A Study of Regularity and Normality
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.
Mitchell Jaeger
Certificate of Approval:
Narendra Govil
Professor
Mathematics & Statistics
Michel Smith, Chair
Professor
Mathematics & Statistics
Geraldo De Souza
Professor
Mathematics & Statistics
George T. Flowers
Dean
Graduate School
A Study of Regularity and Normality
Mitchell Jaeger
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 9th, 2009
A Study of Regularity and Normality
Mitchell Jaeger
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
Mitchell John Jaeger, son of Richard & Vicke Jaeger, was born in Green Bay, Wis-
consin on June 13th, 1984. After graduating from Oconto Falls High School in 2002, he
entered the University of Wisconsin - Madison as a Nuclear Engineering major. In 2006, he
graduated with a dual degree in Nuclear Engineering and Mathematics. He graduated with
the distinction of distinguished scholar. He then enrolled in Auburn University to pursue a
Master Degree in Mathematics.
iv
Thesis Abstract
A Study of Regularity and Normality
Mitchell Jaeger
Master of Science, May 9th, 2009
(B.S., University of Wisconsin - Madison, 2006)
32 Typed Pages
Directed by Michel Smith
The paper is a broad overview of the separation axioms, commonly known as the
Hausdorff, Regular and Normal axioms. After a review of basic definitions and lemmas, we
proceed to certain examples and theorems dealing with Paracompact and Lindel?of spaces.
We then progress to metric spaces, separable spaces and countability before we end with
several examples of spaces that satisfy certain separation axioms and not others, and the
properties that these spaces satisfy.
v
Acknowledgments
I would like to thank my advisor, Dr. Michel Smith, for guiding my graduate career
from beginning to end. I would also like to thank my committee members Dr. Narendra
Govil, Dr. Geraldo De Souza and Dr. George Flowers. Furthermore, I would like to thank
all of my friends that I have made in the mathematics department here at Auburn. I would
not have been able to do this without them.
I would like to thank my family for their love and support. I would also like to thank
my parents for believing in me and encouraging me in whatever I chose to pursue as a
career. I can not say enough for the foundation they have laid for me.
Finally, I thank the Lord for all the blessings He has bestowed upon me. I would be
nothing without Him.
vi
Style manual or journal used Journal of Approximation Theory (together with the style
known as ?aums?). Bibliography follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (specifically LATEX)
together with the departmental style-file aums.sty.
vii
Table of Contents
1 Introduction 1
2 Background Definitions 2
3 Paracompact and Lindel?of Spaces 6
4 Metric Spaces 9
5 Separability and Countability 13
6 regularity 15
7 Normality 18
Bibliography 24
viii
Chapter 1
Introduction
Topology is one of the major areas of Abstract Mathematics. In essence, it is the study
of geometric spaces and continuity. The Euclidean plane is a topological space, but in fact
any set X may be a topological space if it satisfies certain requirements. Namely, if there
exists a collection T of subsets of X such that (1) both the empty set and X are in T, (2)
the union of the elements of any subcollection of T is in T, and (3) the intersection of the
elements of any finite subcollection of T is in T. If these three criteria are met, then T is
called a topology on X, and X together with T form a topological space.
These topological spaces allow the formal definition of concepts such as convergence,
connectedness, and continuity. They appear in virtually every branch of modern mathe-
matics and are a central unifying notion. In this thesis we will take a closer look at some
properties held by topological spaces, namely Regularity and Normality, and the types of
spaces that have them, and those that don?t.
1
Chapter 2
Background Definitions
If X is a set, a basis for a topology on X is a collection B of subsets of X (called basis
elements) such that: (1) For each x ? X, there is at least one basis element B containing
x and (2) If x belongs to the intersection of two basis elements B1 and B2, then there is a
basis element B3 containing x such that B3 ? (B1?B2). If B satisfies these two conditions,
then we define the topology T generated by B as follows: A subset U of X is said to be open
in X (or to be an element of T) if for each x ? U, there is a basis element B ? B such that
x ? B and B ? U. Then B is a basis for the topology on X and each member of B is called
a basic open set of X.
A topological space X is called a Hausdorff space if for each pair a,b of distinct points
of X, there exist disjoint open sets containing a and b respectively. We sometimes refer to
this as the Hausdorff axiom, one of the separation axioms.
A space X is said to be regular if for each pair consisting of a point x and a closed set
B disjoint from x, there exist disjoint open sets containing x and B, respectively. The space
X is said to be normal if for each pair A, B of disjoint closed sets of X, there exist disjoint
open sets containing A and B respectively. These are the other two separation axioms to
be dealt with in this thesis.
Suppose that X is a space, M ? X and G is a collection of open sets which covers the
set M. Then the collection Gprime is said to be a refinement of G covering M if and only if for
each point x ? M there is an element of Gprime that contains x and lies in some element of G.
2
Suppose that X is a space and G is a collection of subsets of X. Then G is said to be
locally finite if and only if for each point x ? X there is an open set U containing x so that
U intersects only finitely many elements of G.
The topological space X is said to be paracompact if and only if it is Hausdorff and
every collection of open sets that covers X has a locally finite refinement that covers X.
The topological space X is said to be Lindel?of if and only if for every collection G of
open sets that covers X, there is a countable subcollection Gprime ? G that covers X.
A space X is said to be completely separable if it has a countable basis.
A subset A of X is said to be dense in X if any open set of X intersects A.
A space X is said to be separable if it has a countable dense subset.
A metric on a set X is a function d : X ?X ??R having the properties:
d(x,y) ? 0 for all x,y ? X; equality holds if and only if x = y.
d(x,y) = d(y,x) for all x,y ? X.
(Triangle inequality) d(x,y) +d(y,z) ? d(x,z), for all x,y,z ? X.
Given r > 0, the r-Ball is defined as B(x,r) = {y : d(x,y) < r}. If these r-balls form a
basis for a topology over X, then we define this topology to be the metric topology.
A space X is first countable if it has a countable basis for each point x ? X.
3
LetW be the collection of all half-open real intervals of the form [a,b) = {x : a ? x < b},
where a < b. Then the topology generated by W on the real line is called the Sorgenfrey
line.
Let S be the sorgenfry line. Let X = S ? S be the product space of the Sorgenfry
line crossed with itself (called the Sorgenfrey Plane) in which a basic open set containing a
point (x,y) is a square containing only the border below and the border to the left of the
rectangle with the point (x,y) as its lower left vertex. Thus we will denote a basic open
square of length epsilon1 containing a point a to be S(a,epsilon1).
Lemma 1.1 (Pasting Lemma) Let X = A?B, where A and B are closed in X. Let
f : A ? Y and g : B ? Y be continuous. If f(x) = g(x) for every x ? A?B, then f and g
combine to give a continuous function h : X ? Y, defined by setting h(x) = f(x) if x ? A
and h(x) = g(x) if x ? B.
Example 1.2 Let X = {(x,y)|y ? 0} be the space given by the upper half plane
including the x-axis. Define all basic open sets containing a point (x,y) above the x-
axis to be the product of all open intervals U = (a,b) (such that x ? U) and all open
V = (c,d),c > 0 (such that y ? V). Define all basic open sets containing a point (x,0) on
the x-axis to be the union of the point (x,0) with the product of all open intervals U = (a,b)
(such that x ? U) and all open intervals V = (0,d). We define this space to be the tangent
rectangle topology.
Example 1.3 Let X = {(x,y) : x,y ? R,y > 0} be the upper half-plane with the usual
Euclidean topology Tx. Let Y be the real line that represents the x-axis. Let Ty be the
topology generated by all sets of the form {x}? D where x ? Y and D is an open disc
in X that is tangent to Y at x. Let Z = X ? Y and let the topology Tz on Z be the
4
topology generated by Tx together with the topology generated by Ty. We shall refer to
this topological space as the Tangent Disc Topology.
A space X is perfectly normal if X is normal and every closed subset of X is the
common part of countably many open sets.
Suppose that X is a metric space with metric d. Then the sequence {pi} is said to be
a Cauchy Sequence if and only if for each epsilon1 > 0 there exists and integer N so that if n and
m are integers larger than N, then d(pn,pm) < epsilon1.
Suppose that X is a metric space with metric d. Then the metric is said to be a
complete metric for X if and only if every Cauchy sequence (with respect to the metric d)
in X converges to a point in X. A space is said to be a complete metric space if and only
if there is a complete metric for the topology of X.
A Baire space is a topological space in which every countable intersection of dense open
sets is non-empty. Often an equivalent definition is used that states a topological space is
a Baire Space if given any countable collection of nowhere dense closed sets, their union is
not the whole space.
5
Chapter 3
Paracompact and
Lindel?of Spaces
Theorem 3.1 If M is a closed subspace of the paracompact space X then it is also
paracompact.
Proof: Let M be a closed subspace of the paracompact space X. Let A be a covering
of M by sets open in M. For each ? ? A, we pick an open set ?prime of X such that ?prime?M = ?.
Then we cover X by the open sets ?prime and the open set X ?M. We then let B be a locally
finite open refinement of this covering that covers X. Then R = {? ?M : ? ? B} is a
locally finite open refinement of A, and thus M is paracompact. ?
Theorem 3.2 Every paracompact Hausdorff space is regular.
Proof: Let X be a paracompact space. Also let x ? X and let H ? X be closed such
that x /? H. Take an h ? H, then there exists an Ah with h ? Ah and a Bh with x ? Bh
such that Ah ?Bh = ?. We can then see that {Ah : h ? H}?{X ?H} covers X. Then,
by paracompactness, there exists a locally finite refinement R which covers X. We define
a new set C of elements of R such that C = {S ? R : S ?H negationslash= ?}. If U = uniontextS?C S, then
obviously U is open as the union of open sets that contains H. Now, since R is locally
finite, there exists an open O with x ? O that only meets a finite number of sets in R.
We call these sets A1,A2,...,An. Each Ai is contained in some Ahi or X ?H. If we define
B := Ointersectiontext(intersectiontextni=1 Bhi) then B is open with x ? B and U ?B = ?. ?
Theorem 3.3 Every paracompact Hausdorff space is normal.
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Proof: Let X be a paracompact space. From the previous theorem we know that X is
regular. So let A and B be closed sets of X such that A?B = ?. Since X is regular, we
can choose, for each b ? B an open set Ub containing b and an Vb containing A such that
Ub ?Vb = ?. We then cover X with the open sets Ub and the open set X ?B. Since X
is paracompact, we take a locally finite open refinement C that covers X. We then form
the subcollection D of C that consists of all the elements of C that intersect B, that is
D = {S ? C : S ?B negationslash= ?}. Then D covers B. We then see that for every a ? A there exists
some open Oa containing a that intersects only finitely many elements of the locally finite
refinement C. Let these elements that Oa intersects be denoted by Sa1,...,San. Each of
these Sai is contained in some element Ubai of the original cover of X. Each of these Ubai
has a corresponding Vbai that contains a but misses Ubai. Therefore, Oa ?Vbai contains a
but misses Ubai and thus misses Sai. So, we see that the open set Wa = Oa?Vba1 ?...?Vban
contains a but misses ?b?BUb the open set containing B. Such a open Wa can be found for
every a ? A. So ?a?AWa is an open set that contains A and misses ?b?BUb, the open set
containing B and thus X is normal. ?
Theorem 3.4 A completely separable space is Lindel?of.
Proof: Let X be a completely separable space and let B = {Bn}?n=1 be a countable
basis of X and C and open cover of X. For each natural number n, let cn ? C be an
element of C that contains Bn if one exists. If one does not exist, then define cn = ?. We
then define Cprime = {cn : cn negationslash= ?}. Notice that Cprime is countable. We then claim that this Cprime
covers X. We show this by letting x ? X. Then there is some c ? C that contains x since C
covers X. Then there exists a Bm ? B with x ? Bm ? c. Thus we see there is some cn ? Cprime
with Bm ? cn. Therefore x ? cn which implies Cprime covers X, and thus X is Lindel?of. ?
Theorem 3.5 A separable paracompact space is Lindel?of.
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Proof: Let X be a separable paracompact space. Let I be an index set such that
{U?}??I is an open cover of X. By the paracompactness of X, we can find a locally finite
open refinement {V?}??J for some index set J, of {U?}??I where we assume V? negationslash= ? ? ?.
Since X is separable, it has a countable dense subset {xi : i ?Z}, each V? contains at least
one xi. We suppose towards a contradiction that {V?} is an uncountable collection. Then
there exists some xi contained in uncountably many V?. But this contradicts the local
finiteness of {V?}. Thus we conclude that {V?} is in fact a countable collection. So, for
each V?, we choose some U?(?) ? V?. Then braceleftbigU?(?)bracerightbig is a countable subcover of {U?}, and
thus X is Lindel?of. ?
8
Chapter 4
Metric Spaces
Theorem 4.1 A metric space is normal
Proof: Let X be a metric space and let d be the metric on X. Let A and B be
two disjoint closed sets in X. We need to show there are two disjoint open sets U and V
containing A and B respectively in X. For each element a ? A, choose an ra such that the
open ball B(a,ra) around a with radius ra does not intersect the closed set B. Similarly, for
each b ? B, we choose an rb such that B(b,rb) does not intersect A. Then we define the open
sets U and V to be U = ?a?AB(a,ra/2) and V = ?b?BB(b,rb/2). It is clear that U contains
the closed set A and that V contains the closed set B. We claim that U?V = ?. We assume
towards a contradiction that this is not the case. Then there is some element x ? U ?V.
This would imply that for some a ? A and some b ? B, x ? B(a,ra/2)?B(b,rb/2). Then
the triangle inequality tells us that d(a,b) < (ra +rb)/2. So if ra ? rb, then d(a,b) < rb we
see that B(b,rb) contains a ? A. But this is not possible by the definition of B(b,rb). On
the other hand, if rb ? ra, then d(a,b) < ra. This then implies that B(a,ra) conains b ? B.
But again, this is not possible by the definition of B(a,ra). Therefore, we see that U and
V are disjoint, and therefore X is normal. ?
Let A be an index set. For every a ? A let Ia = (0,1]?{a}. Then define the space X
to be X = uniontexta?A Ia ?(0,0). Furthermore, define a metric on X to be:
d((x,a),(y,b)) = |x?y| if a = b
d((x,a),(y,b)) = |x|+|y| if a negationslash= b
9
Notice that if m and n are two elements in X, then d(m,n) ? 0, and if l,m,n ? X
then d(l,m) + d(m,n) = |l ? m| + |m ? n| ? |l ? n| = d(l,n). Thus d((x,a),(y,b)) is a
metric on X. The space X together with this metric d is known as the hedgehog space based
on the index set A.
Theorem 4.2 The hedgehog space based on the uncountable index set A is a non-
separable metric space.
Proof: Let H be the hedgehog space which we know to be metric, and let D be any
dense set in H. Define Ux to be the open subset Ux = (1/3,2/3)?{x} of one of the ?spines?.
Since D is dense in H, there exists some d ? D such that d ? Ux. However, since there
is a spine Ix for every a ? A, there exists uncountably many disjoint open sets Ux which
must each contain distinct dx ? D. Then, since the reals are uncountable, D must also be
uncountable. Therefore, there can not exist a countable dense subset of H, and thus H is
non-separable. ?
Theorem 4.3 A separable metric space is completely separable.
Proof: Let X be a separable metric space, and let A = {a0,a1,...} be its countably
dense subset. Then take the open balls centered at each ai with radius epsilon1 = d(ai,aj) for
j negationslash= i so that there are countably many balls centered at countably many points, thus leaving
countably many open sets we claim to be a basis for X. First we show these open sets cover
X. It is clear that these open sets cover A, so we must only look at some element p ? X?A
and show that it is contained in one of the open sets. Fix some ai ? A and let r = d(p,ai)/2.
Then let U = B(p,r) be an open ball of radius r centered at p. Since A is dense, this open
U must contain some aj ? A. We then notice that d(p,aj) < d(p,ai)/2 which implies that
2d(p,aj) < d(p,ai). Moreover, by the triangle inequality, we see that d(p,aj) + d(aj,ai) ?
d(p,ai). This implies that d(aj,ai) ? d(p,ai)?d(p,aj) > 2d(p,aj)?d(p,aj) = d(p,aj) since
2d(p,aj) < d(p,ai). Thus the open ball B(aj,d(ai,aj)) contains the point p, and thus the
10
open sets described above over X. To show that this cover is a basis of X, suppose we have
an open set O containing some x ? X. Then there exists some r 0 such that B(x,r) ? O.
Then suppose we take B(x,r/8) which must contain some an from the countable dense
subset. Also, we can look at the open set B(x,3r/8) minus the closure of B(x,r/4). This
is an open set, being an open set minus a closed set. Since this open set is disjoint from
B(x,r/8), it must contain another distinct element am of the countable dense subset. Now,
by construction, r/8 < d(an,am) < r/2. This then implies that B(an,d(an,am)) contains x
and lies completely in B(x,r) ? O. Thus the constructed cover is a basis for the topology
on X. ?
Baire Category Theorem (4.4) If X is a complete metric space, then X is a Baire
Space.
Proof: Let X be a complete metric space with metric d and let {Un} be a countable
collection of dense open sets in X. Let xo ? X and epsilon1o > 0 be given. Since U1 is dense
and open, we can choose epsilon11 > 0 and x1 ? U1 such that d(xo,x1) < epsilon1o/2, epsilon11 < epsilon1o/2 and
B(x1,epsilon11) ? U1. Then we can choose epsilon12 > 0 and x2 ? U2 such that d(x1,x2) < epsilon11/2, epsilon12 <
epsilon11/2 and B(x2,epsilon12) ? U1. By induction, we can construct a sequence xn with xn ? Un for
each n and a sequence epsilon1n such that d(xn?1,xn) < epsilon1n?1/2, epsilon1n < epsilon1n?1/2 and B(xn,epsilon1n) ? Un.
Therefore for some positive integer j less than n, we have
d(xj,xn) < epsilon1j(1/2+...+1/2n?j) < epsilon1j ? epsilon1o/2j.
Thus the sequence xn, n = 1,2,3,... is a cauchy sequence and converges by hypothesis
to some x ? X. We also know that for every n, d(x,xn) ? epsilon1n. Moreover, it follows that
d(x,xn) ? d(x,xn+1) +d(xn,xn+1) < epsilon1n+1 +epsilon1n/2 < epsilon1n.
11
This then implies that d(x,xn) < epsilon1n for every n, implying d(x,xo) < epsilon1o. Thus x ? Un
for all n = 1,2,3..., and X is therefore a Baire space. ?
12
Chapter 5
Separability and Countability
Example 5.1 The Sorgenfrey Line is separable. This can be easily seen since the
rational numbers are dense in the topology generated by the Sorgenfrey Line. Let W be
the Sorgenfrey Line, and let r be any real number. Then [r,?) is an open set containing
r. Thus, any basis for the space W must contain some basic open set Ur that contains r
that is contained in [r,?). Also note that if r1 < r2 then r1 cannot lie in Ur2. Since this is
true for all real numbers r ? R, we see that there must at least be uncountably many basic
open sets for any basis of W. ?
Example 5.2 The Sorgenfrey line is first countable. Let X be the Sorgenfrey line and
let x ? X. Then the set of all basis elements of the form [x,x+ 1/n) for all natural numbers
n is a basis for each point x in X. And since this set has countably many elements, X is
first countable. ?
Example 5.3 The Sorgenfrey line is not completely separable. Let S be the Sorgenfrey
line. Let B be a basis for S. Choose for each x ? S an element Bxn of B that contains x
such that Bxn ? [x,x + 1/n). Thus x = inf(Bxn|n ? N), and Bxn = Byn implies x = y.
For any y such that x < y, Bxn does not contain y. Thus for each x ? S we must have a
local basis. Since S has uncountably many elements, we see that B must be uncountable.
?
Example 5.4 The Tangent Rectangle Topology is separable since the subset of the
rationals crossed with the positive rationals are countable and dense in X.?
13
Example 5.5 The Tangent Rectangle Topology X is not completely separable. Let L
be the x-axis and let x ? L. Clearly L ? X. If we choose a basic open set U in the Tangent
Rectangle Topology containing x, then L?U = x. Since L countains uncountably many
elements, this implies that there would necessarily have to be uncountably many basic open
sets to cover all of L, and thus X can not be completely separable. ?
14
Chapter 6
regularity
Theorem 6.1 A space X is regular if and only if given a point x ? X and an open
set U containing x, there is an open set V ? U such that x ? V and the closure of V is
contained in U.
Proof: First suppose X is a regular space and that x ? X. Suppose there is an open
set U ? X such that x ? U. Let A = X ?U. Then A is a closed subset of X and x /? A.
Thus, since X is regular, we can find two disjoint open sets V and W such that x ? V and
A ? W. Now we see that the closure of V misses A, since if this were not true A would
contain a limit point of V. If y ? A, then W is open and contains y but misses V. Thus y
is not a limit point of V, and thus the closure of V misses A and is contained in U.
Conversely, suppose there exists a point x ? X and a closed set A ? X such that
x /? A. Let U = X?A. Thus U is open and contains x. By hypothesis, we can find an open
V that contains x and whose closure is contained in U. If we let W be the complement in
X of the closure of V, then W is open and contains A since the closure of V is a proper
subset of U and thus completely misses A. Therefore, V ? W = ? and x ? V, A ? W,
implying that X is regular. ?
Theorem 6.2 A metric space is regular.
Proof: By Theorem 4.1, we know that a metric space is normal. And since normality
implies regularity, we know that a metric space is regular. ?
Example 6.3 The Tangent Rectangle Topology is Hausdorff but not regular.
15
Proof: First we show that X is Hausdorff. Let p,q ? X such that p negationslash= q. Then
p = (xp,yp), q = (xq,yq). If both yp,yq > 0, then we have two choices. Since p negationslash= q, we
know that either xp negationslash= xq or yp negationslash= yq. Suppose first that xp negationslash= xq, then we can find two
disjoint open intervals Up and Uq on the x-axis that correspond to basic open sets in X
such that xp ? Up, xq ? Uq and Up ?Uq = ? since the real line is Hausdorff. If xp = xq,
then we know that yp negationslash= yq. In this case we can find disjoint open intervals Vp and Vq that
correspond to basic open sets in X on the positive y-axis such that yp ? Vp and yq ? Vq. If
yp = 0 and yq > 0, then we can find two disjoint open intervals on the y-axis Vp = (0,a) and
Vq = (b,c) such that b > a that correspond to disjoint basic open sets in X containing p and
q respectively. A similar argument follows if yq = 0 and yp > 0. Now if both yp = yq = 0,
then we know that xp negationslash= xq. We choose two disjoint open intervals on the x-axis Up and Uq
that correspond to disjoint basic open sets in X containing p and q respectively. Therefore,
we conclude that X is Hausdorff.
To show that X is not regular, we take a point p on the horizontal axis. Then we take
a basic open set U in the tangent rectangle topology containing p. We know that this open
set is of the form {(a,b)?(0,c)} ?{p} where a < p < b. Now we choose another basic
open set V containing p such that V ? U. Then V is of the form (d,e)?(0,f) ?p, where
a < d < p < e < b and f < c. But now the closure of V in not fully contained in U. This is
because the closure of V contains all points on the horizontal axis between (d,0) and (e,0),
which are explicitly not contained in U. And thus X can not be regular. ?
Example 6.4 The Tangent Disc Topology is regular. Suppose we have a point x in
Z. There are two possibilities for x. Either it is in X (the open upper half plane) or it is
in Y (the horizontal axis). First let us suppose x ? X. We can then find an r > 0 and
the basic open ball B(x,r). If we take the open ball B(x,r/2) which contains x and whose
closure is properly contained in B(x,r), then we have regularity by the Theorem 6.1. In
16
the case where x is in Y, we can then find an r > 0 and the basic open disc tangent to the
horizontal axis at x of diameter r. We then consider the open disc tangent to x of diameter
r/2. The closure of this disc is properly contained in the disc of radius r, and hence again
by theorem 6.1, we have regularity of Z. ?
17
Chapter 7
Normality
Example 7.1 The Sorgenfrey Plane is not normal. We first look at the diagonal across
the space with negative slope, that is A = {(x,y) : y = ?x}. Clearly A is closed in X and
is discrete since we can find open neighborhoods S(a,epsilon1) of each point a ? A such that
S(a,epsilon1) ? A = {a}. Now let Q be the rationals and let B = {(b,?b) : b /? Q}. It is also
easy to see that B and A?B are closed in X. We claim that we cannot find disjoint open
sets in X containing B and A?B respectively. Let U and V be disjoint opens sets such
that B ? U and A ? B ? V. For each irrational real number b, there exists some open
square S(b,rb) with (b,?b) as its lower left vertex lying in U. We then define the subset
Bn ? B to be Bn = {(b,?b) ? B : rb > 1/n}. Thus we form the countable collection of
subsets {Bn}. Since A?B is clearly countable (as it contains all of the elements (q,?q),
where q is a rational number), and since {Bn} contains all (b,?b), b irrational, we form a
countable covering of A in the usual Euclidean topology. Now since A is nonmeager in the
usual Euclidean topology (it is not the countable union of nowhere dense sets), we see that
at least one of the Bn fails to be nowhere dense. Thus there must be some interval J in A
such that {(x,?x) ? A : rx > 1/m} is dense in the usual Euclidean topology. This means
that any open set containing some (q,?q), q rational, in the interval J must intersect the
open set V. Thus U and V cannot be disjoint, and therefore X is not normal. ?
Theorem 7.2 If X is normal, then for each closed set A in X and any open set O
containing A, there exists and open set U such that A ? U ? ?U ? O.
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Proof: Let X be a normal space and let A be a closed subset of X. Suppose O is open
in X such that A ? O. Then Oc is closed and Oc ?A = ?. Thus there exists disjoint open
sets U and V such that A ? U and Oc ? V. Since U?V = ?, U ? V c which in turn implies
that ?U ? V c because V is open. But we know that Oc ? V which implies that V c ? O and
so we have A ? U ? ?U ? O. And thus for each closed set A and any open set O containing
A, there exists an open set U such that A ? U ? ?U ? O. ?
Theorem 7.3 X is normal if and only if for each pair of disjoint closed sets A and B,
there exists a continuous function from X into the unit interval [0,1] such that f(A) = 0
and f(B) = 1.
Proof: Let f : X ?? [0,1] be a continuous function from X into [0,1] such that
f(A) = 0 and f(B) = 1. Let U be an open subset of [0,1]. Since f is continuous, the
inverse image of U under f is open. If we let U = [0,a) and V = (b,1] be two open sets
of [0,1] such that a < b, then U ? V = ?. Also, since U and V are open, we know that
f?1(U) and f?1(V) are open, and because 0 ? U and 1 ? V, we have f?1(0) ? f?1(A) and
f?1(1) ? f?1(V). Furtermore, since U ?V = ?, we get f?1(U)?f?1(V) = ? and X is thus
normal.
Conversely, let A and B be disjoint closed subsets of a normal space X. Set U1 = Bc.
Then U1 is an open set that contains A. Now since X is normal, we can use Theorem
7.2 to find an open set U1/2 such that A ? U1/2 ? ?U1/2 ? U1. Similarly, by normality
of X, there exists open sets U1/4 and U3/4 such that A ? U1/4 ? ?U1/4 ? U1/2 ? ?U1/2 ?
U3/4 ? ?U3/4 ? U1. Continuing in this way, we define an open set Ud for every d in the set
D = {m/(2n) : m,n ? Z;m ? 2n} such that A ? Ud ? ?Ud ? U1 and ?Ud ? Uc if d < c. Now,
if we define the function f to be:
f(x) = inf {d : x ? Ud,d ? D}
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f(x) = 1 if x /? Ud if d ? D
Then it is clear that f(A) = 0 and f(B) = 1, and we need only check that f is
continuous. Take some basic open set V in [0,1]. Then V = (a,b) for some a,b ?
[0,1],a < b. This also means that V = [0,b) ? (a,1], the intersection of two open sub-
sets in [0,1]. Then f?1(V) = f?1([0,b)?(a,1]) = {x ? X : 0 ? f(x) < b and a < f(x) ?
1 = {x ? X : 0 ? f(x) < b}?{a < f(x) ? 1} = f?1([0,b))?f?1((a,1]).
However, f?1([0,b)) = {x ? X : 0 ? f(x) < b} = ?d<b{Ud : d ? D} and thus is open
being the union of open sets. Furthermore, for a ? [0,1), f(x) > a if and only if x /? ?Ud
for some d > a. Then {x : f(x) > b} = ?d>abraceleftbig ?Udc : d ? Dbracerightbig is again open. Thus the inverse
image of an open set under f is again open, implying f is continuous, thereby finishing the
proof. ?
Example 7.4 The set X of all countable ordinals together with the first uncountable
ordinal is normal, but it is not perfectly normal. X is clearly normal. With the usual
notation, if we let the first uncountable ordinal to be ?1, then the set {?1} is closed in X,
but it cannot be the countable common part of open sets in X. To see this, take any open
set that contains ?1. Since all the other elements in the set are countable, the compliment of
the open set must have only countably many elements in it. This leaves uncountably many
elements in the open set containing ?1, and this will be true for any open set containing ?1.
And thus we would need {?1} to be the common part of at least uncountably many open
sets, and thus X is not perfectly normal. ?
Theorem 7.5 X is perfectly normal if and only if for each pair A,B of disjoint closed
subsets of X, there exists a continuous function f from X into [0,1] such that the inverse
image of 0 under f is A and the inverse image of 1 under f is B.
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Proof: First suppose there is a continuous function f(x) such that f(x) = 0 if and only
if x ? A for some closed set A ? X. Let Gn = {x ? X : f(x) < (1/n)}. Then intersectiontext?n=1
{x ? X : f(x) < (1/n)} = {x ? X : f(x) = 0} = A, and thus the closed set A is the common
part of countably many open sets.
Now suppose the space X is perfectly normal. That is, suppose X is normal and that
every closed subset of X is the common part of countably many open sets. Let A and B
be two disjoint closed subsets of X. We need to define a continuous function h : X ? [0,1]
satisfying the conditions set forth in the theorem. Since X is perfectly normal, we know
that the closed set A can be written as G1 ?G2 ?G3 ?... where each Gn is open. Assume
Gn ? B = ?, otherwise replace Gn with Gn ? B. Since X is normal, we know that for
each n there exists a continuous function fn : X ? [0,1] such that fn(x) = 0 for x ? A
and fn(x) = 1 for x ? X ?Gn. We then define f(x) = summationtext?n=1(1/2n)fn(x). We claim that
this function is continuous. We first notice that f(x) is the limit of the partial sums of the
infinite summation. Each of these finite partial sums (call them gN = summationtextNk=1(1/2k)fk(x))
is clearly continuous being the finite sum of continuous functions. Since each of the fn(x)
maps all of X into the unit interval, and since each fn(x) is multiplied by 1/2n, the sequence
of the partial sums is a uniformly convergent sequence. Thus we let epsilon1 > 0 and let xo be a
given point of X. Since the sequence is uniformly convergent, we know there exists a M
such that for any x, we have |gM(x) ? f(x)| < epsilon1/3, and since gM(x) is continuous at all
x ? X, it is also continuous at xo. Thus there is an open set O containing xo such that
|gM(x)?gM(xo)| < epsilon1/3 for each x ? O, implying that |f(x)?f(xo)| < epsilon1. And thus we see
that xo is a point of continuity of f, and since the choice of xo was arbitrary, we see that
the function f is continuous at all x ? X, that is f is continuous.
Now notice that for each n , B ? X ? Gn. This implies that fn(x) = 1 for x ? B.
So we know there exists a continuous f : X ? [0,1] such that x ? A ? f(x) = 0 (since
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if x /? A and x /? Gm for some m, then x ? X ? Gm implying fm(x) = 1) and x ? B
implies f(x) = 1. Now let D = {x ? X : f(x) < 1/2} , E = {x ? X : f(x) = 1/2} , and
F = {x ? X : f(x) > 1/2}. Then we see that D ? E and E ? F are closed and that
(D?E)?B = ?. Thus we have two closed disjoint sets in X. Now by a similar argument
as above, we can find a new continous function g : X ? [1/2,1] such that x ? (D ? E)
implies that g(x) = 1/2 and x ? B if and only if g(x) = 1. So if we define a new function
h : X ? [0,1] to be such that h(x) = f(x) for x ? D ?E and h(x) = g(x) for E ?F, the
we see that h is continuous (by the pasting lemma) since (D ?E)?(E ?F) = E and for
x ? E, f(x) = 1/2 = g(x) and (D?E)?(E ?F) = X. ?
Example 7.6 The tangent open disc topological space is not normal. First it should
be noted that the horizontal axis Y is closed in Z. This is because each basic open set
containing some point z ? Y contains at most only that one point of Y. Therefore the
rationals Q and the irrationals I in Y are disjoint closed subsets of Z. Suppose U and V
are open sets in Z such that Q ? U and I ? V. Then for each z ? I there exists some
open disc Dz ? V of radius rz which is tangent to Y at z. Now we define the subset In of
I to be In = {z ? I : rz > 1/n}. Then we form the countable collection of sets {In}. Since
Q is countable and {In} clearly contains all the irrationals, they form a countable covering
of Y under the usual Euclidean topology. We know that in the usual Euclidean topology,
the real line is not the countable union of nowhere dense sets. We thus claim that at least
one of the In fails to be nowhere dense in the usual Euclidean topology. We know that
each element of Q is nowhere dense, and thus we know for some fixed positive integer m,
there is an interval [e,f] in Y such that the set {z ? I : rz > 1/m} is dense in [e,f]. This
means that every open set containing an element of the rationals in the interval (e,f) must
intersect the open set V. But this in turn means that U and V cannot be disjoint. And
thus we have two closed sets that can not be contained in two disjoint open sets, and thus
the tangent open disc topological space is not normal. ?
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Bibliography
[1] K. Kuratowski, Topology, vol. II, Academic Press, New York, 1968.
[2] John G. Hocking and Gail S. Young Topology, Addison-Wesley Publishing Co., New
York, 1961
[3] James R. Munkres, Topology, Second Edition, Prentice Hall Inc., Upper Saddle River,
NJ, 1975
[4] Class notes from Topology course work, completed during the 2006-2007 school year
at Auburn University
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