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 nicerespecially 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 with Y B − Y C Â Y,
there exists an open set with ,
or Z 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
there exists an open set with
and Z B Â Y ß C − Z
3) X#-space (or, Hausdorff space) if, whenever B Á C − \, there exist disjoint open sets Y
and in such that Z \ 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 aBÁaC that is, different points
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 % !eitherB Â F ÐCÑ% orC 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 iffaB œaC.
Let 1 À \ Ä \Î µ œ ] by 1ÐBÑ œ ÒBÓÞGive the quotient topology. Then is continuous, onto,] 1
open (not automatic for a quotient map!) and the quotient is a X! space:
If is open in S \, 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 in C S× œ S.
If ÒBÓ Á ÒCÓ − ], then is not equivalent to , so there is an open set B C S © \ with
(say) B − S and C Â S. Since is open, 1 1ÒSÓ is open in and ] ÒBÓ − 1ÒSÓ. Moreover,
ÒCÓ Â 1ÒSÓ or else would be C 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 is one-1
to-one and is a homeomorphism. Applied to a 1 X! space, the X!-identification accomplishes nothing.
If Ð\ß .Ñis a pseudometric space, the X!-identification is the same as the metric identification
discussed in Example VI.5.6 because, in that case,aBœaC if and only if .ÐBß CÑ œ !Þ
5) For 3 œ !ß "ß # À if Ð\ß Ñg is a space and X3 g w ªg is a new topology on , then \ Ð\ßg wÑ
is also a space.X3
Example 2.3
1) (Exercise) It is easy to check that a space \is aX" space
iff for each B − \ ÖB×, is closed
iff for each B − \ ÖB× œ, +ÖS À Sis 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 ∞Ñis an open
set that contains and not ; but there is no open set that contains and not .C B B C
5) With the cofinite topology, is X" but not X# because, in an infinite cofinite space, any
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 -space is aX3 X3-space
b) If \ œ#α−E\ Á gα , then is a -space iff each \ X3 \α is a -space.X3
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 Yw is an open set in \ containing but not ,B C
then Y œ Y ∩ Ew is an open set in E containing but not . Similarly we can find an open set in B C Z E
containing but not Therefore C BÞ Eis 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 B Á Cα α
for some . Pick an open set α Yα in \α containing Bα but not Cα. Then Y œ + Y α is an open set
in \containing but not . Similarly, we find an open set B C Z in containing but not . Therefore\ C B
\ is a X"-space. ñ
Exercise 2.5 Is a continuous image of a -space necessarily a -space? How about a quotient?X3 X3
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 is a closed set and J B Â J ß
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 is an open set containing , then there exists an open set S B Y © \ such that
B − Y ©clY © S
iii) at each point B − \ there exists a neighborhood base consisting of closed
neighborhoods.
Proof i)Êii) Suppose \ is regular and is an open set with S 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 ©clY © S (since clY © \ Z).
ii)Êiii) If R −aB, then B − S œintR. By ii), we can find an open set so thatY
B − Y ©clY © S © R Þ Since clY is a neighborhood of , the closed neighborhoods of form aB B
neighborhood base at .B
iii)Êi) Suppose is closed and J B Â J. By ii), there is a closed neighborhood O of suchB
that B − O © \ J. We can choose Y œintO 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 JandJ is closed. We have
a continuous function 0 ÐBÑ œ .ÐBß J Ñ for which 0 Ð+Ñ œ - ! and 0 lJ œ !ÞThis 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
To bring things into line, we make the following definition.
Definition A topological space is called a \ X$-space if is regular \ andX".
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$”).
Example 2.11 X Ê X# Î $. We will put a new topology on the set \ œ‘#. At each point : − \, let a
neighborhood base U: consist of all sets of the formR
a finite number of straight lines through for some
R œ F Ð:Ñ Ð% :Ñ ∪ Ö:× % !Þ
(Check that the conditions in the Neighborhood Base Theorem III.5.2 are satisfied.) With the
resulting topology, \ is called the slotted plane. Note that F Ð:Ñ −% U: (because “ ” is a finite!
number), so each F Ð:Ñ% is among the basic neighborhoods in U:so the slotted plane topology on ‘#
contains the usual Euclidean topology. It follows that is \ X#.
The set J œ ÖÐBß !Ñ À B Á !× œ“the -axis with the origin deleted” is a closed set in (B \ why?).
If is any open set containing the origin Y Ð!ß !Ñ, then there is a basic neighborhood withR
Ð!ß !Ñ − R © Y ÞUsing the in the definition of , we can choose a point% R : œ ÐBß !Ñ − Jwith
! + B + Þ% Every basic neighborhood set of must intersect (: R why?) and therefore must intersect
Y. It follows that Ð!ß !Ñand cannot be separated by disjoint open sets, so the slotted plane is notJ
Note: The usual topology in ‘#is regular. This example shows that an “enlargement” of a regular (or X$) topology may not be regular (or X$). Although the enlarged topology has more open sets to work with, there are also more “point/closed set pairs Bß J” that need to be separated. By contrast, it is easy to see that an “enlargement” of a topology X3 Ð3 œ !ß "ß #Ñ is still X Þ3
Example 2.12 The Moore plane (Example III.5.6) is clearly > X Þ# In fact, at each point, there is a
neighborhood base of closed neighborhoods. The figure illustrates this for a point Ton the -axis andB
a point Uabove the -axis. Therefore is B > X$.
Theorem 2.13 a) A subspace of a regular (X Ñ$ space is regular (X$).
b) Suppose \ œ#α−E\ Á g \α . is regular (X$) iff each \αis regular (X$).
Proof a) Let E © \ where is regular. Suppose \ + − E and that is a closed set J in E that does
not contain . There exists a closed set + Jw in \ such that J ∩ E œ J Þw Choose disjoint open sets
Yw and Z w in with \ + − Y w and J © Z Þw w Then Y œ Y ∩ Ew andZ œ Z ∩ Ew are open in ,E
disjoint, + − Y, and J © Z. Therefore is regular.E
b) If \ œ#α−E\ Á gα is regular, then part a) implies that each \α is regular because eachß
\α is homeomorphic to a subspace of . Conversely, suppose each \ \αis regular and that
Y œ + Y ß ÞÞÞß Yα" α8 is a basic open set containing . For each B α3, we can pick an open setZα3in
\α3such that B − Z ©α3 α3 clZ © Y Þα3 α3 Then B − Z œ + Z ß ÞÞÞß Zα" α8 ©clZ
© +clZ ß ÞÞÞßα" clZα8 © Y Þ (Why is the last inclusion true?) Therefore is regular.\
Since the X" property is hereditary and productive, a) and b) also hold for X$-spaces ñ
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 , in with Y Z \ E © Y and F © Z Þ \ is called a X%-space if \ is
Example 2.15 a) Every pseudometric space Ð\ß .Ñis normal (so every metric space is X Ñ% .
In fact, if and are disjoint closed sets, we can defineE F 0 ÐBÑ œ .ÐBßEÑ < .ÐBßFÑ.ÐBßEÑ . Since the
denominator cannot be , is continuous and ! 0 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 and respectively. Therefore is normal.E F \ 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 and ?E F
b) Let have the right ray topology ‘ g œ ÖÐBß ∞Ñà B − × ∪ Ögß ×‘ ‘ . Ð ß Ñ‘ g is
normal because the only possible pair of disjoint closed sets is and and we can separate theseg \
using the disjoint open sets Y œ g and Z œ \Þ Also, Ð ß Ñ‘ g is not regular: for example is not in"
the closed set J œ Ð ∞ß !Ó, but every open set that contains also contains . So J " normal
ÊÎ regular. But Ð ß Ñ‘ g is not X" and therefore notX Þ%
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 is a closed set not containing , then J B ÖB× and are disjoint closedJ
sets. By normality, we can find disjoint open sets separating ÖB×and . It follows that is regularJ \
and therefore X$. ñ
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 of \ Y closed subsets such that every
closed set is an intersection of sets from . Clearly, is a base for the closed sets in iffJ Y Y \
Uœ ÖS À S œ \ J ß J −Y× is a base for the open sets in .\
For a polynomialT in real variables, define the 8 zero set of asT
^ÐT Ñ œ ÖÐB ß B ß ÞÞÞß B Ñ −" # 8 ‘8À T ÐB ß B ß ÞÞÞß B Ñ œ !×" # 8
a) Prove that Ö^ÐT Ñ À T a polynomial in real variables is the base for the closed sets of a8 ×
topology (called the Zariski topology) on ‘8Þ
b) Prove that the Zariski topology on ‘8is but not .
" #
X X
c) Prove that the Zariski topology on is the cofinite topology, but that if ‘ 8 ", the Zariski
topology on ‘8is not the cofinite topologyÞ
Note: The Zariski topology arises in studying algebraic geometry. After all, the sets ^ÐT Ñare rather special geometric objects—those “surfaces” in ‘8 which can be described by polynomial equations T ÐB ß B ß ÞÞÞß B Ñ œ !" # 8 .
E3. A space \ is a X&Î# space if, whenever B Á C − \, there exist open sets Y and Zsuch that
B− Y C − Z, and clY ∩clZ œ g. Clearly, T ( $ÊX&Î# ÊT#Þ) a) Prove that a subspace of a X&Î# space is a X&Î#space.
b) Suppose \ œ#\ Á gα . Prove that is \ X&Î# iff each \α isX&Î# .
c) Let W œ ÖÐBß CÑ −‘#À C !×and P œ ÖÐBß CÑ − W À C œ !×ÞDefine a topology on withW
the following neighborhood bases:
if :− W P À U: œ ÖF Ð:Ñ ∩ W À !×% %
if : − P: U: œ ÖF Ð:Ñ ∩ ÐW PÑ ∪ Ö:× À !×% %
You may assume that these U:'s satisfy the axioms for a neighborhood base.
E4. Suppose E © \. Define a topology on by\
g œ ÖS © \ À S ª E× ∪ Ög×
Decide whether or not Ð\ß Ñg is normal.
E5. A function 0 À \ Ä ] is called perfect if is continuous, closed, onto, and, for each 0 C − ]ß
0" ]
$
ÐCÑ is compact. Prove that if \ is regular and is perfect, then 0 is regular and that if à \ is X ,
the is also ] X Þ$
E6. a) Suppose \ is finite. Prove that Ð\ß Ñg is regular iff there is a partition of U \ 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 is closed in a J X$-space \ÞProve that
i) Prove that J œ+ÖS À S is open and J © S×Þ
ii) Define B µ C iff B œ C or Bß C − J Þ Prove that the quotient space \Î µ is Hausdorff.
d) Suppose is an infinite subset of a F X$-space . Prove that there exists a sequence of open sets\
Y8 such that each Y ∩ F Á g8 and that clY ∩8 clY7 œ g whenever 8 Á 7.
e) Suppose each point in a space has a neighborhood such that clC ] Z Z is regular. Prove
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 can be defined on a 0 metric space Ð\ß .Ñusing 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$-space can sometimes be very deficient in continuous real-valued functionsin 1946, Hewett gave
an example of a infinite X$-space on which the L 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 \
is bounded the collection of
G Ð\Ñ œ Ö0 − GÐ\Ñ À 0‡ × œ 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 œmaxÖ0 ÐBÑß 1ÐBÑ×
min
Ð0 • 1ÑÐBÑ œ 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#
Definition 3.2 A space is called \ completely regular if whenever Jis a closed set and + Â J ßthere
exists a function 0 − GÐ\Ñ such that 0 Ð+Ñ œ ! and 0 lJ œ "Þ
Informally, “completely regular” means that “ and can be separated by a continuous real-valued+ J 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 , there is also a continuous 0 1 À \ Ä Ò!ß "Ósuch 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 "Óis 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 9À 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 œ9‰ 1 À \ Ä Ò<ß =Ó and depending on how you chose ß 9ß 2Ð+Ñ œ <and 2lJ œ , or vice-versa.)
Putting these observations together, we see Definition 3.2 is equivalent to:
Definition 3.2 w A space is called \ completely regular if whenever Jis 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 < and0 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 possible to contrive a purely internal definition for “completely regular,”is
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 and J + Â J ÞThen
0 ÐBÑ œ .ÐBßJ Ñ.Ð+ßJ Ñis 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$", we can choose 0 − GÐ\Ñwith
#
0 Ð+Ñ œ !and 0 lJ œ "Þ Then Y œ 0"ÒÐ ∞ß ÑÓ"# and Z œ 0"ÒÐ ß ∞ÑÓ"# are disjoint open sets with
+ − Y ß J © Z Þ Therefore is regular. Since is \ \ X", \ is X Þ ñ$
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$" (the example, in 9, is
# Ann. Math., 47(1946) 503-50
rather complicated ). For that purpose, it is a little easier. but still nontrivialto find a X$ 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\
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 in comments later in this section.L
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$" .
# #
b) Suppose \ œ#α−E\ Á g \α . is completely regular (X Ñ$" iff each \αis
#
completely regular (X Ñ$" .
#
Proof Suppose + Â J © E © \, where \is completely regular and is a closed set J in E. Pick a
closed set O in \ such that O ∩ E œ J and an 0 − GÐ\Ñ such that 0 Ð+Ñ œ !and 0 lO œ ". Then
1 œ 0 lE − GÐEÑß 1Ð+Ñ œ ! and 1lJ œ "Þ Therefore is completely regular.E
If g Á \ œ#α−E\α is completely regular, then each \α is homeomorphic to a subspace of
\so each \α is completely regular. Conversely, suppose each \α is completely regular and that isJ
a closed set in not containing . There is a basic open set such that\ + Y
+ − Y œ + Y ß ÞÞÞYα" α8 © \ J
For each 3 œ "ß ÞÞÞß 8 we can pick a continuous function 0 À \ Ä Ò!ß "Óα3 α3 with 0 Ð+ Ñ œ !α3 α3 and
0 Ð\ Y Ñ œ "Þα3| α3 α3 Define 0 À \ Ä Ò!ß "Ó by
max max
0 ÐBÑ œ ÖÐ0 ‰α3 1α3ÑÐBÑ À 3 œ "ß ÞÞÞß 8× œ Ö0 ÐB Ñ À 3 œ "ß ÞÞÞß 8×α3 α3
Then is continuous and 0 0 Ð+Ñ œmaxÖ0 Ð+ Ñ À 3 œ "ß ÞÞÞß 8× œ !Þα3 α3 If B − J, then for some 3ß
B  Yα3 α3 and 0 ÐB Ñ œ "α3 α3 , so 0 ÐBÑ œ ". Therefore 0 lJ œ " and \is completely regular.
.
Since the X" property is both hereditary and productive, the statements in a) and b) also hold
for X Þ$" ñ
#
Corollary 3.8 For any cardinal , the “cube” 7 Ò!ß "Ó7 and all its subspaces are X .
$" #
Since a Tychonoff space \ is defined using functions in GÐ\Ñ, we expect that these functions will
have a close relationship to the topology on . We want to explore that connection.\
Definition 3.9 Suppose 0 − GÐ\Ñ. Then ^Ð0 Ñ œ 0"ÒÖ!×Ó œ ÖB − \ À 0 ÐBÑ œ !× is called the
zero set of . If 0 E œ ^Ð0 Ñ for some 0 − GÐ\Ñ, we call a zero set in . The complement of a zeroE \
set in is called a \ cozero set: cozÐ0 Ñ œ \ ^Ð0 Ñ œ ÖB − \ À 0 ÐBÑ Á !×Þ
A zero set ^Ð0 Ñin \ is closed because is continuous. In addition, 0 ^Ð0 Ñ œ+8œ"∞ S8, where
S œ ÖB − \ À l0 ÐBÑl +8 8"×. Each S8 is open. Therefore a zero set is always a closed K$-set.
For 0 − GÐ\Ñ, let 1 œ Ð " ” 0 Ñ • " − G Ð\ÑÞ‡ Then ^Ð0 Ñ œ ^Ð1Ñ. Therefore GÐ\Ñand G Ð\ч
produce the same zero sets in (and therefore also the same cozero sets).\
Example 3.10
1) A closed set in a pseudometric space J 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 UBœ 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 EÑsatisfies
E © \ Ö:×, so Eis 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 is a closed set not containing , then either J B J © \ Ö:×or ÖB× © \ Ö:×.
So one of the sets or J 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$set in , so \ Ö:×is not a zero set in .\
Suppose : −+8œ"∞ S8 where S8 is open. For each , there is a basic neighborhood 8 F8
of such that : : − F © S ß8 8 so \ S © \ F8 8is countable. Therefore
is countable. Since is uncountable, we conclude
\ +8œ"∞ S œ8 -8œ"∞ Ð\ S Ñ8 \
that Ö:× Á+8œ"∞ S Þ8
Even when is both closed J and a K$ set, might not be a zero set. We will see examplesJ 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
where ^ œ ÖB À 0 ÐBÑ œ <× œ ^Ð1Ñß 1ÐBÑ œ 0 ÐBÑ < 0 , where ^ œ ÖB À 0 ÐBÑ × œ ^Ð1Ñ 1ÐBÑ œ 0 ÐBÑ l0 ÐBÑl 0 , where ^ œ ÖB À 0 ÐBÑ Ÿ × œ ^Ð1Ñ 1ÐBÑ œ 0 ÐBÑ < l0 ÐBÑl , where ^ œ ÖB À 0 ÐBÑ r× œ ^Ð1Ñ 1ÐBÑ œ Ð0 ÐBÑ <Ñ l0 ÐBÑ <l , where ^ œ ÖB À 0 ÐBÑ Ÿ r× œ ^Ð1Ñ 1ÐBÑ œ 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 where ^Ð1Ñ œ ÖB À 0 ÐBÑ œ <× 0 ÐBÑ œ 1ÐBÑ < < 0 where ^Ð1Ñ œ ÖB À 0 ÐBÑ × 0 ÐBÑ œ l1ÐBÑl 0 where ^Ð1Ñ œ ÖB À 0 ÐBÑ Ÿ × 0 ÐBÑ œ l1ÐBÑl where ^Ð1Ñ œ ÖB À 0 ÐBÑ r× 0 ÐBÑ œ < l1ÐBÑl where ^Ð1Ñ œ ÖB À 0 ÐBÑ Ÿr× 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 is closed in J \Þ If \ is regular, we can find disjoint
open sets and Y Z with B − Y and J © Z Þ But if is completely regular, we can separate and \ B J
with “special” open sets and Y Z cozero sets! Just choose 0 − GÐ\Ñ with 0 ÐBÑ œ ! and 0 lJ œ ",
then
and
B − Y œ ÖB À 0 ÐBÑ + ×" J © Z œ ÖB À 0 ÐBÑ ×"
# #
In fact this observation characterizes completely regular spacesthat is, if a regular space fails to be
completely regular, it is because there is a “shortage” of cozero setsbecause 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 and g 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 Ñ GÐ\Ñg , and G Ð\ч induce the same weak topology g on \,
A
and a base for gA is the collection of all cozero sets in .\
Proof A subbase for gAconsists of all sets of the form 0"ÒY Ó, where is open in and Y ‘ 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 forgA
In fact, the cozero sets are actually a base because cozÐ0 Ñ ∩ cozÐ1Ñ œ cozÐ0 1Ñ: the intersection of
two cozero sets is a cozero set.
The same argument, with G Ð\ч replacing GÐ\Ñß shows that the cozero sets of G Ð\ч are a base for
the weak topology on \generated by G Ð\ÑÞ‡ But GÐ\Ñ and G Ð\ч produce the same cozero sets in
\, and therefore generate the same weak topology gAon . \ ñ
Now we can now see the close connection between \ and GÐ\Ñ in completely regular spaces. For
any space Ð\ß Ñg the functions in GÐ\Ñ certainly are continuous with respect to g (by definition of
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 Ñg , 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\ч )
d) GÐ\Ñ (equivalently, G Ð\ч ) separates points from closed sets.
Proof The preceding theorem shows that b) and c) are equivalent.
a)Êb) Suppose + − S where is open Let S Þ 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)Êd) Suppose is a closed set not containing . By b), we can choose J + 0 − GÐ\Ñ so that
+ −cozÐ0 Ñ © \ J Þ Then 0 Ð+Ñ œ < Á !, so 0 Ð+Ñ Âcl0 ÒJ Ó œ Ö!×Þ Therefore GÐ\Ñ separates points and closed sets.
d)Êa) Suppose is a closed set not containing . There is some J + 0 − GÐ\Ñfor which
0 Ð+Ñ Âcl0 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\ч by 1ÐBÑ œminÖl0 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\ч (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\ßg Ñ is completely regularÞ Ð\ßg ) is Tychonoff is Yseparates points.
Y Y
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 ©gYÞ
On the other hand: Y © GÐ\Ñby definition of gY, and the larger collection of functions GÐ\Ñ
generates a (potentially) larger weak topology. Therefore gY ©gAÞ
Example 3.14
1) If Y œ Ö0 −‘‘ À 0 is nowhere differentiable , then the weak topology × g on generated‘
Y
by is completely regular.Y
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\ч , we have ranÐ0 Ñ © Ò+ ß ,0 0]œ M0 for some + + , − Þ0 0 ‘ The evaluation map / À \ Ä#ÖM À 0 − G Ð\Ñ×0 ‡ is an embedding.
Proof \ is X", the 's are continuous and the collection of 's 0 0 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, #ÖM À 0 − G Ð\Ñ×0 ‡ is homeomorphic to Ò!ß "Ó ß7 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 Ð Ñl‡ ‘ , even though we can embed ‘ in Ò!ß "Ó œ Ò!ß "Ó Þ" The
following theorem improves the value for 7in 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 of cardinality . Then can be embedded in\ U 7 \ 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
\ © Ò!ß "Ó © Ò!ß "Ó Þtop top 7
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 ßZÐBÑ + "# for all B − Y and
0Y ßZÐBÑ œ " for all B − \ Z Þ Clearly, Y © Z for a distinguished pair ÐY ß Z ÑÞ For each
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 separates points and closed sets:Y
Suppose is a closed set not containing . Choose a basic set J + Z −Uwith + − Z © \ J Þ
There is a distinguished pair ÐY ß Z Ñ with + − Y © Z © \ J Þ Then 0Y ßZÐ+Ñ œ < + "# and
0Y ßZlJ œ ", so 0Y ßZÐ+Ñ Âcl0Y ßZÒJ Ó œ Ö"×Þ
By Corollary VI.4.11, / À \ Ä Ò!ß "Ól lY is an embedding. Since 7 is infinite,
l l Ÿ l ‚ l œ 7 œ 7Y U U # . ñ
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”) A second countable Tychonoff space \ is metrizable.
Proof By Theorem 3.17, \ © Ò!ß "Ótop i!. 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.
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 Àif 0 Ð:Ñ œ ! then there is a neighborhood of R :
such that 0 ± R ´ !
E10. Let 3 À ‘Ä‘ be the identity map and let
for some }.
Ð3Ñ œ Ö0 − GÐ Ñ À 0 œ 13‘ 1 − GÐ Ñ‘
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 Ð‘ or, more generally, GÐ\ÑÑ is a commutative ring The constant function is the zeroÞ ! 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 Ð!Ñw exists .×
b) Exhibit two functions 0 ß 1 in GÐ Ñ‘ for which 0 1 − Ð3Ñ yet 0 Â Ð3Ñ and 1 Â 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 nor is identically 0 on .0 1 \
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 and in the ring 0 1 GÐ\Ñ are called “zero divisors.”
d) Prove that there are exactly two functions 0 − GÐ Ñ‘ for which 0 œ 0# . (M8 GÐ\Ñßthe
notation 0 ÐBÑ# means 0 ÐBÑ † 0 ÐBÑ, not 0 Ð0 ÐBÑÑ.)
e) Prove that there are exactly functions in ( ) for which c 0 G 0# œ 0.
An element in GÐ\Ñthat equals its own square is called an idempotent. Part d) shows that GÐ Ñ‘ and GÐ Ñ are not isomorphic rings since they have different numbers of idempotents. Is eitherGÐ Ñ‘ or GÐ Ñ isomorphic to GÐ Ñ ?
One classic part of general topology is to explore the relationship between the space \ and the rings GÐ\Ñ and G Ð\ч . For example, if \ is homeomorphic to ], then GÐ\Ñ is isomorphic toGÐ] Ñ. This necessarily implies (why?) that G Ð\ч is isomorphic to G Ð] ч . The question “when does
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 HÐ Ñ‘
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 is a multiple of in 0 1 GÐ\Ñßthat 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 is not a multiple of in 0 1 GÐ\Ñ.
E13. Let \ be a Tychonoff space with subspaces and J Eßwhere is closed and is countable.J E
Prove that if J ∩ E œ g, then is disjoint from some zero set that contains .E J
E14. A space \ is called pseudocompact if every continuous 0 À \ Ä‘is bounded, that is, if
GÐ\Ñ œ G Ð\ч (see Definition IV.8.7). Consider the following condition (*) on a space :\
(*) Whenever Z ª Z ª" # ...ª Z ª ÞÞÞ8 is a decreasing sequence of nonempty open sets,
then +8œ"∞ clZ Á g8 .
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 pseudocompactnessthat is, a characterization that makes no explicit reference to .‘
4. Normal and
X
%-Spaces
We now return to a topic in progress: normal spaces and Ð X%-spaces . Even though normal spaces areÑ
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% $"” as an immediate corollary.
#
To begin, the following technical variation on the definition of normality is very useful.
Lemma 4.1 A space is normal \ iff wheneverE is closedß S is open and E © S, there exists an open set U
with E © Y ©cl Y © SÞ
Proof Suppose \ is normal and that Sis an open set containing the closed set E. Then E and
F œ \ Sare disjoint closed sets. By normality, there are disjoint open sets and with Y Z E © Y
and F © Z Þ Then E © Y ©clY © \ Z © SÞ
Conversely, suppose satisfies the stated condition and that \ Eß F are disjoint closed sets.
Then E © S œ \ Fß so there is an open set with Y E © Y ©clY © \ FÞ Let Z œ \ cl .Y
Theorem 4.2 a) A closed subspace of a normal (X%) space is normal (X%).
b) A continuous closed image of a normal (X%) space normal (X%).
Proof a) Suppose is a closed subspace of a normal space J \ and let and be disjoint closedE F
sets in J. Then Eß F are also closed in \ so we can find disjoint open sets Yw and Z w in \
containing and respectively. Then E F Y œ Y ∩ Jw and Z œ Z ∩ Jw are disjoint open sets in thatJ
contain and , so is normal.E F J
b) Suppose \ is normal and that 0 À \ Ä ] is continuous, closed and onto. If and E Fare
disjoint closed sets in ], then 0"ÒEÓand 0"ÒFÓ are disjoint closed sets in \. Pick Y w and Z w
disjoint open sets in with \ 0"ÒEÓ © Y w and 0"ÒFÓ © Z Þw Then Y œ ] 0 Ò\ Y Ów and
Z œ ] 0 Ò\ Z Ów are open sets in .]
If C − Y, then C  0 Ò\ Y Ów . Since is onto, 0 C œ 0 ÐBÑfor some B − Y © \ Z Þw w Therefore
C − 0 Ò\ Z Ów so C  Z. Hence Y ∩ Z œ gÞ
If C − E, then 0"ÒÖC×Ó © 0"ÒEÓ © Y w, so 0"ÒÖC×Ó ∩ Ð\ Y Ñ œ gw . Therefore C  0 Ò\ Y Ów so
C − ] 0 Ò\ Y Ó œ Yw . Therefore E © Yand, similarly, F © Z so is normal.]
Since the X" property is hereditary and is preserved by closed onto maps, the statements in a) and b)
hold for X% as well as normality. ñ
The next theorem gives us more examples of normal (and X%) spaces.
Theorem 4.3 Every regular Lindelo¨f space is normal (and therefore every Lindelo¨f \ X$-space is X%).
Proof Suppose and E Fare disjoint closed sets in . For each \ B − E, use regularity to pick an open
set YB such that B − Y ©B clY © \ FÞB Since the Lindelo¨f property is hereditary on closed
subsets, a countable number of the YB's cover : relabel these as E Y ß Y ß ÞÞÞß Y ß ÞÞÞ" # 8 . For each , we8
have clY ∩ F œ gÞ8 Similarly, choose a sequence of open sets Z ß Z ß ÞÞÞß Z ß ÞÞÞ" # 8 covering such thatF
clZ ∩ E œ g8 for each .8
We have that -8œ"∞ Y ª E8 and -8œ"∞ Z ª F8 , but these unions may not be disjoint. So we define
cl cl Y œ Y "‡ Z Z œ Z "‡ Y " " " " cl cl cl cl Y œ Y Ð Z ∪#‡ Z Ñ Z œ Z Ð Y ∪#‡ Y Ñ # " # # " # ã ã cl cl cl cl cl cl Y œ Y Ð Z ∪8‡ Z ∪ ÞÞÞ ∪ Z Ñ Z œ Z Ð Y ∪8‡ Y ∪ ÞÞÞ ∪ Y Ñ 8 " # 8 8 " # 8 ã ã Let Y œ-∞8œ"Y8‡and Z œ-8œ"∞ Z8‡.
If B − E, then B Âcl Z8 for all . But 8 B − Y5 for some , so 5 B − Y © Y Þ5‡ Therefore E © Yand,
similarly, F © Z Þ
Then B − Y5‡ for some 5ß so B ÂclZ ∪" clZ ∪ ÞÞÞ ∪# clZ ß5
so B Â Z ∪ Z ∪ ÞÞÞ ∪ Z" # 5
so B Â Z ∪ Z ∪ ÞÞÞ ∪ Z"‡ #‡ ‡ 5
so B Â Z8‡ for any 8 Ÿ 5 Þ
Since B − Y5‡, then B − Y Þ So, if 8 5, then B  Z œ Z Ð Y ∪ ÞÞÞ ∪‡ cl clY ∪ ÞÞÞ ∪clY Ñ
5 8 8 " 5 8
So B  Z8‡for all , so 8 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 is Lindelo¨f, so is normal.+ W W
Since is W X", we have that is W 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 be any dense subset of U ‘Þ Suppose open
sets Y © \< have been defined, one for each < − U, in such a way that:
i) \ œ-<−UY< and +<−UY œ g<
ii) if <ß = − Uand < + =, then clY © Y< =.
For B − \, define 0 ÐBÑ œinfÖ< − U À B − Y ×< . Then 0 À \ Ä‘is continuous.
We will use this Lemma only once, with U œÞ 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
From the definition of , we get that for 0 <ß = − Uß
a) if B −clY<, then B − Y= for all = < so 0 ÐBÑ Ÿ <
b) if 0 ÐBÑ + =, then B − Y=.
We want to prove is continuous at each point 0 + − \. Since is dense in ,U ‘
and
ÖÒ<ß =Ó À <ß = − U < + 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 of such that Y + 0 ÒY Ó © Ò<ß =ÓÞ
Since 0 Ð+Ñ + =, we have + − Y=, and 0 Ð+Ñ <gives us that + ÂclY<. Therefore Y œ Y = clY< is
an open neighborhood of . If + D − Y, then D − Y ©= clY ß= so 0 ÐDÑ Ÿ =; and D Âcl 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 Urysohn's Lemma Ð Ñ 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 exists, we say0
that and are E F completely separated.)
Note: Notice that if and happen to be disjoint E F 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
Then is continuous, and .
0 ÐBÑ œ 1 ÐBÑ < 2 ÐBÑ1 ÐBÑ 0 0 lE œ ! 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 and were zero sets inE F
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 exists,0
then Y œ ÖB À 0 ÐBÑ + ×" and Z œ ÖB À 0 ÐBÑ ×" are disjoint open sets (in fact, cozero sets)
# #
containing and respectively. It is the other half of Urysohn's Lemma for which Urysohn deservesE F
credit.
Let and be disjoint closed sets in a normal space . We will define sets open sets E F \ Y < −< ( Ñin
Enumerate the remaining rationals in ∩ Ò!ß "Ó as < ß < ß ÞÞÞß < ß ÞÞÞ" # 8 , 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
cl
E © Y ©<# Y © Y<# <" œ \ F
Then ! œ < + < + < œ "# $ " , and we use normality to pick an open set Y<$ so that
cl cl
E © Y ©<# Y © Y ©<# <$ 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 < + < + < Ð3ß 4ß 5 Ÿ 8Ñß3 4 5 then
clY © Y ©<3 <4 clY © Y<4 <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
the largest among that is smaller than and
< œ5 < ß < ß ÞÞÞß <" # 8 <8<"ß
the smallest among that is larger than
< œ6 < ß < ß ÞÞÞß <" # 8 <8<"Þ
By the induction hypothesis, we already have clY © Y Þ<5 <6 Then use normality to pick an open set
Y<8<" so that
clY © Y<5 <8<" ©clY<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 œinfÖ< −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 we can replace it, if we like, by 0 1 œ Ð! ” 0 Ñ • " so that and areE F completely separated by a function 1 − G Ð\ÑÞ‡ It is also clear that we can modify further to get an1 2 − G Ð\ч 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%ÊX$" #.
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).
Lemma 5.4 (Weierstrass Q-Test) Let \ be a topological space. Suppose 0 À \ Ä8 ‘is
continuous for each 8 − and that l0 ÐBÑl Ÿ Q8 8 for all B − \. If Q + ∞8 , then
8œ" ∞ 0 ÐBÑ œ0 ÐBÑ B 0 À \ Ä 8œ" ∞
8 converges (absolutely) for all and ‘is continuous.
Proof For each , B l0 ÐBÑl ŸQ + ∞, so 0 ÐBÑ converges (absolutely) by the Comparison
8œ" 8œ" 8œ"
∞ ∞ ∞
8 8 8
Test.
Suppose + − \ and % !. Choose R so that Q + Each 0 is continuous, so for
8œR <" ∞
8 %%. 8
8 œ "ß ÞÞÞß R we can pick a neighborhood Y8 of such that for + B − Y ß l0 ÐBÑ 0 Ð+Ñl +8 8 8 #R% . Then
Y œ+8œ"R Y8 is a neighborhood of , and for + B − Y we get l0 ÐBÑ 0 Ð+Ñl
œ lÐ0 ÐBÑ 0 Ð+ÑÑ < Ð0 ÐBÑ 0 Ð+ÑÑl Ÿl0 ÐBÑ 0 Ð+Ñl < l0 ÐBÑ 0 Ð+Ñl 8œ" 8œ" R ∞ R ∞ 8 8 8 8 8 8 8 8 8œR <" 8œR <" Ÿ l0 ÐBÑ 0 Ð+Ñl < l0 ÐBÑl < l0 Ð+Ñl + R † < #Q + < œ Þ 8œ" R ∞ ∞ 8 8 8 8 8 8œR <" 8œR <" % % % #R # # % Therefore is continuous at . 0 + ñ
Lemma 5.5 Let be a closed set in a normal space and let be a positive real number. SupposeE \ + 2 À E Ä Ò <ß <Ó is continuous. Then there exists a continuous 9À \ Ä ß Ó[ < < such that
$ $
l2ÐBÑ ÐBÑl Ÿ9 #<$ 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 , so E E"and F" are closed in . By Urysohn's Lemma, there exists a\
continuous function 9À \ Ä Ò ß Ó$ $< < such that 9lE œ " <$ and 9lF œ" <$.
If B − E", then < Ÿ 2ÐBÑ Ÿ <$ and 9ÐBÑ œ <$, so l2ÐBÑ ÐBÑl Ÿ l < Ð Ñl9 <$
œ #<$à and similarly if B − F ß l2ÐBÑ ÐBÑl Ÿ" 9 #<$Þ If B − E ÐE ∪ F Ñ" " , then 2ÐBÑ and 9ÐBÑ are
both in Ò ß Ó<$ $< so l2ÐBÑ ÐBÑl Ÿ9 #<$. 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\Ñsuch that 1lE œ 0 Þ
Note: if is a closed subset of E \ 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 , theM
endpoints are in , where is already defined. If E 0 M œ Ð+ß ,Ñ then extend the definition of 0
over by using a straight line segment to join M Ð+ß 0 Ð+ÑÑand Ð,ß 0 Ð,ÑÑ on the graph of . If0
M œ Ð+ß ∞ÑÞ then extend the graph of over using a horizontal right ray at height 0 M 0 Ð+Ñà if
M œ Ð ∞ß ,Ñß then extend the graph of over using a horizontal left ray at height 0 M 0 Ð,ÑÞ
As with Urysohn's Lemma, half of the proof is easy. The significant part of theorem is proving the
Proof (É) Suppose and are disjoint closed sets in . and are clopen E F \ E F 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\Ñsuch that 1lÐE ∪ FÑ œ 0 Þ Then
Y œ ÖB À 1ÐBÑ + ×" Z œ ÖB À 1ÐBÑ ×"
# and # are disjoint open sets (cozero sets, in fact) that contain
E and respectively. Therefore is normal.F \
(Ê) The idea is to find a sequence of functions 1 − GÐ\Ñ3 such that
l0 ÐBÑ 1 ÐBÑl Ä ! 8 Ä ∞ Ð 1 ÐBÑ
3œ" 3œ"
8 8
3 as for each B − E where is defined 0 ). The sums 3 are
defined on all of \ and as 8 Ä ∞ we can think of them as giving better and better approximations to
the extension that we want. Then we can let 1 1ÐBÑ œ lim 1 ÐBÑ œ 1 ÐBÑÞ The details follow.
8Ä∞ 3œ" 3œ"
8 ∞
3 3
We proceed in three steps, but the heart of the argument is in Case I.
Case I Suppose is continuous and that 0 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" 9À \ 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#$ #. So0 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" # $ #() œ Ð Ñ#$ $.
So 0 Ð1 < 1 < 1 Ñ À E Ä Ò " # $ #( #()ß )ÓÞ
We continue, using induction, to find for each a continuous function3
1 À \ Ä Ò 3 #$ ß#$ Ó l0 ÐBÑ 1 ÐBÑl Ÿ Ð#Î$Ñ3 B − E
3œ" 8
8
3" 3"
3 3 such that for .
Since the series converges (absolutely) for every
3œ" 3œ" 3œ"
∞ ∞ ∞
3 #$ 3
l1 ÐBÑl Ÿ 3"3 + ∞ß 1ÐBÑ œ 0 ÐBÑ
B − \, and is continuous by the Weierstrass 1 Q-Test. Since l1ÐBÑl œ l1 ÐBÑl
3œ" ∞ 3 Ÿl1 ÐBÑl Ÿ œ "ß 1 À \ Ä Ò "ß "ÓÞ 3œ" 3œ" ∞ ∞ 3 #$ 3" 3 we have
Finally, for B − Eß l0 ÐBÑ 1ÐBÑl œliml0 ÐBÑ 1 ÐBÑl Ÿ limÐ Ñ œ !, so 1lE œ 0 and
8Ä∞ 3œ" 8Ä∞
8
3 #$ 8
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 "Ówith J lE œ 0 Þ To get , we merely make a slight modification to1 J to get a that still extends but where has all its values in 1 0 1 Ð "ß "ÑÞ
Let F œ ÖB − \ À J ÐBÑ œ „ "× E. and are disjoint closed sets in , so byF \
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 "Ñwith 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\ч 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 is normal (W 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‘Hl 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 is continuous, that is, E ‘Eœ GÐEÑ and so lGÐEÑl œ - œ # Þ- - If W ‚ Wwere 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‘El œ - œ # -- - . which is false. Therefore normality is not
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 -axis in is> > B >
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 "Ómust
be normal. In the 1950's, Dowker proved that \ ‚ Ò!ß "Óis normal iff is normal and “countably\
paracompact.”
However, this result was unsatisfyingbecause 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 "Óis 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 is W X%, so is W X$" and therefore the Sorgenfrey plane W ‚ W is
#
also X$". So W ‚ W is an example showing that X$" X%.
# # does not imply
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 is a subgroup of and E \ 0 À E Ä F is an isomorphism, can be extended to a homomorphism 0 1 À \ Ä F ?
If we let 3 À E Ä \ be the injection3Ð+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.”
Specific extension theorems impose conditions on and , and usually we want to share someE \ 1
property of such as continuity. Here are some illustrations, without further details.0
1) Extension theorems that generalize of Tietze's Theorem: by putting stronger hypotheses
on , we can relax the hypotheses on .\ F
Suppose is closed in and E \ 0 À E Ä F is continuous.
If
is normal and (Tietze's Theorem)
is normal and
is collectionwise normal** and is a separable Banach s
Ú Ý Ý Û Ý Ý Ü \ F œ \ F œ \ F ‘ ‘8 pace*
is paracompact** and is a Banach space*
\ F
then has a continuous extension 0 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 0 ÐBÑÑ
" # 8
where each 0 À E Ä3 ‘. By Tietze's Theorem, there exists for each a continuous3
extension 1 À \ Ä3 ‘ with 1 lE œ 03 3. If we let 1ÐBÑ œ Ð1 ÐBÑß ÞÞÞß 1 ÐBÑÑ" 8 , then
1 À \ Ä‘8 and 1lE œ 0. In other words, we separately extend the coordinate
functions in order to extend . And in this example, could even be an infinite0 8
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 axiomsfor 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, ‘8À the usual norm lÐB ß B ß ÞÞÞß B Ñl œ" # 8 ÈB < B < ÞÞÞ < B#" ## 8# produces the usual metric, which is complete. So ‘8is 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
paracompact collectionwise normal normal metric or compact Ú Û Ü X Ê Ê Ê #
Therefore, in the theorems cited above, a continuous map defined on a closed subset of a0 metric space (or, compact X#space) and valued in a Banach space can be continuouslyF extended a function 1 À \ Ä F.
