Chapter 6 Products and Quotients - Department of Mathematics

1. Introduction
Chapter VI
Products and Quotients
In Chapter III we defined the product of two topological spaces \ and ] and developed some simple
properties of the product. ( See Examples III.5.10-5.12 and Exercise IIIE20. ) Using simple induction,
that work generalized easy to finite products. It turns out that looking at infinite products leads to
some very nice theorems. For example, infinite products will eventually help us decide which
topological spaces are metrizable.
2. Infinite Products and the Product Topology
The set \‚] was defined as ÖÐBßCÑÀ B−\ßC −]× . How can we define an “infinite product” set
\œÖ\ α Àα −E× ? Informally, we want to say something like
\œÖ\ α Àα −EלÖÐBαÑÀBα −\ α×
so that a point B in the product consists of “coordinates B α” chosen from the \ α's.
But what exactly
does a symbol like BœÐBÑmean if there are “uncountably many coordinates?”
α
We can get an idea by first thinking about a countable product. For sets \ " ß \ # ß ÞÞÞß \ 8,...
we can
informally define the product set as a certain set of sequences: \œ ∞ 8œ" \ 8 œÖÐB8ÑÀB8 −\ 8×
.
But if we want to be careful about set theory, then a legal definition of \ should have the form
\œÖÐB8Ñ−Y ÀB8 −\ 8 × . From what “pre-existing set” Y will the sequences in \ be chosen? The
answer is easy: we write
\œ ∞ \ œÖB−Ð ∞
\Ñ ÀBÐ8ÑœB−\×Þ 8œ" 8 8œ" 8 8 8
∞ 8œ"
Thus the elements of \ 8 are certain functions (sequences) defined on the index set .
This idea
generalizes naturally to any productÞ
Definition 2.1 Let Ö\ α À α − E× be a collection of sets. We define the product set
\œÖ\ À −EלÖB− E
α α \ α ÀBÐαÑ−\ α× . The \ α's
are called the factors of \ . For each
α, the function 1α À Ö\ α À α − E× Ä \ α defined by 1αÐBÑ
œ Bα
is called the α - .
th projection map
For B−\ , we write more informally B œBÐαÑœthe α - coordinate of Band write BœÐB).
th
α α
Note: the index set E might not be ordered. So, even though we use the informal notation B œ ÐBα Ñ,
such phrases as “the first coordinate of B,” “the next coordinate in B after Bα,”
and “the coordinate
in B preceding Bα” may not make sense. The notation ÐBαÑis handy but can lead you into errors if
you're not careful.
239

α α
A point B in Ö\ À α − E× is, by definition, a function that “chooses” a coordinate B from each set
in the collection Ö\ α À α − E× . To say that such a “choice function” must exist if all the \ α's
are
nonempty is precisely the Axiom of Choice. ( See the discussion following Theorem I.6.8.)
Theorem 2.2 The Axiom of Choice (AC) is equivalent to the statement that every product of
nonempty sets is nonempty.
Note: In ZF set theory, certain special products can be shown to be nonempty without using AC.
∞
For example, if \ 8 œ ß then
8œ" \ 8 œ ÖB − À BÐ8Ñ − × œ . Without using AC, we can
precisely describe a point ( œ function) in the product for
example, the identity function
3œÖÐ7ß8Ñ− ‚ À7œ8×so œ \ Ág.
∞ 8œ" 8
We will often write Ö\ α À α − E× as
α−E\
α.
If the indexing set E is clearly understood, we may
simply write \α.
Example 2.3
1) If E œ g, then Ö\ À − E× œ ÖÐ E
α \ Ñ À BÐαÑ− \ × œ Ög×Þ
α α−E
α α
2) Suppose \ α œ g for some α − E . Then α \ α œ ÖÐ α \ αÑ À BÐαÑ − \ α×Þ
! ! E
−E −E
Since BÐ Ñ − \ is impossible, \ œ g.
α! α! αα−E
α
3) Strictly speaking, we have two different definitions for a finite product \ " ‚\ # :
i) \ " ‚\ # œÖÐBßBÑÀB " # " −\ßB " # −\× # (a set of ordered pairs)
Ö"ß#×
" # " # 3
ii) \ ‚\ œ ÖB − Ð\ ∪\ Ñ À BÐ3Ñ − \ × (a set of functions)
But there is an obvious way to identify these two sets: the ordered pair ÐB" ß B# Ñ corresponds to the
function B œ ÖÐ"ß B" Ñß Ð#ß B# Ñ× − Ð\ " ∪ \ # Ñ Þ Ö"ß#×
4) If Eœ, then Ö\ À8− ל
∞ \ œÖB−Ð ∞
8 8œ" 8 8œ" \ 8ÑÀB8 −\ 8×
œ ÖÐB" ß B# ß ÞÞÞß B8ßÞÞÞß Ñ À B8 − \ 8× œ the set of all sequences ÐB8Ñwhere B8 − \ 8Þ
5) Suppose the \ α's are identical, say \ α œ ] for all α − E. Then Ö\
α À α − E×
œÖB−Ð E E E
α−E\ αÑÀBÐα Ñ−\ αלÖB−] ÀBα −]ל] . If lElœ7,
we will sometimes
7 write this product simply as ] œ “the product of 7 copies of ] ” because the number of factors 7 is
often important while the specific index set E is not.
Now that we have a definition of the set Ö\ αÀ α − E× , we can think about a product topology.
We
begin by recalling the definition and a few basic facts about the “weak topology.” (See Example
III.8.6.)
240

Definition 2.4 Let \ be a set. For each α − E, suppose Ð\ αßgαÑ is a topological space and that
0α À \ Ä \ α. The weak topology on \ generated by the collection Y œ Ö0αÀ α − E× is the smallest
topology on that makes all the 's continuous.
\ 0α
Certainly, there is at least one topology on \ that makes all the 0α's
continuous: the discrete topology.
Since he intersection of a collection of topologies on \ is a topology ( why? ), the weak topology
exists we can describe it “from the top down” as { g : g is a topology on \ making all the 0α's
continuous×Þ
However, this slick description of the weak topology doesn't give a useful description of what sets are
open. Usually it is more useful to describe the weak topology on \ “from the bottom up.” to make all
the 0α's
continuous it is necessary and sufficient that
for each α −Eand for each open set Y ©\ , the set 0 ÒYÓmust be open.
α α α α
"
Therefore the weak topology g is the smallest topology that contains all such sets 0αÒYαÓ and that is
" the topology for which Æ œÖ0 ÒYÓÀ α −EßY open in \ × is a subbase. ( See Example III.8.6. )
α α α α
Therefore a base for the weak topology consists of all finite intersections of sets from Æ.
A typical
" " "
basic open set has form 0α ÒYα ] ∩0α ÒYα Ó∩... ∩0α ÒYα Ówhere each α − E and each Yα
is
" " # # 8 8 3
3
open in \α3.
To cut down on symbols, we will use a special notation for these subbasic and basic open
sets:
We will write Yα for 0 ÒYαÓÞ " " "
A typical basic open set is then Y œ 0 ÒY ] ∩0 ÒY Ó∩... ∩0 ÒY Ó
α "
α" α α# α α8
α
which we further abbreviate as Y œ Yα" ß Yα# ß ÞÞÞß Yα8
241
" # 8
So B − Y œ Y ß Y ß ÞÞÞY iff 0 ÐBÑ − Y for each 3 œ "ß ÞÞÞß 8Þ
α α α α α
" # 8 3 3
This notation is not standard but it should be because it's very handy. You should verify that to get a
base for the weak topology g on \ , it is sufficient to use only the sets Yα ß Yα ß ÞÞÞß Yα where
" # 8
each Y is a basic (or even subbasic) open set in \ .
α α
3
Example 2.5 Suppose Ð\ßgÑ is any topological space and E©\ÞLet 3ÀEÄ\ be the inclusion
map 3Ð+Ñ œ +Þ Then the subspace topology on E is the same as the weak topology generated by
"
Y œÖ3×ÞTo see this, just note that a base for the weak topology is Ö3 ÒYÓÀY open in \× and that
" 3 ÒYÓ œ Y ∩E which is a typical open set in the subspace topology.
The following theorem tells us that a map 0 into a space \ with the weak topology is continuous iff
each composition 0 ‰0 is continuous.
α
Theorem 2.6 For α −E, suppose 0α À\ÄÐ\ αßgαÑ and that \ has the weak topology generated by
the functions 0α. Let ^ be a topological space and 0 À ^ Ä \ . Then 0 is continuous iff
α
0 ‰0 À ^ Ä \
α α is continuous for every .
Proof If 0 is continuous, then each composition 0 α ‰0 is continuous. Conversely, suppose each
"
0 α ‰0 is continuous. To show that 0 is continuous, it is sufficient to show that 0 ÒZÓ is open in ^
whenever Z is a subbasic open set in \ ( why? ). So let Z œ Yα with Y α open in \ α.
Then
" " " "
0 ÒZ Ó œ 0 Ò0 ÒY ÓÓ œ Ð0 ‰ 0Ñ ÒY Ó which is open since 0 ‰ 0 is continuous.
ñ
α α α α α
3
"

Definition 2.7 For each α −Eßlet Ð\ ßg Ñbe
a topological space. The g on the
set \ is the weak topology generated by the collection of projection maps Y œ Ö1 À α − E×Þ
α α product topology
α α
The product topology is sometimes called the “Tychonoff topology.” We always assume that a
product space has the product topology unless something else is explicitly stated.
\α
Since the product topology is a weak topology, a subbase for the product topology consists of all sets
"
of form Yα œ1αÒYαÓ, where α −E and Y α is open in \ αÞA
base consists of all finite
intersections of such sets À
" "
α α α α α α
Y ∩ ÞÞÞ ∩ Y œ 1 ÒY ] ∩ ... ∩ 1 ÒY Ó
" 8 " " 8 8
α α α
œ ÖB − \ À B − Y for each 3 œ "ß ÞÞÞß 8×
3 3
α−E α α α
" # 8
œ Y where Y œ \ for α Á α ß α ß ÞÞÞß α (*)
( It is sufficient to use only Y ' s which are basic (or even subbasic) open sets in \ Why? )
α αÞ
A basic open set in “depends on only finitely many coordinates” in the following sense:
\α
B − Yα" ß Yα# ß ÞÞÞß Yα8 iff B satisfies the finitely many restrictions Bα3 − Yα3 Ð3 œ "ß ÞÞÞß 8Ñ.
Since (basic) open sets containing B are the “standard” for measuring closeness to B,
we can
say, roughly, that in the product topology “closeness depends on only finitely many
coordinates.”
For a finite index set E œ Ö"ß #ß ÞÞÞß 8× , the product topology on
α−E\
α is just the one for which the
basic open sets are the open boxes
α−E
α:
Y
8
Y œ Y œ Y ß Y ß ÞÞÞß Y œ Y ‚ Y ‚ ÞÞÞ ‚ Y
α−E α 3œ" 3 α" α# α8
" # 8
Then condition (*) that “ Yα œ \ α for all but finitely many α'
s” is satisfied automatically so, for finite
products the product topology is exactly as we defined it in Chapter III.
Definition 2.7 is probably not what was expected. Someone's “first guess” would more likely be to
use all “boxes”
α−EYα ( Y α open in \ αÑ
as a base in defining the product topology. But in Definition
2.7, a “box” of the form
α−EYα
might not be open because (*) may not hold. It is perfectly possible,
of course, to define a different topology on the set using all boxes of the form
α−E\ α α−EYαas
a
base. This alternate topology on
α−E
\ α is called the box topology.
But we will see that our
definition of the product topology is the “right” definition to use. (Of course, the box topology and the
product topology coincide for finite products. )
Theorem 2.8 Each projection map 1α À ÖÖ\
α À α − E× Ä \ α is continuous and open.
If Ö\ α À α − E× Á g, each 1α
is onto. A function 0 À ^ Ä \
α is continuous iff
1α ‰0 À^ Ä\ α is continuous for every α.
Proof By definition, the product topology is one that makes all the 1α's
continuous. and clearly, each
1α is onto if the product is nonempty. To show that 1α
is open, it is sufficient to show that the image
of a basic open set Y œ Y ß Y ß ÞÞÞß Y is open. This is clearly true if Y œ g.
α α α
" # 8
242

Yα3 For Y Á g, we get 1αÒY Ó œ 1αÒYα"
ß Yα# ß ÞÞÞß Yα8 Ó œ
\ α
if α œ α3
..
if α Á α" ß ÞÞÞß α8
Either way, 1αÒYα" ß Yα# ß ÞÞÞß Yα8 Ó is open.
Because the product has the weak topology generated by the projections, Theorem 2.6 gives
that 0 is continuous iff each composition ‰0 is continuous. ñ
1α
Generally, projection maps are not closed. For example, if JœÖÐBßCÑ−‘ ÀCœ ßB!× ,
# "
B
then 1 "ÒJ Ó œ the projection of J on the B-axis ‘ is not closed in ‘ Þ
Example 2.9
243
" "
1) A subbasic open set in ‘ ‚ ‘ has form 1 " ÒYÓ œ Y ‚ ‘ or 1 # ÒZÓ œ ‘ ‚Z, where Y and
Z are open in ‘ Þ ( We get a subbase even if we restrict ourselves just to using basic open intervals
YßZ in ‘ Þ)
. So basic open sets have form ÐY ‚ ‘ Ñ ∩ Ð ‘ ‚ Z Ñ œ Y ‚ Z . Therefore the product
#
topology on ‘ ‚ ‘ is the usual topology on ‘ .
# #
The function 0À‘ Ä ‘ ‚ ‘ given by 0Ð>ÑœÐ>ß sin >Ñ is continuous because the
# #
compositions Ð 1 ‰ 0ÑÐ>Ñ œ > and Ð 1 ‰ 0ÑÐ>Ñ œ sin > are both continuous functions from ‘ to ‘ .
" #
i!
2) Let \œ œ“the
product of countably many copies of .”
A base for the product
topology consists of all sets YœY8 where finitely many Y8's
are singletons and all the others are
equal to Þ Each Y is infinite ( in fact lY l œ - why? ), so every nonempty open set is infinite. In
particular, \ has no isolated points Þ More generally, an infinite product of (nonempty) discrete spaces
is not discrete.
The box topology on \ is the discrete topology. For each point B œ Ð5" ß 5# ß ÞÞÞß 58, ... Ñ − \ ,
the set ÖB× œ ∞ 8œ" Ö5 8×
is open in the box topology. ( For a finite product, the box and product
topologies care the same: a finite product of discrete spaces is discrete. )
‘ ‘
< <
3) Let ‘ œ Ö\ À < − ‘ × , where \ œ ‘ . Each point 0 in ‘ is a function 0 À ‘ Ä ‘ and
1< Ð0Ñ œ 0Ð

∞
"
3
3œR"
# 3 3 #R
#
#
R
3œ"
3 3
#
∞
3œR"
3 3
# "Î#
# %
#R
# %
#
"Î#
# #
Pick R so that % %
Þ If ÐB C Ñ for each 3 œ "ß ÞÞÞß R,
then we have
.ÐBß CÑ œ Ð ÐB C Ñ ÐB C Ñ Ñ ÐR † Ñ œ % Þ
244
This is the natural
metric topology on the product L, and we see here that we can make “ C close to B”
by
requiring “closeness” in just finitely many coordinates "ß ÞÞÞß RÞ This is just what the product
topology does and, in fact, the product topology on L turns out to be the topology g. Þ.
In Ð!ß "Ñ we can see a similar phenomenon quite clearly: for two points B œ !ÞB" ÞÞÞB8ÞÞÞ and
C œ !ÞC" ÞÞÞC8ÞÞÞ ß “ C will be close to B” if C and B agree in, say, the first 8 decimal places.
Roughly speaking, “closeness” depends on only “finitely many coordinates.”
A handy “rule of thumb” that has proved true every time I've used it is that if a topology on a
product set makes “closeness” depend on only a finite number of coordinates, then that
topology is the product topology.
3) The bottom line: a mathematical definition justifies itself by the fruit it bears. The
definition of the product topology will lead to some beautiful theorems. For example, we will
see that compact Hausdorff spaces are (topologically) nothing other than the closed subspaces
7
of cubes Ò!ß "Ó with the product topology ( 7 may be infinite). For the time being, you will
need to accept that things work out nicely down the road, and that by contrast, the box
topology turns out to be rather ill-behaved. ( See Exercise E11.)
As a simple example of nice behavior, the following theorem is exactly what one would hope for and
the proof depends on having the “correct” definition for the product topology. The theorem says that
convergence of sequences in a product is “coordinatewise convergence” : that is, in a product,
>2 >2
ÐB8ÑÄBiff for all α , the α coordinate of B 8converges
(in \ αÑ
to the α coordinate of B.
The
product topology is sometimes called the “topology of coordinatewise convergence.”
Theorem 2.11 Suppose ÐB8Ñ is a sequence in \ œ Ö\ αÀ α − E× . Then ÐB8ÑÄB− \ iff
( 1 ÐB Ñ) Ä 1 ÐBÑ in \ for all α − E.
α 8 α α
Proof If ÐB8ÑÄB, then ( 1αÐB8Ñ) Ä 1αÐBÑ because each 1α
is continuous.
Conversely, suppose ( 1αÐB8Ñ) Ä 1αÐBÑ in \ α for each α and consider any basic open let
Y œ Yα" ß Yα# ß ÞÞÞß Yα that contains B œ ÐBαÑ. For each 3 œ "ß ÞÞÞß 5, we have Bα − Y
5 3 α3.
Since
( 1α ÐB ) 1α , we have 1α
α for some . If max , then for
3 8ÑÄ3ÐBÑ ÐB Ñ − Y 8 R R œ ÖR ß ÞÞÞß R ×
3 8 3
3 " 5
8 R we have 1α ÐB Ñ − Yα for every 3 œ "ß ÞÞÞß 5. This means that B − Y for 8 R,
so
3 8 3
8
ÐB8 Ñ Ä B. ñ
In the proof, R is the max of a finite set. If \ has the box topology, the basic open set Y œ Yα might involve infinitely many open sets Yα Á \ α. For each such α,
we could pick Rα,
just as in the
proof. But the set of all Rα's might not have a max R so the proof would collapse. Can you create a
specific example with the box topology where this happens?
< <
‘ Example 2.12 Consider ‘ œ Ö\ À < − ‘ × , where \ œ ‘ . Each point 0 in the product is a
‘
function 0À‘ Ä‘ Þ Suppose that Ð0Ñ 8 is a sequence of points in ‘ . By Theorem 2.11, Ð0ÑÄ0 8
iff Ð08 Ð

Question: if ‘ is given the box topology, is convergence of a sequence simply
‘ Ð08Ñ uniform convergence (as defined in analysis)?
The following “theorem” is stated loosely. You can easily create variations. Any reasonable version
of the statement is probably true.
“Theorem 2.13” Topological products are associative in any “reasonable” sense: for example, if
EœF∪Gwhere Fand Gare
disjoint indexing sets, then
Ö\ Àα −E× Ö\ À" −F× ‚ Ö\
À# −G×
α " #
“Proof” A point B−Ö\ α Àα −E× is a function BÀEÄ
α−E\
α.
Define
0ÀÖ\ α Àα −E×ÄÖ\ " À" −F× ‚ Ö\
# À# −G× by 0ÐBÑœÐBlFßBlGÑ.
Clearly 0 is one-to-one and onto.
For convenience, let ] œ Ö\ " À " − F× and ^ œ Ö\
# À # − G× so that
1 ‰0 ÀÖ\ Àα −E×Ä] and 1 ‰0 ÀÖ\ Àα −E×Ä^Þ
" α #
α
0 is a mapping into a product Y ‚^, so 0 is continuous iff 1" ‰0 and 1#
‰0 are both continuous.
But 1" ‰0 ÀÖ\ α Àα −E×Ä] œ Ö\
" À −F× is also a map into a product ß so 1"
‰0 is
continuous iff 1" ‰ 1" ‰0 À Ö\
α À α − E× Ä \ " is continuous for all " − FÞ
But 1 ‰ 1 ‰0 œ 1 À Ö\
À α − E× Ä \ which is continuous.
" " " α "
The proof that 1# ‰0 is continuous is completely similar.
" 0 À ] ‚^ Ä "
Ö\ α À α − E× is given by B œ 0 ÐCßDÑ œ C∪D ( the union of two functions)
and
" "
0 is continuous iff 1α ‰0 À ] ‚^ Ä \ α is continuous for each α − E œ F ∪GÞ
"
Suppose α − F : then 1α ‰ 0 ÐCß DÑ œ 1αÐBÑ, where B œ C ∪ DÞ Since α − Fß Bα
"
œ Ð 1α ‰ 1" ÑÐCß DÑß so 1α ‰ 0 œ 1α ‰ 1" , which is continuous. The case where α − G is completely
similar.
Therefore 0 is a homeomorphism. ñ
The question of topological commutativity for products only makes sense when the index set E is
ordered in some way. Even in that case, if we view a product as a collection of functions, the question
of commutativity is irrelevant the question reduces to the fact that set theoretic unions are
Ö"ß#×
commutative. For example, \ " ‚\ # œ ÖB − Ð\ " ∪\ # Ñ À BÐ3Ñ − \ 3 for 3 œ "ß#×
Ö"ß#×
œ ÖB − Ð\ # ∪ \ " Ñ À BÐ3Ñ − \ 3 for 3 œ "ß #× œ \ # ‚ \ " Þ
So viewed as sets of functions, \ " ‚\ # and \ # ‚\ " are exactly the same set! The same observation
applies to any product viewed as a collection of functions.
But we might look at an ordered product in another way: for example, thinking of \ " ‚\ # and
\ # ‚\ " as sets of ordered pairs. Then generally \ " ‚\ # Á\ # ‚\ " . From that point of view, the
topological spaces \ 1 ‚\ 2 and \ 2 ‚\ 1 are not literally identical, but there is a homeomorphism
between them: 0ÐBßBÑœÐBßBÑÞ
" # # " So the products are still topologically identical. We can make a
similar argument whenever any ordered product is “commuted” by permuting the index set.
The general rule of thumb is that “whenever it makes sense, topological products are commutative.”
245

Exercise 2.14 Recall that for a space \ and cardinal 7 , \ denotes the product of 7 copies of the
7 8 78 space \ . Prove that Ð\ Ñ is homeomorphic to \ . ( Hint: Look at the proof of Theorem I.14.7;
the bijection 9 given there is a homeomorphism. )
Notice that “cancellation properties” may not be true. For example, ‚ Ö!× and ‚
Ö!ß "× are
homeomorphic (both are countable discrete spaces) but topologically you can't “cancel the ” À Ö!× is
not homeomorphic to Ö 0,1 × !
Here are a few results which are quite simple but very handy to remember. The first states that
singleton factors are topologically irrelevant in a product.
Lemma 2.15 \ ‚ Ö: × \
α −E α " −F " α−E
α
Proof
" −FÖ: " × is itself a one-point space Ö: }, so we only need to prove that
\ ‚Ö:× \ . The map 0ÐBß:Ñ œ Bis clearly a homeomorphism Þ ñ
α−E α α−E
α
Lemma 2.16 Suppose . For any ,
α−E\ α Á g F © E α−F\
α is homeomorphic to a subspace ^
of that is, can be embedded in
α−E\ α α−F\ α α−E\
αÞ In fact, if all the \ α's
are X"
-spaces,
then \ is homeomorphic to a closed subspace ^ of \ .
α−F α α−E
α
α
α−E
α
Proof Pick a point :œÐ: Ñ− \ . Then by Lemma 2.15,
\ \ ‚ Ö: × œ ^ © \
α−F α α−F α α−EF α α−E
α
Now suppose all the \ α's are X" . If C − Ð
α−E\
α Ñ^, then for some # − EFß C# Á : # Þ Since
\ is a X space, there is an open set Y in \ that contains C but not : . Then C − Y and
# "
# # # # #
Y ∩Ð \ ‚ Ö: ×Ñœg.
# α−F α α−EF
α
Therefore ^ is closed in
α−E\
α.
ñ
Note: 1) Assume all the \ α 's œ ‘ . Go through the preceding proof step-by-step when E œ Ö"ß ÞÞÞß 5×
and when Eœ and
2) In the case FœÖ α! × , Lemma 2.16 says that each factor \ α! is homeomorphic to subspace
of
α−E \ α (a closed subspace if all the \ α's
are X"
).
3) But Lemma 2.16 does not say that if all the \ α 's are X"
, then every embedded copy of
\ α \
! α α in is closed: only that there a closed homeomorphic copy. It is very easy to show
−E exists (
# #
a copy of ‘ embedded in ‘ that is not closed in ‘ , for example ... ?)
α α " #
Lemma 2.17 Suppose \œ Ö\ À −E×Ág . Then \ is a X space (or, X space) iff every
factor \ is X (or, X ).
α " #
246
7

Proof for Hausdorff Suppose all the \ α's are Hausdorff. If B Á C − \ , then Bα! Á Cα!
for some
α!. Pick disjoint open sets Yα and Z α in \ α containing Bα and Cα
. Then
! ! ! ! !
Y and Z are disjoint (basic) open sets in \
that contain Band C.
α α α
! !
Conversely, suppose \Ág . By Lemma 2.16, any factor \ α is homeomorphic to a subspace of \ .
Since a subspace of a Hausdorff space is Hausdorff ( why? Ñ , \ is Hausdorff. ñ
Exercise 2.18 Prove Lemma 2.17 if “Hausdorff” is replaced by “ X" .” ( The proof is similar but
easier.)
Theorem 2.19 The product of countably many 2-point discrete spaces is homeomorphic to the Cantor
set G.
Proof We show that if \ 8 œ Ö!ß #× , then ∞
8œ" \ 8 G.
To construct G we defined, for each sequence ÐB" ß B# ß ÞÞÞß B8ßÞÞÞÑ − Ö!ß #× , a descending
sequence of closed sets J B" ª J B" B# ª ÞÞÞ ª J B" B# ÞÞÞB8
ª ÞÞÞ in Ò!ß "Ó whose intersection gave a unique
point :−GÀ Ö:ל∞ 8œ" JB" B# ÞÞÞB8
(s ee Section IV.10).
For each 8, we can write Gas
a union of
8 # disjoint clopen sets: G œ
ÐB ßÞÞÞßB Ñ−Ö!ß#× 8 ÐG ∩JBBÞÞÞB Ñ.
" 8
" # 8
Define 0 À G Ä ∞ 8œ" \ 8 by 0Ð:Ñ œ B œ ÐB" ß B# ß ÞÞÞß B8ßÞÞÞÑÞ Clearly 0 is one-to-one and
onto. To show that 0 is continuous at : − G, it is sufficient to show that for each 8,
the function
18 ‰ 0 À G Ä Ö!ß #× is continuous at : , so pick any 8. One of the clopen sets G ∩ J B" B# ÞÞÞB8
contains :
(call it Y) and Ð 18 ‰0ÑlY œ B 8. Thus, 18 ‰ 0 is constant on a neighborhood of : in G , so 18
‰0 is
continuous at : .
By Lemma 2.17, ∞ 8œ" \ 8 is Hausdorff. Since 0 is a continuous bijection of a compact space
onto a Hausdorff space, 0 is a homeomorphism ( why? ). ñ
Corollary 2.20 Ö!ß #× i! is compact.
This corollary is a very special case of the Tychonoff Product Theorem which states that any product
of compact spaces is compact. The Tychonoff Product Theorem is much harder and will be proved in
Chapter IX.
Corollary 2.21 The Cantor set is homeomorphic to a product of countably many copies of itself.
i i i i †i i
Proof By Exercise 2.14 above, G
8 G G for 8 − is similar. ñ
! ÐÖ!ß #× ! Ñ ! Ö!ß #× ! ! Ö!ß #× ! GÞ The case of
Example 2.22 Convince yourself that each assertion is true:
1) If \ is the Sorgenfrey line, then \ ‚\ is the Sorgenfrey plane ( see Examples
III.5.3 and III.5.4).
" # "
2) Let W be the unit circle in ‘ . Then W ‚ Ò!ß "Ó is homeomorphic to the
cylinder ÖÐBß Cß DÑ − ‘ À B C œ "ß D − Ò!ß "Ó×Þ
$ # #
" "
3) W ‚W is homeomorphic to a torus ( œ “the surface of a doughnut”).
247
α

Exercises
E1. Does it ever happen that \‚Ö!× open in \‚‘ ? If so, what is a necessary and sufficient
condition on \ for this to happen?
E2. a) Suppose \ and ] are topological spaces and that E © \ß F © ]Þ Prove that
int \‚] ( E‚FÑœint\ E‚ int ] F:
that is, “the interior of the product is the product of the interiors.”
( By induction, the same result holds for any finite product.)
Give an example to show that the
statement may be false for infinite products.
b) Suppose E © \ for all α − E . Prove that in the product \ œ \
,
α α α
cl( EÑœcl E.
α α
Note: When the 's are closed, this shows that is closed: so “any product of closed sets is
Eα Eα
closed.” Can you see any plausible reason why products of closures are better behaved than products
of interiors?
c) Suppose \œ \ α Ágand that E α ©\ α. Prove that E
α is dense in \ iff Eα
is dense in
\α for each α. Note: Part c) implies that a finite product of separable spaces is separable.
Part c) doesn't tell us whether or not an infinite product of separable spaces is separable: why not?
d) For each α , let ; α − \ α. Prove that F œ ÖB − \
α À Bα œ ; α for all but at most finitely many
α× is dense in
\ α. ( Note: Suppose \œ‘ œ
8−\ 8 where each \ 8 œ‘ Þ Suppose each ; 8 is
chosen to be a rational say ; 8 œ ! . What does d) say about ‘ ? )
e) Let α! −E . Prove that ] α œÖB−
! \ αÀBα œ: α for all α Á α!
× is homeomorphic
to \α! . Note: So any factor of a product has a “copy” of itself inside the product in a “natural” way.
For example, in ‘ , the set of points where all coordinates except the first are is homeomorphic to
8 !
the first factor, ‘ . Þ
f) Give an example of infinite spaces \ß] ß^ such that \ ‚ ] is homeomorphic to \ ‚ ^ but ]
is not homeomorphic to ^.
E3. Let \ be a topological space and consider the “diagonal” set
? œÖÐBßBÑÀB−\ש\‚\Þ
a) Prove that ? is closed in \ ‚\ iff \ is Hausdorff.
b) Prove that ? is open in \ ‚\ iff \ is discrete.
E4. Suppose \ is a Hausdorff space and that \ α © \ for each α − E..
Show that
] œ Ö\ À α − E× is homeomorphic to a closed subspace of the product Ö\
À α − E×.
α α
248

E5. For each 8−, suppose H is a countable dense set in Ð\ ßg ÑÞ
8 8 8
a) Prove that HœH is dense in \
.
8 8
b) Prove that \8 is separable. ( Caution: H may not be countable (when is H countable?
. Hint: In a product “closeness depends on only finitely many coordinates.” )
E6. a) Suppose a3ß4 − Ö!ß #× for all 3ß 4 − . Prove that there exists a sequence Ðn5Ñin such that,
for each 3, lim a38
5Ä∞
exists.
, 5
Hint: “picture” the + 3ß4 in an infinite matrix. For each fixed 4,
the “j-th column" of the matrix is a
i!
point in the Cantor set G œ Ö!ß #× , and G is a compact metric space.
∞
8
8œ"
∞
8œ"
8 w
w
8
∞
w
8 or 8 is called a subseries of
8œ"
WœÖ=−‘ À=
8.
∞
+× 8
‘
8œ"
∞
8œ"
8 w
i i
b) Let + be an absolutely convergent sequence of real numbers. A series +
where each
+ œ + + œ ! +
Prove that is the sum of a subseries of is closed in .
Hint: “Absolute convergence” just guarantees that every subseries converges. Each subseries +
can be associated in a natural way with a point B − Ö!ß "× Þ Consider the mapping 0 À Ö!ß "× Ä
given by 0ÐBÑœ ∞
+ − ‘ Þ Must 0 be a homeomorphism?
8œ"
8 w
249
! ! ‘
c) Suppose K is an open dense subset of the Cantor set G. Must Fr K be countable?
Hint: Consider ÖÐ+ß ,Ñ − G ‚ G À , Á !×Þ
E7. Let have the cofinite topology.
a) Does the product have the cofinite topology? Does the answer depend on ?
7 7
b) Prove is separable (
7 Hint: When 7 is infinite, consider the simplest possible points in
the product.Ñ
Part b) implies that an arbitrarily large product of X" spaces with more than one point can
be separable. According to Theorem 3.8, this statement is false for X#
spaces.

E8. In any set \ , we can define a topology by choosing a nonempty family of subsets Y and defining
closed sets to be all sets which can be written as an intersection of finite unions of sets from Y .
Y is called a subbase for the closed sets of \ . ( This construction is “complementary” to generating a
topology on \ by using a collection of sets as a subbase for the open sets. )
a) Verify that this procedure does give a topology on \ .
b) Suppose \ α is a topological space. Give \
α the topology for which collection of “closed
boxes” Y œÖJα ÀJ αclosed in \ α×
is a subbase for the closed sets. Is this topology the product
topology?
E9. Prove or disprove:
i i
! !
There exists a bijection 0 À \ œ Ö!ß "× Ä œ ] such that for all B − \ and for all 8,
the first 8 coordinates of C œ 0ÐBÑare determined by the first 7 coordinates of B.
Here, 7 depends on 8 and B.
More formally, we are asking whether there exists a bijection
0 such that:
aB − \ a8 − b7 − such that changing B4 for any 4 7 does not change C5
for 5Ÿ8
(Hint: Think about continuity and the definition of the product topology.)
250

3. Productive Properties
We want to consider how some familiar topological properties behave with respect to products.
Definition 3.1 Suppose that, for each α −E , the space \ α has a certain property T.
We say that
productive
if \
α must have property T
the property T is countably productive if \
α must have property T when Eis
countable
finitely
productive if \ must have property T when Eis
finite
For example, Lemma 2.17 shows that the X" and X#
properties are productive.
α
Some topological properties behave very badly with respect to products. For example, the Lindelof ¨
property is a “countability property” of spaces, and we might expect the Lindelof ¨ property to be
countably productive. Unfortunately, this is not the case.
Example 3.2 The Lindelof ¨ property is not finitely productive; in fact if \ is a Lindelof ¨ space, then
\‚\ may not be Lindelof. ¨ Let \ be the set of real numbers with the topology for which a
neighborhood base at + is U+ œ ÖÒ+ß ,Ñ À +ß , − W , , +× . ( Recall that \ is called the Sorgenfrey
Line: see Example III.5.3.) We begin by showing that f is Lindelof. ¨
It is sufficient to show that a collection h of basic open sets covering \ has a countable
subcover. Given such a cover h, Let i œÖÐ+ß,ÑÀ Ò+ß,Ñ− h× and define E œi.
For a
moment, we can think of E as a subspace of ‘ with its usual topology. Then E is Lindelof ¨
w
( why? ), and i is a covering of E by usual open sets, so there is a countable subfamily i with
i w œE.
Replace the left endpoints of the intervals in i to get h œÖÒ+ß,ÑÀÐ+ß,Ñ− i ש h. If h
covers \ we are done, so suppose \ w Á g. For each B − \ w
h h , pick a set Ò+ß,Ñ in
h that contains B. In fact, B must be the left endpoint of Ò+ß ,Ñ for if B − Ò+ß ,Ñ − h and
BÁ+ , then B−i œ i © h Þ So, for each BÂh we can pick a set [ B ) −h.
251
w w w w
w w w , ,B
h w
If B and C are distinct points not in , then ÒB, , B) and ÒC,
, C)
must be disjoint ( why? ). But
w w
there can be at most countably many disjoint intervals ÒBß , BÑÞ
So h ∪ ÖÒB, ,B)
À B Â h
} is
a countable subcollection of h that covers W.
However the Sorgenfrey plane W‚Wis not Lindelof. ¨ If it were, then its closed subspace
HœÖÐBßCÑÀBCœ"× would also be Lindelof ¨ (Theorem III.7.10). But that is impossible
since H is uncountable and discrete in the subspace topology. ( See the figure on the following
page.)

Fortunately, many other topological properties interact more nicely with products. Here are several
topological properties T to which the “same” theorem applies. We combine these into one large
theorem for efficiency.
Theorem 3.3 Suppose \œÖ\ α Àα −E×Ág. Let T be any one of the properties “first
countable,” “second countable,” “metrizable,” or “completely metrizable.” Then \ has property T iff
1) all the \ α's
have property T,
and
2) there are only countably \α's
that do not have the trivial topology.
This theorem “essentially” is a statement about countable products because:
1) The nontrivial \α's
are the “interesting” factors, and 2) says there are only countably
many of them. In practice, one hardly ever works with trivial spaces; and if we completely exclude
trivial spaces from the discussion, then the theorem just states that \ has property T iff \ is a
countable product of spaces with property T .
2) A nonempty X " space \ α has the trivial topology iff l\ αl
œ "Þ So, if we deal
only with X " spaces (as is most often the case) the theorem says that \ has property T iff all the \ α's
have property T and all but countably many of the \α's
are “topologically irrelevant” singletons.
Of course, in the case of metrizability (or complete metrizability), the X" condition is automatically
satisfied.
Proof
Throughout, let Fœthe set of “interesting” indices œÖα−EÀ \ α is a nontrivial space × .
We begin with the case Tœ“first
countable.”
Suppose 1) and 2) hold and B−\. We need to produce a countable neighborhood base at B.
8
For each α −F, let ÖY À8− ×
be a countable open neighborhood base at B −\ Þ Let
α α α
U œÖYÀYis a finite intersection of sets of the form Y ßα −Fß8− ×
B
Since F is countable, UB is a countable collection of open sets containing B and we claim that
UB is a neighborhood base at B − \. To see this, suppose Z œ Zα ß ÞÞÞß Zα is a basic
" 5
open set containing B. ( We may assume that all α3's
are in FÀwhy?
). For each 3 œ "ß ÞÞÞß 5ß
83 83 8" 8 8
#
5
pick Y so that B − Y © Z Þ Then Y œ Y ß Y ß ÞÞÞß Y − U
and B − Y © Z Þ
α α α α α α α
3 3 3
3 " #
252
k
8
α
B

Conversely, suppose \ is first countable.
We need to prove that 1) and 2) hold.
Since \Ág , Lemma 2.16 gives that each \ α is homeomorphic to a subspace of \ . Therefore
each \α is first countable, so 1) holds.
We prove the contrapositive for 2): assuming F is uncountable, we find a point B − \ at
which there cannot be a countable neighborhood base. Pick a point : œÐ: αÑ−\
For each " −Fßpick an open set S " ©\ " for which gÁS" Á\ " . Choose
and Define using the coordinates
B −S C ÂS Þ BœÐB Ñ−\
" " " " α
B" if α œ " − F
Bαœ
: − \ if α Â F
α α
Suppose UBis a countable collection of neighborhoods of B. For each R − UBchoose
a basic
open set Y with B − Y œ Yα" ß ÞÞÞß Yα © R.
There are only finitely many α3's
involved
5
in the expression for the chosen Y, and there are only countably many R's in UB.
So, since F
is uncountable, we can pick a " −Fthat is not one of the α3's
involved in the expression for
any of the chosen sets Y . Then
i) S" is an open set that contains Bbecause B" −S"
ii) for all R−UB ßR©Î S" : to see this, consider the point A œ ÐAαÑ which is the same as B except in the " coordinate:
Bαif α Á "
Aαœ
C if α œ "
"
For each R − FB, we chose Y œ Yα" ß ÞÞÞß Yα © R. Since B − Y and A has the same
5
α" ,..., α8
coordinates as B, we have A−Y also. But AÂ S" because A" œC" ÂS" Þ
Since Y©Î S , certainly R ©Î S Þ So is not a neighborhood at B.
" " UB base
We are now able to see immediately that if the product \œ \ α has any of the other properties T,
then conditions 1) and 2) must hold.
If \ is second countable, metrizable or completely metrizable, then \ is first countable and,
by the first part of the proof, condition 2) must hold.
If \ is second countable or metrizable then every subspace has these same properties in
particular each \ α is second countable or metrizable respectively. If \ is completely
metrizable, then \ and all the subspaces \ α are X" . By Lemma 2Þ 16, \ α is homeomorphic to
a closed therefore complete subspace of \ . Therefore \αis
completely metrizable.
It remains to show that if Tœ“second
countable,” “metrizable,” or “completely metrizable” and
conditions 1) and 2) hold, then \œ \ α
also has property T.
253

Suppose Tœ“ second countable.”
For each α in the countable set F, let UαœÖSαßSαßÞÞÞß
SαßÞÞÞ } be a countable base for the
open sets in \α and let
254
" # 8
U œÖSÀSis a finite intersection of sets of form S , where α −Fand 8−}
U is countable and we claim U is a base for the product topology on \ .
Suppose B − Z œ Zα ß ÞÞÞß Zα , a basic open set in \ . For each 3 œ "ß ÞÞÞß 5ß
" 5
83
Bα3 − Z α3 so we can choose a basic open set in \ α3 such that Bα3 − S α © Z 3 α3.
Then
8" 8# 85 8" 8# 85
B − Sα ß Sα ß ÞÞÞß Sα © Z and Sα ß Sα ß ÞÞÞß Sα − U . Therefore Z can be
" # 5 " #
5
written as a union of sets from U, so U is a base for \ .
Suppose Tœ“ metrizableÞ”
Since all the \ α's are X " , condition 2) implies that all but countably many \ α's
are singletons
and dropping the singleton factors from the product does not change \ topologically.
Therefore it is sufficient to prove that if each space \ " ß \ # ß ÞÞÞß \ 8ßÞÞÞ
is metrizable, then
∞
\œ \ 8 is metrizable.
8œ"
Let . 8 be a metric for \ 8ß where without loss of generality, we can assume each . 8 Ÿ "
( why? ). For points BœÐBÑ, CœÐCÑ−\ , define
8 8
∞ .ÐBßCÑ
.ÐBßCÑ œ 8 8 8
# 8
8œ"
Then . is a metric on \ ( check ), and we claim that g. is the product topology g . Because \
is a countable product of first countable spaces, g is first countable, so g can be described
using sequences: so it is sufficient to show that ÐD5ÑÄDin Ð\ß g Ñ iff ÐD5 Ñ Ä D in Ð\ß g.
ÑÞ
But ÐD5ÑÄDin Ð\ß g Ñ iff the ÐD5Ñconverges “coordinatewise.” Therefore it is sufficient to
show that:
( D5 Ñ Ä D in Ð\ß g.
Ñ iff a8 ÐD5Ð8ÑÑ Ä DÐ8Ñ in Ð\ 8ß . 8Ñ,
that is,
a8 . 8ÐD5Ð8Ñß DÐ8ÑÑ Ä !
i) Suppose .ÐD ß DÑ Ä ! . Let % ! and consider a particular 8 . We can choose O
5
5
∞
8œ"
#
!
.ÐDÐ8ÑßDÐ8ÑÑ
8 5
# 8
%
8!
. 8 ÐD Ð8 ÑßDÐ8 ÑÑ
! 5 ! !
#
%
# 8 5 ! !
so that 5 O implies .ÐDßDÑœ . Therefore, if 5 O
certainly 8 ! 8 , and therefore so . ÐD Ð8 Ñß DÐ8 ÑÑ % .
! !
So . ÐD Ð8 Ñß DÐ8 ÑÑ Ä ! .
8 5 ! !
!
ii) On the other hand, suppose . ÐD Ð8Ñß DÐ8ÑÑ Ä ! for every 8. Let ! . Choose
∞
"
# 8 #
8œR"
α 8
8 5 %
R so that %
and then choose O so that if 5 O
. " ÐD5Ð"Ñß DÐ"ÑÑÎ#
# %
. # ÐD5Ð#Ñß DÐ#ÑÑÎ# #R
ÞÞÞ
R
. ÐD ÐRÑß DÐRÑÑÎ#
R 5
Then for 5 O we have .ÐDßDÑœ
5
"
%
#R
%
#R
∞
8œ"
.ÐDÐ8ÑßDÐ8ÑÑ
8 5
# 8
.

R
.ÐDÐ8ÑßDÐ8ÑÑ
∞
.ÐDÐ8ÑßDÐ8ÑÑ
8œ"
# 8
8œR"
# 8
#R
œ 8 5
8 5
% % R † œ % Þ
Therefore .ÐD ß DÑ Ä ! .
Suppose Tœ“ completely metrizableÞ”
5
Just as for Tœ “metrizable,” we can assume \œ \ 8 and that . 8 is a complete metric
on \ 8 with . 8 Ÿ " . Using these . 8 's, we define . in the same way. Then g.
is the product
topology on \ . We only need to show that Ð\ß.Ñis
complete.
255
∞ 8œ"
Suppose ÐD5Ñ is a Cauchy sequence in Ð\ß .Ñ . From the definition of . it is easy to see that
ÐD5Ð8ÑÑ is Cauchy in Ð\ 8ß . 8Ñfor each 8, so that ÐD5Ð8ÑÑ Ä some point + 8 − \ 8Þ
Let +œÐ+ 8Ñ−\ . Since ÐD5Ð8ÑÑÄ+ 8 for each 8, we have ÐD5ÑÄ+ in the product topology
œ g.. Therefore Ð\ß.Ñ is complete. ñ
What is the correct formulation and proof of the theorem for Tœ“pseudometrizable”
?
We might wonder why Tœ“separable”
is not included with Theorem 3.3. Since separability is a
“countability property,” we might hope that separability is preserved in countable products although
our experience Lindelof ¨ spaces could make us hesitate. The reason for the omission is that separability
is better behaved for products than the other properties. ( You should try to prove directly that a
countable product of separable spaces is separable remembering that in the product topology,
“closeness depends on finitely many coordinates.” If necessary, first look at finite products.) But
surprisingly, the product of as many as - separable spaces must be separable, and the product of more
than - nontrivial separable spaces can sometimes be separable.
We begin the treatment of separability and products with a simple lemma which is really nothing but
set theory.
Lemma 3.4 Suppose lEl Ÿ -Þ There exists a countable collection e of subsets of E with the
following property: given distinct α" , α# , ... , α8 − E, there are disjoint sets E" ß E# ß ÞÞÞß E8 − e such
−E 3
that for each .
α3 3
Proof (Think about how you would prove the theorem when Eœ‘ Þ The general case is just a
“carry over” on the same idea.)
"
Since lElŸ-, there is a one-to-one map 9 ÀEÄ‘ . Let e œÖ9 ÒÐ+ß,ÑÓÀ+ß,− ×Þ
For distinct α" , α# ,..., α8 − E, 9Ðα" Ñ, 9Ðα# Ñ, ..., 9Ðα8Ñare distinct real numbers; we can choose
+ 3, , 3 − so that 9Ðα3Ñ − Ð+ 3ß, 3Ñ and so that the intervals Ð+ 3ß, 3Ñ
are pairwise disjoint. Then the sets
" E œ 9 ÒÐ+ ß, ÑÓ − e are the ones we need. ñ
3 3 3
Theorem 3.5
α−E α α
1) Suppose \œ \ Ág . If \ is separable, then each \ is separable.
α
α−E
α
2) If each \ is separable and lEl Ÿ - , then \ œ \ is separable.
Part 2) of Theorem 3.5 is attributed (independently) to several people. In a slightly more general
version, it is sometimes called the “Hewitt-Marczewski-Pondiczery Theorem.” Here is an amusing
#

sidelight, written by topologist Melvin Henriksen online in the Topology Atlas.
A few words have
been modified to conform with our notation:
Most topologists are familiar with the Hewitt-Marczewski-Pondiczery theorem. It states that if
7
7 is an infinite cardinal, then a product of # topological spaces, each of which has a dense
set of cardinality Ÿ7ß also has a dense set with Ÿ7points. In particular, the product of -
separable spaces is separable (where - is the cardinal number of the continuum). Hewitt's
proof appeared in [Bull. Amer. Math. Soc. 52 (1946), 641-643], Marczewski's proof in [Fund.
Math. 34 (1947), 127-143], and Pondiczery's in [Duke Math. 11 (1944), 835-837]. A proof
and a few historical remarks appear in Chapter 2 of Engelking's General Topology.
The
spread in the publication dates is due to dislocations caused by the Second World War; there
is no doubt that these discoveries were made independently.
Hewitt and Marczewski are well-known as contributors to general topology, but who was (or
is) Pondiczery? The answer may be found in Lion Hunting & Other Mathematical Pursuitsß
edited by G. Alexanderson and D. Mugler, Mathematical Association of America, 1995. It is a
collection of memorabilia about Ralph P. Boas Jr. (1912-1992), whose accomplishments
included writing many papers in mathematical analysis as well as several books, making a lot
of expository contributions to the American Mathematical Monthly, being an accomplished
administrator (e.g., he was the first editor of Mathematical Reviews (MR) who set the tone for
this vitally important publication, and was the chairman for the Mathematics Department at
Northwestern University for many years and helped to improve its already high quality), and
helping us all to see that there is a lot of humor in what we do. He wrote many humorous
articles under pseudonyms, sometimes jointly with others. The most famous is “A Contribution
to the Mathematical Theory of Big Game Hunting” by H. Petard that appeared in the Monthly
in 1938. This book is a delight to read.
In this book, Ralph Boas confesses that he concocted the name from Pondicheree (a place in
India fought over by the Dutch, English and French), changed the spelling to make it sound
Slavic, and added the initials E.S. because he contemplated writing spoofs on extra-sensory
perception under the name E.S. Pondiczery. Instead, Pondiczery wrote notes in the Monthly,
reviews for MR, and the paper that is the subject of this article. It is the only one reviewed in
MR credited to this pseudonymous author.
One mystery remains. Did Ralph Boas have a collaborator in writing this paper? He certainly
had the talent to write it himself, but facts cannot be established by deduction alone. His son
Harold (also a mathematician) does not know the answer to this question...
Proof 1) Let H be a countable dense set in \ . For each ß ÒHÓ is countable and dense in \
(because \ œ1 Ò\Óœ1 Òcl HÓ© cl 1 ÒHÓ ). Therefore each \ is separable.
α α α α α
256
α 1α α
1 2 8
2) Choose a family e as in Lemma 3.4 and for each α, let Hα œ ÖBα, BαßÞÞÞß BαßÞÞÞ × be a
countable dense set in \α.
Define a countable set f by
f œ ÖÐE ß ÞÞÞß E ß 6 ß ÞÞÞß 6 Ñ À 8 − ß6−, E − e with the E 's pairwise disjoint×Þ
" 8 " 8 3 3 3
Pick a point : α − \ α for each α. For each #8 -tuple = œ ÐE" ß ÞÞÞß E8 ß 6" ß ÞÞÞß 68Ñ−f, define a point
B= − \ with coordinates

6
3 B if − E3
BÐ = αÑœ
α α
: if α Â E
α
3
Let HœÖB= À=− f×
. His countable and we claim that H is dense in \ . To see this, consider any
nonempty basic open set Y œ Y ß ÞÞÞß Y À we will show that Y ∩ H Á gÞ
For Y œ Y ß ÞÞÞß Y Á g,
α α
" 5
α α
" 5
i) Choose disjoint sets E ß ÞÞÞß E in e so that α − E ß ÞÞÞß α − E
" 5
257
" " 5 5
8
α α α α
3
ii) For each 3 œ "ß ÞÞÞß 5ß H is dense in \ and we can pick a point Bα in H ∩ Y
3
" # 8
œ ÖB , B ß ÞÞÞß B ß ÞÞÞ × ∩ Y Þ
α α α α
3 3 3 3
3 3 3 3
3
Let = œ ÐE" ß ÞÞÞß E3ßÞÞÞß E5 ß 8" ß ÞÞÞ83ßÞÞÞß 85Ñ−fÞBecause α3 − E3, we have B= Ðα3ÑœBα−Yα. 3
3
Therefore B − Y ∩H. ñ
=
Example 3.6 The rather abstract construction of a dense set H in the proof of Theorem 3.5 can be
‘
nicely illustrated with a concrete example. Consider ‘ Ðœ

Theorem 3.8 shows that for Hausdorff spaces, a nonempty product with more than - factors cannot be
separable.
Example 3.7 For each α −E , let \ α be a set with l\ αl"Þ Choose : α −\ α and let
gα œ ÖS © \ α À : α − S× ∪ Ög×Þ The singleton set Ö: α× is dense in Ð\ αßgαÑ , so \ α is separable.
If :œÐ: αÑ−\œ
α−E\
α,
then singleton set Ö:× is dense ( why? look at a nonempty basic
open set Y . ) So \ is separable, no matter how large the index set E is.
The \ α 's in this example are not X"
. But Exercise E7 shows that an arbitrarily large
product of separable X" spaces can turn out to be separable.
α α α
Theorem 3.8 Suppose \œ −E \ Ág where each \ is a X#
space with more than one point.
If \ is separable, then lEl Ÿ -Þ
Proof For each α, we can pick a pair of disjoint, nonempty open sets Yαand Z α in \ αÞ
Let H be a
countable dense set in \ and let Hα œ Yα ∩H for each α. If α Á " − Eß there is a point
:− Yα ßZ" ∩Hbecause H is dense. Then :−H α but :ÂH" œ Y" ∩Hsince
B"  Y" À therefore Hα Á H" . Therefore the map 9 À E Ä cÐHÑ given by 9ÐαÑ œ Hαis
one-toi!
one, so lElŸlcÐHÑlœ# œ-Þñ We saw in Corollary V.2.19 that a finite product of connected spaces is connected. In fact, the
following theorem shows that connectedness behaves very nicely with respect to any product. The
proof of the theorem is interesting because, unlike our previous proofs about products, this proof uses
the theorem about finite products to prove the general case.
α−E α α
Theorem 3.9 Suppose \œ \ Ág . \ is connected iff each \ is connected.
Proof \ is nonempty so each map 1α À \ Ä \ α is onto. If \ is connected, then each \ α is a
continuous image of a connected space so \α is connected (see Example V.2.6).
Conversely, suppose each \ α is connected Þ For each α, pick a point : α − \ α.
For each finite
set J©E , let \ J œ \ ‚
−J −EJ Ö:× . \ J is homeomorphic to the finite product
α α α α α−J
\ α,
so each \ J is connected. Let H œ Ö\
J À J is a finite subset of E× . Each \ J contains the point
:œÐ: α Ñ, so Corollary V.2.10 tells us that His connected. Unfortunately, HÁ\ (except in trivial
cases; why?).
But we claim that H is dense in \ . If Y œ Yα ß ÞÞÞß Yα is any nonempty basic open set in \ ,
" 8
we need to show that H ∩ Y Á g. Choose a point Bα3 − Yα3 for each 3 œ "ß ÞÞÞß 8, and define B − \
having coordinates
B if α œ α
BÐαÑ œ
: if α Á α , α ,..., α
α3
3
α
" # 8
If we let J œ Ö+ ß α ß ÞÞÞß α × , then B − \ ∩ Y © H ∩ Y , so H ∩ Y Á gÞ
" # 8 J
Therefore \œcl His connected (Corollary V.2.20). ñ
Question: is the analogue of Theorem 3.9 true for path connected spaces?
Just for reference, we state one more theorem here. We will not prove Theorem 3.10 until Chapter IX,
but we may use it in examples. (Of course, the proof in Chapter IX will not depend on any of these
258

examples! )
Theorem 3.10 (Tychonoff Product Theorem) Suppose \œ
α−E\
α Ág . Then \ is compact iff
each \α is compact.
The proof “ ” in Tychonoff's Theorem is quite easy ( And the easy theorem that a
Ê why?). finite
product of compact spaces is compact was in Exercise IV.E.26. .
259

Exercises
E10. Let \œÒ!ß"Ó have the product topology.
Ò!ß"Ó
a) Prove that the set of all functions in \ with finite range ( sometimes called step functions)
is
dense in \ .
in \ .
b) By Theorem 3.5, \ is separable. Describe a countable set of step functions which is dense
c) Let E œ ÖB − \ À 0 is the characteristic function of a singleton set Ö

4. Embedding Spaces in Products
It is often possible to embed a space \ in a product \ αÞ
Such embeddings will give us some nice
theorems for example, we will see that there is a separable metric space \ that contains all other
separable metric spaces topologically a “universal” separable metric space.
To illustrate the general method we will use, consider two maps 0" À Ò!ß"Ó Ä ‘ and 0# À Ò!ß"Ó Ä ‘
# B #
given by 0ÐBÑœB " and 0ÐBÑœ/Þ # Using these maps we can define /ÀÒ!ß"ÓÄ ‘ ‚ ‘ œ‘
by
# B
using 0" and 0 # as “coordinate functions”: /ÐBÑ œ Ð0" ÐBÑß 0# ÐBÑÑ œ ÐB ß / ÑÞ This map / is called the
evaluation map defined by the set of functions Ö0" , 0 # × . In this example, / is an embedding that is,
# # / is a homeomorphism of Ò!ß "Ó into ‘ so that Ò!ß "Ó ran Ð/Ñ © ‘ ( see the figure).
An evaluation map does not always give an embedding: for example, the evaluation map
# / À Ò!ß "Ó Ä ‘ defined by the family Ö cos # 1Bß sin # 1B×
is not a homeomorphism between Ò!ß "Ó and
ran Ð/Ñ © ‘ ( )
# why? what is ran Ð/Ñ?
We want to generalize the idea of an evaluation map / into a product and to find some conditions under
which / will be an embedding.
Definition 4.1 Suppose \ and \ α ( α − EÑare topological spaces and that 0α À \ Ä \ α for each αÞ
The evaluation map defined by the family Ö0 À α − E× is the function / À \ Ä \ given by
α
α−E
α
/ÐBÑÐαÑ œ 0αÐBÑ α α α
th
Then /ÐBÑ is the point in the product
coordinate notation, /ÐBÑ œ Ð0αÐBÑÑÞ \ whose coordinate function is 0 ÐBÑ:\.
In more informal
α α α
Exercise 4.2 Suppose \œ −E\ÞFor each α there is a projection map 1 À\Ä\ . What is
α
the evaluation map defined by the family Ö1 À α − E× ?
α
261

Definition 4.3 Suppose \ and \ α( α − EÑare topological spaces and that 0α À \ Ä \ αÞ
We say that
the family Ö0α À α − E× separates points if, for each pair of points B Á C − \ , there exists an α − E
for which 0ÐBÑÁ0ÐCÑ.
α α
Clearly, the evaluation map /À\Ä
α−E\
α is one-to-one Í for all BÁC−\ß/ÐBÑÁ/ÐCÑ
Í for all BÁC−\ there is an α −E for which /ÐBÑÐαÑœ0αÐBÑÁ0αÐCÑœ/ÐCÑÐαÑ Í the family Ö0 À − E× separates points.
α α
Theorem 4.4 Suppose \ has the weak topology generated by the maps 0α À \ Ä Ð\ αßgαÑ and that
the family Ö0α À α − E× separates points. Then / is an embedding that is, / a homeomorphism
between \ and /Ò\Ó © \ α.
Proof Since Ö0 α×
separates points, / is one-to-one and / is continuous because each composition
1α ‰/œ0α is continuous.
/ preserves unions and (since / is one-to-one) intersections. So to check that / is an open map
from \ to /Ò\Ó , it is sufficient to show that / takes subbasic open sets in \ to open sets in /Ò\Ó.
Because \ has the weak topology, a subbasic open set has the form Y œ 0αÒZÓ, where Z is open in
"
" "
\ α. But then /ÒY Ó œ /Ò0α ÒZ ÓÓ œ 1α ÒZ Ó ∩ /Ò\Ó is an open set in /Ò\ÓÞ ñ
( Note: /ÒY Ó might not be open in \α,
but that is irrelevant. See the earlier example where
# B /ÐBÑ œ ÐB ß / )Þ
The converse of Theorem 4.4 is also true: if e is an embedding, then the 0 α's
separate points and \
has the weak topology generated by the 0α's.
However, we do not need this fact and will omit the proof
(which is not very hard).
Example 4.5
Let Ð\ß .Ñ be a separable metric space. We can assume, without loss of generality, that . Ÿ "Þ \ is
second countable so there is a countable base U œ ÖY" ß ÞÞÞß Y8ßÞÞÞ× for the open sets. For each 8,
let
08ÐBÑ œ .ÐBß \ Y8ÑÞ Then 08 À \ Ä Ò!ß "Ó is continuous and, since \ Y8
is closed, we have
08ÐBÑ! iff B−Y8ÞIf BÁC−\ , there is an 8such that B−Y8 and CÂY8ÞThen 08ÐCÑœ!Á08ÐBÑ, so 08's
separate points.
We claim that the topology g. on \ is actually just the weak topology gAgenerated
by the 08's.
Because the functions 0 8 are continuous if \ has the topology g. , we know that g. ª gAÞ
To show
g. © gA , suppose B−Y −g. Þ For some 8, we have B−Y 8 ©Yand therefore 08ÐBÑœ-!ÞBut " -
Z8 œ08 ÒÐ# ß"ÓÓis a subbasic open set in the weak topology and B−Z 8 ©Y 8 ©YÞTherefore
Y−gA.
i i
! !
By Theorem 4.4, /À\ ÄÒ!ß"Ó is an embedding, so \ /Ò\Ó©Ò!ß"Ó Þ We sometimes write this
top
i! i!
as \ © Ò!ß "Ó . From Theorems 3.3 and 3.5, we know that Ò!ß "Ó is itself a separable metrizable
space and therefore all its subspaces are separable and metrizable. Putting all this together, we get
that topologically, the separable metrizable spaces are nothing more and nothing less than the
subspaces of Ò!ß "Ói! .
We can view this fact with a “half-full” or “half-empty” attitude:
262

i) separable metric spaces must not be very complicated since topologically they are nothing
more than the subspaces of a single very nice space: the “cube” Ò!ß "Ó i!
ii) separable metric spaces can get quite complicated, so the subspaces of a cube Ò!ß "Ó i!
are more complicated than we imagined.
i
∞ "
! Since Ò!ß "Ó
8œ" Ò!ß 8 Ó œ L (the “Hilbert cube”) © j#
, we can also say that topologically
the separable metrizable spaces are precisely the subspaces of j# . This is particularly amusing
because of the almost “Euclidean” metric on :
. j#
∞
8 8
8œ"
.ÐBß CÑ œ Ð ÐB C Ñ Ñ
In some sense, this elegant “Euclidean-like” metric is adequate to describe the topology of any
separable metric space.
We can summarize by saying that each of L and j# is a “universal separable metric space” ( But they
are not homeomorphic: L is compact but j# is not. )
Example 4.6
Suppose Ð\ß .Ñ is a metric space, not necessarily separable, and that ÖY À − E× is a base for the
topology g. , where 7 œ lElÞ An argument completely analogous to the reasoning in Example 4.5
(just replace “ 8 ” everywhere with “ α ”) shows that \ © Ò!ß"Ó . Therefore topologically
top
7 every metric
7 space is a subspace of some sufficiently large “cube.” Of course, if 7 i! , the cube Ò!ß "Ó is not
itself metrizable ( why? ); in general this cube will have many subspaces that are nonmetrizable. So the
result is not quite as dramatic as in the separable case.
263
" # #
α α
The weight AÐ\Ñ of a topological space Ð\ß g Ñ is defined as min ÖlUlÀ U is a base for g × i! Þ
We are assuming here that the “min” in the definition exists: see Example 5.22 in Chapter VIII.
J or some very simple spaces, the “min” could be finite in which case the “ i! ” guarantees
that AÐ\Ñ i Ðconvenient
for purely technical reasons that don't matter in these notes).
!
The density $ Ð\Ñ is defined as min ÖlHl À H is dense in Ð\ß g Ñ× i! Þ For a metrizable space, it is
not hard to prove that AÐ\Ñ œ $ Ð\Ñ.
The proof is just like our earlier proof (in Theorem III.6.5) that
separability and second countability are equivalent in metrizable spaces. Therefore we have that for
any metric space Ð\ß .Ñß
top
AÐ\Ñ $ Ð\Ñ
\ © Ò!ß "Ó œ Ò!ß "Ó (*)
Notice that, for a given space Ð\ß .Ñ,
the exponents in this statement are not necessarily the smallest
top
AБ Ñ i!
possible. For example, (*) says that ‘ © Ò!ß "Ó œ Ò!ß "Ó , but in fact we can do much better than
"
the exponent i À ‘ Ð!ß "Ñ © Ò!ß "Ó œ Ò!ß "Ó !
!
We add one additional comment, without proof: For a given infinite cardinal 7,
it is possible to define
a metrizable space L7with weight 7 such that every metric space \ with weight 7 can be
i! i!
embedded in H 7Þ In other words, L7 is a metrizable space which is “universal for all metric spaces
with AÐ\Ñ œ 7.” The price of metrizability, here, is that we need to replace Ò!ß "Ó by a more
complicated space L7.

Without going into all the details, you can think of L7as a “star” with 7 different copies of
Ò!ß "Ó (“rays”) all placed with ! at the center of the star. For two points Bß C on the same “ray”
of the star, .ÐBß CÑ œ lB Cl; if Bß C are on different rays, the distance between them is
measured “via the origin” À .ÐBß CÑ œ lBl lClÞ
The condition in Theorem 4.4 that “ \ has the weak topology generated by a collection of maps
0α À \ Ä \ α” is sometimes not so easy to check. The following definition and theorem can
sometimes help.
Definition 4.7 Suppose \ and \ α( α − EÑare topological spaces and that, for each α,
0α À \ Ä \ αÞ
We say that the collection Y œÖ0αÀ α −E× separates points from closed sets if whenever Jis
a
closed set in \ and B Â J , there is an such that 0 ÐBÑ Â cl 0 ÒJ ÓÞ
α α α
Example 4.8 Let \œ‘ and Y œGБ Ñ. Suppose J is a closed set in ‘ and

Note: the more open (closed) sets there are in \ , the harder it is for a given family Ö0 À − E× to
succeed in separating points and closed sets. In fact ß the lemma shows that a family of continuous
functions Ö0α À α − E× succeeds in separating points and closed sets only if g is the smallest
topology that makes the 0α's
continuous.
"
Proof Suppose U is a base. Then the sets 0αÒZÓ in U are open, so the 0α's
are continuous and
i) holds.
To prove ii), suppose J is closed in \ and B Â J. For some α and some Z open in \ α,
we
" "
have B − 0αÒZÓ © \ JÞ Then 0αÐBÑ − Z and Z ∩0αÒJÓ œ g Ðsince 0αÒZÓ∩J œ gÑ.
Therefore 0ÐBÑÂcl 0ÒJÓso
ii) also holds.
α α
Conversely, suppose i) and ii) hold. If B−Sand S is open in \ , we need to find a set
" "
0α ÒZÓ − U such that B − 0α ÒZÓ © SÞ Since B  J œ \ S,
condition ii) gives us an α for which
" "
0αÐBÑ Â cl 0αÒJ Ó. Then 0αÐBÑ − Z œ \ α cl 0αÒJ ÓÞ Then B − 0α ÒZ Ó and we claim 0α ÒZ Ó © S:
"
Suppose D − 0α ÒZ Ó so 0αÐDÑ−Zœ \ α cl 0αÒJ ÓÞ
But if D−J, then 0ÐDÑ−0ÒJÓœ0Ò α α α cl JÓ© cl 0ÒJÓÞ α
Therefore D−\JœSÞñ Theorem 4.10 Suppose 0 À \ Ä \ is continuous for each α − E. If the collection Ö0 À α − E×
265
α α
α α α
i) separates points and closed sets, and
ii) separates points
then the evaluation map /À\Ä \ α is an embedding.
Proof Since the 0 α's
are continuous, Lemma 4.9 gives us that \ has the weak topology. Then
Theorem 4.4 implies that / is an embedding. ñ
If the space \ we are trying to embed is a X" -space ( as is most often the case),
then
i) Ê ii) in Theorem 4.10 , so we have the simpler statement given in the following corollary.
Corollary 4.11 Suppose 0α À \ Ä \ α is continuous for each α − E . If \ is a X"
-space and
Ö0 À α − E× separates points and closed sets, then evaluation map / À \ Ä \
is an embedding.
α α
Proof By Theorem 4.10, it is sufficient to show that the 0α's separate points, so suppose B Á C − \ .
Since B  the closed set ÖC× , there is an α for which 0αÐBÑ Â cl 0αÒÖC×Ó. Therefore 0αÐBÑ Á 0αÐCÑß so
Ö0α À α − E× separates points.
ñ

Exercises
E14. A space Ð\ß g Ñ is called a X! space if whenever B Á C − \ , then aB Á aC
(equivalently, either
there is an open set Y containing B but not C, or vice-versa). Notice that the X! condition is weaker
than X ( see example III.2.6.4). Clearly, a subspace of a X -space is X .
" ! !
α α ! α !
a) Prove that a nonempty product \œ Ö\ À −E× is X iff each \ is X.
b) Let W be “Sierpinski space” that is, W œ Ö!ß "× with the topology g œ Ögß Ö 1 ×ß Ö!ß "×× . Use
7
the embedding theorems to prove that a space \ is X ! iff \ is homeomorphic to a subspace of W for
some cardinal 7. ( Nearly all interesting spaces are X! , so nearly all interesting spaces are
7 (topologically) just subspaces of W for some cardinal 7.
)
Hint Ê : for each open set Y in \ , let ;Y be the characteristic function of YÞ Use an embedding
theorem.
E15. a) Let Ð\ß .Ñ be a metric space. Prove that GÐ\Ñ ( œ the collection of continuous real-valued
functions on \ ) separates points from closed sets Þ ( Since \ is X" , GÐ\Ñ therefore also separates
points).
b) Suppose \ is any X! topological space for which GÐ\Ñ separates points and closed sets.
Prove that \ can be embedded in a product of copies of ‘ .
E16. A space \ satisfies the countable chain condition (CCC) if every collection of nonempty
pairwise disjoint open sets must be countable. ( For example, every separable space satisfies CCC. )
Suppose that \α is separable for each α −E . Prove that \œÖ\ α:
α −E× satisfies CCC.
( There isn't much to prove when | A | Ÿ c; why? )
Hint: Let ÖU > À > − X× be any such collection. We can assume all the U> 's are basic open sets (why?).
Prove that if W© T and | W| Ÿ c, then | W| Ÿ i! and hence Xmust
be countable.)
E17. There are several ways to define dimension for topological spaces. One classical method is the
following inductive definition.
Define dim gœ 1.
For :−\ , we say that \ has dimension Ÿ0 at p if there is a neighborhood base at p
consisting of sets with 1 dimensional (that is, empty) frontiers. Since a set has empty frontier iff
it is clopen, \ has dimension Ÿ 0 at : iff : has a neighborhood base consisting of clopen sets.
We say \ has dimension Ÿ 8 at : if : has a neighborhood base consisting of sets with n 1
dimensional frontiers.
We say \ has dimension Ÿ n, and write dimÐ\Ñ Ÿ n, if \ has dimension Ÿ 8 at p for each
p − X and that dim(X) œ n if dim( \ ) Ÿ 8 but dim Ð\Ñ ŸÎ8" . We say dimÐ\Ñ
œ ∞ if
dim Ð\Ñ Ÿ 8 is false for every 8 − Þ
While this definition of dim Ð\Ñ makes sense for any topological space \ , it turns out that
“dim” produces a nicely behaved dimension theory only for separable metric spaces. The
dimension function “dim” is sometimes called small inductive dimension to distinguish it from
other more general definitions of dimension. The classic discussion of small inductive dimension
is in Dimension Theory (
Hurewicz and Wallman).
266

It is clear that “dim Ð\Ñ œ 8 ” is a topological property of \ . There is a theorem stating that
8 8 7
dim( ‘ ) œ8, from which it follows that ‘ is not homeomorphic to ‘ if m Án.
In proving the
8 8
theorem, showing dim( Б Ñ Ÿ n is easy; the hard part is showing that dim Б Ñ ŸÎ n 1.
a) Prove that dim( ‘ ) œ" .
b) Let G be the Cantor set. Prove that dim( G)
œ !Þ
c) Suppose Ð\ß .Ñ is a ! -dimensional separable metric space. Prove that \ is homeomorphic
to a subspace of C. ( Hint: Show that \ has a countable base of clopen sets. View G as Ö!ß #× i!
and apply the embedding theorems.)
E18. Suppose \ and ] are X ! -spaces. Part a) outlines a sufficient condition enabling ] to be
top
embedded in a product of \ 's, i.e., that ] © \ for some cardinal m. Parts b) and c) look at some
7
corollaries.
a) Theorem Let \ and ] be X ! spaces. Then ] can be topologically embedded in \ for
some cardinal 7 if, for every closed set J ©] and every point C ÂJ,
there exists a
continuous function 0À]Ä\ such that f( y) cl f[ F] (here, n is finite and depends on f).
8 Â
Proof Let X œ Ö> œ ÐCßJÑ À J is closed in ] and C  J× and, for each such pair
8
>œÐCß0Ñ, let 0 > be the function given in the hypothesis. Let \ > œ\ (the space
containing the values of 0> ). Then clearly
7
Ö\ > À > − X× is homeomorphic to \ for some
7 , so it suffices to show ] can be embedded in Ö\ > À > − X× . Let
2À] Ä Ö\ > À>−X× be defined as follows: for C−] , 2Ð>Ñ has for its > -th coordinate
0> ÐCÑß i.e., 2ÐCÑÐ>Ñ œ 0> ÐCÑÞ
i) Show 2 is continuous.
ii) Show 2 is one-to-one.
iii) Show that 2 is a closed mapping onto its range 2Ò]Ó to complete the proof that 2 is
+ homeomorphism between ] and 2Ò]Ó © Ö\ > À > − X× . ( Note: the converse of the
theorem is also true. Both the theorem and converse are due to S. Mrowka.)
b) Let F denote Sierpinski space {{0,1},{ ,{0},{0,1}}. Use the theorem to show that every T
7
space Y can be embedded in F for some m.
g !
c) Let D denote the discrete space Ö 0,1 × . Use the theorem to show that every X " -space ]
7
satisfying dim Ð] Ñ œ ! (see Exercise E14) can be embedded in D for some m.
Parts b) and c) are due to Alexandroff.
Mrowka also proved that there is no T" -space \ such that every T" -space ] can be
7
embedded in X for some m.
7
E19 Þ Let \ œ Ö!ß "× and consider the product space \ , where 7 is an infinite cardinal.
267
7

7
a) Show that \ contains a discrete subspace of cardinality 7.
7 b) Show that AÐ\ Ñ œ 7 Ðsee
Example 4.6).
E20. A space \ is called totally disconnected every connected subset E satisfies lEl Ÿ "Þ Prove that
a totally disconnected compact Hausdorff space is homeomorphic to a closed subspace of Ö!ß "× for 7
some 7. ( Hint: see Lemma V.5.6).
E21. Suppose \ is a countable space that does not have a countable neighborhood base at the point
+−\Þ ( For instance, let +œÐ!ß!Ñin the space P,
Example III.9.8. )
i! Let E4 œ ÖB − \ À B3 œ + for 3 4× and E œ ∞ i!
4œ! E 4 © \ . Prove that no point in the
(countable) space E has a countable neighborhood base. ( Note: it is not necessary that \ be
countable. That condition simply forces E to be countable and makes the example more dramatic. )
268

5. The Quotient Topology
Suppose that for each α −Ewe have a map 1α À\ α Ä] , where \ α is a topological space and ] is a
set. Certainly there is a topology for ] that will make all the 1α's
continuous: for example, the trivial
topology on ] . But what is the largest topology on ] that will do this? Let
"
α
g œÖS©] Àfor all α −E, 1 ÒSÓ is open in \ ×
It is easy to check that g is a topology on ] . For each α,
by definition, each set 1αÒSÓ is open in \ α
so g makes all the 1α's continuous. Moreover, if F © ] and F Â g , then for at least one α,
" 1αÒFÓ is not open in \ α so adding F to g would “destroy” the continuity of at least one 1αÞ Therefore g is the largest possible topology on ] making all the 1α's
continuous.
Definition 5.1 Suppose 1α À \ α Ä ] for each α − E.
The strong topology g on ] generated by the
"
maps 1 α is the largest topology on ] making all the maps 1α continuous ß and S − g iff 1αÒSÓis open
in \α for every α.
The strong topology generated by a collection of maps Ö1 À − E× is “dual” to the weak topology in
α α
the sense that it involves essentially the same notation but with “all the arrows pointing in the opposite
direction.” For example, the following theorem states that a map 0 out of a space with the strong
topology is continuous iff each map 0‰1αis continuous; but a map 0 into a space with the weak
topology generated by mappings 0αis continuous iff all the compositions 0 α ‰0 are continuous ( see
Theorem 2.6)
Theorem 5.2 Suppose ] has the strong topology generated by a collection of maps Ö1αÀ α − E×Þ If
^is a topological space and 0À] Ä^, then 0 is continuous iff 0‰1α À\ α Ä^ is continuous for
each α −E.
1α0 Proof For each α −E , we have \ α Ä] Ä^, and the 1 α's
are continuous since ] has the strong
topology.
If 0 is continuous, so is each composition 0 ‰1αÞ Suppose each 0‰1αis continuous and that Yis open in ^. We want to show that 0 ÒYÓ
" " "
is open in ] . But ] has the strong topology, so 0 ÒY Ó is open in ] iff each 1α Ò0 ÒY ÓÓ
" " "
is open in \ αÞBut 1α Ò0 ÒYÓÓœ Ð0 ‰1αÑÒYÓwhich is open because 0 ‰1α
is
continuous. ñ
We introduced the idea of the strong topology as a parallel to the definition of weak topology.
However, we are going to use the strong topology only in a special case:
when there is only one map
1α œ 1 and 1 is onto.
Definition 5.3 Suppose Ð\ß g Ñ is a topological space and 1 À \ Ä ] is onto.
The strong topology on
] generated by 1 is also called the quotient topology
on ] . If ] has the quotient topology from 1,
we
269
α
"
"

say that 1À\Ä] is a quotient mapping and we say ] is a quotient of \ . We also say the ] is a
quotient space of \ and sometimes as ] œ \Î1.
From our earlier discussion we know that if \ is a topological space and 1 À \ Ä ] , then the quotient
" "
topology on ] is g œ ÖY À 1 ÒY Ó is open in \× . Thus Y − g if and only if 1 ÒY Ó is open in \ :
notice in this description that
“only if” guarantees that 1 is continuous and
“if” guarantees that g is the largest topology on ] making 1 continuous.
Quotients of \ are used to create new spaces ] by “pasting together” (“identifying”) several points of
\ to become a single new point. Here are two intuitive examples:
i) Begin with \œÒ!ß"Ó and identify ! with " (that is, “paste” ! and " together to become a
" "
single point). The result is a circle, WÞ This identification is exactly what the map 1ÀÒ!ß"ÓÄW
given by 1ÐBÑ œ Ð cos # 1Bß sin # 1BÑ accomplishes. It turns out ( see below)
that the usual topology on
" W is the same as quotient topology generated by the map 1. Therefore we can say that 1 is a quotient
"
map and that W is a quotient of Ò!ß "Ó
"
ii) If we take the space \œW and use a mapping 1to
“identify” the north and south poles
together, the result is a “figure-eight” space ] . The usual topology on ] (from ‘ ) turns out to be the
#
same as quotient topology generated by 1 ( see below).
Therefore we can say that 1 is a quotient
mapping and the the “figure-eight” is a quotient of W . "
Suppose we are given an onto map 1ÀÐ\ßg ÑÄÐ]ßg Ñ. How can we tell whether 1is
a quotient
w
map that is, how can we tell whether g is the quotient topology? By definition, we must check that
w
w "
Y−g iff 1 ÒYÓ−g Þ Sometimes it is fairly straightforward to do this. But the following theorem
can sometimes make things much easier.
Theorem 5.4 Suppose 1ÀÐ\ßg ÑÄÐ]ßg Ñis continuous and onto. If 1is
open (or closed), then
w
g w is the quotient topology, so 1 is a quotient map. In particular,
if \ is compact and ] is Hausdorff,
1 is a quotient mapping.
Note: Whether the map 1À\Ä] is continuous depends, of course, on the topology g , but if g
makes 1 continuous, then so would any smaller topology on ] . The theorem tells us that if 1 is
w w
both continuous and open (or closed), then g is completely determined by 1:
g is the largest
topology making 1 continuous.
270
w " w "
w w
Proof Suppose Y © ] Þ W / must show Y − g iff 1 ÒY Ó − g Þ If Y − g , then 1 ÒY Ó − g
" "
because 1 is continuous. On the other hand, suppose 1 ÒY Ó − g . Since 1 is onto, 1Ò1 ÒY ÓÓ œ Y and
" w
so, if 1 is open, Y œ 1Ò1 ÒY ÓÓ − g Þ
" "
For J © ] ß 1 Ò] J Ó œ \ 1 ÒJ ÓÞ It follows easily that J is closed in the quotient
"
space if and only if 1 ÒJÓ is closed in \Þ With that observation, the proof that a continuous, closed,
onto map 1is a quotient map is exactly parallel to the proof when 1 is open.
If \ is compact and ] is X , then 1 must be closed, so 1 is a quotient mapping. ñ
#
"
Note: Theorem 5.4 implies that the map 1ÐBÑ œ Ð cos # 1Bß sin # 1 BÑ from Ò!ß "Ó to W is a quotient
"
map, but 1 is not open. The same formula 1 can be used to define a quotient map 1 À ‘ Ä W
which is not closed (why?). Exercise E25 gives examples of quotient maps 1 that are neither

open nor closed.
Suppose µ is an equivalence relation on a space \ . For each B − \ , the equivalence class of B is
ÒBÓœÖD−\ÀDµB× . The equivalence classes partition \ into a collection of nonempty pairwise
disjoint sets. (Conversely, it is easy to see that any partition of \ is the collection of equivalence
classes for some equivalence relation namely, B µ D iff Band D are in the same set of the partition. )
The set of equivalence classes, ] œÖÒBÓÀB−\× , is sometimes denoted \ÎµÞ There is a natural
onto map 1À\Ä] given by 1ÐBÑœÒBÓ . We can think of the elements of ] as “new points” which
arise from “identifying together as one” all the members of each equivalence class in \ . If \ is a
topological space, we can give the set of equivalence classes ] the quotient topology. Conversely,
whenever 1À\Ä] is any onto mapping, we can think of ] as the set of equivalence classes for
"
some equivalence relation on \ namely B µ C Í C − 1 ÐBÑ Í 1ÐCÑ œ 1ÐBÑ .
Example 5.5 For + , , − ,
define + µ , iff ,+ is even. There are two equivalence classes
Ò!Ó œ ÖÞÞÞß %ß #ß !ß #ß %ß ÞÞÞ× and Ò"Ó œ ÖÞÞÞ $ß "ß "ß $ß ÞÞÞ× so Î
µ œ ÖÒ!Óß Ò"Ó×Þ
Define 1À Äε by 1Ð+ÑœÒ+Óand give ε
the quotient topology. A set Y is
" open in ε iff 1 ÒYÓis open in , and that is true for every Y © ε because is discrete.
Therefore the quotient Î µ is a two point discrete space.
Example 5.6 Let Ð\ß .Ñ be a pseudometric space and define a relation µ in \ by B µ D iff
.ÐBßDÑœ! . Let ] œ\ε , define 1À\Ä] by 1ÐBÑœÒBÓ . Give ] the quotient topologyÞ
Points at distance ! in \ have been “identified with each other” to become one point (the equivalence
class) in ]Þ
w w
For ÒBÓß ÒDÓ − ] , define . ÐÒBÓß ÒDÓÑ œ .ÐBß DÑ . In order to see that . is well-defined,
we need to check that the definition is independent of the representatives chosen from the equivalence
classes:
w w w w
If ÒBÓœÒBÓand ÒDÓœÒDÓ , then .ÐBßBÑœ! and .ÐDßDÑœ!ÞTherefore w w w w w w
.ÐBßDÑ Ÿ .ÐBßB Ñ .ÐB ß D Ñ .ÐD ß DÑ œ .ÐB ß D Ñ,
and similarly
w w w w w w w w
.ÐB ß D Ñ Ÿ .ÐBß DÑ . Thus .ÐB ß D Ñ œ .ÐBß DÑ so . ÐÒB Óß ÒD ÓÑ œ . ÐÒBÓß ÒDÓÑÞ
w w w
It is easy to check that . is a pseudometric on ] . In fact, . is a metric:
if . ÐÒBÓßÒDÓÑ œ ! , then
.ÐBß DÑ œ ! , which means that B µ D and ÒBÓ œ ÒDÓÞ
We now have two definitions for topologies on ] : the quotient topology g and the metric topology
g w. We claim that g œ g wÞ
To see this, first notice that
. .
w . w .
% %
ÒCÓ − F ÐÒBÓÑ iff . ÐÒCÓß ÒBÓÑ % iff .ÐCß BÑ % iff C − F ÐBÑ
w w
" . . . .
% % % %
Therefore 1 ÒF ÐÒBÓÑÓ œ F ÐBÑ and 1 ÒF ÐBÑÓ œ F ÐÒBÓÑ.
But then we have
Y−g. w iff Y is a union of . -balls
" iff 1 ÒYÓ is a union of . -balls
" iff 1 ÒYÓ is open in \
iff Y−g Þ
271
w

w The metric space Ð] ß . Ñ is called the metric identification of the pseudometric space Ð\ß .Ñ.
In effect,
we turn the pseudometric space into a metric space by agreeing that points in \ at distance ! are
“identified together” and treated as one point.
Note: In this particular example, it is easy to verify that the quotient mapping 1À\Ä] is open,
so 1 would be a homeomorphism if only 1 were "" . If the original pseudometric . is actually a
metric, then 1 is "" and a homeomorphism: the metric identification of a metric space Ð\ß.Ñ is
itself.
Example 5.7 Exactly what does it mean if we say “identify the endpoints of Ò!ß "Ó and get a circle”?
Of course, one could simply take this to be the definition of a (topological) “circle.” Or, it could mean
that we already know what a circle is and are claiming that a certain quotient space is homeomorphic to
a circle. We take the latter point of view.
" Let 1ÀÒ!ß"ÓÄW be given by 1ÐBќРcos # 1Bß sin # 1BÞ ) This map is onto and "" except
that 1Ð!Ñœ1Ð"Ñ, so 1 corresponds to the equivalence relation on \ for which !µ" (and there are no
other equivalences except that BµB for every BÑÞ We can think of the equivalence classes as
corresponding in a natural way to the points of W . "
" "
Here W has its usual topology and 1 is continuous. However, since \ is compact and W is Hausdorff,
" Theorem 5.4 gives that the usual topology on W is the quotient topology and 1is
a quotient map.
When it “seems apparent” that the result of making certain identifications produces some familiar
space ] , we need to check that the familiar topology on ] is actually the quotient topology. Example
5.7 is reassuring: if we believed, intuitively, that the result of identifying the endpoints of Ò!ß "Ó should
" "
be W but then found that the quotient topology on the set W turned out to be different from the usual
topology, we would be inclined to think that we had made the “wrong” definition for a “quotient.”
Example 5.8 Suppose we take a square Ò!ß "Ó and identify points on top and bottom edges using the
#
equivalence relation ÐBß !Ñ µ ÐBß "ÑÞ
We can schematically picture this identification as
272

The arrows indicate that the edges are to be identified as we move along the top and bottom edges in
# $
the same direction. We have an obvious map 1 from Ò!ß "Ó to a cylinder in ‘ which identifies points
in just this way, and we can think of the equivalence classes as corresponding in a natural way to the
points of the cylinder.
The cylinder has its usual topology from and the map is (clearly) continuous and onto. Again,
‘ $ 1
Theorem 5.4 gives that the usual topology on the cylinder is, in fact, the quotient topology.
273

Example 5.9 Similarly, we can show that a torus (the “surface of a doughnut”) is the result of the
#
following identifications in Ò!ß"Ó : ÐBß!ѵÐBß"Ñand Ð!ßCѵÐ"ßCÑ
Thinking in two steps, we see that the identification of the two vertical edges produces a cylinder; the
circular ends of the cylinder are then identified (in the same direction) to produce the torus.
The two circles darkly shaded on the surface represent the identified edges.
We can identify the equivalence classes naturally with the points of this torus in and just as before
‘ $
we see that the usual topology on the torus is in fact the quotient topology.
274

Example 5.10 Define an equivalence relation µ in Ò!ß "Ó by setting ÐBß !Ñ µ Ð" Bß "ÑÞ Intuitively,
the idea is to identify the points on the top and bottom edges with each other as we move along the
edges in opposite directions. We can picture this schematically as
Physically, we can think of a strip of paper and glue the top and bottom edges together after making a
“half-twist.” The quotient space \Î µ is called a Mobius ¨ strip.
We can take the quotient \Î µ as the definition of a Mobius ¨ strip, or we can consider a “real” Mobius ¨
strip Q in ‘ and define a map 1 À Ò!ß "Ó Ä Q that accomplishes the identification we have in mind.
$ #
In that case there is a natural way to associate the equivalence classes to the points of the torus in ‘ $
and again Theorem 5.4 guarantees that the usual topology on the Mobius ¨
strip is the quotient topology.
275
#

Example 5.11 If we identify the vertical edges of Ò!ß "Ó (to get a cylinder) and then identify its
#
circular ends with a half-twist (reversing orientation): Ð!ß CÑ µ Ð"ß CÑ and ÐBß !Ñ µ Ð" Bß "Ñ.
We get
a quotient space which is called a Klein bottle.
It turns out that a Klein bottle cannot be embedded in ‘ the physical construction would require a
$
“self-intersection” (additional points identified) which is not allowed. A pseudo-picture looks like
In these pictures, the thin “neck” of the bottle actually intersects the main body in order to re-emerge
“from the inside” in a “real” Klein bottle (in ‘ , say), the slef-intersection would not happen.
%
In fact, you can imagine the Klein bottle as a subset of ‘ using color as a 4 dimension. To each
% >2
point ÐBßCßDÑon the “bottle” pictured above, add a 4th coordinate to get ÐBßCßDßÑwhere BßCßD depend on time > and > is recorded as a 4 -coordinate. A point on the
surface then has coordinates of form ÐBßCßDß>Ñ. At “a point” where we see a self-intersection in ‘ ß $
there are really two different points (with different time coordinates >
).
276

Example 5.12
" "
1) In W, identify antipodal points that is, in vector notation, TµTfor each T−W.
" "
Convince yourself that the quotient Wε is W.
# #
2) Let H be the unit disk ÖT − ‘ À lTl Ÿ "× . Identify antipodal points on the boundary of
# " #
H : that is, T µ T if T − W © H Þ The quotient HÎ µ is called the projective plane,
a space
which, like the Klein bottle, cannot be embedded in ‘ . $
3) For any space \ , we can form the product \ ‚ Ò!ß "Ó and let ÐBß "Ñ µ ÐCß "Ñ for all
Bß C − \ . The quotient Ð\ ‚ Ò!ß "ÓÑÎ µ is called the cone over \ . ( Why? )
4) For any space \ , we can form the product \‚Ò"ß"Óand define ÐBß"ѵÐCß"Ñand
ÐBß"ѵÐCß"Ñfor all BßC−\ . The quotient Ð\‚Ò"ß"ÓÑε is called the suspension of \ .
( Why? )
There is one other very simple construction for combining topological spaces. It is often used in
conjunction with quotients.
Definition 5.13 For each α −E, let Ð\ αßgαÑ be a topological space, and assume that the sets \ α are
pairwise disjoint. The topological sum (or “free sum”) of the \ α's is the space Ð
α−E\
αßgÑ
where
g œÖS©
α−E \ α ÀS∩\ α is open in \ α for every α −E}.
We denote the topological sum by
\ . In the case lEl œ # , we use the simpler notation \ \ Þ
α−E
α
277
" #
In \ , each \ is a clopen subspace. Any set open (or closed) in \ is open (or closed) in the sum.
α−E
α α α
The topological sum \ can be pictured as a union of the disjoint pieces \ , all “far apart” from
α−E
α α
each other so that there is no topological “interaction” between the pieces.
Example 5.14 In , let E" and E # be open disks with radius " and centers at Ð!ß !Ñ and Ð$ß !ÑÞ
Then toppological sum E" E# is the same as E" ∪E#
with subspace topology
Let F " be an open disk with radius " centered at Ð!ß!Ñ and let F#
be a closed disk
with radius " centered at Ð#ß !Ñ. Then F" F# is not the same as F" ∪ F#
with the subspace topology
(why? ÑÞ
‘ #
Exercise 5.15 Usually it is very easy to see whether properties of the \α's
do or do not carry over to
\ . For example, you should convince yourself that:
α−E
α
1) If the \ 's are nonempty and separable, then \ is separable iff lEl Ÿ i .
α α
α−E
2) If the \ 's are nonempty and second countable, then \ is second countable iff lEl Ÿ i .
α α
α−E
3) A function 0À \ Ä^is continuous iff each 0l\ is continuous.
α−E
α α
!
!

4) If . Ÿ 1 is a metric for \ , then the topology on \ is the same as g where
α α α
α−E
. ÐBß CÑ if Bß C − \
.ÐBßCÑ œ
" otherwise
α α
so \ is metrizable if all the \ 's are metrizable. You should be able to find similar statements for
α−E
α α
other topological properties such as first countabe, second countable, Lindelof, ¨ compact, connected,
path connected, completely metrizable, ... .
Definition 5.16 Suppose \ and ] are disjoint topological spaces and 0 À E Ä ] , where E © \ . In
the sum \] , define BµC iff Cœ0ÐBÑÞIf we form Ð\]Ñε , we say that we have attached
\ to ] with 0 and write this space as \ ] .
0
For each : − 0ÒEÓß its equivalence class under µ is Ö:× ∪ 0 ÒÖ:×ÓÞ You may think of the function
“attaching” the two spaces by repeatedly selecting a group of points in \ , identifying them together,
and “sewing” them all onto a single point in ]just
as you might run a needle and thread through
several points in the fabric \ and then through a point in ] and pull everything tight.
Example 5.17
1) Consider disjoint cylinders \ and ] . Let E be the circle forming one end of \ and F the
circle forming one end of ] . Let 0ÀEÄ0ÒEÓœF©] be a homeomorphism. Then 0 “sews
together” \ and ] by identifying these two circles. The result is a new cylinder.
2) Consider a sphere W © ‘ . Excise from the surface W two disjoint open disks H" and H#
and let E" ∪E # be union of the two circles that bounded those disks. Let ] be a cylinder whose
ends are bounded by the union of two circles, F" ∪F# . Let 0 be a homeomorphism carrying the
points of E" and E" clockwise onto the points of F" and F#
respectively..
# Then W ] is a “sphere with a handle.”
0
278
"
# $ #
# $ #
3) Consider a sphere W © ‘ . Excise from the surface W an open disk and let E be the
#
circular boundary of the hole in the surface W . Let Q be a Mobius ¨ strip and let F be the curve that
"
bounds it. Of course, FW . Let 0ÀEÄFbe a homeomorphism. The we can use 0 to join the
spaces by “sewing” the edge of the Mobius ¨ strip to the edge of the hole in W . The result is a “
# sphere
with a crosscap.”
There is a very nice theorem, which we will not prove here, which uses all these ideas. It is a
“classification” theorem for certain surfaces.
Definition 5.18 A Hausdorff space \ is a 2- manifold if each B − \ has an open neighborhood Y that
is homeomorphic to ‘ . Thus, a 2-manifold looks “locally” just like the Euclidean plane. A is
# surface
a Hausdorff 2-manifold.
Theorem 5.19 Let \ be a compact, connected surface. The \ is homeomorphic to a sphere W or to #
W # with a finite number of handles and crosscaps attached.
You can read more about this theorem and its proof in (William
Algebraic Topology: An Introduction
Massey).
.

Exercises
E22. a) Let µ be the equivalence relation on given by ÐBßCѵÐBßCÑ " " # # iff C" œC#
. Prove that
# ‘ ε is homeomorphic to ‘ .
‘ #
b) Find a counterexample to the following assertion: if µ is an equivalence relation on a
space \ and each equivalence class is homeomorphic to the same space ] , then Ð\Î µ Ñ‚] is
homeomorphic to \ .
Why might someone ever wonder whether this assertion might be true? In part a), we have
\œ ‘ ‚ ‘ , each equivalence class is homeomorphic to ‘ and ( \Î µ Ñ‚ ‘ ‘ ‚ ‘ \ . In this
example, you “divide out” equivalence classes that all look like ‘ , then “multiply” by ‘ , and you're
back where you started.
# # # #
c) Let 1À‘ Ä‘ be given by 1ÐBßCÑœB C. Then the quotient space ‘ Î1 is
homeomorphic to what familiar space?
d) On ‘ , define an equivalence relation ( iff . Prove
# # #
BßCѵÐBßCÑ " " # # B" C" œB# C#
that ‘ / is homeomorphic to some familiar space.
# µ
E23. For Bß C − Ò!ß "Ó, define B µ C iff B C is rational. Prove that the corresponding quotient
space Ò!ß "ÓÎ µ is trivial.
E24. Prove that a 1-1 quotient map is a homeomorphism.
E25. a) Let ] " œ Ò!ß "Ó with its usual topology and ] # œ Ò!ß "Ó with the discrete topology. Define
1À] " ] # Ä] by letting 1 be “the identity map” on both ] " and ]Þ # Prove that 1is
a quotient map
that is neither open nor closed.
# b) Let ‘ have the topology g for which a subbase consists of all the usual open sets together
#
with the singleton set ÖÐ!ß !Ñ× . Let ‘ have the usual topology and let 0 À ‘ Ä ‘ be the projection
0ÐBßCÑœ B. Prove that 0 is a quotient map which is neither open nor closed.
E26. State and prove a theorem of the form:
“for two disjoint subsets Eand F of ‘ , EFE∪Fiff
... ”
#
#
E27. Let ‘ have the topology g for which a subbasis consists of all the usual open sets together with
#
the singleton set ÖÐ!ß !Ñ× . Let ‘ have the usual topology and define 0À ‘ Ä ‘ by 0ÐBß CÑ œ B.
Prove that 0 is a quotient map which is neither open nor closed.
E28. Let ] œ Ð ‚
Ð!ß "ÑÑ and let ] œ ] Ò!ß "Ñ . Prove ] is not homeomorphic to ] but
" # " " #
279

that each is a continuous one-to-one image of the other
E29. Show that no continuous image of ‘ can be represented as a topological sum \] , where
\ß] Á g. How can this be result be strengthened?
E30. Suppose \= (s −W ) and ] > ( >−X ) are pairwise disjoint spaces. Prove that \ = ‚ ]
> is
homeomorphic to Ð\ ‚ ] ÑÞ
=−W >−X
=−Wß>−X
= >
E31. This problem outlines a proof (due to Ira Rosenholtz) that every nonempty compact metric space
\ is a continuous image of the Cantor set G . From Example 4.5, we know that \ is homeomorphic to
i!
a subspace of Ò!ß "Ó Þ
a) Prove that the Cantor set G©‘ consists of all reals of the form 4
where
each + 4 œ ! or 2.
280
∞
+
$
4œ!
∞ ∞
+ 4
b) Prove that Ò!ß "Ó is a continuous image of G. Hint: Define 1Ð Ñ œ + 4
Þ
$ #
4œ! 4œ!
4
4 4
i!
c) Prove that the cube Ò!ß "Ó is a continuous image of G.
i! i! . i!
Hint: By Corollary 2.21, G Ö!ß #× G Use 1 from part b) to define 0 À G Ä Ò!ß "Ó by i!
0ÐB" ß B# ß ÞÞÞß ÞÞÞÑ œ Ð1ÐB" Ñß 1ÐB# Ñß ÞÞÞß ÞÞÞÑ
d) Prove that a closed set O©Gis a continuous image of G.
(Hint: G is homeomorphic to the “middle two-thirds" set G consisting of all reals of the form
w
∞
, 4
'
4œ!
w G w BßC−G
BC
#
w ÂG Þ w O w G
w w
4 . has the property that if , then If is closed in , we can map
G Ä O by sending each point B to the point in O nearest to B.)
e) Prove that every nonempty compact metric space \ is a continuous image of G.
.

Chapter VI Review
Explain why each statement is true, or provide a counterexample.
i i
! !
1. Ð!ß "Ñ is open in Ò!ß "Ó Þ
2. Suppose J is a closed set in Ò!ß "Ó ‚ ‘ . Then 1 ÒJ Ó is a closed set in ‘ Þ
3. is discrete
i!
#
4. If G is the Cantor set, then there is a complete metric on G which produces the product topology.
i!
‘ ‘
8 8
5. Let ‘ have the box topology. A sequence Ð0 Ñ Ä 0 − ‘ iff Ð0 Ñ Ä 0 uniformly.
8 Ò!ß"Ó
8 8 8
6. Let 0 À Ò!ß "Ó Ä Ò!ß "Ó be given by 0 ÐBÑ œ B . The sequence Ð0 Ñ has a limit in Ò!ß "Ó .
‘ #
7. Let 1−‘ be defined by 1ÐBÑœBfor all B−‘ . Give an example of a sequence Ð08Ñ of distinct
functions in ‘ that converges to .
‘ 1
8. Let G be the Cantor set. Then G is homeomorphic to the topological sum G G.
9. The projection maps 1 and 1 from ‘ ‘ separate points from closed sets.
B C
# Ä
10. The letter N is a quotient of the letter M .
11. Suppose µ is an equivalence relation on \ and that for B − \ß ÒBÓ represents its equivalence
class. If B is a cut point of \ , then ÒBÓ is a cut point of the quotient space \Î µ .
12. If 1À \ Ä ] is a quotient map and ] is compact T # , then ] is compact T # .
13. Suppose E is infinite and, that in each space \ α ( α − E)
there is a nonempty proper open subset
Sα. Then S α is not a basic open set of the product \ œ \ α, and, in fact, Sα
cannot even be
open in the product.
14. If \œ∞ E , where the E 's are disjoint clopen sets in \ , then \z∞ 8œ" 8 8 8œ" E8 ( œthe
topological sum of the 's).
E8
15. Let \ œ Ö!ß "× and ] œ with their usual topologies. Then \ is homeomorphic to ]
Þ
8 8 8 8
8œ" 8œ"
281
∞ ∞
∞
16. Let \ 8 œ Ö!ß "× and ] 8 œ with their usual topologies. Then
8œ" \ 8 is homeomorphic to
]Þ
∞ 8œ" 8
$ # # $
17. Let E œ ÖÐBß Cß DÑ − ‘ À B C #CD D lsin ÐBCDÑl × , and let 0À ‘ Ä ‘ be given by
0ÐBßCßDÑœ B# . Then 0 ÒEÓis
open but not closed in ‘
.

18. Suppose E 8 is a connected subset of \ 8Ð Á gÑ and that ∞ E 8 is dense in ∞
8œ" 8œ" \ 8Þ
Then each
\8 is connected.
19. In ‘ , every neighborhood of the function sin contains a step function (that is, a function with
‘
finite range).
20. Let X be an uncountable set with the cocountable topology. Then { ÐBß BÑ À B − \× is a closed
subset of the product \‚\Þ.
"
8
21. Let EœÖ À8− ש ‘ , and let 7be any cardinal number. E is a closed set in ‘ Þ
22. If Gis the Cantor set, then G‚G‚G‚G is homeomorphic to G‚G.
" " 33. W ‚W is homeomorphic to the “infinity symbol”: ∞
282
5 5
34. 31. Let T be the set of all real polynomials in one variable, with domain restricted to Ò!ß "Ó,
for
Ò!ß"Ó
which ranÐT Ñ © Ò!ß "ÓÞ Then T is dense in Ò!ß "Ó Þ
35. Every metric space is a quotient of a pseudometric space.
37. A separable metric space with a basis of clopen sets is homeomorphic to a subspace of the Cantor
set.
38. Let { \ α À α − E } be a nonempty product space. Each \ α is a quotient of the product space
Ö\ α À α − E×
39. Suppose \ does not have the trivial topology. Then \ cannot be separable.
#-
40. Every countable space \ is a quotient of
41. ‚ is homeomorphic to the sum of i! disjoint copies of .
42. Suppose BœÐBÑ−int E, where E© \ . Then for every α, B −int 1 ÒEÓ.
α α α α
"
α α α
43. The unit circle, W , is homeomorphic to a product −E \ , where each \ © Ò!ß"Ó
" (i.e., W can be “factored” into a product of subspaces of Ò!ß "Ó ).
i!
44. is homeomorphic to ‘ .