic ideas about different topological properties. From here I was given eight different
spaces to look into. All that was given to me was a list of eight sets and a base for each.
In the beginning I did not know what any of these spaces were called, or even know that
they had a name. What you are about to read are the conjectures and proofs that I
came up with for two of these spaces.
Many of the theorems that you will see were inspired by work I had done in classes
and from colloquia I have attended. During the research portion of my study, I assumed
the well known properties of the real line with the usual topology without proof. Other
than that, any ideas that I may have needed, I personally proved. For the sake of the
reader, many of the proofs of the easier facts have been left out.
Chapter 2 may be used as a reference chapter. In it the reader will find common
definitions and statements. Included with these will be proofs of theorems that will be
used throughout the other two chapters. Many of these came from the initial fundamental
research I did, along with general assignments given in different courses. I believe that
some of the ideas used in these proofs are interesting concepts in themselves.
1
In chapter 3, the reader will encounter a brief investigation of the Sorgenfrey topol-
ogy on the real line. For this space, I was forced to break away from the preconceived
notions that I had had about the real line and intervals in general. Even though many
properties are shared by the usual topology and the Sorgenfrey topology, I feel that the
interesting part is found when the two are contrasted.
The Lexicographic Order is the subject of chapter 4. I would say that out of the
eight spaces given to me, this one is my favorite. I began completely lost on any idea
as to what to do with this space. I struggled throughout my research to come to many
of the conclusions that are contained in this chapter, yet I feel that everything that I
learned has come together to describe this space.
As you read through this paper, I hope that you will find something that you enjoy
and maybe something that you find unfamiliar.
2
Chapter 2
Fundamental Definitions and Theorems
The following are definitions and proofs of statements which will be needed in other
parts of the paper. The first two theorems were given as homework problems in my
undergraduate level general topology course taught by Dr. Jo Heath.
Definition 2.1. A space X is said to be regular if for each x ? X and each closed set,
M ? X such that x negationslash? M, there exist disjoint open sets U and V such that x ? U and
M ? V.
Theorem 2.1. A space X is regular if and only if for each x ? X and U open in X
such that x ? U, there exists V open in X such that x ? V ? V ? U.
Proof. Suppose X is a regular topological space. Let x ? X and U be open in X such
that x ? U. Then Uc is a closed subset of X such that x negationslash? Uc. Thus we can find U0 and
V0 open subsets of X such that x ? U0, Uc ? V0, and U0 ?V0 = ?. Let V = U0. Then
x ? V ? V ? V c0 ? U.
Now we must prove the reverse implication. So, suppose that for all x ? X and U
open in X such that x ? U, there exists V open in X such that x ? V ? V ? U. Let
x ? X and M be a closed subset of X such that x negationslash? M. Then x ? Mc and Mc is open,
which implies that there exists V open in X such that x ? V ? V ? Mc. Let U = V c.
Then U, and V are open sets such that M ? U, x ? V, and U ?V = V c ?V = ?.
Definition 2.2. Nonempty subsets A and B of a space X separate X if X = A?B
and A?B = B ?A = ?. A space is connected if there do not exist subsets A and B
which separate X.
3
Theorem 2.2. A space is not connected iff it contains a proper clopen subset.
Proof. Let X be a topological space. If we assume that X contains a proper clopen
subset, K, then K and Kc are both nonempty clopen subsets of X. Also, X = K ?Kc.
So K and Kc separate X. Hence X is not connected.
Conversely, let?s assume that X is not connected. Then there exist A and B,
nonempty subsets of X such that X = A?B and A?B = ? and B ?A = ?. Hence
X = A?B = A?B, which implies that Bc = A. So B is open. Similarly, X = A?B
implies that B = B. Thus B is closed. Therefore B is a proper clopen subset of X.
The next six theorems are just a few of the things that were proven during my
preliminary research. They give alternate definitions for some topological terms based
on basic open sets, and extend certain properties to subspaces. Let us first define a basis
for a space.
Definition 2.3. Let X be a set, and let B be a collection of subsets of X. Then B is
called a basis, or base, for X if B satisfies the following conditions.
1. B covers X (i.e., X = uniontextB).
2. Whenever B1 ?B2 negationslash= ? for B1, B2 ? B, there exists Bx ? B for each x ? B1 ?B2
such that x ? Bx ? B1 ?B2.
If X is a topological space, then we need a third condition.
3. Each open set U can be written as a union of elements of B.
An element of B is called a basic open set.
Definition 2.4. A space is said to be zero-dimensional if it has a basis consisting of
clopen sets.
4
Theorem 2.3. A collection of open sets B is a base for a space X iff for each x ? X
and for each open neighborhood U of x, there exists Bx ?B such that x ? Bx ? U.
Proof. Let X be a topological space, and let B be a collection of open sets. Suppose
that B is a base for X. Let x ? X, and let U be an open neighborhood of x. Then
U = uniontexti?I Bi where Bi ? B for each i ? I. So there exists i ? I such that x ? Bi, and
Bi ? U.
Conversely, suppose that for each x ? X and for each open neighborhood U of x,
there exists Bx ? B such that x ? Bx ? U. Then obviously B covers X and for each
U open U = uniontextx?U Bx. Now suppose that x ? B1 ?B2 for some B1 and B2 ? B. Then
B1 ?B2 is open which implies that there exists Bx ? B such that x ? Bx ? B1 ?B2.
Therefore B is a base for X.
Definition 2.5. A space is called a Hausdorff space if for each pair of distinct points
x and y, there exist disjoint open sets U and V such that x ? U and y ? V.
Theorem 2.4. A space X is a Hausdorff space iff for every pair of distinct points, x
and y ? X, there exist disjoint basic open sets Bx and By such that x ? Bx and y ? By.
Proof. Let X have a basis B, and let x and y ? X be distinct. Suppose that X is a
Hausdorff space. Then there exist disjoint open sets U and V such that x ? U and
y ? V. We know that U = uniontexti?I Bi and V = uniontextj?J Bj where Bi and Bj ? B for each
i ? I and j ? J. Thus there exists an i ? I and a j ? J such that x ? Bi and y ? Bj.
Since U and V are disjoint, Bi and Bj are also disjoint.
Now suppose that for every pair of distinct points, x and y ? X, there exist disjoint
basic open sets Bx and By such that x ? Bx and y ? By. Since Bx and By are open, we
clearly have that X is a Hausdorff space.
5
Definition 2.6. A space is compact provided that every open cover of the space has a
finite subcover. A space is Lindel?of provided that every open cover of the space has a
countable subcover.
Theorem 2.5. A space X is compact (respectively Lindel?of) iff for every open cover, G,
consisting of basic open sets, there is a finite (respectively countable) subcover of G.
Proof. Let X have a basis B. Suppose X is compact, and let G be an open cover
consisting of basic open sets. Then by the definition of compact, there exists a finite
subcover of G.
Now suppose that for every open cover, G, consisting of basic open sets, there is
a finite subcover of G. Let H be an arbitrary open cover of X. Then for each x ? X,
choose Ux ?H such that x ? Ux. Since B is a base for X, there exists Bx ?B such that
for each x ? X, x ? Bx ? Ux. Let GB = {Bx : x ? X}. Then there exists N ? N such
that hatwideGB = {Bxn : n ? N} is a subcover of GB. Let hatwideG = {Uxn : n ? N}. Then hatwideG is a
subcover of G. Therefore X is compact.
This same proof can be used to show that a space is Lindel?of with a few alterations.
When showing compactness, the following property is very useful.
Definition 2.7. A collection of sets, C, is said to have the finite intersection property
provided that every finite subcollection has nonempty intersection.
Theorem 2.6. A space X is compact iff every collection of closed subsets with the finite
intersection property has nonempty intersection.
Proof. Suppose that X is compact. Let C be a collection of closed subsets of X with
the finite intersection property. Suppose that intersectiontextCG = ?. Define a collection G = {Kc :
6
K ? C}. Notice that G is an open cover of X. Hence there exists n ? N and some Kc1,
..., Kcn ? G such that uniontextni=1 Kci = X. Thus intersectiontextni=1 Ki = (uniontexti = 1nKci)c = ?. This is a
contradiction because C has the finite intersection property.
Now conversely assume that every collection of closed subsets of X with the finite
intersection property has nonempty intersection. Let G be an open cover of X. Suppose
that it does not have a finite subcover. Define the collection C = {Uc : U ? G}. Then
C is a collection of closed subsets with the finite intersection property. So we have that
intersectiontextC negationslash= ?. Thus uniontextG = (intersectiontextC)c negationslash= X, a contradiction since G covers X.
Theorem 2.7. A sequence {xn}n?N converges to a point x in a space X iff for each
basic open set Bx containing x, Bx also contains xn for all but finitely many n ?N.
Proof. Let X be a space with a basis B. Suppose that the sequence {xn}n?N converges
to a point x ? X. Then for every open set U such that x ? U, xn ? U for all but finitely
many n ? N. Thus obviously for each Bx ? B such that x ? Bx, xn ? Bx for all but
finitely many n ? N.
Conversely, suppose that for each basic open set Bx containing x, Bx contains xn
for all but finitely many n ? N. Let U be an open neighborhood of x. Then there exist
Bx ? B such that x ? Bx ? U. So by the hypothesis, xn ? Bx ? U for all but finitely
many n ?N. Therefore {xn}n?N converges to x.
Definition 2.8. A function f : X ? Y from a space X to a space Y is continuous if
f?1(V) is open in X for every V open in Y.
Theorem 2.8. Let f : X ? Y be a function from a space X to a space Y. Then f is
continuous iff for each basic open set BY ? Y, f?1(BY ) is open in X.
7
Proof. Let f : X ? Y be a function from a space X to a space Y. Suppose that f is
continuous. Then obviously for each basic open set BY ? Y, f?1(BY ) is open in X.
Conversely, suppose that for each basic open set BY ? Y, f?1(BY ) is open in X.
Let V be open in Y. Then V = uniontexti?I Bi where Bi is a basic open set in Y for each
i ? I. Then f?1(V) = f?1(uniontexti?I Bi) = uniontexti?I f?1(Bi) which is open in X. Therefore f is
continuous.
Definition 2.9. A subspace of a space X is a subset A of X with the subspace topology,
T|A defined by T|A = {U ?A : U open in X}.
Definition 2.10. A property C is said to be hereditary provided that every subspace
has property C whenever the whole space has property C. A space is said to have a
property hereditarily, if every subspace has the property.
Theorem 2.9. The properties of regularity and Hausdorff are hereditary.
Proof. Assume that X is a Hausdorff space and that A is a subspace of X. Let a and
b ? A be distinct. Since A ? X, there exist U and V disjoint open sets in X such that
a ? U and b ? V. Let hatwideU = U ?A and hatwideV = V ?A. Then hatwideU and hatwideV are disjoint open
subsets of A such that a ? hatwideU and b ? hatwideV. Therefore A is a Hausdorff space.
Now let X be regular, and let M be a nonempty closed subset of A. Let a ? A
such that a negationslash? M. Then notice that A\M is open. So there exists U open in X such
that A\M = U ?A. Then M = A\(A\M) = A?(U ?A)c = A?Uc. Let K = Uc.
Now notice M = K ?A, K is closed in X, and a negationslash? K. This implies that there exists
U and V disjoint open subsets of X such that a ? U and K ? V. Let hatwideU = U ? A
and hatwideV = V ? A. Then hatwideU and hatwideV are disjoint open subsets of A such that a ? hatwideU and
M = K ?A ? V ?A = hatwideV. Therefore A is regular.
8
The next two theorems were very beneficial in looking into normality for the follow-
ing two chapters. They were introduced as homework in Dr. Gary Gruenhage?s course,
Set Theoretic Topology.
Lemma 2.1. Let K and M be disjoint subsets of a space X. Suppose there exist collec-
tions {Un : n ?N} and {Vn : n ?N} of open sets satisfying the following:
1. M ? uniontextn?NUn and K ? uniontextn?NVn;
2. for each n ?N, Un ?K is empty; and,
3. for each n ?N, V n ?M is empty.
Then there exist disjoint open sets U and V such that M ? U and K ? V.
Proof. Let K and M be disjoint subsets of a space X. Suppose there exist collections
{Un : n ?N} and {Vn : n ?N} of open sets satisfying the following:
1. M ? uniontextn?NUn and K ? uniontextn?NVn;
2. for each n ?N, Un ?K is empty; and,
3. for each n ?N, V n ?M is empty.
Let hatwideUn = Un \(uniontextk?n V k) and hatwideVn = Vn \(uniontextk?n Uk). Notice that for each n ?N, hatwideUn and
hatwideVn are open.
Now we claim that for each i and j ?N, hatwideUi and hatwideVj are disjoint. Suppose that i = j.
Notice that hatwideUi ?Vi is empty. Thus hatwideUi ? hatwideVi is empty. Now suppose that i < j and that
hatwideUi ? hatwideVj is nonempty. Then there exists a point x ? hatwideUi ? hatwideVj which implies that x ? hatwideUi
and x ? hatwideVj. Thus x ? Ui and x negationslash? Ui, a contradiction. Hence hatwideUi ? hatwideVj is empty. Similarly
if i > j, we have that hatwideUi ? hatwideVj is empty.
9
Let U = uniontextn?N hatwideUn and V = uniontextn?N hatwideVn. By the preceding claim, U and V are disjoint
open sets. Since M ? uniontextn?NUn, we have that M ? U. Similarly, K ? V.
Definition 2.11. A space X is normal provided that for each pair of disjoint closed
sets H and K, there exist disjoint open sets U and V such that H ? U and K ? V.
Theorem 2.10. Every regular Lindel?of space is normal.
Proof. Assume that X is a regular Lindel?of space. Let H and K be nonempty disjoint
closed subsets of X. Then for each h ? H, there exists Uh and Vh disjoint open subsets
of X such that h ? Uh and K ? Vh. Notice that this implies that Uh?K is empty. Now
{Uh : h ? H} is an open cover of H. So there exist hn ? H for each n ? N such that
H ? uniontextn?NUhn. Recall that for each n ?N, Uhn ?K is empty.
Similarly, we can find open sets Vkn for some point kn ? K for each n ?N such that
K ? uniontextn?NVkn and V kn ?H is empty. By Lemma 2.1, there exist U and V disjoint open
subsets of X such that H ? U and K ? V. Therefore X is normal.
We have finished proving the general statements that will be used throughout this
paper. Let us define a few more terms with which the reader may not be familiar.
Definition 2.12. A set is called degenerate if it contains only a single element.
Definition 2.13. A subset D of a space X is dense in X provided that D = X. In
other words, D is dense in X if for each open set U, U ?X is nonempty.
Definition 2.14. A space is separable if it contains a countable dense subset.
Definition 2.15. Let X be a set. A function d : X ?X ? [0,?) is a metric provided
that for each x, y, and z ? X
10
1. d(x,y) = 0 iff x = y,
2. d(x,y) = d(y,x), and
3. d(x,z) ? d(x,y) +d(y,z).
Define an open ball centered at x with radius e, denoted B(x,e), as the collection of all
elements, y ? X, such that d(x,y) < e. A space (X,?) is said to be metrizable if there
exists a metric on X which can be used to build a basis consisting of open balls which
generates ?.
Definition 2.16. Let f : X ? Y be a continuous bijection from a space X to a space
Y. Then f is called a homeomorphism if f?1 is also continuous. We say that a space
X is homeomorphic to a space Y if there exists a homeomorphism f : X ? Y.
Now that we have gotten familiar with a few terms and general theorems, let us
move on to our first space.
11
Chapter 3
The Sorgenfrey Line
What would happen to the real line if we gave it a finer topology than the usual
topology? Would it lose any of its properties? Would it gain any? These are the
questions we will explore in this chapter.
Let S be the real line. Define a collection B by B = {(a,b] ? S : a < b}. Notice that
this collection of intervals covers S. Also if we note that the intersection of two intervals
of the form (a,b] is another interval of that form or the empty set, then it is easy to
see that B is a basis for S. Let T be the topology generated by B. In other words, let
T be the collection of all unions of elements of B. We call the topological space (S,T)
the Sorgenfrey line. In particular, since the intervals are closed on the right, T is known
as the right Sorgenfrey topology. If instead we use intervals closed on the left, then it
would be called the left Sorgenfrey topology.
We will discover that (S,T) satisfies most of the separation properties. Let us first
show that this is a finer topology than the usual topology. For one topology, ?1, to be
finer than another topology, ?2, we must show that ?2 is a subset of ?1.
Theorem 3.1. Any open interval in the usual topology on S is open in the Sorgenfrey
topology.
Proof. Let (a,b) be a nonempty subset of S. Then (a,b) = uniontextn?Nparenleftbiga,b? 1nbracketrightbig is an element
of T since it is a union of basic open sets.
One of the bases for the usual topology is the collection of all open intervals in S.
By Theorem 3.1, each open interval is open in the Sorgenfrey topology, T. So T is a
12
finer than the usual topology. What does this tell us? The properties of regularity and
of being a Hausdorff space are preserved by finer topologies. Thus (S,T) is a regular
Hausdorff space. Notice that a half closed interval such as (0,1] cannot be written as
the union of open intervals. So the Sorgenfrey topology on S is strictly finer than the
usual topology.
We will now show that any dense subset of S with the usual topology, is also dense
in S with the Sorgenfrey topology. In particular, the rationals would be dense in (S,T).
So this would imply that (S,T) is separable.
Theorem 3.2. Let D be a dense subset of S with the usual topology. Then D is dense
in S with the Sorgenfrey topology.
Proof. Let ? denote the usual topology on S, and let D be dense in (S,?). Choose an
arbitrary interval, (a,b] ?B. Now take note that every basic open set (a,b] ?B contains
the open interval (a, a+b2 ). So since D is dense in (S,?), there exists an element of D
which we will call d ? D ? (a, a+b2 ). Hence d ? (a,b] which implies that D ? (a,b] is
nonempty. Therefore D is dense in (S,T).
Next we will consider the countability properties. Obviously the two spaces share
the same cardinality, since their underlying sets are the same. What about the first and
second countability properties? We know that the usual topology has both properties.
How about the Sorgenfrey topology?
Theorem 3.3. (S,T) is first countable.
Proof. To show that the Sorgenfrey line is first countable, we must show that each point
has a countable local base. Suppose x ? S. Let Gx = {(p,x] : p < x, p ?Q}. Then Gx is
13
countable. Now we will show that Gx is a local base at x. Let U be an open neighborhood
of x. Then there is a basic open set, (a,b] ?B such that x ? (a,b] ? U. Thus a < x ? b
which gives us a p ? Q such that a < p < x ? b. Hence x ? (p,x] ? (a,b] ? U and
(p,x] ? Gx.
Theorem 3.4. The Sorgenfrey line is not second countable.
Proof. Suppose that (S,T) is second countable. Then (S,T) has a countable basis which
we will call A. We will use the usual basic open sets with irrational endpoints to con-
tradict the fact that A is countable. Let V = {(x ? 1,x] : x ? Qc} ? T. Since A is
a basis, for each x ? Qc we can choose Ax ? A such that x ? Ax ? (x ? 1,x]. Let
hatwideA = {Ax : x ?Qc}?A.
Now we will show that there exists an injective function from the irrationals to hatwideA.
Define a function f : Qc ? hatwideA by f(x) = Ax. Suppose that f is not injective. Then
there are points x, y ? Qc such that x negationslash= y and f(x) = f(y). We may assume that
x < y. Notice that f(x) = f(y) implies that Ax = Ay, but x < y implies that y negationslash? Ax; a
contradiction since y ? Ay. Therefore f is injective.
Since there is an injective function from Qc to hatwideA and hatwideA ? A, we have that |Qc| ?
| hatwideA|?|A|. Thus A is uncountable, a contradiction.
Interestingly enough, the Sorgenfrey line does not share the second countability
property with the usual topology. This will give rise to a very important difference
between the two spaces. The reals with the usual topology is a metric space. We will see
that the Sorgenfrey Line is not metrizable, using the fact that it is separable, but not
second countable. First we will show that a separable metric space is second countable.
Lemma 3.1. Every separable metric space is second countable.
14
Proof. Assume X is a separable metric space. Let d be a metric on X, and D be a
countable dense subset of X. Let B = {B(x,e) : x ? X, e > 0} which is a basis for a
metric space, and let B0 = {B(x,e) : x ? D, e ? Q, e > 0}. Then B0 is countable, and
B0 ? B. Now we will show that B0 is a basis for X. Let U be an open neighborhood of
x ? X. Then there is a B(x0,e0) ? B such that x ? B(x0,e0) ? U. Thus there exists
e1 > 0 such that B(x,e1) ? B(x0,e0). Since D is dense in X, there is a p ? D such that
p ? Bparenleftbigx, e14 parenrightbig. We know there is an e ? Q such that e14 < e < 3e14 . Hence B(p,e) ? B0.
To show that B0 is a basis, we will show that x ? B(p,e) and that B(p,e) ? U.
Claim 1: x ? B(p,e)
d(p,x) = d(x,p) < e14 < e ? x ? B(p,e).
Claim 2: B(p,e) ? B(x,e1)
Let y ? B(p,e). Then d(p,y) < e, and
d(x,y) ? d(x,p) +d(p,y) < e14 +e < e14 + 3e14 = e1.
Thus y ? B(x,e1) ? B(p,e) ? B(x,e1).
So by claims 1 and 2, x ? B(p,e) ? B(x,e1) ? B(x0,e0) ? U. Hence B0 is a countable
basis for X, and this implies X is second countable.
Corollary 3.1. (S,T) is not metrizable.
Since (S,T) is not second countable, either it is not separable or it is not metrizable.
We have shown that it is separable, and thus it is nonmetrizable.
Now we will show an interesting property of the Sorgenfrey line, not shared with
the reals with the usual topology.
15
Lemma 3.2. (S,T) is zero-dimensional. In other words, (S,T) has a basis of clopen
sets.
Proof. We will show that the basis B is made up of clopen sets. Let (a,b] ? B. Then
clearly (a,b] is open. We now will show that its complement is open. Notice that
(a,b]c = (??,a]?(b,?) = (uniontextn?N(a?n,a])?(uniontextn?N(b,b+n]) which is open.
This can be a very useful property, because it would automatically give us regularity.
The natural thing to do now, would be to prove normality. Instead we will con-
sider uncountable subsets of the reals. This property and the resulting theorems were
introduced to me by Dr. Gary Gruenhage in a class on Set-Theoretic Topology. This
property uses the fact that the usual topology on the reals is second countable, and will
be advantageous to prove that (S,T) is hereditarily normal.
Lemma 3.3. If H is an uncountable subset of the reals, then there exists x ? H such
that for each ? > 0, the intersection (x??,x)?H is nonempty.
Proof. Let H be an uncountable subset of R. Suppose that for each x ? H, there exists
?x > 0 such that (x??x,x)?H is empty. Then for each x ? H we can choose a rational
number, px ? (x??x,x). We will use this set of rational numbers to contradict the fact
that H is uncountable.
Let f : H ?Q be defined by f(x) = px. Suppose that f(x) = f(y) for some x and
y ? H. Then px = py. So px ? (x ? ?x,x) ? (y ? ?y,y). Since neither interval meets
H, x negationslash? (y ??y,y) and y negationslash? (x??x,x). This gives us x = y. Thus f is injective, which
implies that |H|?|Q|, a contradiction.
16
Notice that by altering the proof slightly, we can also show that there exists a point,
y ? H such that for each ? > 0, the intersection (y,y + ?)?H is also nonempty. This
will come in handy when we begin to discuss compactness.
Now Lemma 3.3 will give us the property of hereditarily Lindel?of. It tells us that
even though we can?t find a countable basis for (S,T), we can at least find countable
subcovers for open covers anywhere in the space.
Theorem 3.5. The Sorgenfrey line is hereditarily Lindel?of.
Proof. Note that whenever we have a point x ? S, and an open neighborhood, U, of x,
we can find a basic open set of the form (b,x] ?B such that (b,x] ? U.
Let X be a subset of S. Now consider an open cover of X of the form, G =
{(bx,x] : x ? X}. By the preceding remark, we can always find a collection of this form
where each element is contained in an element of our cover for any open cover. Let
M = {(bx,x) : (bx,x)?X negationslash= ?}, and let K be the set of all elements of X that are not
covered by M.
We would like to show that K is countable. To do this let us first suppose that
K is not countable. Then we know that there exists y ? K such that for each ? > 0,
(y ? ?,y) ? K negationslash= ? implying that (by,y) ? K negationslash= ?. Hence there exists z ? (by,y) ? K,
which means that z ? (by,y)?X since K ? X. Then z is covered by M, which tells us
that it is not in K, a contradiction. Therefore K must be countable.
Notice that M covers X \K in the usual topology on R. If we recall that the usual
topology is second countable and thus hereditarily Lindel?of, then we see that M has a
countable subcover, hatwiderM. Let hatwideG = {(bx,x] : (bx,x) ? hatwiderM}?{(by,y] : y ? K}. Then hatwideG is
17
the union of two countable subcollections of G which together covers all of X. Thus X
is Lindel?of.
We have shown that each subspace of (S,T) is both regular and Lindel?of which
implies that each is normal by theorem 2.10.
Corollary 3.2. (S,T) is hereditarily normal.
In the usual topology, every interval is connected and all closed bounded subsets are
compact. Since the Sorgenfrey topology is a finer topology on S, we have added more
open sets. This gives more opportunity for the subspaces to be disconnected. It also
lets us add more open sets to covers which gives the subspaces less of a chance to be
compact.
We will start by looking for non-trivial connected subspaces for (S,T). The best
place to begin is naturally with the whole space itself.
Theorem 3.6. (S,T) is not connected.
This can easily be seen by considering any element of B. Each element is a proper
clopen subset of S.
The next lemma will extend our use of the clopen basis to subspaces of (S,T).
Lemma 3.4. Let K be clopen in a space X and Y ? X, then K ?Y is clopen in the
subspace topology on Y.
Proof. Let X be a topological space, and let Y ? X. Let K be a clopen subset of X.
Then K and Kc are both open in X. Hence by the definition of subspace, K ?Y and
Y \K = Kc ?Y are both open in Y. Therefore K is clopen in Y.
18
Now that we know that the space (S,T) is not connected one may wonder if there is
any nontrivial connected subspaces of (S,T). The next theorem answers this question.
Theorem 3.7. The only nonempty connected subspaces of the Sorgenfrey line are the
degenerate sets.
Proof. Let X be a nonempty subspace of (S,T). If we suppose that X is degenerate,
then clearly X is connected. Suppose that X is nondegenerate. Then there exist distinct
points a and b ? X. We may assume that a < b. By the preceding lemma, (a,b]?X is
a proper clopen subset of X. Thus X is not connected.
Earlier we discussed briefly how a finer topology could affect the compact subsets
of a space. The next theorems deal with this issue. First we will consider the subsets of
S containing intervals.
Theorem 3.8. If X, a subset of S, contains an interval, then X is not compact.
Proof. Suppose that X is a subset of S which contains an interval. Then there exist
a and b ? X such that [a,b] ? X. Let C = {(a, (2n?1)a+b2n ] : n ? N}. Then C is a
collection of closed sets in the subspace X with the finite intersection property. Notice
that intersectiontextC = ?. Hence X is not compact.
Already with this one theorem we have shown that many of the compact subspaces
of S with the usual topology are not compact with the Sorgenfrey topology. This even
excludes the space itself from being compact since it contains the interval (0,1).
Corollary 3.3. (S,T) is not compact.
Obviously any finite subset of (S,T) is compact, but are there others? We will look
into unbounded subsets, and then we will take a look at uncountable subsets.
19
Theorem 3.9. Unbounded subsets of S are not compact.
Proof. Let X be an unbounded subset of (S,T). Suppose that X is compact. Let
hatwideG = {(?n,n] : n ? N}. Notice that hatwideG is an open cover of S, and thus it is an open
cover of X. Then hatwideG has a finite subcover of X. So there exists an N ? N such that
{(?ni,ni] : i ? N} covers X.
Let M = max{ni : i ? N}. Then X ? uniontexti?N(?ni,ni] = (?M,M]. So this shows
that X is bounded by M and ?M, a contradiction. Therefore X cannot be compact.
To deal with uncountable subsets, we will use some earlier theorems. In particular,
we will use the fact that a space X is compact if and only if for each collection of closed
subsets of X with the finite intersection property, C, the intersection intersectiontextC is nonempty.
The reader can find the proof of this statement in Cchapter 2.
Theorem 3.10. Uncountable subsets of S are not compact.
Proof. Let X be an uncountable subset of S. We may assume that X does not contain
an interval by Theorem 3.9. By Lemma 3.3, there exists a point in X such that for each
? > 0, the intersection X ? (x,x + ?), is nonempty. Let C = {(x,x + ?] ?X : ? > 0}.
Notice that since the subsets in C are nested and nonempty, this collection has the finite
intersection property. By Lemmas 3.2 and 3.4 , C is a collection of closed sets in X. If
we look at intersectiontextC, we realize that this intersection is empty. Hence X is not compact.
So far we only have the trivially compact subsets. By now one may be thinking that
this is all there is. Surprisingly enough, it?s not. Let us digress to look at convergent
sequences. Once we show the existence of a nontrivial convergent sequence, we will use
it to find an infinite compact subset of the Sorgenfrey line.
20
Theorem 3.11. If {xn}n?N is a sequence converging to some x ? R with the usual
topology, and for all but finitely many n, xn ? x, then {xn}n?N converges to x in (S,T).
Proof. Let ? be the usual topology on S. Let {xn}n?N be a sequence in S which is
convergent to some point x in (S,?). Suppose this sequence also satisfies the condition
that for all but finitely many n, xn ? x. Now let us show that {xn}n?N converges to
x in (S,T). Let U be an open neighborhood of x in (S,T). Then there exists a basic
open set (a,b] ? U such that x ? (a,b]. Since {xn}n?N converges to x in (S,?), there
exists N1 ?N such that xn ? (a,b) ? (a,b] for all n ? N1. Now notice that there exists
N2 ?N such that xn ? x for all n ? N2. So we have that xn ? (a,x] ? (a,b] ? U for all
n ? N1 +N2. Thus {xn}n?N converges to x in (S,T).
Now we know that sequences such as nn+1n?N converge. Now that we have convergent
sequences, let?s see how we can use it.
Theorem 3.12. If {xn}n?N converges to x in (S,T) from the left then {xn}n?N ?{x}
is a compact subspace of (S,T).
Proof. Let {xn}n?N converge to a point x in (S,T) from the left. Let X = {xn}n?N?{x}.
Suppose that G is an open cover of X. Then there exists U ? G which is an open
neighborhood of x. Since {xn}n?N converges to x, we have that for some N ? N, the
points, xn ? U for all n ? N. Since G covers X, there exists Un an open neighborhood
of xn for each n ? N. Let hatwideG = {Un : n ? N}?{U}. Then hatwideG is an open cover of X.
Therefore X is compact.
That will conclude our study of the Sorgenfrey line. In summary, by taking a basis
which adds an endpoint to the open intervals in R, we have altered the usual topology
21
on the reals into a much finer space, the Sorgenfrey line. Even though many of the
nicer qualities of the usual topology have been lost such as metrizability, the Sorgenfrey
topology does share some properties of the usual topology, while gaining a few more.
22
Chapter 4
The Lexicographic Order on R?[0,1]
To organize our papers and files, we normally alphabetize them. In other words, we
order them by the first letter in the name, and then by the consecutive letters. This is
a very common order that can be extended to coordinate systems. Let us define a set
L = R?[0,1]. Elements of this set are expressed as ordered pairs. It is now natural to
order the first coordinates and then the second coordinates. So define an order ? < ? on
L by (x,xprime) < (y,yprime) provided that either x < y or, x = y and xprime < yprime. This is called the
lexicographic order on L. Clearly this is a linear order on L.
We can now use this ordering to construct ?open intervals.? Define ((a,aprime),(b,bprime))
for (a,aprime), (b,bprime) ? L as the set of points, {(x,xprime) ? L : (a,aprime) < (x,xprime) < (b,bprime)} where
(a,aprime) < (b,bprime). Notice that the intersection of two open intervals is either another open
interval or the empty set. Thus if we let B be the collection of all open intervals on L,
then it is easy to see that B is a basis for L. Let R be the topology generated by B.
We start by showing that L is a Hausdorff space.
Theorem 4.1. (L,R) is a Hausdorff space.
Proof. Let (a,aprime), and (b,bprime) be two distinct points in L. We may assume that (a,aprime) <
(b,bprime). Then either a < b, or a = b and aprime < bprime.
Case 1: Suppose a < b. Since a, b ? R there exists c ? R such that a < c < b. Let
U = ((a ? 1,aprime),(c,aprime)) and V = ((c,aprime),(b + 1,bprime)). Then U and V are open,
(a,aprime) ? U, (b,bprime) ? V, and U ?V = ?.
23
Case 2: Suppose a = b. Then aprime < bprime. Since aprime, bprime ? [0,1] and [0,1] ?R, there exists cprime ?
[0,1] such that aprime < cprime < bprime. Let U = ((a?1,aprime),(a,cprime)) and V = ((b,cprime),(b+1,bprime)).
Then, again, U and V are open, (a,aprime) ? U, (b,bprime) ? V, and U ?V = ?.
Therefore (L,R) is a Hausdorff space.
We prove regularity using the definition of a regular space. Let H? denotes the
interior of H which is an open set.
Theorem 4.2. (L,R) is a regular space.
Proof. Let M be a closed subset of L, and let (x,xprime) ? L such that (x,xprime) /? M. Then
Mc is open and (x,xprime) ? Mc implies that there exists a basic open set ((a,aprime),(b,bprime))
such that (x,xprime) ? ((a,aprime),(b,bprime)) ? Mc. Then there exists (c,cprime), (d,dprime) ? L such that
(a,aprime) < (c,cprime) < (x,xprime) < (d,dprime) < (b,bprime). Let U = ((c,cprime),(d,dprime)) and V = (Uc)?. Then
U and V are open, (x,xprime) ? U, M ? ((a,aprime),(b,bprime))c ? (Uc)? = V, and U ? V = ?.
Hence (L,R) is regular.
Now it is easy to see that L is uncountable, but is it separable, first countable, or
second countable? We will first show that the space L is not separable. From this it will
easily follow that the space is not second countable, since if it were, we could construct a
countable dense set by choosing a point from each basic open set in the countable basis.
Theorem 4.3. (L,R) is not separable.
Proof. Suppose (L,R) is separable. Then there is a dense countable subset of L. Let
D = {(xn,yn) : n ? N} be this subset. Now we will construct an open set which does
not intersect D. Since R is uncountable, we may choose x ? L such that x /? {xn :
24
n ? N}. Then for each n ? N, (xn,yn) /? ((x,0),(x,1)). Hence D is not dense in L, a
contradiction. Therefore (L,R) is not separable.
Corollary 4.1. (L,R) is not second countable.
Even though we cannot find a countable basis for the whole space, we will prove
that we can find a countable local basis at each point. This proves that (L,R) is first
countable.
Theorem 4.4. (L,R) is first countable.
Proof. Let (a,aprime) ? L. Then either aprime ? (0,1), aprime = 1, or aprime = 0. For each point in each
case we will construct a countable local base.
Case 1: Suppose aprime ? (0,1). Let G(a,aprime) = {((a,p),(a,q)) : p, q ? Q? [0,1] and p <
aprime < q}. Let U be an open neighborhood of (a,aprime). Then there is a basic open
set, ((b,bprime),(c,cprime)) such that (a,aprime) ? ((b,bprime),(c,cprime)) ? U. Notice that there exists
p and q ? Q? [0,1] such that (b,bprime) < (a,p) < (a,aprime) < (a,q) < (c,cprime). Then
((a,p),(a,q)) is an element of G(a,aprime), a subset of U, and it contains (a,aprime). Hence
G(a,aprime) is a local base of (a,aprime).
Case 2: Suppose aprime = 0. Let G(a,0) = {((p,0),(a,q)) : p ?Q and q ?Q?(0,1] and p <
a} Again, let U ? T such that (a,0) ? U. Then there is a basic open set,
((b,bprime),(c,cprime)) such that (a,0) ? ((b,bprime),(c,cprime)) ? U. Notice that there exists p
and q ? Q ? [0,1] such that (b,bprime) < (p,0) < (a,aprime) < (a,q) < (c,cprime). Then
((p,0),(a,q)) is an element of G(a,aprime), is a subset of U, and it contains (a,0). Hence
G(a,aprime) is a local base of (a,0).
25
Case 3: Now we are left with the case when aprime = 1. We will construct a countable local
base for (a,aprime) in much the same way as in Case 2. Let G(a,1) = {((a,p),(q,1)) :
p ?Q?[0,1) and q ?Q and q > a}. Again, let U ? T such that (a,1) ? U. Then
there is a basic open set ((b,bprime),(c,cprime)) such that (a,1) ? ((b,bprime),(c,cprime)) ? U. In
much the same way as before, there is an element of G(a,1) which contains (a,1)
and is a subset of ((b,bprime),(c,cprime)). Thus G(a,1) is a local base for (a,1).
Therefore there is a local countable base at each point, which implies that (L,R) is first
countable.
The following theorems deal with specific subspaces and convergent sequences of
(L,R). We wish to categorize the subspaces we mention. To do this we will find a
homeomorphism between the subspace and a familiar space.
Theorem 4.5. Let x ?R. Then [(x,0),(x,1)] with the subspace topology is homeomor-
phic to [0,1] with the usual topology.
Proof. Let X = [(x,0),(x,1)] for some x ?R, and let T|X be the subspace topology on
X. Let [0,1] have the usual topology. Define a function f : X ? [0,1] by f((x,xprime)) = xprime
for each (x,xprime) ? X. Then it is easy to see that f is a well defined function which is also
a bijection. Now we need to show that f is a homeomorphism.
First we will prove that f is continuous by using Theorem 5 in the Chapter 2. This
means that we need to show that the preimage of a basic open set is open. Choose A
to be a basic open set in [0,1]. Then A = (a,b)?[0,1] for some a, b ? R. Notice that
f?1(A) = ((x,a),(x,b))?X which is open in X. Hence f is continuous.
26
Now we need to show that f?1 is continuous. Let V = ((a,aprime),(b,bprime))?X for some
(a,aprime), (b,bprime) ? L. Then we have one of four situations; either V = ((x,aprime),(x,bprime)), or
V = [(x,0),(x,bprime)), or V = ((x,aprime),(x,1)], or V = X.
Case 1: Assume V = ((x,aprime),(x,bprime)). Then f(V) = (aprime,bprime) which is open in [0,1].
Case 2: Assume V = ([(x,0),(x,bprime)). Then f(V) = [0,bprime) = (?bprime,bprime)?Y which is open
in [0,1].
Case 3: Assume V = ((x,aprime),(x,1)]. Then f(V) = (aprime,1] = (aprime,aprime + 1)?[0,1] which is
open in [0,1].
Case 4: Assume V = X. Since f is a bijection, f(V) = f(X) is [0,1] which is definitely
open.
Hence f?1 is continuous. Therefore X is homeomorphic to [0,1] with the usual topology.
Now we have seen what a projection onto the second coordinate looks like, for any
fixed first coordinate. What does a projection onto the first coordinate look like? We
will see that it depends on which second coordinate we fix.
Theorem 4.6. For any y ? [0,1] let Ky = R?{y}, with the subspace topology.
1. If y ? (0,1), then Ky is homeomorphic to the reals with the discrete topology.
2. If y = 0, then Ky is homeomorphic to the reals with the Right Sorgenfrey topology.
3. If y = 1, then Ky is homeomorphic to the reals with the Left Sorgenfrey topology.
27
Proof. Let y ? [0,1], and let Ky = R?{y}. Define f : Ky ?R by f((x,y)) = x for each
(x,y) ? Ky. Then it is obvious that f is a well-defined bijection. We will go on to show
that when we change the value of y and put the corresponding topology on the reals, f
is a homeomorphism.
Lets begin by assuming that y ? (0,1) and that R has the discrete topology. Then
obviously f?1 is continuous since R is discrete. Thus for this case, we only need to show
that f is continuous.
Let U be a basic open set in R. Then U = {x} for some x ? R. So f?1(U) =
{(x,y)} = ((x,0),(x,1)) ? Ky which is open. Hence f is continuous. Therefore Ky is
homeomorphic to the reals with the discrete topology.
Now suppose that y = 0. Let TR denote the Right Sorgenfrey topology on R. So
we need to show that f and its inverse are both continuous. Let (a,b] ? R be a basic
open set in (R,TR). Then f?1((a,b]) = (a,b]?{0} = ((a,1),(b,1))?K0 which is open
in K0. Hence f is continuous.
Next let A be a basic open set in K0. Then A = ((a,aprime),(b,bprime)) ? K0 for some
((a,aprime),(b,bprime)) ? B which impies that either a = (a,b)?{0} or, in the case that bprime = 0,
A = (a,b)?{0}. Then f(A) = (a,b] ? TR or f(A) = (a,b) ? TR since TR is finer than
the usual topology on the reals. Hence f?1 is continuous, and so f is a homeomorphism
between K0 and R with the Right Sorgenfrey topology.
If we let y = 1 and give R the Left Sorgenfrey topology, then it is easy to see that f
is a homeomorphism between K1 and R with the Left Sorgenfrey topology by using an
argument similar to the y = 0 case.
28
We will now discuss how a few interesting sequences converge to some very unex-
pected points.
Theorem 4.7. Let x ?R, y ? [0,1], and let {(xn,yn)}n?N be a sequence in L.
1. If {yn}n?N converges to y in the usual topology on [0,1] and xn = x for all but
finitely many n ?N, then {(xn,yn)}n?N converges to (x,y) in L.
2. If {xn}n?N converges to x in the usual topology and xn < x for all but finitely many
n ?N, then {(xn,yn)}n?N converges to (x,0) in L.
3. If {xn}n?N converges to x in the usual topology and xn > x for all but finitely many
n ?N, then {(xn,yn)}n?N converges to (x,1) in L.
Proof. Let x ?R, y ? [0,1], and let {(xn,yn)}n?N be a sequence in L.
Case 1: Suppose yn ? y in the usual topology on [0,1] and xn = x for all but finitely
many n ?N. Let U be an open neighborhood of (x,y), and let X = [(x,0),(x,1)].
Then U ? X is a nonempty open neighborhood of (x,y) in X. Now, by the hy-
pothesis, (xn,yn) ? X for all but finitely many n ? N. Since yn ? y in the usual
topology on [0,1] and X is homeomorphic to [0,1], {(xn,yn)}n?N converges to (x,y)
in X. Then (xn,yn) ? U ?X for all but finitely many n ? N which implies that
(xn,yn) ? U for all but finitely many n ? N. Hence the sequence {(xn,yn)}n?N
converges to (x,y) in L.
Case 2: Assume that x ? R, and {(xn,yn)}n?N ? L such that {xn}n?N converges to
x in the usual topology on R and xn < x for all but finitely many n ? N. Let
((a,aprime),(b,bprime)) be a basic open set in (L,R) such that (x,0) ? ((a,aprime),(b,bprime)). Then
(a,aprime) < (x,0) < (b,bprime). Thus either a < x < b or a < x = b.
29
Suppose that a < x < b. Then x ? (a,b), and this implies that xn ? (a,b) for all
but finitely many n ? N. Thus (xn,yn) ? ((a,aprime),(b,bprime)) for all but finitely many
n ?N.
Now suppose that a < x = b. Then x ? (a,b] which implies that x ? (a,b + 1).
Hence there exist N1 ? N such that xn ? (a,b + 1) for all n ? N1. Since xn < x
for all n ? N2 for some N2 ?N, xn ? (a,b) for all n ? N1 +N2. Hence (xn,yn) ?
((a,aprime),(b,bprime)) for all n ? N1 +N2. Therefore (xn,yn) ? (x,0).
Case 3: Suppose that {xn}n?N converges to x in the usual topology and xn > x for
all but finitely many n ? N. Then similarly to Case 2, {(xn,yn)}n?N converges to
(x,1).
This idea of using the subspaces to prove properties about L is not isolated to
showing sequence convergence. We will use this idea to show that L is a Lindel?of space.
Then using Theorem 2.10, we have that L is normal as a corollary.
Theorem 4.8. (L,R) is Lindel?of.
Proof. Let G be an open cover of L consisting of basic open sets. Then G is an open
cover of K0 = R?{0} with the subspace topology. Since K0 is homeomorphic to the
right Sorgenfrey topology on the reals, by Theorem 3.6, G has a countable subcover of
K0. Let G0 = {((an,aprimen),(bn,bprimen)) : n ? N} be such a countable subcover of K0. Let A0
be the collection of first coordinates of both endpoints of each interval in G0. In other
words, let A0 = {an : n ?N}?{bn : n ?N}. Notice that G0 covers Ac0 ?[0,1], and that
A0 is countable.
30
We can use a similar argument to find a countable subcover G1 of K1 = R?{1}.
Let A1 be the collecton of first coordinates of both endpoints of each interval in G1.
Again notice that G1 covers Ac1 ? [0,1]. Also G0 ?G1 covers (A0 ?A1)c ? [0,1]. Note
that so far we have a countable subcover of all but those points with first coordinates in
the countable set A1 ?A0.
Now let a ? A0 ? A1. Then Xa = [(a,0),(a,1)] is homeomorphic to the closed
interval [0,1] with the usual topology. Hence for each a ? A0 ?A1, G has a countable
subcover, Ga, of Xa.
Let hatwideG = G0 ?G1 ?(uniontexta?A0?A1 Ga). Then obviously hatwideG ?G. Notice that G0 and G1
are countable. Also A0 ?A1 is countable and, for each a ? A0 ?A1, Ga is countable.
Thus we have that hatwideG is a countable union of countable collections. Now to show that hatwideG
is a cover of L, consider the fact that L = ((A0 ?A1)?[0,1])?((A0 ?A1)c ?[0,1]) =
(uniontexta?A0?A1(uniontextGa))?(uniontextG0)?(uniontextG1) = uniontext hatwideG. Hence hatwideG is a countable subcover of G.
Corollary 4.2. The space L is normal.
Let us now look a little closer at our underlying set, L. To show that our space is
connected and that certain subsets are compact, it would help to have the least upper
bound property for L.
Lemma 4.1. Bounded subsets of L have the least upper bound property.
Proof. Let X be a bounded subset of L. Then there exist points (a,aprime), (b,bprime) ? L such
that for each point (x,xprime) ? X, (a,aprime) < (x,xprime) < (b,bprime). So if we let E = {x : (x,xprime) ?
X}, or in other words let E be the set of all first coordinates of points in X, then E is
bounded on the real line by a and b. Thus E has a least upper bound which we will call
c.
31
Suppose that c ? E. Then X ?[(c,0),(c,1)] has a least upper bound, (c,cprime), since
[(c,0),(c,1)] is homeomorphic to the closed interval [0,1]. Now we claim that in this
case, (c,cprime) is the least upper bound of X. Let (x,xprime) ? X. Then either x < c or x = c
and xprime < cprime. Either way (x,xprime) ? (c,cprime). Now suppose that (d,dprime) is an upper bound of
X. Then either c < d or d = c and cprime ? dprime. Thus (c,cprime) ? (d,dprime). Hence (c,cprime) is the
least upper bound of X.
Now suppose that c negationslash? E. Then for each (x,xprime) ? X, x < c. We claim that in this
case (c,0) is the least upper bound of X. First of all, for each (x,xprime) ? X, x < c implies
that (x,xprime) < (c,0). Suppose (d,dprime) is an upper bound of X. Then either c < d or c = d.
If c < d, then obviously (c,0) < (d,dprime). If d = c, then 0 ? dprime and (c,0) ? (d,dprime). So (c,0)
is the least upper bound of X.
Earlier we described an open interval by the subset ((a,aprime),(b,bprime)) = {(c,cprime) ? L :
(a,aprime) < (c,cprime) < (b,bprime)}. This is very similar to the definition of an open interval on
the real line. The open interval, (a,b), on the real line is the set, {c ? R : a < c < b}.
In any linearly ordered space, this is the form of an open interval. When we talk about
any interval, we are talking about subsets of a linearly ordered set with the same form
as intervals on the real line.
So how does the least upper bound property help us? It is an important element in
showing that intervals are connected.
Theorem 4.9. Every interval in L is connected.
Proof. Let I be an interval in L. Suppose that I is not connected. Then there exist
nonempty disjoint open sets, U and V such that I = U ?V. Hence there exist elements
(a,aprime), (b,bprime) ? I such that (a,aprime) ? U and (b,bprime) ? V. Without loss of generality we
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will assume that (a,aprime) < (b,bprime). Let hatwideU = U ?[(a,aprime),(b,bprime)] and hatwideV = V ?[(a,aprime),(b,bprime)].
Notice that hatwideU and hatwideV are open subsets of [(a,aprime),(b,bprime)] that separate [(a,aprime),(b,bprime)].
Now we will find a point of [(a,aprime),(b,bprime)] that is in neither hatwideU nor hatwideV. Let (c,cprime) be
the least upper bound of hatwideU. Then (c,cprime) ? (x,xprime) for each (x,xprime) ? hatwideU and for any upper
bound, (y,yprime) of hatwideU, (c,cprime) ? (y,yprime). Since (a,aprime) ? hatwideU and (b,bprime) is an upper bound on hatwideU,
we have that (c,cprime) ? [(a,aprime),(b,bprime)].
Suppose that (c,cprime) ? hatwideU. Then there is some basic open set BU such that (c,cprime) ?
BU ?[(a,aprime),(b,bprime)]. Since (b,bprime) negationslash? hatwideU, we know that BU ?[(a,aprime),(b,bprime)] is either of the
form ((d,dprime),(e,eprime)) for some (d,dprime), (e,eprime) ? [(a,aprime),(b,bprime)] or of the form [(a,aprime),(e,eprime))
for some (e,eprime) ? ((a,aprime),(b,bprime)]. Either way ((c,cprime),(e,eprime)) ? BU ?[(a,aprime),(b,bprime)]. How-
ever we know that there is a point (x,xprime) such that (c,cprime) < (x,xprime) < (e,eprime) which implies
that (x,xprime) ? hatwideU, a contradiction since (c,cprime) is the least upper bound. Hence (c,cprime) negationslash? hatwideU.
So (c,cprime) must be in hatwideV. Again there is a basic open set BV such that (c,cprime) ?
BV ?[(a,aprime),(b,bprime)], but this time BV ?[(a,aprime),(b,bprime)] is either of the form ((d,dprime),(e,eprime))
for some (d,dprime), (e,eprime) ? [(a,aprime),(b,bprime)] or of the form ((d,dprime),(b,bprime)] for some (d,dprime) ?
[(a,aprime),(b,bprime)). Thus ((d,dprime),(c,cprime)) ? BV ? [(a,aprime),(b,bprime)], yet there exists a point
(x,xprime) ? ((d,dprime),(c,cprime)) which is not in hatwideU. This makes (x,xprime) an upper bound on hatwideU
which is less than (c,cprime), a contradiction.
So we have that (c,cprime) negationslash? hatwideU ? hatwideV meaning that [(a,aprime),(b,bprime)] is not separated by hatwideU
and hatwideV. Therefore I must be connected.
If we recall that any union of connected sets with nonempty intersection is also
connected, then we get the following corollary.
Corollary 4.3. (L,R) is connected.
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Proof. Let G = {((?n,0),(n,1)) : n ?N}. Then G is a collection of connected sets, and
intersectiontextG = ((?1,0),(1,1)) which is obviously nonempty. So uniontextG = L is connected.
Are there any more connected subsets of L? We will next show that there are not.
Theorem 4.10. If X is a proper nondegenerate subset of L which is not an interval,
then X is not connected.
Proof. Let X be a proper subset of L. Suppose that X is not an interval. Then there
are points (a,aprime), (b,bprime) ? X and (c,cprime) negationslash? X such that (a,aprime) < (c,cprime) < (b,bprime). Let
U = (uniontextn?N((a?n,aprime),(c,cprime)))?X = (??,(c,cprime))?X, and let V = (uniontextn?N((c,cprime),(b +
n,bprime)))?X = ((c,cprime),?)?X. Then U and V are nonempty disjoint open subsets of X
such that X = U ?V.
It may have been apparent that the arguments for connectedness worked in much
the same way as they work for the real line with the usual topology. We will now see
that the same holds for compact subsets of L.
Theorem 4.11. Unbounded subsets of L are not compact.
Proof. Let X be an unbounded subset of L. Suppose that X is compact. Let G =
{((?n,0),(n,1)) : n ? N}. Then G is an open cover of X. Since we assume that X is
compact, there exists N ? N such that GN = {((?ni,0),(ni,1)) : i ? N} is a subcover
of L. Then X ? uniontextGN. Let m = max{ni : i ? N}. Then X is bounded by (?m,0) and
(m,1), a contradiction. Therefore X is not compact.
We know that L itself is unbounded, so it is clear to see that L is not compact.
What type of subsets are compact? If we keep with the idea that the compact
subsets of L resemble the compact subsets of R with the usual topology, then why don?t
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we look into closed bounded subsets of L? Lets first show that closed intervals are
compact.
The following proof was inspired by a proof that the interval [0,1] is compact in the
usual topology on R by Dr. John O?Connor [2].
Theorem 4.12. If X is a closed interval in L, then X is compact.
Proof. Let X = [(a,aprime),(b,b)prime] for some (a,aprime), (b,bprime) ? L. Let G be an open cover of X
consisting of basic open sets. Let
hatwideX = {(x,xprime) ? X : [(a,aprime),(x,xprime)] can be covered by finitely many elements of G}. Let
(c,cprime) be the least upper bound of hatwideX.
We claim that (c,cprime) = (b,bprime). Suppose that (c,cprime) < (b,bprime). Then there exists
((d,dprime),(e,eprime)) ? G such that (c,cprime) ? ((d,dprime),(e,eprime)). Choose (y,yprime) ? ((d,dprime),(c,cprime)).
Then (y,yprime) must be a point in hatwideX. So there exists a finite subcover hatwideG of [(a,aprime),(y,yprime)].
Now choose (z,zprime) ? ((c,cprime),(e,eprime)). Then hatwideG ?{((d,dprime),(e,eprime))} is a finite subcover of
[(a,aprime),(z,zprime)] which implies that (z,zprime) ? hatwideX. However since (c,cprime) is an upper bound
on hatwideX and (c,cprime) < (z,zprime), we have that (z,zprime) negationslash? hatwideX, a contradiction. Also, (c,cprime) ? (b,bprime)
because (b,bprime) is an upper bound on hatwideX. Hence (c,cprime) = (b,bprime).
Now choose ((d,dprime),(e,eprime)) ?G such that (b,bprime) ? ((d,dprime),(e,eprime)). Then we know that
(d,dprime) ? hatwideX by the prior claim. So there exists a finite subcover, Gd, of [(a,aprime),(d,dprime)]. No-
tice that Gd?((d,dprime),(e,eprime)) is a finite subcover of [(a,aprime),(b,bprime)]. Therefore [(a,aprime),(b,bprime)]
is compact.
We can now expand this idea to bounded closed subsets of L.
Corollary 4.4. Every bounded closed subset of L is compact.
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Proof. Let X be a closed subset of L, and let X be bounded by (a,aprime), (b,bprime) ? L.
By Theorem 4.12,[(a,aprime),(b,bprime)] is compact. Since X is closed in L, X is closed in
[(a,aprime),(b,bprime)]. Now recall that closed subsets of a compact space are compact.
Since we have shown that L is a Hausdorff space, every compact subset of L must
be closed, and by Theorem 4.11, it must be bounded. So these are the only compact
subsets of L.
This will conclude our exploration of the Lexicographic order on R?[0,1], and in
essence this paper.
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Bibliography
[1] Herbert S. Gaskill and P. P. Narayanaswani, ?Elements of Real Analysis,? Prentice
Hall,1998.
[2] John O?Connor, ?Lecture Notes for MT4522,? http://www-history.mcs.st-
and.ac.uk/ john/MT4522/Lectures/L21.html, 2004.
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