Download Chapter 7 Separation Properties

Chapter VII
Separation Properties
1. Introduction
“Separation” refers here to whether objects such as points or disjoint closed sets can be enclosed in
disjoint open sets. In spite of the similarity of terminology, “separation properties” have no direct
connection to the idea of “separated sets” that appeared in Chapter 5 in the context of connected
spaces.
We have already met some simple separation properties of spaces: the X! ß X" and X# (Hausdorff)
properties. In this chapter, we look at these and others in more depth. As hypotheses for “more
separation” are added, spaces generally become nicer and nicer especially when “separation” is
combined with other properties. For example, we will see that “enough separation” and “a nice base”
guarantees that a space is metrizable.
“Separation axioms” translates the German term Trennungsaxiome used in the original literature.
Therefore the standard separation axioms were historically named X! , X" ,X# , X$ , and X% , each one
stronger than its predecessor in the list. Once these had become common terminology, another
separation axiom was discovered to be useful and “interpolated” into the list: X$ "# Þ It turns out that the
X$ "# spaces (also called Tychonoff spaces) are an extremely well-behaved class of spaces with some
very nice properties.
2. The Basics
Definition 2.1 A topological space \ is called a
1) X! -space if, whenever B Á C − \ , there either exists an open set Y with B − Y , C Â Y
or there exists an open set Z with C − Z , B Â Z
2) X" -space if, whenever B Á C − \ , there exists an open set Y with B − Y ß C Â Z
and there exists an open set Z with B Â Y ß C − Z
3) X# -space (or, Hausdorff space) if, whenever B Á C − \ , there exist disjoint open sets Y
and Z in \ such that B − Y and C − Z .
It is immediately clear from the definitions that X# Ê X" Ê X! Þ
Example 2.2
1) \ is a X! -space if and only if: whenever B Á C , then aB Á aC that is, different points
in \ have different neighborhood systems.
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2) If \ has the trivial topology and l\l ", then \ is not a X! -space.
3) A pseudometric space Ð\ß .Ñ is a metric space in and only if Ð\ß .Ñ is a X! -space.
Clearly, a metric space is X! . On the other hand, suppose Ð\ß .Ñ is X! and that B Á CÞ
Then for some % ! either B Â F% ÐCÑ or C Â F% ÐBÑ. Either way, .ÐBß CÑ %, so . is
a metric.
4) In any topological space \ we can define an equivalence relation B µ C iff aB œ aC .
Let 1 À \ Ä \Î µ œ ] by 1ÐBÑ œ ÒBÓÞ Give ] the quotient topology. Then 1 is continuous, onto,
open (not automatic for a quotient map!) and the quotient is a X! space:
If S is open in \ , we want to show that 1ÒSÓ is open in ] , and because ] has the
quotient topology this is true iff 1" Ò1ÒSÓÓ is open in \ . But 1" Ò1ÒSÓÓ
œ ÖB − \ À 1ÐBÑ − 1ÒSÓ× œ ÖB − \ À for some C − S, 1ÐBÑ œ 1ÐCÑ×
œ ÖB − \ À B is equivalent to some point C in S× œ S.
If ÒBÓ Á ÒCÓ − ] , then B is not equivalent to C, so there is an open set S © \ with
(say) B − S and C Â S. Since 1 is open, 1ÒSÓ is open in ] and ÒBÓ − 1ÒSÓ. Moreover,
ÒCÓ Â 1ÒSÓ or else C would be equivalent to some point of S implying C − SÞ
] is called the X! -identification of \ . This identification turns any space into a X! -space by
identifying points that have identical neighborhoods. If \ is a X! -space to begin with, then 1 is oneto-one and 1 is a homeomorphism. Applied to a X! space, the X! -identification accomplishes nothing.
If Ð\ß .Ñ is a pseudometric space, the X! -identification is the same as the metric identification
discussed in Example VI.5.6 because, in that case, aB œ aC if and only if .ÐBß CÑ œ !Þ
5) For 3 œ !ß "ß # À if Ð\ß g Ñ is a X3 space and g w ª g is a new topology on \ , then Ð\ß g w Ñ
is also a X3 space.
Example 2.3
1) (Exercise) It is easy to check that a space \ is a X" space
iff for each B − \ , ÖB× is closed
iff for each B − \ , ÖB× œ +ÖS À S is open and B − S×
2) A finite X" space is discrete.
3) Sierpinski space \ œ Ö!ß "× with topology g œ Ögß Ö"×ß Ö!ß "××Ñ is X! but not X" : Ö"× is
an open set that contains " and not !; but there is no open set containing ! and not 1.
4) ‘, with the right-ray topology, is X! but not X" : if B + C − ‘, then S œ ÐBß ∞Ñ is an open
set that contains C and not B; but there is no open set that contains B and not C .
5) With the cofinite topology, is X" but not X# because, in an infinite cofinite space, any
two nonempty open sets have nonempty intersection.
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These separation properties are very well-behaved with respect to subspaces and products.
Theorem 2.4 For 3 œ !ß "ß #:
a) A subspace of a X3 -space is a X3 -space
b) If \ œ #α−E \α Á g, then \ is a X3 -space iff each \α is a X3 -space.
Proof All of the proofs are easy. We consider here only the case 3 œ ", leaving the other cases as an
exercise.
a) Suppose + Á , − E © \ , where \ is a X" space. If Y w is an open set in \ containing B but not C,
then Y œ Y w ∩ E is an open set in E containing B but not C . Similarly we can find an open set Z in E
containing C but not BÞ Therefore E is a X" -space.
b) Suppose \ œ #α−E \α is a nonempty X" -space. Each \α is homeomorphic to a subspace of \ ,
so, by part a), each \α is X" Þ Conversely, suppose each \α is X" and that B Á C − \ . Then Bα Á Cα
for some α. Pick an open set Yα in \α containing Bα but not Cα . Then Y œ + Yα is an open set
in \ containing B but not C . Similarly, we find an open set Z in \ containing C but not B. Therefore
\ is a X" -space. ñ
Exercise 2.5 Is a continuous image of a X3 -space necessarily a X3 -space? How about a quotient?
A continuous open image?
We now consider a slightly different kind of separation axiom for a space \ À formally, the definition
is “just like” the definition of X# , but with a closed set replacing one of the points.
Definition 2.6 A topological space \ is called regular if whenever J is a closed set and B Â J ß
there exist disjoint open sets Y and Z such that B − Y and J © Z .
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There are some easy equivalents of the definition of “regular” that are useful to recognize.
Theorem 2.7 The following are equivalent for any space \ :
i) \ is regular
ii) if S is an open set containing B, then there exists an open set Y © \ such that
B − Y © cl Y © S
iii) at each point B − \ there exists a neighborhood base consisting of closed
neighborhoods.
Proof i) Ê ii) Suppose \ is regular and S is an open set with B − SÞ Letting J œ \ S, we use
regularity to get disjoint open sets Y ß Z with B − Y and J © Z as illustrated below:
Then B − Y © cl Y © S (since cl Y © \ Z ).
ii) Ê iii) If R − aB , then B − S œ int R . By ii), we can find an open set Y so that
B − Y © cl Y © S © R Þ Since cl Y is a neighborhood of B, the closed neighborhoods of B form a
neighborhood base at B.
iii) Ê i) Suppose J is closed and B Â J . By ii), there is a closed neighborhood O of B such
that B − O © \ J . We can choose Y œ int O and Z œ \ O to complete the proof that \ is
regular. ñ
Example 2.8 Every pseudometric space Ð\ß .Ñ is regular. Suppose + Â J and J is closed. We have
a continuous function 0 ÐBÑ œ .ÐBß J Ñ for which 0 Ð+Ñ œ - ! and 0 lJ œ !Þ This gives us disjoint
open sets with + − Y œ 0 " ÒÐ #- ß ∞ÑÓ and J © Z œ 0 " ÒÐ ∞ß #- ÑÓ. Therefore \ is regular.
At first glance, one might think that regularity is a stronger condition than X# . But this is false: if
Ð\ß .Ñ is a pseudometric space but not a metric space, then \ is regular but not even X! .
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To bring things into line, we make the following definition.
Definition A topological space \ is called a X$ -space if \ is regular and X" .
It is easy to show that X$ Ê X# ( Ê X" Ê X! ): suppose \ is X$ and B Á C − \ . Then J œ ÖC× is
closed so, by regularity, there are disjoint open sets Y , Z with B − Y and C − ÖC× © Z .
Caution Terminology varies from book to book. For some authors, the definition of “regular”
includes X" À for them, “regular” means what we have called “X$ .” Check the definitions when
reading other books.
Exercise 2.10 Show that a regular X! space must be X$ (so it would have been equivalent to use
“X! ” instead of “X" ” in the definition of “X$ ”).
Î X$ . We will put a new topology on the set \ œ ‘# . At each point : − \ , let a
Example 2.11 X# Ê
neighborhood base U: consist of all sets R of the form
R œ F% Ð:Ñ Ða finite number of straight lines through :Ñ ∪ Ö:× for some % !Þ
(Check that the conditions in the Neighborhood Base Theorem III.5.2 are satisfied.) With the
resulting topology, \ is called the slotted plane. Note that F% Ð:Ñ − U: (because “!” is a finite
number), so each F% Ð:Ñ is among the basic neighborhoods in U: so the slotted plane topology on ‘#
contains the usual Euclidean topology. It follows that \ is X# .
The set J œ ÖÐBß !Ñ À B Á !× œ “the B-axis with the origin deleted” is a closed set in \ (why?).
If Y is any open set containing the origin Ð!ß !Ñ, then there is a basic neighborhood R with
Ð!ß !Ñ − R © Y Þ Using the % in the definition of R , we can choose a point : œ ÐBß !Ñ − J with
! + B + %Þ Every basic neighborhood set of : must intersect R (why?) and therefore must intersect
Y . It follows that Ð!ß !Ñ and J cannot be separated by disjoint open sets, so the slotted plane is not
regular (and therefore not X$ ).
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Note: The usual topology in ‘# is regular. This example shows that an “enlargement” of a regular
(or X$ ) topology may not be regular (or X$ ). Although the enlarged topology has more open sets to
work with, there are also more “point/closed set pairs Bß J ” that need to be separated. By contrast,
it is easy to see that an “enlargement” of a X3 topology Ð3 œ !ß "ß #Ñ is still X3 Þ
Example 2.12 The Moore plane > (Example III.5.6) is clearly X# Þ In fact, at each point, there is a
neighborhood base of closed neighborhoods. The figure illustrates this for a point T on the B-axis and
a point U above the B-axis. Therefore > is X$ .
Theorem 2.13 a) A subspace of a regular (X$ Ñ space is regular (X$ ).
b) Suppose \ œ #α−E \α Á g. \ is regular (X$ ) iff each \α is regular (X$ ).
Proof a) Let E © \ where \ is regular. Suppose + − E and that J is a closed set in E that does
not contain +. There exists a closed set J w in \ such that J w ∩ E œ J Þ Choose disjoint open sets
Y w and Z w in \ with + − Y w and J w © Z w Þ Then Y œ Y w ∩ E and Z œ Z w ∩ E are open in E,
disjoint, + − Y , and J © Z . Therefore E is regular.
b) If \ œ #α−E \α Á g is regular, then part a) implies that each \α is regularß because each
\α is homeomorphic to a subspace of \ . Conversely, suppose each \α is regular and that
Y œ + Yα" ß ÞÞÞß Yα8 is a basic open set containing B. For each α3 , we can pick an open set Zα3 in
\α3 such that Bα3 − Zα3 © cl Zα3 © Yα3 Þ Then B − Z œ + Zα" ß ÞÞÞß Zα8 © cl Z
© + cl Zα" ß ÞÞÞß cl Zα8 © Y Þ (Why is the last inclusion true?) Therefore \ is regular.
Since the X" property is hereditary and productive, a) and b) also hold for X$ -spaces
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The obvious “next step up” in separation is the following:
Definition 2.14 A topological space \ is called normal if, whenever Eß F are disjoint closed sets in
\ , there exist disjoint open sets Y ,Z in \ with E © Y and F © Z Þ \ is called a X% -space if \ is
normal and X1 Þ
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Example 2.15 a) Every pseudometric space Ð\ß .Ñ is normal (so every metric space is X% Ñ.
.ÐBßEÑ
In fact, if E and F are disjoint closed sets, we can define 0 ÐBÑ œ .ÐBßEÑ < .ÐBßFÑ . Since the
denominator cannot be !, 0 is continuous and 0 lE œ !, 0 lF œ "Þ The open sets Y œ ÖB À 0 ÐBÑ + "# ×
and Z œ ÖB À 0 ÐBÑ "# × are disjoint that contain E and F respectively. Therefore \ is normal.
Note: the argument given is slick and clean. Can you show Ð\ß .Ñ is normal by directly constructing
a pair of disjoint open sets that contain E and F ?
b) Let ‘ have the right ray topology g œ ÖÐBß ∞Ñà B − ‘× ∪ Ögß ‘×. Ð‘ß g Ñ is
normal because the only possible pair of disjoint closed sets is g and \ and we can separate these
using the disjoint open sets Y œ g and Z œ \Þ Also, Ð‘ß g Ñ is not regular: for example " is not in
the closed set J œ Ð ∞ß !Ó, but every open set that contains J also contains ". So normal
Î regular. But Ð‘ß g Ñ is not X" and therefore not X% Þ
Ê
When we combine “normal < X" ” into X% , we have a property that fits perfectly into the separation
hierarchy.
Theorem 2.16 X% Ê X$ Ð Ê X# Ê X" Ê X! Ñ
Proof Suppose \ is X% . If J is a closed set not containing B, then ÖB× and J are disjoint closed
sets. By normality, we can find disjoint open sets separating ÖB× and J . It follows that \ is regular
and therefore X$ . ñ
.
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Exercises
E1. \ is called a door space if every subset is either open or closed. Prove that if a X# -space \
contains two points that are not isolated, then \ is not a door space, and that otherwise \ is a door
space.
E2. A base for the closed sets in a space \ is a collection of Y of closed subsets such that every
closed set J is an intersection of sets from Y . Clearly, Y is a base for the closed sets in \ iff
U œ ÖS À S œ \ J ß J − Y × is a base for the open sets in \ .
For a polynomial T in 8 real variables, define the zero set of T as
^ÐT Ñ œ ÖÐB" ß B# ß ÞÞÞß B8 Ñ − ‘8 À T ÐB" ß B# ß ÞÞÞß B8 Ñ œ !×
a) Prove that Ö^ÐT Ñ À T a polynomial in 8 real variables× is the base for the closed sets of a
topology (called the Zariski topology) on ‘8 Þ
b) Prove that the Zariski topology on ‘8 is X" but not X# .
c) Prove that the Zariski topology on ‘ is the cofinite topology, but that if 8 ", the Zariski
topology on ‘8 is not the cofinite topologyÞ
Note: The Zariski topology arises in studying algebraic geometry. After all, the sets ^ÐT Ñ are rather
special geometric objects—those “surfaces” in ‘8 which can be described by polynomial equations
T ÐB" ß B# ß ÞÞÞß B8 Ñ œ !.
E3. A space \ is a X&Î# space if, whenever B Á C − \ , there exist open sets Y and Z such that
B − Y , C − Z and cl Y ∩ cl Z œ g. (Clearly, T$ Ê X&Î# Ê T# Þ)
a) Prove that a subspace of a X&Î# space is a X&Î# space.
b) Suppose \ œ #\α Á g. Prove that \ is X&Î# iff each \α is X&Î# .
c) Let W œ ÖÐBß CÑ − ‘# À C !× and P œ ÖÐBß CÑ − W À C œ !×Þ Define a topology on W with
the following neighborhood bases:
if : − W P À
if : − P:
U: œ ÖF% Ð:Ñ ∩ W À % !×
U: œ ÖF% Ð:Ñ ∩ ÐW PÑ ∪ Ö:× À % !×
You may assume that these U: 's satisfy the axioms for a neighborhood base.
Prove that S is T&Î# but not T$ .
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E4. Suppose E © \ . Define a topology on \ by
g œ ÖS © \ À S ª E× ∪ Ög×
Decide whether or not Ð\ß g Ñ is normal.
E5. A function 0 À \ Ä ] is called perfect if 0 is continuous, closed, onto, and, for each C − ] ß
0 " ÐCÑ is compact. Prove that if \ is regular and 0 is perfect, then ] is regularà and that if \ is X$ ,
the ] is also X$ Þ
E6. a) Suppose \ is finite. Prove that Ð\ß g Ñ is regular iff there is a partition U of \ that is a base
for the topology.
b) Give an example to show that a compact subset O of a regular space \ need not be closed.
However, show that if \ is regular then cl O is compact.
c) Suppose J is closed in a X$ -space \Þ Prove that
i) Prove that J œ +ÖS À S is open and J © S×Þ
ii) Define B µ C iff B œ C or Bß C − J Þ Prove that the quotient space \Î µ is Hausdorff.
d) Suppose F is an infinite subset of a X$ -space \ . Prove that there exists a sequence of open sets
Y8 such that each Y8 ∩ F Á g and that cl Y 8 ∩ cl Y 7 œ g whenever 8 Á 7.
e) Suppose each point C in a space ] has a neighborhood Z such that cl Z is regular. Prove
that ] is regular.
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3. Completely Regular Spaces and Tychonoff Spaces
The X$ property is well-behaved. For example, we saw in Theorem 2.13 that the X$ property is
hereditary and productive. However, the X$ property is not sufficiently strong to give us really nice
theorems.
For example, it's very useful if a space has many (nonconstant) continuous real-valued functions
available to use. Remember how many times we have used the fact that continuous real-valued
functions 0 can be defined on a metric space Ð\ß .Ñ using formulas like 0 ÐBÑ œ .ÐBß +Ñ or
0 ÐBÑ œ .ÐBß J Ñ; when l\l ", we get many nonconstant real functions defined on Ð\ß .ÑÞ But a
X$ -space can sometimes be very deficient in continuous real-valued functions in 1946, Hewett gave
an example of a infinite X$ -space L on which the only continuous real-valued functions are the
constant functions.
In contrast, we will see that the X% property is strong enough to guarantee the existence of lots of
continuous real-valued functions and, therefore, to prove some really nice theorems (for example, see
Theorems 5.2 and 5.6 later in this chapter). The downside is that X% -spaces turn out also to exhibit
some very bad behavior: the X% property is not hereditary (explain why a proof analogous to the one
given for Theorem 2.13b) doesn't work) and it is not even finitely productive. Examples of such bad
behavior are a little hard to find right now, but later they will appear rather naturally.
These observations lead us to look first at a class of spaces with separation somewhere “between X$
and X% .” We want a group of spaces that is well-behaved, but also with enough separation to give us
some very nice theorems. We begin with some notation and a lemma.
Recall that
GÐ\Ñ œ Ö0 − ‘\ À 0 is continuous× œ the collection of continuous real-valued
functions on \
G ‡ Ð\Ñ œ Ö0 − GÐ\Ñ À 0 is bounded× œ the collection of continuous bounded
real-valued functions on \
Lemma 3.1 Suppose 0 ß 1 − GÐ\Ñ. Define real-valued functions 0 ” 1 and 0 • 1 by
(0 ” 1ÑÐBÑ œ max Ö0 ÐBÑß 1ÐBÑ×
Ð0 • 1ÑÐBÑ œ min Ö0 ÐBÑß 1ÐBÑ×
Then 0 ” 1 and 0 • 1 are in GÐ\ÑÞ
Proof We want to prove that the max or min of two continuous real-valued functions is continuous.
But this follows immediately from the formulas
(0 ” 1ÑÐBÑ œ
0ÐBÑ < 1ÐBÑ
#
<
l 0ÐBÑ 1ÐBÑ l
#
Ð0 • 1ÑÐBÑ œ
0ÐBÑ < 1ÐBÑ
#
l 0ÐBÑ 1ÐBÑ l
#
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ñ
Definition 3.2 A space \ is called completely regular if whenever J is a closed set and + Â J ß there
exists a function 0 − GÐ\Ñ such that 0 Ð+Ñ œ ! and 0 lJ œ "Þ
Informally, “completely regular” means that “+ and J can be separated by a continuous real-valued
function.”
Note i) The definition requires that 0 lJ œ ", in other words, that J © 0 " ÒÖ"×ÓÞ However, these
two sets might not be equal.
ii) If there is such a function 0 , there is also a continuous 1 À \ Ä Ò!ß "Ó such that 1Ð+Ñ œ !
and 1lJ œ ". For example, we could use 1 œ Ð0 ” !Ñ • " which, by Lemma 3.1, is continuous.
iii) Suppose 1 À \ Ä Ò!ß "Ó is continuous and 1Ð+Ñ œ !, 1lJ œ "Þ The particular values !ß " in
the definition are not important.: they could be any real numbers < + =Þ ÐIf we choose a
homeomorphism 9 À Ò!ß "Ó Ä Ò<ß =Ó, then it must be true that either 9Ð!Ñ œ <, 9Ð"Ñ œ = or 9Ð!Ñ œ =,
9Ð"Ñ œ < why?. Then 2 œ 9 ‰ 1 À \ Ä Ò<ß =Ó andß depending on how you chose 9ß 2Ð+Ñ œ < and
2lJ œ , or vice-versa. )
Putting these observations together, we see Definition 3.2 is equivalent to:
Definition 3.2w A space \ is called completely regular if whenever J is a closed set, + Â J ß and
<ß = are real numbers with < + =ß then there exists a continuous function 0 À \ Ä Ò<ß =Ó for which
0 Ð+Ñ œ < and 0 lJ œ =Þ
In one way, the definition of “completely regular space” is very different the definitions for the other
separation properties: the definition isn't “internal”because an “external” space, ‘ is an integral part of
the definition. While it is possible to contrive a purely internal definition for “completely regular,”
the definition is complicated and seems completely unnatural: it simply imposes some very
unintuitive conditions to force the existence of enough functions in GÐ\Ñ.
Example 3.3 Suppose Ð\ß .Ñ is a pseudometric space with a closed subset J and + Â J Þ Then
.ÐBßJ Ñ
0 ÐBÑ œ .Ð+ßJ Ñ is continuous, 0 Ð+Ñ œ " and 0 lJ œ !. So Ð\ß .Ñ is completely regular, but if . is not a
metric, then this space is not even X! .
Definition 3.4 A completely regular X" -space \ is called a Tychonoff space (or X$ "# -space).
Theorem 3.5 X$ "# Ê X$ Ð Ê X# Ê X" Ê X! Ñ
Proof Suppose J is a closed set in \ not containing +. If \ is X$ "# , we can choose 0 − GÐ\Ñ with
0 Ð+Ñ œ ! and 0 lJ œ "Þ Then Y œ 0 " ÒÐ ∞ß "# ÑÓ and Z œ 0 " ÒÐ "# ß ∞ÑÓ are disjoint open sets with
+ − Y ß J © Z Þ Therefore \ is regular. Since \ is X" , \ is X$ Þ ñ
Hewitt's example of a X$ space on which every continuous real-valued function is constant is more
than enough to show that a X$ space may not be X$ "# (the example, in Ann. Math., 47(1946) 503-509, is
rather complicated.). For that purpose, it is a little easier but still nontrivial to find a X$ space \
containing two points :ß ; such that for all 0 − GÐ\Ñß 0 Ð:Ñ œ 0 Ð;ÑÞ Then : and Ö;× cannot be
separated by a function from GÐ\Ñ so \ is not T$ "# Þ (See D.J. Thomas, A regular space, not
completely regular, American Mathematical Monthly, 76(1969), 181-182). The space \ can then be
used to construct an infinite X$ space L (simpler than Hewitt's example) on which every continuous
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real-valued function is constant (see Gantner, A regular space on which every continuous real-valued
function is constant, American Mathematical Monthly, 78(1971), 52.) Although we will not present
these constructions here, we will occasionally refer to L in comments later in this section.
Note: We have not yet shown that X% Ê X$ "# : this is true (as the notation suggests), but it is not at all
easy to prove: try it! This result is in Corollary 5.3.
Tychonoff spaces continue the pattern of good behavior that we saw in preceding separation axioms,
and they will also turn out to be a rich class of spaces to study.
Theorem 3.7 a) A subspace of a completely regular (X$ "# Ñ space is completely regular (X$ "# Ñ.
b) Suppose \ œ #α−E \α Á g. \ is completely regular (X$ "# Ñ iff each \α is
completely regular (X$ "# Ñ .
Proof Suppose + Â J © E © \ , where \ is completely regular and J is a closed set in E. Pick a
closed set O in \ such that O ∩ E œ J and an 0 − GÐ\Ñ such that 0 Ð+Ñ œ ! and 0 lO œ ". Then
1 œ 0 lE − GÐEÑß 1Ð+Ñ œ ! and 1lJ œ "Þ Therefore E is completely regular.
If g Á \ œ #α−E \α is completely regular, then each \α is homeomorphic to a subspace of
\ so each \α is completely regular. Conversely, suppose each \α is completely regular and that J is
a closed set in \ not containing +. There is a basic open set Y such that
+ − Y œ + Yα" ß ÞÞÞYα8 © \ J
For each 3 œ "ß ÞÞÞß 8 we can pick a continuous function 0α3 À \α3 Ä Ò!ß "Ó with 0α3 Ð+α3 Ñ œ ! and
0α3 |Ð\α3 Yα3 Ñ œ "Þ Define 0 À \ Ä Ò!ß "Ó by
0 ÐBÑ œ max ÖÐ0α3 ‰ 1α3 ÑÐBÑ À 3 œ "ß ÞÞÞß 8× œ max Ö0α3 ÐBα3 Ñ À 3 œ "ß ÞÞÞß 8×
Then 0 is continuous and 0 Ð+Ñ œ max Ö0α3 Ð+α3 Ñ À 3 œ "ß ÞÞÞß 8× œ !Þ If B − J , then for some 3ß
Bα3  Yα3 and 0α3 ÐBα3 Ñ œ ", so 0 ÐBÑ œ ". Therefore 0 lJ œ " and \ is completely regular.
.
Since the X" property is both hereditary and productive, the statements in a) and b) also hold
for X$ "# Þ ñ
Corollary 3.8 For any cardinal 7, the “cube” Ò!ß "Ó7 and all its subspaces are X$ "# .
Since a Tychonoff space \ is defined using functions in GÐ\Ñ, we expect that these functions will
have a close relationship to the topology on \ . We want to explore that connection.
Definition 3.9 Suppose 0 − GÐ\Ñ. Then ^Ð0 Ñ œ 0 " ÒÖ!×Ó œ ÖB − \ À 0 ÐBÑ œ !× is called the
zero set of 0 . If E œ ^Ð0 Ñ for some 0 − GÐ\Ñ, we call E a zero set in \ . The complement of a zero
set in \ is called a cozero set: cozÐ0 Ñ œ \ ^Ð0 Ñ œ ÖB − \ À 0 ÐBÑ Á !×Þ
A zero set ^Ð0 Ñ in \ is closed because 0 is continuous. In addition, ^Ð0 Ñ œ +∞
8œ" S8 , where
"
S8 œ ÖB − \ À l0 ÐBÑl + 8 ×. Each S8 is open. Therefore a zero set is always a closed K$ -set.
Taking complements shows that coz Ð0 Ñ is always an open J5 -set in \ .
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For 0 − GÐ\Ñ, let 1 œ Ð " ” 0 Ñ • " − G ‡ Ð\ÑÞ Then ^Ð0 Ñ œ ^Ð1Ñ. Therefore GÐ\Ñ and G ‡ Ð\Ñ
produce the same zero sets in \ (and therefore also the same cozero sets).
Example 3.10
1) A closed set J in a pseudometric space Ð\ß .Ñ is a zero set: J œ ^Ð0 Ñ, where 0 ÐBÑ œ .ÐBß J ÑÞ
2) In general, a closed set in \ might not be a zero set in fact, a closed set in \ might not even be
a K$ set.
Suppose \ is uncountable and : − \ . Define a topology on \ by letting UB œ ÖÖB×× be a
neighborhood at each point B Á : and letting U: œ ÖF À : − F and \ F is countable× be
the neighborhood base at :. (Check that the conditions of the Neighborhood Base Theorem
III.5.2 are satisfiedÞ )
All points in \ Ö:× are isolated and \ is clearly X" Þ In fact, \ is X% .
If E and F are disjoint closed sets in \ , then one of them (say EÑ satisfies
E © \ Ö:×, so E is clopen. We then have open sets Y œ E and Z œ \ E ª F ,
so \ is normal.
We do not know yet that X% Ê X$ "# in general, but it's easy to see that this space \ is also X$ "# .
If J is a closed set not containing B, then either J © \ Ö:× or ÖB× © \ Ö:×.
So one of the sets J or ÖB× is clopen and he characteristic function of that clopen set
is continuous and works to show that \ is completely regular.
The set Ö:× is closed but Ö:× is not a K$ set in \ , so Ö:× is not a zero set in \ .
Suppose : − +∞
8œ" S8 where S8 is open. For each 8, there is a basic neighborhood F8
of : such that : − F8 © S8 ß so \ S8 © \ F8 is countable. Therefore
-∞
\ +∞
8œ" S8 œ
8œ" Ð\ S8 Ñ is countable. Since \ is uncountable, we conclude
∞
+
that Ö:× Á 8œ" S8 Þ
Even when J is both closed and a K$ set, J might not be a zero set. We will see examples
later.
For purely technical purposes, it is convenient to notice that zero sets and cozero sets can be described
in a many different forms. For example, if 0 − GÐ\Ñ, then we can see that each set in the left column
is a zero set by choosing a suitable 1 − GÐ\Ñ À
^
^
^
^
^
œ ÖB À 0 ÐBÑ œ <×
œ ÖB À 0 ÐBÑ 0×
œ ÖB À 0 ÐBÑ Ÿ 0×
œ ÖB À 0 ÐBÑ r×
œ ÖB À 0 ÐBÑ Ÿ r×
œ ^Ð1Ñß
œ ^Ð1Ñ,
œ ^Ð1Ñ,
œ ^Ð1Ñ,
œ ^Ð1Ñ,
where 1ÐBÑ œ
where 1ÐBÑ œ
where 1ÐBÑ œ
where 1ÐBÑ œ
where 1ÐBÑ œ
295
0 ÐBÑ <
0 ÐBÑ l0 ÐBÑl
0 ÐBÑ < l0 ÐBÑl
Ð0 ÐBÑ <Ñ l0 ÐBÑ <l
Ð0 ÐBÑ <Ñ < l0 ÐBÑ <l
On the other hand, if 1 − GÐ\Ñ, we can write ^Ð1Ñ in any of the forms listed above by choosing an
appropriate function 0 − GÐ\Ñ À
^Ð1Ñ œ ÖB À 0 ÐBÑ œ <×
^Ð1Ñ œ ÖB À 0 ÐBÑ 0×
^Ð1Ñ œ ÖB À 0 ÐBÑ Ÿ 0×
^Ð1Ñ œ ÖB À 0 ÐBÑ r×
^Ð1Ñ œ ÖB À 0 ÐBÑ Ÿ r×
where
where
where
where
where
0 ÐBÑ œ 1ÐBÑ < <
0 ÐBÑ œ l1ÐBÑl
0 ÐBÑ œ l1ÐBÑl
0 ÐBÑ œ < l1ÐBÑl
0 ÐBÑ œ < < l1ÐBÑl
Taking complements, we get the corresponding results for cozero sets: if 0 − GÐ\Ñ
i) the sets ÖB À 0 ÐBÑ Á <×, ÖB À 0 ÐBÑ + !×, ÖB À 0 ÐBÑ !×ß ÖB À 0 ÐBÑ + <×, {B À 0 ÐBÑ <}
are cozero sets, and
ii) any given cozero set can be written in any one of these forms.
Using the terminology of cozero sets, we can see a nice comparison/contrast between regularity and
complete regularity. Suppose B Â J , where J is closed in \Þ If \ is regular, we can find disjoint
open sets Y and Z with B − Y and J © Z Þ But if \ is completely regular, we can separate B and J
with “special” open sets Y and Z cozero sets! Just choose 0 − GÐ\Ñ with 0 ÐBÑ œ ! and 0 lJ œ ",
then
B − Y œ ÖB À 0 ÐBÑ + "# ×
and
J © Z œ ÖB À 0 ÐBÑ "# ×
In fact this observation characterizes completely regular spaces that is, if a regular space fails to be
completely regular, it is because there is a “shortage” of cozero sets because there is a “shortage” of
functions in GÐ\Ñ (see Theorem 3.12, below). For the extreme case of a X$ space L on which the
only continuous real valued functions are constant, the only cozero sets are g and L !
The next theorem reveals the connections between cozero sets, GÐ\Ñ and the weak topology on \Þ
Theorem 3.11 For any space Ð\ß g Ñ, GÐ\Ñ and G ‡ Ð\Ñ induce the same weak topology gA on \ ,
and a base for gA is the collection of all cozero sets in \ .
Proof A subbase for gA consists of all sets of the form 0 " ÒY Ó, where Y is open in ‘ and 0 − GÐ\ÑÞ
Without loss of generality, we can assume the sets Y are subbasic open sets of the form Ð+ß ∞Ñ and
Ð ∞ß ,Ñ, so that the sets 0 " ÒY Ó have form ÖB − \ À 0 ÐBÑ +× or ÖB − \ À 0 ÐBÑ + ,×. But these
are cozero sets of \ , and every cozero set in \ has this form. So the cozero sets are a subbase for gA
In fact, the cozero sets are actually a base because coz Ð0 Ñ ∩ coz Ð1Ñ œ coz Ð0 1Ñ: the intersection of
two cozero sets is a cozero set.
The same argument, with G ‡ Ð\Ñ replacing GÐ\Ñß shows that the cozero sets of G ‡ Ð\Ñ are a base for
the weak topology on \ generated by G ‡ Ð\ÑÞ But GÐ\Ñ and G ‡ Ð\Ñ produce the same cozero sets in
\ , and therefore generate the same weak topology gA on \ . ñ
Now we can now see the close connection between \ and GÐ\Ñ in completely regular spaces. For
any space Ð\ß g Ñ the functions in GÐ\Ñ certainly are continuous with respect to g (by definition of
GÐ\Ñ ). But is g the smallest topology making this collection of functions continuous? In other
296
words, is g the weak topology on \ generated by GÐ\Ñ? The next theorem says that is true precisely
when \ is completely regular.
Theorem 3.12 For any space Ð\ß g Ñ, the following are equivalent:
a) \ is completely regular
b) The cozero sets of \ are a base for the topology on \ (equivalently, the zero sets of \
are a base for the closed sets—meaning that every closed set is an intersection of zero sets)
c) \ has the weak topology from GÐ\Ñ (equivalently, from G ‡ Ð\Ñ )
d) GÐ\Ñ (equivalently, G ‡ Ð\Ñ ) separates points from closed sets.
Proof The preceding theorem shows that b) and c) are equivalent.
a) Ê b) Suppose + − S where S is openÞ Let J œ \ SÞ Then we can choose 0 − GÐ\Ñ
with 0 Ð+Ñ œ ! and 0 lJ œ ". Then Y œ ÖB À 0 ÐBÑ + "# × is a cozero set for which + − Y © SÞ
Therefore the cozero sets are a base for \ .
b) Ê d) Suppose J is a closed set not containing +. By b), we can choose 0 − GÐ\Ñ so that
+ − coz Ð0 Ñ © \ J Þ Then 0 Ð+Ñ œ < Á !, so 0 Ð+Ñ Â cl 0 ÒJ Ó œ Ö!×Þ Therefore GÐ\Ñ separates
points and closed sets.
d) Ê a) Suppose J is a closed set not containing +. There is some 0 − GÐ\Ñ for which
0 Ð+Ñ Â cl 0 ÒJ ÓÞ Without loss of generality (why?), we can assume 0 Ð+Ñ œ !Þ Then, for some % !,
Ð %ß %Ñ ∩ 0 ÒJ Ó œ g, so that for B − J , l0 ÐBÑl %Þ Define 1 − G ‡ Ð\Ñ by 1ÐBÑ œ minÖl0 ÐBÑlß %×Þ
Then 1Ð+Ñ œ ! and 1lJ œ %, so \ is completely regular.
At each step of the proof, GÐ\Ñ can be replaced by G ‡ Ð\Ñ (check!) ñ
The following corollary is curious and the proof is a good test of whether one understands the idea of
“weak topology.”
Corollary 3.13 Suppose \ is a set and let gY be the weak topology on \ generated by any family of
functions Y © ‘\ . Then Ð\ß gY Ñ is completely regularÞ Ð\ß gY ) is Tychonoff is Y separates points.
Proof Give \ the topology the weak topology gY generated by Y . Now \ has a topology, so the
collection GÐ\Ñ makes sense. Let XA be the weak topology on \ generated by GÐ\ÑÞ
The topology gY does make all the functions in GÐ\Ñ continuous, so gA © gY Þ
On the other hand: Y © GÐ\Ñ by definition of gY , and the larger collection of functions GÐ\Ñ
generates a (potentially) larger weak topology. Therefore gY © gA Þ
Therefore gY œ gA . By Theorem 3.12, Ð\ß gY Ñ is completely regular. ñ
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Example 3.14
1) If Y œ Ö0 − ‘‘ À 0 is nowhere differentiable×, then the weak topology gY on ‘ generated
by Y is completely regular.
2) If L is an infinite X$ space on which every continuous real-valued function is constant (see
the comments at the beginning of this Section 3), then the weak topology generated by GÐ\Ñ has for a
base the collection of cozero sets {gß L×. So the weak topology generated by GÐ\Ñ is the trivial
topology, not the original topology on \ .
Theorem 3.12 leads to a lovely characterization of Tychonoff spaces.
Corollary 3.15
Suppose \ is a Tychonoff space.
For each 0 − G ‡ Ð\Ñ, we have
ranÐ0 Ñ © Ò+0 ß ,0 ] œ M0 for some +0 + ,0 − ‘Þ The evaluation map / À \ Ä # ÖM0 À 0 − G ‡ Ð\Ñ× is
an embedding.
Proof \ is X" , the 0 's are continuous and the collection of 0 's Ð œ G ‡ Ð\Ñ Ñ separates points and
closed sets. By Corollary VI.4.11, / is an embedding. ñ
Since each M0 is homeomorphic to Ò!ß "Ó, #ÖM0 À 0 − G ‡ Ð\Ñ× is homeomorphic to Ò!ß "Ó7 ß where
7 œ lG ‡ Ð\Ñl. Therefore any Tychonoff space can be embedded in a “cube.” On the other hand
(Corollary 3.8) Ò!ß "Ó7 and all its subspaces are Tychonoff. So we have:
Corollary 3.16 A space \ is Tychonoff iff \ is homeomorphic to a subspace of a “cube” Ò!ß "Ó7 for
some cardinal number 7.
The exponent 7 œ lG * Ð\Ñl in the corollary may not be the smallest exponent possible. As an extreme
case, for example, we have - œ lG ‡ БÑl, even though we can embed ‘ in Ò!ß "Ó œ Ò!ß "Ó" Þ The
following theorem improves the value for 7 in certain cases (and we proved a similar result for metric
spaces Ð\ß .Ñ: see Example VI.4.5.)
Theorem 3.17 Suppose \ is Tychonoff with a base U of cardinality 7. Then \ can be embedded in
Ò!ß "Ó7 . In particular, \ can be embedded in Ò!ß "ÓAÐ\Ñ .
Proof Suppose 7 is finite. Since \ is X" , ÖB× œ +ÖF À F is a basic open set containing B×Þ Only
finitely many such intersections are possible, so \ is finite and therefore discrete. Hence
top
top
\ © Ò!ß "Ó © Ò!ß "Ó7 Þ
Suppose U is a base of cardinal 7 where 7 is infinite. Call a pair ÐY ß Z Ñ − U ‚ U
distinguished if there exists a continuous 0Y ßZ À \ Ä Ò!ß "Ó with 0Y ßZ ÐBÑ + "# for all B − Y and
0Y ßZ ÐBÑ œ " for all B − \ Z Þ Clearly, Y © Z for a distinguished pair ÐY ß Z ÑÞ For each
distinguished pair, pick such a function 0Y ßZ and let Y œ Ö0Y ßZ À ÐY ß Z Ñ − U ‚ U is distinguished×.
298
We note that if + − Z − U, then there must exist Y − U such that + − Y and ÐY ß Z Ñ is distinguishedÞ
To see this, pick an 0 À \ Ä Ò!ß "Ó so that 0 Ð+Ñ œ ! and 0 l\ Z œ "Þ Then choose Y − U so that
+ − Y © 0 " ÒÒ!ß "# ÑÓ © Z Þ
We claim that Y separates points and closed sets:
Suppose J is a closed set not containing +. Choose a basic set Z − U with + − Z © \ J Þ
There is a distinguished pair ÐY ß Z Ñ with + − Y © Z © \ J Þ Then 0Y ßZ Ð+Ñ œ < + "# and
0Y ßZ lJ œ ", so 0Y ßZ Ð+Ñ Â cl 0Y ßZ ÒJ Ó œ Ö"×Þ
By Corollary VI.4.11, / À \ Ä Ò!ß "ÓlY l
lY l Ÿ lU ‚ U l œ 7# œ 7. ñ
is
an
embedding.
Since
7
is
infinite,
A theorem that states that certain topological properties of a space \ imply that \ is metrizable is
called a “metrization theorem.” Typically the hypotheses of a metrization theorem involve that
1) \ has “enough separation” and 2) \ has a “sufficiently nice base.” The following theorem is a
simple example.
Corollary 3.18 (“Baby Metrization Theorem”)
metrizable.
A second countable Tychonoff space \ is
top
Proof By Theorem 3.17, \ © Ò!ß "Ói! . Since Ò!ß "Ói! is metrizable, so is \ . ñ
In Corollary 3.18, \ turns out to be metrizable and separable (since \ is second countable). On the
other hand Ò!ß "Ói! and all its subspaces are separable metrizable spaces. Thus, the corollary tells us
that the separable metrizable spaces (topologically) are precisely the second countable Tychonoff
spaces.
299
Exercises
E7. Prove that if \ is a countable Tychonoff space, then there is a neighborhood base of clopen sets
at each point. (Such a space \ is sometimes called zero-dimensional.)
E8. Prove that in any space \ , a countable union of cozero sets is a cozero set or, equivalently, that
a countable intersection of zero sets is a zero set.
E9. Prove that the following are equivalent in any Tychonoff space \ :
a) every zero set is open
b) every K$ set is open
c) for each 0 − GÐ\Ñ À if 0 Ð:Ñ œ ! then there is a neighborhood R of :
such that 0 ± R ´ !
E10. Let 3 À ‘ Ä ‘ be the identity map and let
Ð3Ñ œ Ö0 − GÐ‘Ñ À 0 œ 13 for some 1 − GБÑ}.
Ð3Ñ is called the ideal in GÐ‘Ñ generated by the element 3.
For those who know a bit of algebra: if we definite addition and multiplication of functions pointwise,
then GÐ‘Ñ Ðor, more generally, GÐ\ÑÑ is a commutative ringÞ The constant function ! is the zero
element in the ring; there is also a unit element, namely the constant function “".”
a) Prove that Ð3Ñ œ Ö0 − GÐ‘Ñ À 0 Ð!Ñ œ ! and the derivative 0 w Ð!Ñ exists×.
b) Exhibit two functions 0 ß 1 in GÐ‘Ñ for which 0 1 − Ð3Ñ yet 0  Ð3Ñ and 1  Ð3Ñ.
c) Let \ be a Tychonoff space with more than one point. Prove that there are two functions
0 ß 1 − GÐ\Ñ such that 0 1 œ ! on \ yet neither 0 nor 1 is identically 0 on \ .
Thus, there are functions 0 ß 1 − GÐ\Ñ for which 0 1 œ ! although 0 Á ! and 1 Á !Þ In an
algebra course, such elements 0 and 1 in the ring GÐ\Ñ are called “zero divisors.”
d) Prove that there are exactly two functions 0 − GÐ‘Ñ for which 0 # œ 0 . (M8 GÐ\Ñß the
notation 0 # ÐBÑ means 0 ÐBÑ † 0 ÐBÑ, not 0 Ð0 ÐBÑÑ. )
e) Prove that there are exactly c functions 0 in G () for which 0 # œ 0 .
An element in GÐ\Ñ that equals its own square is called an idempotent. Part d) shows that GÐ‘Ñ and
GÐÑ are not isomorphic rings since they have different numbers of idempotents. Is either GÐ‘Ñ or
GÐÑ isomorphic to GÐÑ ?
One classic part of general topology is to explore the relationship between the space \ and the rings
GÐ\Ñ and G ‡ Ð\Ñ. For example, if \ is homeomorphic to ] , then GÐ\Ñ is isomorphic to GÐ] Ñ.
This necessarily implies (why?) that G ‡ Ð\Ñ is isomorphic to G ‡ Ð] Ñ. The question “when does
300
isomorphism imply homeomorphism?” is more difficult. Another important area of study is how the
maximal ideals of the ring GÐ\Ñ are related to the topology of \ . The best introduction to this
material is the classic book Rings of Continuous Functions (Gillman-Jerison).
f) Let HÐ‘Ñ be the set of differentiable functions 0 À ‘ Ä ‘. Are the rings GÐ‘Ñ and HБÑ
isomorphic? Hint: An isomorphism between GÐ‘Ñ and HÐ‘Ñ preserves cube roots.
E11. Suppose \ is a connected Tychonoff space with more than one point. Prove l\l -Þ.
E12. Let \ be a topological space. Suppose 0 ß 1 − GÐ\Ñ and that ^Ð0 Ñ is a neighborhood of ^Ð1Ñ
(that is, ^Ð1Ñ © int ^Ð0 ÑÞ )
a) Prove that 0 is a multiple of 1 in GÐ\Ñß that is, prove there is a function 2 − GÐ\Ñ such
that 0 ÐBÑ œ 1ÐBÑ2ÐBÑ for all B − \ .
b) Give an example where ^Ð0 Ñ ª ^Ð1Ñ but 0 is not a multiple of 1 in GÐ\Ñ.
E13. Let \ be a Tychonoff space with subspaces J and Eß where J is closed and E is countable.
Prove that if J ∩ E œ g, then E is disjoint from some zero set that contains J .
E14. A space \ is called pseudocompact if every continuous 0 À \ Ä ‘ is bounded, that is, if
GÐ\Ñ œ G ‡ Ð\Ñ (see Definition IV.8.7). Consider the following condition (*) on a space \ :
(*) Whenever Z" ª Z# ª ... ª Z8 ª ÞÞÞ is a decreasing sequence of nonempty open sets,
then +∞
8œ" cl Z8 Á g.
a) Prove that if \ satisfies (*), then \ is pseudocompact.
b) Prove that if \ is Tychonoff and pseudocompact, then \ satisfies (*).
Note: For Tychonoff spaces, part b) gives an “internal" characterization of pseudocompactness that
is, a characterization that makes no explicit reference to ‘.
.
301
4. Normal and X% -Spaces
We now return to a topic in progress: normal spaces Ðand X% -spacesÑ . Even though normal spaces are
badly behaved in some ways, there are still some very important (and nontrivial) theorems that we can
prove. One of these will give “X% Ê X$ "# ” as an immediate corollary.
To begin, the following technical variation on the definition of normality is very useful.
Lemma 4.1 A space \ is normal
iff
whenever E is closedß S is open and E © S,
there exists an open set U
with E © Y © cl Y © SÞ
Proof Suppose \ is normal and that S is an open set containing the closed set E. Then E and
F œ \ S are disjoint closed sets. By normality, there are disjoint open sets Y and Z with E © Y
and F © Z Þ Then E © Y © cl Y © \ Z © SÞ
Conversely, suppose \ satisfies the stated condition and that Eß F are disjoint closed sets.
Then E © S œ \ Fß so there is an open set Y with E © Y © cl Y © \ FÞ Let Z œ \ cl Y .
Y and Z are disjoint closed sets containing E and F respectively, so \ is normal. ñ
302
Theorem 4.2
a) A closed subspace of a normal (X% ) space is normal (X% ).
b) A continuous closed image of a normal (X% ) space normal (X% ).
Proof a) Suppose J is a closed subspace of a normal space \ and let E and F be disjoint closed
sets in J . Then Eß F are also closed in \ so we can find disjoint open sets Y w and Z w in \
containing E and F respectively. Then Y œ Y w ∩ J and Z œ Z w ∩ J are disjoint open sets in J that
contain E and F , so J is normal.
b) Suppose \ is normal and that 0 À \ Ä ] is continuous, closed and onto. If E and F are
disjoint closed sets in ] , then 0 " ÒEÓ and 0 " ÒFÓ are disjoint closed sets in \ . Pick Y w and Z w
disjoint open sets in \ with 0 " ÒEÓ © Y w and 0 " ÒFÓ © Z w Þ Then Y œ ] 0 Ò\ Y w Ó and
Z œ ] 0 Ò\ Z w Ó are open sets in ] .
If C − Y , then C  0 Ò\ Y w Ó. Since 0 is onto, C œ 0 ÐBÑ for some B − Y w © \ Z w Þ Therefore
C − 0 Ò\ Z w Ó so C  Z . Hence Y ∩ Z œ gÞ
If C − E, then 0 " ÒÖC×Ó © 0 " ÒEÓ © Y w , so 0 " ÒÖC×Ó ∩ Ð\ Y w Ñ œ g. Therefore C  0 Ò\ Y w Ó so
C − ] 0 Ò\ Y w Ó œ Y . Therefore E © Y and, similarly, F © Z so ] is normal.
Since the X" property is hereditary and is preserved by closed onto maps, the statements in a) and b)
hold for X% as well as normality. ñ
The next theorem gives us more examples of normal (and X% ) spaces.
Theorem 4.3 Every regular Lindelöf space \ is normal (and therefore every Lindelöf X$ -space
is X% ).
Proof Suppose E and F are disjoint closed sets in \ . For each B − E, use regularity to pick an open
set YB such that B − YB © cl YB © \ FÞ Since the Lindelöf property is hereditary on closed
subsets, a countable number of the YB 's cover E: relabel these as Y" ß Y# ß ÞÞÞß Y8 ß ÞÞÞ . For each 8, we
have cl Y8 ∩ F œ gÞ Similarly, choose a sequence of open sets Z" ß Z# ß ÞÞÞß Z8 ß ÞÞÞ covering F such that
cl Z8 ∩ E œ g for each 8.
-∞
We have that -∞
8œ" Y8 ª E and
8œ" Z8 ª F , but these unions may not be disjoint. So we define
Y"‡ œ Y" cl Z"
Y#‡ œ Y# Ðcl Z" ∪ cl Z# Ñ
ã
Y8‡ œ Y8 Ðcl Z" ∪ cl Z# ∪ ÞÞÞ ∪ cl Z8 Ñ
ã
Z"‡ œ Z" cl Y"
Z#‡ œ Z# Ðcl Y" ∪ cl Y# Ñ
ã
Z8‡ œ Z8 Ðcl Y" ∪ cl Y# ∪ ÞÞÞ ∪ cl Y8 Ñ
ã
‡
‡
-∞
Let Y œ -∞
8œ" Y8 and Z œ
8œ" Z8 .
If B − E, then B Â cl Z8 for all 8. But B − Y5 for some 5 , so B − Y5‡ © Y Þ Therefore E © Y and,
similarly, F © Z Þ
To complete the proof, we show that Y ∩ Z œ g. Suppose B − Y .
303
Then B − Y5‡ for some 5ß
so
so
so
so
B Â cl Z" ∪ cl Z# ∪ ÞÞÞ ∪ cl Z5 ß
B Â Z" ∪ Z# ∪ ÞÞÞ ∪ Z5
B Â Z"‡ ∪ Z#‡ ∪ ÞÞÞ ∪ Z5‡
B Â Z8‡ for any 8 Ÿ 5 Þ
Since B − Y5‡ , then B − Y5 Þ So, if 8 5 , then B  Z8‡ œ Z8 Ðcl Y" ∪ ÞÞÞ ∪ cl Y5 ∪ ÞÞÞ ∪ cl Y8 Ñ
So B  Z8‡ for all 8, so B  Z and therefore Y ∩ Z œ gÞ ñ
Example 4.4 The Sorgenfrey line W is regular because the sets Ò+ß ,Ñ form a base of closed
neighborhoods at each point +. We proved in Example VI.3.2 that W is Lindelöf, so W is normal.
Since W is X" , we have that W is X% .
5. Urysohn's Lemma and Tietze's Extension Theorem
We now turn our attention to the issue of “X% Ê X$ "# ”. This is hard to prove because to show that a
space \ is X$ "# , we need to prove that certain continuous functions exist; but the hypothesis “X% ” gives
us no continuous functions to work with. As far as we know at this point, there could even be X%
spaces on which every continuous real-valued function is constant! If X% -spaces are going to have a
rich supply of continuous real-valued functions, we will have to show that these functions can be
“built from scratch” in a X% -space. This will lead us to two of the most well-known classical theorems
of general topology.
We begin with the following technical lemma. It gives a way to use a certain collection of open sets
ÖY< À < − U× to construct a function 0 − GÐ\Ñ. The idea in the proof is quite straightforward, but I
attribute its elegant presentation (and that of Urysohn's Lemma which follows) primarily to Leonard
Gillman and Meyer Jerison.
Lemma 5.1 Suppose \ is any topological space and let U be any dense subset of ‘Þ Suppose open
sets Y< © \ have been defined, one for each < − U, in such a way that:
i) \ œ -<−U Y< and +<−U Y< œ g
ii) if <ß = − U and < + =, then cl Y< © Y= .
For B − \ , define 0 ÐBÑ œ inf Ö< − U À B − Y< ×. Then 0 À \ Ä ‘ is continuous.
We will use this Lemma only once, with U œ Þ So if you like, there is no harm in assuming that
U œ in the proof.
Proof Suppose B − \Þ By i) we know that B − Y< for some <, so Ö< − U À B − Y< × Á g. And by ii),
we know that B Â Y= for some =. For that = À if B − Y< , then (by ii) = Ÿ <, so = is a lower bound for
Ö< − U À B − Y< ×. Therefore Ö< − U À B − Y< × has a greatest lower bound, so the definition of 0
makes sense: 0 ÐBÑ œ inf Ö< − U À B − Y< × − ‘.
304
From the definition of 0 , we get that for <ß = − Uß
a) if B − cl Y< , then B − Y= for all = < so 0 ÐBÑ Ÿ <
b) if 0 ÐBÑ + =, then B − Y= .
We want to prove 0 is continuous at each point + − \ . Since U is dense in ‘,
ÖÒ<ß =Ó À <ß = − U and < + 0 Ð+Ñ + =×
is a neighborhood base at 0 Ð+Ñ in ‘. Therefore it is sufficient to show that whenever < + 0 Ð+Ñ + =,
then there is a neighborhood Y of + such that 0 ÒY Ó © Ò<ß =ÓÞ
Since 0 Ð+Ñ + =, we have + − Y= , and 0 Ð+Ñ < gives us that +  cl Y< . Therefore Y œ Y= cl Y< is
an open neighborhood of +. If D − Y , then D − Y= © cl Y= ß so 0 ÐDÑ Ÿ =; and D Â cl Y< , so D Â Y<
and 0 ÐDÑ <Þ Therefore 0 ÒY Ó © Ò<ß =ÓÞ ñ
Our first major theorem about normal spaces is still traditionally referred to as a “lemma” because it
was a lemma in the paper where it originally appeared. Its author, Paul Urysohn, died at age 26, on the
morning of 17 August 1924, while swimming off the coast of Brittany.
Theorem 5.2 ÐUrysohn's LemmaÑ A space \ is normal iff whenever Eß F are disjoint closed sets
in \ , there exists a function 0 − GÐ\Ñ with 0 lE œ ! and 0 lF œ "Þ (When such an 0 exists, we say
that E and F are completely separated.)
Note:
Notice that if E and F happen to be disjoint zero sets, say E œ ^Ð1Ñ and F œ ^Ð2Ñ, then the
conclusion of the theorem is true in any space, without assuming normality: just let
1# ÐBÑ
0 ÐBÑ œ 1# ÐBÑ < 2# ÐBÑ . Then 0 is continuous, 0 lE œ ! and 0 lF œ ".
The conclusion of Urysohn's Lemma only says that E © 0 " Ð!Ñ and F © 0 " Ð"Ñ: equality
might not be true. In fact, if E œ 0 " Ð!Ñ and F œ 0 " Ð"Ñ, then E and F were zero sets in
the beginning, and the hypothesis of normality would have been unnecessary.
This shows again that zero sets are very special closed sets: in any space, disjoint zero sets are
completely separated. Put another way: given Urysohn's Lemma, we can conclude that every
nonnormal space must contain a closed set that is not a zero set.
Proof The proof of Urysohn's Lemma in one direction is almost trivial. If such a function 0 exists,
then Y œ ÖB À 0 ÐBÑ + "# × and Z œ ÖB À 0 ÐBÑ "# × are disjoint open sets (in fact, cozero sets)
containing E and F respectively. It is the other half of Urysohn's Lemma for which Urysohn deserves
credit.
Let E and F be disjoint closed sets in a normal space \ . We will define sets open sets Y< (< − Ñ in
\ in such a way that Lemma 5.1 applies. To start, let Y< œ g for < + ! and Y< œ \ for < ".
305
Enumerate the remaining rationals in ∩ Ò!ß "Ó as <" ß <# ß ÞÞÞß <8 ß ÞÞÞ , beginning the list with <" œ " and
<# œ !Þ We begin by defining Y<" œ Y" œ \ FÞ Then use normality to define Y<# Ð œ Y! Ñ À since
E © Y<" œ \ F , we can pick Y<# so that
E © Y<# © cl Y<# © Y<" œ \ F
Then ! œ <# + <$ + <" œ ", and we use normality to pick an open set Y<$ so that
E © Y<# © cl Y<# © Y<$ © cl Y<$ © Y<" œ \ F
We continue by induction. Suppose 8 $ and that we have already defined open sets Y<" ß Y<# ß ÞÞÞß Y<8
in such a way that whenever <3 + <4 + <5 Ð3ß 4ß 5 Ÿ 8Ñß then
cl Y<3 © Y<4 © cl Y<4 © Y<5 ЇÑ
We need to define Y<8<" so that Ð‡Ñ holds for 3ß 4ß 5 Ÿ 8 < "Þ
Since <" œ " and <# œ !, and <8<" − Ð!ß "Ñß it makes sense to define
<5 œ the largest among <" ß <# ß ÞÞÞß <8 that is smaller than <8<" ß and
<6 œ the smallest among <" ß <# ß ÞÞÞß <8 that is larger than <8<" Þ
By the induction hypothesis, we already have cl Y<5 © Y<6 Þ Then use normality to pick an open set
Y<8<" so that
cl Y<5 © Y<8<" © cl Y<8<" © Y<6 .
The Y< 's defined in this way satisfy the conditions of Lemma 5.1, so the function 0 À \ Ä ‘ defined
by 0 ÐBÑ œ inf Ö< − À B − Y< × is continuous. If B − E, then B − Y<# œ Y! and B  Y< if < + !, so
0 ÐBÑ œ !Þ If B − F then B  Y" , but B − Y< œ \ for < ", so 0 ÐBÑ œ "Þ ñ
Once we have the function 0 we can replace it, if we like, by 1 œ Ð! ” 0 Ñ • " so that E and F are
completely separated by a function 1 − G ‡ Ð\ÑÞ It is also clear that we can modify 1 further to get an
2 − G ‡ Ð\Ñ for which 2lE œ + and 2lF œ , where + and , are any two real numbers.
With Urysohn's Lemma, the proof of the following corollary is obvious.
Corollary 5.3 X% Ê X$ "# .
There is another famous characterization of normal spaces in terms of GÐ\Ñ. It is a result about
“extending” continuous real-valued functions defined on closed subspaces.
We begin with the following two lemmas. Lemma 5.4, called the “Weierstrass Q -Test” is a slight
generalization of a theorem with the same name in advanced calculus. It can be useful in “piecing
together” infinitely many real-valued continuous functions to get a new one. Lemma 5.5 will be used
in the proof of Tietze's Extension Theorem (Theorem 5.6).
306
Lemma 5.4 (Weierstrass Q -Test)
Let \ be a topological space.
continuous for each 8 − and that l08 ÐBÑl Ÿ Q8
Suppose 08 À \ Ä ‘ is
∞
for all B − \ .
If Q8 + ∞, then
8œ"
∞
0 ÐBÑ œ 08 ÐBÑ converges (absolutely) for all B and 0 À \ Ä ‘ is continuous.
8œ"
∞
∞
∞
8œ"
8œ"
8œ"
Proof For each B, l08 ÐBÑl Ÿ Q8 + ∞, so 08 ÐBÑ converges (absolutely) by the Comparison
Test.
∞
Suppose + − \ and % !. Choose R so that Q8 + %% .
Each 08 is continuous, so for
8œR <"
8 œ "ß ÞÞÞß R we can pick a neighborhood Y8 of + such that for B − Y8 ß l08 ÐBÑ 08 Ð+Ñl +
Y œ +R
8œ" Y8 is a neighborhood of +, and for B − Y we get l0 ÐBÑ 0 Ð+Ñl
R
∞
R
8œ"
R
8œR <"
∞
8œ"
%
#R .
Then
∞
œ l Ð08 ÐBÑ 08 Ð+ÑÑ < Ð08 ÐBÑ 08 Ð+ÑÑl Ÿ l08 ÐBÑ 08 Ð+Ñl < l08 ÐBÑ 08 Ð+Ñl
Ÿ l08 ÐBÑ 08 Ð+Ñl < l08 ÐBÑl < l08 Ð+Ñl + R †
8œ"
8œR <"
Therefore 0 is continuous at +.
8œR <"
%
#R
∞
< #Q8 +
8œR <"
%
#
<
%
#
œ %Þ
ñ
Lemma 5.5 Let E be a closed set in a normal space \ and let + be a positive real number. Suppose
2 À E Ä Ò <ß <Ó is continuous. Then there exists a continuous 9 À \ Ä [ $< ß $< Ó such that
l2ÐBÑ 9ÐBÑl Ÿ #<$ for each B − E.
Proof Let E" œ ÖB − E À 2ÐBÑ Ÿ $< × and F" œ ÖB − E À 2ÐBÑ $< ×. E is closed, and E" and F"
are disjoint closed sets in E, so E" and F" are closed in \ . By Urysohn's Lemma, there exists a
continuous function 9 À \ Ä Ò $< ß $< Ó such that 9lE" œ $< and 9lF" œ $< .
If B − E" , then < Ÿ 2ÐBÑ Ÿ $< and 9ÐBÑ œ $< , so l2ÐBÑ 9ÐBÑl Ÿ l < Ð $< Ñl
œ #<$ à and similarly if B − F" ß l2ÐBÑ 9ÐBÑl Ÿ #<$ Þ If B − E ÐE" ∪ F" Ñ, then 2ÐBÑ and 9ÐBÑ are
both in Ò $< ß $< Ó so l2ÐBÑ 9ÐBÑl Ÿ #<$ . ñ
Theorem 5.6 (Tietze's Extension Theorem) A space \ is normal iff whenever E is a closed set in
\ and 0 − GÐEÑ, then there exists a function 1 − GÐ\Ñ such that 1lE œ 0 Þ
Note: if E is a closed subset of \ œ ‘, then it is quite easy to prove the theorem. In that case,
‘ E is open and can be written as a countable union of disjoint open intervals Mß where
each M œ Ð+ß ,Ñ or Ð ∞ß ,Ñ or Ð+ß ∞Ñ (see Theorem II.3.4). For each of these intervals M , the
endpoints are in E, where 0 is already defined. If M œ Ð+ß ,Ñ then extend the definition of 0
over M by using a straight line segment to join Ð+ß 0 Ð+ÑÑ and Ð,ß 0 Ð,ÑÑ on the graph of 0 . If
M œ Ð+ß ∞ÑÞ then extend the graph of 0 over M using a horizontal right ray at height 0 Ð+Ñà if
M œ Ð ∞ß ,Ñß then extend the graph of 0 over M using a horizontal left ray at height 0 Ð,ÑÞ
As with Urysohn's Lemma, half of the proof is easy. The significant part of theorem is proving the
existence of the extension 1 when \ is normal.
307
Proof ( É ) Suppose E and F are disjoint closed sets in \ . E and F are clopen in the subspace
E ∪ F so the function 0 À E ∪ F Ä Ò!ß "Ó defined by 0 lE œ ! and 0 lF œ " is continuous. Since
E ∪ F is closed in \ , there is a function 1 − GÐ\Ñ such that 1lÐE ∪ FÑ œ 0 Þ Then
Y œ ÖB À 1ÐBÑ + "# × and Z œ ÖB À 1ÐBÑ "# × are disjoint open sets (cozero sets, in fact) that contain
E and F respectively. Therefore \ is normal.
( Ê ) The idea is to find a sequence of functions 13 − GÐ\Ñ such that
8
8
3œ"
3œ"
l0 ÐBÑ 13 ÐBÑl Ä ! as 8 Ä ∞ for each B − E Ðwhere 0 is defined ). The sums 13 ÐBÑ are
defined on all of \ and as 8 Ä ∞ we can think of them as giving better and better approximations to
8
∞
the extension 1 that we want. Then we can let 1ÐBÑ œ lim 13 ÐBÑ œ 13 ÐBÑÞ The details follow.
8Ä∞ 3œ"
3œ"
We proceed in three steps, but the heart of the argument is in Case I.
Case I Suppose 0 is continuous and that 0 À E Ä Ò "ß "Ó. We claim there is a continuous
function 1 À \ Ä Ò "ß "Ó with 1lE œ 0 Þ
Using Lemma 5.5 (with 2 œ 0 ß < œ "Ñ we get a function 1" œ 9 À \ Ä Ò "$ ß "$ Ó such that
for B − E, l0 ÐBÑ 1" ÐBÑl Ÿ #$ . Therefore 0 1" À E Ä Ò #$ ß #$ ÓÞ
Using Lemma 5.5 again (with 2 œ 0 1" ß < œ $# Ñ, we get a function 1# À \ Ä Ò #* ß #* Ó such
that for B − Eß l0 ÐBÑ 1" ÐBÑ 1# ÐBÑl Ÿ *% œ Ð $# Ñ# . So 0 Ð1" < 1# Ñ À E Ä Ò %* ß %* Ó
Using Lemma 5.5 again (with 2 œ 0 1" 1# ß < œ %* Ñ, we get a function
% %
1$ À \ Ä Ò #(
ß #( Ó such that for B − Eß l0 ÐBÑ 1" ÐBÑ 1# ÐBÑ 1$ ÐBÑl Ÿ
) )
So 0 Ð1" < 1# < 1$ Ñ À E Ä Ò #(
ß #( ÓÞ
)
#(
œ Ð #$ Ñ$ .
We continue, using induction, to find for each 3 a continuous function
8
3"
3"
13 À \ Ä Ò # 3 ß # 3 Ó such that l0 ÐBÑ 13 ÐBÑl Ÿ Ð#Î$Ñ8 for B − E.
$
$
∞
∞
3œ"
3œ"
3œ"
∞
Since l13 ÐBÑl Ÿ #$3 + ∞ß the series 1ÐBÑ œ 03 ÐBÑ converges (absolutely) for every
3"
3œ"
∞
B − \ , and 1 is continuous by the Weierstrass Q -Test. Since l1ÐBÑl œ l 13 ÐBÑl
∞
Ÿ l13 ÐBÑl Ÿ
3œ"
3œ"
∞
#3"3
$
3œ"
œ "ß we have 1 À \ Ä Ò "ß "ÓÞ
8
Finally, for B − Eß l0 ÐBÑ 1ÐBÑl œ lim l0 ÐBÑ 13 ÐBÑl Ÿ lim Ð #$ Ñ8 œ !, so 1lE œ 0 and
8Ä∞
3œ"
8Ä∞
the proof for Step I is complete.
Case II Suppose 0 À E Ä Ð "ß "Ñ is continuous. We claim there is a continuous function
1 À \ Ä Ð "ß "Ñ with 1lE œ 0 Þ
Since 0 À E Ä Ð "ß "Ñ © Ò "ß "Ó, we can apply Case I to find a continuous function
J À \ Ä Ò "ß "Ó with J lE œ 0 Þ To get 1, we merely make a slight modification to
J to get a 1 that still extends 0 but where 1 has all its values in Ð "ß "ÑÞ
308
Let F œ ÖB − \ À J ÐBÑ œ „ "×. E and F are disjoint closed sets in \ , so by
Urysohn's Lemma there is a continuous 2 À \ Ä Ò "ß "Ó such that 2lF œ ! and
2lE œ "Þ If we let 1ÐBÑ œ J ÐBÑ2ÐBÑ, then 1 À \ Ä Ð "ß "Ñ and 1lE œ 0 ,
completing the proof of Case II.
Case III (the full theorem) Suppose 0 À E Ä ‘ is continuous. We claim there is a
continuous function 1 À \ Ä ‘ with 1lE œ 0 Þ
Let 2 À ‘ Ä Ð "ß "Ñ be a homeomorphism. Then 2 ‰ 0 À E Ä Ð "ß "Ñ and, by Step
II, there is a continuous J À \ Ä Ð "ß "Ñ with J lE œ 2 ‰ 0 Þ
Let 1 œ 2" ‰ J À \ Ä ‘. Then for B − E we have 1ÐBÑ œ 2" ÐJ ÐBÑÑ
œ 2" ÐÐ2 ‰ 0 ÑÑÐBÑ œ 0 ÐBÑÞ ñ
It is easy to see that G ‡ Ð\Ñ can replace GÐ\Ñ in the statement of Tietze's Extension Theorem.
Example 5.7 We now know enough about normality to see some of its bad behavior. The Sorgenfrey
line W is normal (Example 4.4) but the Sorgenfrey plane W ‚ W is not normal.
To see this, let H œ ‚ ß a countable dense set in W ‚ W . Every continuous real-valued function
on W ‚ W is completely determined by its values on H. (See Theorem II.5.12. The theorem is stated
for the case of functions defined on a pseudometric space, but the proof is written in a way that
applies just as well to functions with any space \ as domain.) Therefore the mapping
GÐW ‚ WÑ Ä GÐHÑ given by 0 È 0 lH is one-to-one, so lGÐW ‚ WÑl Ÿ lGÐHÑ Ÿ l‘H l œ - i! œ -Þ
E œ ÖÐBß CÑ − W ‚ W À B < C œ "× is closed and discrete in the subspace topology, so every function
defined on E is continuous, that is, ‘E œ GÐEÑ and so lGÐEÑl œ - - œ #- Þ If W ‚ W were normal,
then each 0 − GÐEÑ could be extended (by Tietze's Theorem) to a continuous function in GÐW ‚ WÑ.
This would mean that lGÐW ‚ WÑl l‘E l œ - - œ #- - . which is false. Therefore normality is not
even finitely productive.
309
The comments following the statement of Urysohn's Lemma imply that W ‚ W must contain closed
sets that are not zero sets.
A completely similar argument “counting continuous real-valued functions” shows that the
Moore plane > (Example III.5.6) is not normal: use that > is separable and the B-axis in > is
an uncountable closed discrete subspace.
Questions about the normality of products are difficult. For example, it was an open question for a
long time whether the product of a normal space \ with such a nice, well-behaved space as Ò!ß "Ó must
be normal. In the 1950's, Dowker proved that \ ‚ Ò!ß "Ó is normal iff \ is normal and “countably
paracompact.”
However, this result was unsatisfying because no one knew whether a normal space was
automatically “countably paracompact.” In the 1960's, Mary Ellen Rudin constructed a normal space
\ which was not countably paracompact. But this example was still unsatisfying because the
construction assumed the existence of a space called a “Souslin line” and whether a Souslin line
exists cannot be decided in the ZFC set theory! In other words, the space \ she constructed required
adding a new axiom to ZFC.
Things were finally settled in 1971 when Mary Ellen Rudin constructed a “real” example of a normal
space \ whose product with Ò!ß "Ó is not normal. By “real,” we mean that \ was constructed in ZFC,
with no additional set theoretic assumptions. Among other things, this complicated example made use
of the box topology on a product.
Example 5.8 The Sorgenfrey line W is X% , so W is X$ "# and therefore the Sorgenfrey plane W ‚ W is
also X$ "# . So W ‚ W is an example showing that X$ "# does not imply X% .
Extension theorems such as Tietze's are an important topic in mathematics. In general, an “extension
theorem” has the following form:
E © \ and 0 À E Ä F , then there is a function 1 À \ Ä F such that 1lE œ 0 .
For example, in algebra one might ask: if E is a subgroup of \ and 0 À E Ä F is an
isomorphism, can 0 be extended to a homomorphism 1 À \ Ä F ?
If we let 3 À E Ä \ be the injection 3Ð+Ñ œ +, then the condition “1lE œ 0 ” can be rewritten
as 1 ‰ 3 œ 0 . In the language of algebra, we are asking whether there is a suitable function 1
which “makes the diagram commute.”
310
Specific extension theorems impose conditions on E and \ , and usually we want 1 to share some
property of 0 such as continuity. Here are some illustrations, without further details.
1) Extension theorems that generalize of Tietze's Theorem: by putting stronger hypotheses
on \ , we can relax the hypotheses on F .
Suppose E is closed in \ and 0 À E Ä F is continuous.
If
Ú
Ý
Ý \ is normal
\ is normal
Û
Ý
Ý \ is collectionwise normal**
Ü \ is paracompact**
and F œ ‘ (Tietze's Theorem)
and F œ ‘8
and F is a separable Banach space*
and F is a Banach space*
then 0 has a continuous extension 1 À \ Ä FÞ
The statement that ‘8 can replace ‘ in Tietze's Theorem is easy to prove:
If \ is normal and 0 À E Ä ‘8 is continuous, write 0 ÐBÑ œ Ð0" ÐBÑß 0# ÐBÑß ÞÞÞß 08 ÐBÑÑ
where each 03 À E Ä ‘. By Tietze's Theorem, there exists for each 3 a continuous
extension 13 À \ Ä ‘ with 13 lE œ 03 . If we let 1ÐBÑ œ Ð1" ÐBÑß ÞÞÞß 18 ÐBÑÑ, then
1 À \ Ä ‘8 and 1lE œ 0 . In other words, we separately extend the coordinate
functions in order to extend 0 . And in this example, 8 could even be an infinite
cardinal.
* A normed linear space is a vector space Z with a norm l@l ( œ “absolute value”) that
defines the “length” of each vector. Of course, a norm must satisfy certain axioms for
example, l@" < @# l Ÿ l@" l < l@# l. These properties guarantee that a norm can be used to
define a metric: .Ð@" ß @# Ñ œ l@" @# lÞ A Banach space is a normed linear space which is
complete in this metric ..
For example, ‘8 À the usual norm lÐB" ß B# ß ÞÞÞß B8 Ñl œ ÈB#" < B## < ÞÞÞ < B#8 produces
the usual metric, which is complete. So ‘8 is a separable Banach space.
** Roughly, a “collectionwise normal” space is one in which certain infinite collections of
disjoint closed sets can be enclosed in disjoint open sets. We will not give definitions for
“collectionwise normal” (or the stronger condition, “paracompactness”) here, but is true
that
Ú metric
Ê
Û or
Ü compact X#
paracompact Ê collectionwise normal Ê normal
Therefore, in the theorems cited above, a continuous map 0 defined on a closed subset of a
metric space (or, compact X# space) and valued in a Banach space F can be continuously
extended a function 1 À \ Ä F .
311
2) The Hahn-Banach Theorem is another example, taken from functional analysis, of an
extension theorem. A special case states:
Suppose Q is a linear subspace of a real normed linear space \ and that 0 À Q Ä ‘
is linear and satisfies 0 ÐBÑ Ÿ llBll for all B − Q Þ There there is a linear J À \ Ä ‘
such that J lQ œ 0 and J ÐBÑ Ÿ llBll for all B − \Þ
3) Homotopy is usually not discussed in terms of extension theorems, but extensions are really
at the heart of the idea.
Let 0 , 1 À Ò!ß "Ó Ä \ be continuous and suppose that 0 Ð!Ñ œ 1Ð!Ñ œ B! and
0 Ð"Ñ œ 1Ð"Ñ œ B" Þ Then 0 and 1 are paths in \ that start at B! and end at B" Þ Let F
be the boundary of the square Ò!ß "Ó# © ‘# and define J À F Ä \ by
J ÐBß !Ñ œ 0 ÐBÑ
J Ð!ß >Ñ œ B!
J ÐBß "Ñ œ 1ÐBÑ
J Ð"ß >Ñ œ B"
Thus J agrees with 0 on the bottom edge of F and with 1 on the top edge. J is constant
Ð œ B! Ñ on the left edge of F and constant Ð œ B" Ñon the right edge of F . We ask whether J
can be extended to a continuous map defined on the whole square, L À Ò!ß "Ó# Ä \Þ
If such an extension L does exist, then we have
For each > − Ò!ß "Óß restrict L to the line segment at height > to define 0> ÐBÑ œ LÐBß >Ñ. Then
for each > − Ò!ß "Ó, 0> is also a path in \ from B! to B" . As > varies from ! to ", we can think
of the 0> 's as a family of paths in \ that continuously deform 0! œ 0 into 0" œ 1Þ
The continuous extension L (if it exists) is called a homotopy between 0 and 1 with fixed
endpoints, and we say that the paths 0 and 1 are homotopic with fixed endpoints.
312
In the space \ on the left, below, it seems intuitively clear that 0 can be continuously
deformed (with endpoints held fixed) into 1 in other words, that L exists.
However in the space ] pictured on the right, 0 and 1 together form a loop that surrounds a
“hole” in ] , and it seems intuitively clear that the path 0 cannot be continuously deformed
into the path 1 inside the space ] that is, the extension L does not exist.
In some sense, homotopy can be used to detect the presence of certain “holes” in a space, and
is one important part of algebraic topology.
The next theorem shows us where compact Hausdorff spaces stand in the discussion of separation
properties.
Theorem 5.9 A compact X# space \ is X% .
Proof \ is Lindelöf, and a regular Lindelöf space is normal (Theorem 4.3). Therefore it is sufficient
to show that \ is regular. Suppose J is a closed set in \ and B Â J Þ For each C − J we can pick
disjoint open sets YC and ZC with B − YC and C − ZC . J is compact so a finite number of the ZC 's
cover J say ZC" ß ZC# ß ÞÞÞ ß ZC8 Þ Then B − +83œ" YC3 œ Y , J © -83œ" ZC3 œ Z , and Y ß Z are disjoint
open sets. ñ
Therefore, our results line up as:
Ú compact X
(*) compact metric Ê Û ____
Ü metric
#
Ê X% Ê X$ "# Ê X$ Ê X# Ê X" Ê X!
In particular, Urysohn's Lemma and Tietze's Extension Theorem hold in metric spaces and in compact
X# spaces.
313
Notice that
i)
the space Ð!ß "Ñ is X% but not compact X#
ii) the Sorgenfrey line W is X% (see example 5.4) but not metrizable. ÐIf W were metrizable,
then W ‚ W A9?6. be metrizable and therefore X% which is false À see Example 5.7).
iii) Ò!ß "Ó- is compact X# (assuming, for now, the Tychonoff Product Theorem VI.3.10 ) but
not metrizable (why?)
iv) Ð!ß "Ñ is metrizable but not compact.
Combining these observations with earlier examples, we see that none of the implications in
(*) is reversible.
Example 5.10 (See Example 5.7) The Sorgenfrey plane W ‚ W is X$ "# , so W ‚ W can be embedded
in a cube Ò!ß "Ó7 and Ò!ß "Ó7 is compact X# (assuming the Tychonoff Product Theorem). Since W ‚ W
is not normal, we see now that a normal space can have nonnormal subspaces. This example,
admittedly, is not terribly satisfying since we can't visualize how W ‚ W “sits” inside Ò!ß "Ó7 . In
Chapter VIII (Example 8.10), we will look at an example of a X% space in which it's easy to “see” why
a certain subspace isn't normal.
6. Some Metrization Results
Now we have enough information to completely characterize separable metric spaces topologically.
Theorem 6.1 (Urysohn's Metrization Theorem) A second countable X$ -space is metrizable.
Note: We proved a similar metrization theorem in Corollary 3.18, but there the separation hypothesis
was X$ "# rather than X$ .
Proof \ is second countable so \ is Lindelöf, and Theorem 4.3 tells us that a Lindelöf
X$ -space is X% Þ Therefore \ is X$ "# . So by Corollary 3.18, \ is metrizable. ñ
Because a separable metrizable space is second countable and X$ , we have a complete
characterization: \ is a separable metrizable space iff \ is a second countable X$ -space. So, with
hindsight, we now see that the hypothesis “X$ "# ” in Corollary 3.18 was unnecessarily strong. In fact,
we see that X$ and X$ "# are equivalent in a space that is second countable.
Further developments in metrization theory hinged on work of Arthur H. Stone in the late 1940's in
particular, his result that metric spaces have a property called “paracompactness.” This led quickly to a
complete characterization of metrizable spaces that came roughly a quarter century after Urysohn's
work. We state this characterization here without a proof.
314
A family of sets U in Ð\ß g Ñ is called locally finite if each point B − \ has a neighborhood R that has
nonempty intersection witth only finitely many sets in U. The family U is called 5-locally finite if we
can write U œ -8− U8 where each subfamily U8 is locally finite.
Theorem 6.2 (The Bing-Smirnov-Nagata Metrization Theorem) Ð\ß g Ñ is metrizable iff \ is X$
and has a 5-locally finite base U.
Note: If \ is second countable, a countable base U œ ÖS" ß S# ß ÞÞÞß S8 ß ÞÞÞ × is 5-locally
finite because we can write U œ -U8 ß where U8 œ ÖS8 ×Þ Therefore this Metrization Theorem
includes Urysohn's Metrization Theorem as a special case.
The Bing-Smirnov-Nagata Theorem has the typical form of most metrization theorems: \ is
metrizable iff “\ has enough separation” and “\ has a nice enough base.”
315
Exercises
E15. Let Ð\ß .Ñ be a metric space and W © \ . Prove that if each continuous 0 À W Ä ‘ extends to a
continuous 1À \ Ä ‘, then W is closed. (The converse, of course, follows from Tietze's Extension
Theorem.)
E16. Urysohn's Lemma says that in a X% -space disjoint closed sets are completely separated. Part a)
shows that this is also true in a Tychonoff space if one of the closed sets is compact.
a) Suppose \ is Tychonoff and J ß O © \ where J is closed, O is compact. \ and
J ∩ O œ g. Prove that there is an 0 − GÐ\Ñ such that 0 ± O œ ! and 0 ± J œ ". (This is another
example of the rule of thumb that “compact spaces act like finite spaces.” If necessary, try proving
the result first for a finite set OÞ )
b) Suppose \ is Tychonoff and that : − Y , where Y is open in \Þ Prove Ö:× is a K$ set in \
iff there exists a continuous function 0 À \ Ä Ò!ß "Ó such that 0 " Ð"Ñ œ Ö:× and 0 l\ Y œ !.
E17. Suppose ] is a Hausdorff space. Define B µ C in ] iff there does not exist a continuous
function 0 À \ Ä Ò!ß "Ó such that 0 ÐB) Á 0 ÐCÑ. Prove or disprove: ] Î µ is a Tychonoff space.
E18. Prove that a Hausdorff space \ is normal iff for each finite open cover h œ ÖY" , ... , Y8 × of \ ,
there exist continuous functions 03 À \ Ä Ò!ß "Ó Ð3 œ "ß ÞÞÞ, 8Ñ such that 83œ" 03 ÐBÑ = 1 for each B − \
and such that, for each 3, 03 ± \ Y3 ´ !. (Such a set of functions is called a partition of unity
subordinate to the finite cover h.)
Hint Ð Ê Ñ First build a new open cover i œ ÖV" ,...,V8 × that “shrinks” h in the sense that,
Z3 © cl Z3 © Y3 for each 3. To begin the construction, let J" œ \ -3" Y3 Þ Pick an open
Z" so that J" © Z" © cl Z" © Y" Þ Then ÖZ" ß Y# ß ÞÞÞß Y8 × still covers \Þ Continue by
looking at J# œ \ ÐZ" ∪ -3# Y3 Ñ and defining Z# so that ÖZ" ß Z# ß Y$ ß ÞÞÞß Y8 × is still a
cover and Z# © cl Z# © Y# Þ Continue in this way to replace the U3 's one by one. Then use
Urysohn's lemma to get functions 13 which can then be used to define the 03 's. .
E19. Suppose \ is a compact, countable Hausdorff space. Prove that \ is completely metrizable.
Hint: 1) For each pair of points B8 Á B7 in \ pick disjoint open sets Y7ß8 and Z7ß8
containing these points. Consider the collection of all finite intersections of such sets.
2) Or: Since \ is, countable, every singleton Ö:× is a K$ set. Use regularity to find
a descending sequence of open sets Z8 containing : such that +∞
8œ" cl Z8 œ Ö:×Þ Prove that
the Z8 's are a neighborhood base at :.
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E20. A space \ is called completely normal if every subspace of \ is normal. (For example, every
metric space is completely normal).
a) Prove that \ is completely normal if and only if the following condition holds:
whenever Eß F © \ and (cl E ∩ FÑ ∪ ÐE ∩ cl FÑ œ g (that is, each of Eß F is
disjoint from the closure of the other), then there exist disjoint open set= Y and Z
with E © Y and F © Z Þ
b) Recall that the “scattered line” (Exercise IIIE.10) consist of the set \ œ ‘ with the
topology g œ ÖY ∪ Z À Y is open in the usual topology on ‘ and Z © ×. Prove that the
scattered line is completely normal and therefore X% Þ
E21. A X" -space \ is called perfectly normal if whenever E and F are disjoint nonempty closed sets
in \ , there is an 0 − GÐ\Ñ with 0 " Ð!Ñ œ E and 0 " Ð"Ñ œ F .
a) Prove that every metric space Ð\ß .Ñ is perfectly normal.
b) Prove that \ is perfectly normal iff \ is X% and every closed set in \ is a K$ -set.
Note: Example 3.10 shows a X% . space \ that is not perfectly normal.
c) Show that the scattered line (see Exercise E20) is not perfectly normal, even though every
singleton set Ö: × is a K$ -set.
d) Show that the scattered line is X% Þ
Hint: Use the fact that ‘, with the usual topology, is normal. Nothing deeper than Urysohn's
Lemma is required but the problem is a bit tricky.
E22. Prove that a X$ space Ð\ß g Ñ has a locally finite base U iff g is the discrete topology.
(Compare to Theorem 6.2.)
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Chapter VII Review
Explain why each statement is true, or provide a counterexample.
1. Suppose Ð\ß g Ñ is a topological space and let gA be the weak topology on \ generated by GÐ\ÑÞ
Then g © gA Þ
2. If \ is regular and B − cl ÖC×, then C − cl ÖB×.
3. Every separable Tychonoff space can be embedded in Ò!ß "Ói! .
4. If 0 − GÐÑß we say 1 is a square root of 0 if 1 − GÐÑ and 1# œ 0 . If a function f in GÐÑ has
more than one square root, then it has - square roots.
5. In a Tychonoff space, every closed set is an intersection of zero sets.
6. A subspace of a separable space need not be separable, but every subspace of the Sorgenfrey line is
separable.
7. Suppose has the cofinite topology. If E is closed in , then every 0 − GÐEÑ can be extended to
a function 1 − GÐÑÞ
8. For 8 œ "ß #ß ÞÞÞ, let 08 À ‘ Ä ‘ be given by 08 ÐBÑ œ B < 8 and let g be the weak topology on ‘
generated by the 08 's. Then the evaluation map / À ‘ Ä ‘i! given by /ÐBÑÐ8Ñ œ 08 ÐBÑ is an
embedding.
9. Let G be the set of points in the Cantor set with the subspace topology from the Sorgenfrey line W .
Every continuous function 0 À G Ä ‘ can be extended to a continuous function 1 À W Ä ‘Þ
10. If Ð\ß .Ñ is a metric space, then \ is homeomorphic to a dense subspace of some compact
Hausdorff space.
11. Suppose J © ‘# Þ J is closed iff J is a zero set.
12. Suppose O is a compact subset of the Hausdorff space \ ‚ ] . Let E œ 1\ ÒOÓ. Then E is X% Þ
13. Every space is the continuous image of a metrizable space.
14. Let Y œ Ö0 − ‘‘ À 0 is not continuous×. The weak topology on ‘ generated by Y is the
discrete topology.
55. A compact X# space is metrizable if and only if it is second countable.
16. Suppose J and O are disjoint subsets of a Tychonoff space \ , where J is closed and O is
compact. There are disjoint cozero sets Y and Z with J © Y and O © Z Þ
17. Every X% space is homeomorphic to a subspace of some cube Ò!ß "Ó7 Þ
18.. Suppose \ is a X% -space with nonempty pairwise disjoint closed subspaces J" ß ÞÞÞß J8 . There is an
0 − GÐ\Ñ such that 0 lJ3 œ 3 for all 3 œ "ß ÞÞÞß 8Þ
318