Chapter VI Products and Quotients
1. Introduction
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 ÖÐBßCÑÀB−\ßC −]× . How can we define an “infinite product” set ? Informally, we want to say something like \œ Ö\α Àα −E×
\œ Ö\αααα Àα −EלÖÐBÑÀB −\ × so that a point BB\ in the product consists of “coordinates αα ” chosen from the 's. But what exactly does a symbol like BœÐBÑα mean if there are “uncountably many coordinates?”
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: ∞ . \œ8œ" \8888 œÖÐBÑÀB −\ × But if we want to be careful about set theory, then a legal definition of \ should have the form \œÖÐBÑ−Y888 ÀB −\ ×. From what “pre-existing set” Y will the sequences in \ be chosen? The answer is easy: we write
∞∞ \œ8œ" \8888 œÖB−Ð 8œ" \Ñ ÀBÐ8ÑœB −\×Þ Thus the elements of ∞ are certain functions (sequences) defined on the index set . This idea 8œ"\8 generalizes naturally to any productÞ
Definition 2.1 Let Ö\α Àα − E× be a collection of sets. We define the product set E \œ Ö\αααα Àαα −EלÖB− \ ÀBÐÑ−\ × . The \ 's are called the factors of \ . For each th α1α, the function ααÀÖ\À−E×Ä\ α defined by 1 α ÐBÑœB α is called the α -projection map . th For B−\ , we write more informally Bαα œBÐα Ñœthe α -coordinate of B and write BœÐB ).
Note: the index set EBœÐBÑ might not be ordered. So, even though we use the informal notation α , such phrases as “the first coordinate of BBB ,” “the next coordinate in after α ,” and “the coordinate in BB preceding αα ” may not make sense. The notation ÐBÑ is handy but can lead you into errors if you're not careful.
239 A point in is, by definition, a function that “chooses” a coordinate from each set BÖ\À−E× α α Bα in the collection Ö\α Àα − 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 then ∞ . Without using AC, we can \œ88 ß8œ" \œÖB− ÀBÐ8Ñ−ל precisely describe a point (œ function) in the product for example, the identity function so ∞ . 3œÖÐ7ß8Ñ− ‚ À7œ8× œ8œ" \8 Ág We will often write as . If the indexing set is clearly understood, we may Ö\αα Àα − E×α−E \ E simply write . \α
Example 2.3
1) If , then E E œ g Ö\ααα Àαα − E× œ ÖÐα−E \ Ñ À BÐ Ñ − \ × œ Ög×Þ
2) Suppose \œg for some αα −E . Then \œÖÐ \ÑÀBÐÑ−\×ÞE αααα! ! αα−E −E Since BÐα Ñ − \ is impossible, \ œ g . ! αα! αα−E
3) Strictly speaking, we have two different definitions for a finite product\‚\"# :
i) \"# ‚\ œÖÐBßBÑÀB "#""## −\ßB −\× (a set of ordered pairs)
Ö"ß#× ii) \‚\œÖB−Ð\∪\Ñ"# "# ÀBÐ3Ñ−\× 3 (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 Ö"ß#× function B œ ÖÐ"ß B"# Ñß Ð#ß B Ñ× − Ð\ "# ∪ \ Ñ Þ
4) If , then ∞∞ Eœ Ö\88888 À8− ל8œ" \ œÖB−Ð 8œ" \ Ñ À B −\ × œ ÖÐB"# ß B ß ÞÞÞß B 8 ß ÞÞÞß Ñ À B 8 − \ 8 × œthe set of all sequences ÐB 8 Ñ where B 8 − \ 8 Þ
5) Suppose the 's are identical, say for all . Then \\œ]−EÖ\À−E×αααα α EEE. If , we will sometimes œÖB−Ðα−E \αα Ñ ÀBÐα Ñ−\ לÖB−] ÀB α −]ל] lElœ7 write this product simply as ]œ7 “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 , we can think about a product topology . We Ö\α Àα − E× begin by recalling the definition and a few basic facts about the “weak topology.” (See Example III.8.6.)
240 Definition 2.4 Let \−EÐ\ßÑ be a set. For each αg , suppose αα is a topological space and that 0À\Ä\αα. The weak topology on \ generated by the collection Y œÖ0Àα α −E× is the smallest topology on \0 that makes all the α 's continuous.
Certainly, there is at least one topology on \0 that makes all the α '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 { : is a topology on making all the 's \0 gg α 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 −E and for each open set Yαα ©\ , the set 0α ÒYÓ α must be open.
" Therefore the weak topology g is the smallest topology that contains all such sets 0ÒYÓα α and that is " the topology for which ÆαœÖ0α ÒYÓÀααα −EßY open in \ × is a subbase . (See Example III.8.6. )
Therefore a base for the weak topology consists of all finite intersections of sets from Æ . A typical basic open set has form " ] " ... " where each and each is 0ÒY∩0ÒYÓ∩∩0ÒYÓαα"#αα"# α 8 α 8α3 −E Y α 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 Yαα for 0α ÒY ÓÞ A typical basic open set is then " ] " ... " Yœ0ÒY∩0ÒYÓ∩∩0ÒYÓαα"#αα"# α 8 α 8
which we further abbreviate as Y œ Yαα"# ß Y ß ÞÞÞß Y α 8
So B − Y œ Yαα"# ß Y ß ÞÞÞY α 8 iff 0 α 3 ÐBÑ − Y α 3 for each 3 œ "ß ÞÞÞß 8Þ
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 Yαα"# ß Y ß ÞÞÞß Y α 8 where each Y\ is a basic (or even subbasic) open set in . αα3 3
Example 2.5 Suppose Ð\ßg Ñ is any topological space and E©\Þ Let 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 3ÒYÓœY∩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‰0α is continuous.
Theorem 2.6 For αg−E , suppose 0ααα À\ÄÐ\ß Ñ and that \ has the weak topology generated by the functions 0^α . Let be a topological space and 0À^Ä\0 . Then is continuous iff 0‰0À^Ä\αα is continuous for every α .
Proof If 00‰0 is continuous, then each composition α is continuous. Conversely, suppose each " 0‰0α is continuous. To show that 0 is continuous, it is sufficient to show that 0 ÒZÓ is open in ^ whenever Z\ZœYY\ is a subbasic open set in (why? ). So let αα with open in α . Then " " " " 0 ÒZ Ó œ 0 Ò0α ÒYαα ÓÓ œ Ð0 ‰ 0Ñ ÒY α Ó which is open since 0 α ‰ 0 is continuous. ñ
241 Definition 2.7 For each αg−Eß let Ð\ßαα Ñ be a topological space. The product topology g on the set is the weak topology generated by the collection of projection maps \œÖÀ−E×Þα Y1αα 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 Yαα œ1αα ÒYÓ , where −E and Y α is open in \ αÞ A base consists of all finite intersections of such sets À
"] ... " Yαααα"8"8 ∩ ÞÞÞ ∩ Y œ11αα"8 ÒY ∩ ∩ ÒY Ó
œ ÖB − \ À B − Y for each 3 œ "ß ÞÞÞß 8× αα33 α where for (*) œα−E Yααα Y œ \ααα Á"# ß ß ÞÞÞß α 8
(It is sufficient to use only Y\αα ' s which are basic (or even subbasic) open sets in Þ Why?)
A basic open set in “depends on only finitely many coordinates” in the following sense: \α
B − Yαα"# ß Y ß ÞÞÞß Y α 8 iff B satisfies the finitely many restrictions B α33 − Y α Ð3 œ "ß ÞÞÞß 8Ñ . Since (basic) open sets containing BB are the “standard” for measuring closeness to , we can say, roughly, that in the product topology “closeness depends on only finitely many coordinates.”
For a finite index set , the product topology on is just the one for which the E œ Ö"ß #ß ÞÞÞß 8×α−E \α basic open sets are the open boxes : α−EYα Y œ8 Y œ Y ß Y ß ÞÞÞß Y œ Y ‚ Y ‚ ÞÞÞ ‚ Y α−Eαααα 3œ" 3"#8"# 8
Then condition (*) that “Yœ\αα 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” ( open in as a base in defining the product topology. But in Definition α−EYYαα \Ñ α 2.7, a “box” of the form might not be open because (*) may not hold. It is perfectly possible, α−EYα of course, to define a different topology on the set using all boxes of the form as a αα−E\Yαα −E base. This alternate topology on is called the box topology . But we will see that our α−E\α 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 is continuous and open. 1αααÀ ÖÖ\ À − E× Ä \ α If , each is onto. A function is continuous iff Ö\αα Àα1 − E× Á g 0 À ^ Ä \ α 1ααα‰0 À^ Ä\ is continuous for every .
Proof By definition, the product topology is one that makes all the 1α 's continuous. and clearly, each 11αα is onto if the product is nonempty. To show that is open, it is sufficient to show that the image of a basic open set Y œ Yαα"# ß Y ß ÞÞÞß Y α 8 is open. This is clearly true if Y œ g .
242 Yœα3 if αα3 For Y Á g , we get 11ααααα ÒY Ó œ Ò Y"# ß Y ß ÞÞÞß Y 8 Ó œ .. \α if αα Á"8 ß ÞÞÞß α
Either way, 1ααααÒ Y"# ß Y ß ÞÞÞß Y 8 Ó is open. Because the product has the weak topology generated by the projections, Theorem 2.6 gives that0‰0ñ is continuous iff each composition 1α is continuous.
# " Generally, projection maps are not closed. For example, if JœÖÐBßCÑ−‘ ÀCœB ßB!× , then 1‘‘"ÒJ Ó œ the projection of J on the B -axis is not closed in Þ
Example 2.9
" " 1) A subbasic open set in ‘‘‚ÒYÓœY‚ÒZÓœ‚ZY has form 1"# ‘1 or ‘ , where and ZÞ are open in ‘ (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 function0À‘‘‘ Ä ‚ given by 0Ð>ÑœÐ>ß## sin >Ñ is continuous because the ## compositions Ð11"# ‰ 0ÑÐ>Ñ œ > and Ð ‰ 0ÑÐ>Ñ œ sin > are both continuous functions from ‘‘ to .
2) Let \œi! œ “the product of countably many copies of .” A base for the product topology consists of all sets where finitely many 's are singletons and all the others are Yœ Y88 Y equal to ÞY Each is infinite (in fact lY l œ - why? ), so every nonempty open set is infinite. In particular, \Þ has no isolated points 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 5 8 , ... Ñ − \ , the set ∞ is open in the box topology. (For a finite product, the box and product ÖB× œ8œ" Ö58 × topologies care the same: a finite product of discrete spaces is discrete.)
3) Let ‘‘ , where . Each point in is a function and ‘‘‘‘œÖ\À<−× << \œ 0 0ÀÄ ‘‘ 1<Ð0Ñ œ 0Ð<ÑÞ A basic open set containing 0 is Y œ Y<<"# ß Y ß ÞÞÞß Y < 8 ß where the Y < 3 's are basic open in ‘%% and 0Ð<3< Ñ − Y33 Þ If each Y < is an open interval Ð0Ð< 3333 Ñ ß 0Ð< Ñ Ñ , then ‘ Y œ Ö1 −‘% À l 1Ð<333 Ñ 0Ð< Ñl ß 3 œ "ß ÞÞÞß 8×. In other words, 1 − Y iff the vertical distance in # ‘% between the graphs of 10 and is3 at each of the finitely many points <3 .
Why is the product topology the “correct” topology for set ? Of course there is no “right” \α answer, but a few examples should make it seem a good choice.
Example 2.10
1) Our intuition can completely comprehend only finite objects; we always run risks when we apply it to infinite collections. For finite products, the product and box topologies are exactly the same, and our intuition gives us no solid reason to prefer one over the other in the infinite case.
2) (See Example II.2.6.6 and Exercise II.E10 ) Let L be the “Hilbert cube” ∞ Ò!ß""" Ó œ Ò!ß "Ó ‚ Ò!ß Ó ‚ ÞÞÞ ‚ Ò!ß Ó ‚ ÞÞÞ © j, where j has its usual metric, . . 8œ" 8#8## Suppose Bß C − L and let %% ! . What condition on C will guarantee that .ÐCß BÑ ?
243 ∞ " %%### Pick R so that 3## Þ If ÐB33 C Ñ #R for each 3 œ "ß ÞÞÞß R , then we have 3œR" R∞ # # "Î#%%## "Î# .ÐBß CÑ œ Ð ÐB33 C Ñ ÐB 33 C Ñ Ñ ÐR †#R # Ñ œ% Þ This is the natural 3œ"3œR" metric topology on the product LCB , and we see here that we can make “ close to ” 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 LÞ turns out to be the topology g. . In Ð!ß "Ñ we can see a similar phenomenon quite clearly: for two points B œ !ÞB"8 ÞÞÞB ÞÞÞ and C œ !ÞC" ÞÞÞC 8ÞÞÞ 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 of cubes Ò!ß "Ó7 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 ÐB88 Ñ Ä B iff for all αα , the coordinate of B converges (in \α Ñ to the α coordinate of B . The product topology is sometimes called the “topology of coordinatewise convergence.”
Theorem 2.11 Suppose is a sequence in . Then iff ÐB88 Ñ \ œ Ö\α Àα − E× ÐB Ñ Ä B − \ (11αααÐB8 Ñ ) Ä ÐBÑ in \ for all α − E .
Proof If ÐB88 Ñ Ä B , then (11αα ÐB Ñ ) Ä ÐBÑ because each 1 α is continuous. Conversely, suppose (11αααÐB8 Ñ ) Ä ÐBÑ in \ for each α and consider any basic open let
Y œ Yαα"# ß Y ß ÞÞÞß Y α5 that contains B œ ÐB α Ñ . For each 3 œ "ß ÞÞÞß 5 , we have B α3 − Y α3 . Since
(11αα333ÐB883"5 Ñ ) Ä3 ÐBÑ , we have 1 αα ÐB Ñ − Y for 8 some R . If R œ max ÖR ß ÞÞÞß R × , then for
8 R we have 1αα33 ÐB88 Ñ − Y for every 3 œ "ß ÞÞÞß 5 . This means that B − Y for 8 R , so ÐB8 Ñ Ä B. ñ
In the proof, is the max of a finite set. If has the box topology, the basic open set R\ YœY α might involve infinitely many open sets YÁ\αα . For each such α , we could pick R α , just as in the proof. But the set of all RRα 's might not have a max so the proof would collapse. Can you create a specific example with the box topology where this happens?
‘ Example 2.12 Consider ‘‘‘œÖ\À<−×<< , where \œ . Each point 0 in the product is a ‘ function 0À‘‘ Ä Þ Suppose that Ð0Ñ8 is a sequence of points in ‘ . By Theorem 2.11, Ð0ÑÄ08 iff Ð08 Ð<ÑÑ Ä 0Ð<Ñ for each < − ‘ . With the product topology, convergence of a sequence of functions in ‘‘ is called (in analysis) pointwise convergence .
244 ‘ Question: if ‘ is given the box topology, is convergence of a sequence Ð08 Ñ simply 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∪G where F and G are disjoint indexing sets, then
Ö\α"# Àα"# −E׶ Ö\ À −Fׂ Ö\ À −G× “Proof” A point is a function . Define B− Ö\αα Àα −E× BÀEÄα−E \ by . 0À Ö\α"# Àα"# −E×Ä Ö\ À −Fׂ Ö\ À −G× 0ÐBÑœÐBlFßBlGÑ Clearly 0 is one-to-one and onto.
For convenience, let and so that ] œ Ö\"# À"# − F× ^ œ Ö\ À − G× and 1α"#‰0À Ö\Àαα −E×Ä] 1α ‰0À Ö\À −E×Ä^Þ
0 is a mapping into a product Y ‚^ , so 0 is continuous iff 11"# ‰0 and ‰0 are both continuous. But is also a map into a product so is 1α" ‰0ÀÖ\À−E×Ä]œα" Ö\À−F× ß 1" ‰0 continuous iff is continuous for all 11"α"‰‰0ÀÖ\À−E×Ä\" α " −FÞ But which is continuous. 11""α"‰‰0œÀÖ\À−E×Ä\" 1 α
The proof that 1# ‰0 is continuous is completely similar.
" " 0À]‚^ÄÖ\À−E×α α is given by Bœ0ÐCßDÑœC∪D (the union of two functions ) and " " 0‰0À]‚^Ä\−EœF∪GÞis continuous iff 1ααα is continuous for each " Suppose α11− F : then αα ‰ 0 ÐCß DÑ œ ÐBÑ , where B œ C ∪ DÞ Since α − Fß B α " œ Ð11ααα ‰"" ÑÐCß DÑß so 1 ‰ 0 œ 11 ‰ , which is continuous. The case where α − G is completely similar.
Therefore 0ñ is a homeomorphism.
The question of topological commutativity for products only makes sense when the index setE 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, \‚\œÖB−Ð\∪\Ñ"# "# ÀBÐ3Ñ−\ 3 for 3œ"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 \‚\12 and \‚\ 21 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 \7\7 and cardinal , 7 denotes the product of copies of the space \Ð\Ñ\ . Prove that 78 is homeomorphic to 78 . (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 α"−E\‚α" −F Ö:׶ α −E \ α
Proof is itself a one-point space }, so we only need to prove that "−FÖ:" × Ö: . The map is clearly a homeomorphism αα−E\‚Ö:׶αα −E \ 0ÐBß:ÑœB Þñ
Lemma 2.16 Suppose . For any , is homeomorphic to a subspace αα−E\Ágαα F©E −F \ ^ of that is, can be embedded in In fact, if all the 's are -spaces, αα−E\αα −F \ α −E \ αÞ" \ α X then is homeomorphic to a closed subspace of . αα−F\^\αα −E
Proof Pick a point . Then by Lemma 2.15, :œÐ:Ñ−ααα−E \
ααα−F\¶αα −F \‚ −EF Ö:ל^© α α −E \ α Now suppose all the 's are . If , then for some Since \XC−Ð\Ñ^αα" α−E # −EFßCÁ: ##Þ \X###### is a " space, there is an open set Y\ in that contains C but not : . Then C−Y and
. Y#αα ∩Ðαα−F \ ‚ −EF Ö:×Ñœg Therefore is closed in . ^\ñα−E α
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 FœÖα! × , Lemma 2.16 says that each factor \α! is homeomorphic to subspace of (a closed subspace if all the 's are ). α−E\\Xαα" 3) But Lemma 2.16 does not say that if all the \Xα 's are " , then every embedded copy of \\ in is closed: only that there exists a closed homeomorphic copy. ( It is very easy to show αα! α−E a copy of ‘‘ embedded in ## that is not closed in ‘ , for example ... ?)
Lemma 2.17 Suppose . Then is a space (or, space) iff every \œ Ö\α Àα −E×Ág \ X"# X factor \Xα is "# (or, X ).
246 Proof for Hausdorff Suppose all the \BÁC−\BÁCααα 's are Hausdorff. If , then !! for some
α!. Pick disjoint open sets YZ\ααα!!! and in containing BC αα !! and . Then Y and Z are disjoint (basic) open sets in \ that contain B and C . αα!! 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\ , α 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 \88 œ Ö!ß #× , then 8œ" \ ¶ G . To construct G we defined, for each sequence ÐB"# ß B ß ÞÞÞß B 8 ß ÞÞÞÑ − Ö!ß #× , a descending sequence of closed setsJB""# ª J B B ª ÞÞÞ ª J B "#8 B ÞÞÞB ª ÞÞÞ in Ò!ß "Ó whose intersection gave a unique ∞ point :−GÀ Ö:ל8œ" JB"# B ÞÞÞB 8 (see Section IV.10 ). For each 8 , we can write G as a union of 8 #GœÐG∩JÑ disjoint clopen sets: 8 B B ÞÞÞB . ÐB"8 ßÞÞÞßB Ñ−Ö!ß#× "# 8 Define ∞ by Clearly is one-to-one and 0 À G Ä8œ" \8"#8 0Ð:Ñ œ B œ ÐB ß B ß ÞÞÞß B ß ÞÞÞÑÞ 0 onto. To show that 0:−G is continuous at , 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 ∩ JB"# B ÞÞÞB 8 contains : (call it YЉ0ÑlYœB ) and 11888 . Thus, ‰0 is constant on a neighborhood of :G‰0 in , so 1 8 is continuous at : . By Lemma 2.17, ∞ is Hausdorff. Since is a continuous bijection of a compact space 8œ"\08 onto a Hausdorff space, 0ñ is a homeomorphism (why? ).
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.
Proof By Exercise 2.14 above, Giiii†ii!!!!!! ¶ ÐÖ!ß #× Ñ ¶ Ö!ß #× ¶ Ö!ß #× ¶ GÞ The case of G¶G8 for 8− is similar. ñ
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 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©\ßF©]Þ Prove that int\‚] (E‚FÑœ int \ 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 for all . Prove that in the product , E©\ααα −E \œ\ α cl( cl. EÑœαα E 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 and that . Prove that is dense in iff is dense in \œ \αααα Ág E ©\ E \ E α \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 ;−\αα . Prove that FœÖB−\ÀBœ; ααα for all but at most finitely many α×\ is dense in α . (Note: Suppose \œ‘‘ œ8− \88 where each \ œ Þ Suppose each ; 8 is chosen to be a rational;œ! say 8 . What does d) say about ‘ ?)
e) Let ααα−E . Prove that ] œÖB− \ À B œ: for all Á × is homeomorphic ! αααα! ! to \α! . Note: So any factor of a product has a “copy” of itself inside the product in a “natural” way. For example, in ‘8 , the set of points where all coordinates except the first are ! is homeomorphic to 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 \\©\−E is a Hausdorff space and that α for each α .. Show that is homeomorphic to a closed subspace of the product . ] œ Ö\α Àαα − E× Ö\α À − E×
248 E5. For each 8−g , suppose H888 is a countable dense set in Ð\ß ÑÞ
a) Prove that is dense in . Hœ H88 \ b) Prove that is separable. (Caution: may not be countable (when is countable? \8 H H . Hint: In a product “closeness depends on only finitely many coordinates.” )
E6. a) Suppose an3ß4 − Ö!ß #× for all 3ß 4 − . Prove that there exists a sequence Ð5 Ñ in such that, for each 3 , lima38, exists. 5Ä∞ 5 Hint: “picture” the +43ß4 in an infinite matrix. For each fixed , the “j-th column" of the matrix is a point in the Cantor set G œ Ö!ß #×i! , and G is a compact metric space.
∞ ∞ w b) Let ++8 be an absolutely convergent sequence of real numbers. A series 8 where each 8œ"8œ" ∞ w w +œ+8888 or +œ! is called a subseries of + . 8œ" ∞ Prove thatWœÖ=−‘‘ À= is the sum of a subseries of +×8 is closed in . 8œ" ∞ w Hint: “Absolute convergence” just guarantees that every subseries converges. Each subseries +8 8œ" can be associated in a natural way with a point B − Ö!ß "×ii!! Þ Consider the mapping 0 À Ö!ß "× Ä ‘ ∞ w given by 0ÐBÑ œ +−Þ8 ‘ Must 0 be a homeomorphism? 8œ"
c) Suppose KGK is an open dense subset of the Cantor set . Must Fr be countable?
Hint: Consider ÖÐ+ß ,Ñ − G ‚ G À , Á !×Þ
E7. Let have the cofinite topology.
a) Does the product 7 have the cofinite topology? Does the answer depend on 7 ?
b) Prove 7 is separable (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.
249 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 is a topological space. Give the topology for which collection of “closed \\αα boxes” closed in is a subbase for the closed sets. Is this topology the product Y œÖ Jαα ÀJ \ α × topology?
E9. Prove or disprove:
There exists a bijection 0 À \ œ Ö!ß "×ii!! Ä œ ] such that for all B − \ and for all 8 , the first 8Cœ0ÐBÑ7B coordinates of are determined by the first coordinates of .
Here, 78B depends on and . More formally, we are asking whether there exists a bijection 0 such that:
aB − \ a8 − b7 − such that changing B45 for any 4 7 does not change C 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 must have property \Tα the property T is countably productive if \TE must have property when is countable α finitely productive if \TE must have property when is finite α
For example, Lemma 2.17 shows that the XX"# and properties are productive.
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 , Let and define . For a hiœÖÐ+ß,ÑÀ Ò+ß,Ñ− h × E œ i moment, we can think of EE as a subspace of ‘ with its usual topology. Then is Lindelof¨ (why? ), and ii is a covering of E by usual open sets, so there is a countable subfamily w with w . i œE Replace the left endpoints of the intervals in ihww to get œÖÒ+ß,ÑÀÐ+ß,Ñ− ihh w ש . If w covers we are done, so suppose ww . For each , pick a set in \\ÁgB−\Ò+ß,Ñhh h that contains B . In fact, B must be the left endpoint of Ò+ß ,Ñ for if B − Ò+ß ,Ñ − h and , then ww So, for each w we can pick a set [, ) . BÁ+ B−ii œ © h Þ B h B,B − h If and are distinct points not in w,, then ) and , ) must be disjoint ( why? ). But B Ch ÒB,,BC ÒC there can be at most countably many disjoint intervals So ww, ) } is ÒBß ,B ÑÞhh ∪ ÖÒB,B À B  a countable subcollection of h that covers W .
However the Sorgenfrey plane W‚W is 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.)
251
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 . Let be any one of the properties “first \œ Ö\α Àα −E×Ág T countable,” “second countable,” “metrizable,” or “completely metrizable.” Then \T has property iff
1) all the \Tα 's have property , 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 \T\ has property iff is a countable product of spaces with property T .
2) A nonempty X\" spaceαα has the trivial topology iff l\lœ"Þ So, if we deal only with X\T\" spaces (as is most often the case) the theorem says that has property 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 Fœ the 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 ÖYα À8− × be a countable open neighborhood base at Bαα −\ Þ Let
8 UαB œÖYÀY is a finite intersection of sets of the form Yα ß −Fß8− ×
Since FB is countable, UB is a countable collection of open sets containing and we claim that
UB is a neighborhood base at B −\. To see this, suppose Z œ Zαα" ß ÞÞÞß Z5 is a basic open set containing B . (We may assume that all α3 's are in FÀ why? ). For each 3 œ "ß ÞÞÞß 5ß 88 8885 pick Y33 so that B − Y © Z Þ Then Y œ Y "# ß Y ß ÞÞÞß Yα −U and B − Y © Z Þ αα33αα33 αα"# k B
252 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 FB−\ is uncountable, we find a point at which there cannot be a countable neighborhood base. Pick a point :œÐ:Ñ−\α
For each " −Fß pick an open set S"" ©\ for which gÁS "" Á\ . Choose B"" −S and C "" ÂS Þ Define BœÐBÑ−\ α using the coordinates
Bœ−F" if α" Bœα :−\ααif α ÂF
Suppose UUB is a countable collection of neighborhoods of BR− . For each B choose a basic
open setY with B − Y œ Yαα" ß ÞÞÞß Y5 © R . There are only finitely many α3 's involved in the expression for the chosen YRF , and there are only countably many 's in UB . So, since is uncountable, we can pick a "α−F that is not one of the 3 's involved in the expression for any of the chosen sets Y . Then
i) S""" is an open set that contains B because B −S
ii) for all R−UB ßR©ÎS"α : to see this, consider the point AœÐAÑ which is the same as B except in the " coordinate:
BÁα if α" Aœα Cœ" if α"
For each R − FB , we chose Y œ Yαα" ß ÞÞÞß Y5 © R . Since B − Y and A has the same αα"8,..., coordinates as B , we have A−Y also. But A S"""" because A œC ÂS Þ Since Y©ÎS"" , certainly R©ÎSÞ So UB is not a neighborhood base at B .
We are now able to see immediately that if the product has any of the other properties , \œ \α 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 \\XÞ\ and all the subspaces αα are " . By Lemma 2 16, is homeomorphic to a closed \\ therefore complete subspace of . Therefore α is completely metrizable.
It remains to show that if Tœ “second countable,” “metrizable,” or “completely metrizable” and conditions 1) and 2) hold, then also has property . \œ \α T
253 Suppose Tœ“second countable .”
"# 8 For each αU in the countable set F , let α œ ÖSαα ß S ß ÞÞÞß S α ß ÞÞÞ } be a countable base for the open sets in \α and let
8 UαœÖSÀS is a finite intersection of sets of form Sα , where −F and 8− }
UU is countable and we claim is a base for the product topology on \ .
Suppose B − Z œ Zαα" ß ÞÞÞß Z5 , a basic open set in \ . For each 3 œ "ß ÞÞÞß 5ß so we can choose a basic open set in such that 83 . Then B−Zαα33 \ α333 B−S©Z αα3 α 88"# 855 and 88 "# 8 . Therefore can be B − Sαα"# ß S ß ÞÞÞß S α55 © Z S αα "# ß S ß ÞÞÞß S α −U Z written as a union of sets from UU , so is a base for \ .
SupposeTœ “metrizableÞ ”
Since all the \Xα 's are " , 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 \ 8 ß ÞÞÞ is metrizable, then ∞ is metrizable. \œ8œ" \8
Let .\ß88 be a metric for where without loss of generality, we can assume each .Ÿ"8 (why? ). For points BœÐBÑ88 , CœÐCÑ−\ , define ∞ .ÐBßCÑ888 .ÐBß CÑ œ #8 8œ" Then .\ is a metric on (check ), and we claim that gg. is the product topology . Because \ is a countable product of first countable spaces, gg is first countable, so can be described using sequences: so it is sufficient to show that ÐD55. Ñ Ä D in Ð\ßgg Ñ iff ÐD Ñ Ä D in Ð\ß ÑÞ But ÐD55 Ñ Ä D in Ð\ßg Ñ iff the ÐD Ñ converges “coordinatewise.” Therefore it is sufficient to show that:
(D5. Ñ Ä D in Ð\ßg Ñ iff a8 ÐD 5 Ð8ÑÑ Ä DÐ8Ñ in Ð\ 88 ß . Ñ , that is, a8 .85 ÐD Ð8Ñß DÐ8ÑÑ Ä !
i) Suppose .ÐD5! ß DÑ Ä ! . Let % ! and consider a particular 8 . We can choose O ∞ so that implies .ÐDÐ8ÑßDÐ8ÑÑ85 % . Therefore, if 5 O .ÐDßDÑœ5 #8 #8! 5 O 8œ" .ÐDÐ8ÑßDÐ8ÑÑ certainly 85!! ! % , and therefore so . #8! #8! .85!! ÐD Ð8 Ñß DÐ8 ! ÑÑ %
So .ÐDÐ8ÑßDÐ8ÑÑÄ!85!! ! .
ii) On the other hand, suppose .85 ÐD Ð8Ñß DÐ8ÑÑ Ä ! for every 8 . Let % ! . Choose ∞ " % R so that ##8 and then choose O5 O so that if 8œR" " % ."5 ÐD Ð"Ñß DÐ"ÑÑÎ# #R # % .#5 ÐD Ð#Ñß DÐ#ÑÑÎ# #R ÞÞÞ R % .R5 ÐD ÐRÑß DÐRÑÑÎ# #R . ∞ .ÐDÐ8ÑßDÐ8ÑÑ85 Then for 5 O we have .ÐDßDÑœ5 #8 8œ"
254 R∞ .ÐDÐ8ÑßDÐ8ÑÑ85 .ÐDÐ8ÑßDÐ8ÑÑ 85 % % œ###R88 R†œÞ# % 8œ"8œR" Therefore .ÐD5 ß DÑ Ä !.
Suppose Tœ“completely metrizableÞ ”
Just as for “metrizable,” we can assume ∞ and that is a complete metric Tœ \œ8œ" \88 . on \.Ÿ"88 with . Using these . 8 's, we define . in the same way. Then g . is the product topology on \Ð\ß.Ñ . We only need to show that is complete.
Suppose ÐD5 Ñ is a Cauchy sequence in Ð\ß .Ñ . From the definition of . it is easy to see that ÐD588588 Ð8ÑÑ is Cauchy in Ð\ ß . Ñ for each 8 , so that ÐD Ð8ÑÑ Ä some point + − \ Þ Let +œÐ+Ñ−\8585 . Since ÐDÐ8ÑÑÄ+ for each 8 , we have ÐDÑÄ+ in the product topology œÐ\ß.Ññg.. Therefore is complete.
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 αα"# , , ... , α 8− E , there are disjoint sets E " ß E # ß ÞÞÞß E 8 − e such that α33−E for each 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‘e9 ÀEÄ . Let œÖ" ÒÐ+ß,ÑÓÀ+ß,− ×Þ For distinct αα"# , ,..., α 8−EÐÑ , 9α " , 9α ÐÑ # , ..., 9α ÐÑ 8 are distinct real numbers; we can choose +,−33,9α so that Ð 3 Ñ−Ð+ß,Ñ 33 and so that the intervals Ð+ß,Ñ 33 are pairwise disjoint. Then the sets " Eœ3339e ÒÐ+ß,ÑÓ− are the ones we need. ñ
Theorem 3.5
1) Suppose . If is separable, then each is separable. \œα−E \αα Ág \ \ 2) If each is separable and , then is separable. \lElŸ-\œ\ααα−E 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
255 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# is an infinite cardinal, then a product of 7 topological spaces, each of which has a dense set of cardinality Ÿ7ß also has a dense set with Ÿ7 points. 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\ßÒHÓ\ be a countable dense set in . For each α1 α is countable and dense in α (because \αα œ11 Ò\Óœ α Ò cl HÓ© cl 1 α ÒHÓ ). Therefore each \ α is separable.
12 8 2) Choose a family eα as in Lemma 3.4 and for each , let Hα œ ÖBαα , B ß ÞÞÞß B α ß ÞÞÞ × be a countable dense set in \α . Define a countable set f by
feœ ÖÐE"8"8 ß ÞÞÞß E ß 6 ß ÞÞÞß 6 Ñ À 8 − ß 6 3 − , E 3 − with the E 3 's pairwise disjoint ×Þ
Pick a point :αα − \ for each αf . For each #8 -tuple = œ ÐE"8"8 ß ÞÞÞß E ß 6 ß ÞÞÞß 6 Ñ − , define a point B−\= with coordinates
256 B−E63 if α BÐα Ñœ α 3 = if :ÂEα α 3
Let HœÖB= À=−f × . H is countable and we claim that H is dense in \ . To see this, consider any nonempty basic open set Y œ Yαα" ß ÞÞÞß Y5 À we will show that Y ∩ H Á gÞ
For Y œ Yαα" ß ÞÞÞß Y5 Á g,
i) Choose disjoint sets E" ß ÞÞÞß E5 in eα so that "" − E ß ÞÞÞß α 55 − E
ii) For each is dense in and we can pick a point 83 in 3 œ "ß ÞÞÞß 5ß Hαα33 \ Bα3 H αα 33 ∩ Y "#, 8 œ ÖBαα33 B ß ÞÞÞß B α 3 ß ÞÞÞ × ∩ Yα3 Þ
Let Because , we have 83 . = œ ÐE"35"35 ß ÞÞÞß E ß ÞÞÞß E ß 8 ß ÞÞÞ8 ß ÞÞÞß 8 Ñ −fα Þ 33 − E B= Ð α 3 Ñ œ Bα3 − Yα3 Therefore B−Y∩Hñ= .
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 ‘ , where each Choose ‘‘Ðœ<−‘ \<< \ œ ÑÞ e to be the collection of all open intervals Ð+ß ,Ñ with rational endpoints, and make a list these intervals "# 8 as E"8 ß ÞÞÞß E ß ÞÞÞ . In each \ < ß choose H < œ œ Ö; ß ; ß ÞÞÞ; ß ÞÞÞ×Þ (Since all the H< 's are identical, we can omit the subscript “<8 ” on the points; but just to stay consistent with the notation i the proof, we still index the ; 's using superscripts .) For each <:œ!−\Þ , (arbitrarily) pick <<
One example of a 6-tuple in the collection f is =œÐEßE'#& ßE ß#ß(ß%Ñ , where E " ß E #$ ßE are disjoint ‘ open intervals with rational endpoints. The corresponding point B−==‘‘‘ is the function BÀ Ä with
# ;<−Efor ' ( ;<−Efor # BÐ<Ñœ= ;<−E% for & ! for <ÂE ∪E ∪E '&#
The dense set HB! consists of all step functions (such as = ) that are outside a finite union
E∪E∪ÞÞÞ∪E88"# 85 of disjoint open intervals with rational endpoints and which have a constant rational value on each E3 .
Caution: In Example 2.12 we saw that the product topology on ‘‘ is the topology of pointwise ‘‘ convergence that is, Ð088 Ñ Ä 0 in ‘‘‘ iff Ð0 Ð<ÑÑ Ä 0Ð<Ñ for each < − . But is not first countable (why? ) so we cannot say that sequences are sufficient to describe the topology. In particular, if 0−‘‘ ß then 0− cl H but we cannot say that there must be sequence of step functions from H that converges pointwise to 0Þ
Since ‘‘‘‘ is not first countable, is an example of a separable space that is neither second countable nor metrizable.
In contrast to the properties discussed in Theorem 3.3, an arbitrarily large product of nontrivial separable spaces can sometimes turn out to be separable, as the next example shows. However,
257 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 \αααα be a set with l\l"Þ Choose : −\ and let ggαααœ ÖS © \ À : − S× ∪ Ög×Þ The singleton set Ö: α × is dense in Ð\ ααα ß Ñ , so \ is separable. If , then singleton set is dense (why? look at a nonempty basic :œÐ:Ñ−\œααα−E \ Ö:× open set Y .) So \E is separable, no matter how large the index set is. The \Xα 's in this example are not " . But Exercise E7 shows that an arbitrarily large product of separable X" spaces can turn out to be separable.
Theorem 3.8 Suppose where each is a space with more than one point. \œα−E \αα Ág \ X# If \lElŸ-Þ is separable, then
Proof For each α , we can pick a pair of disjoint, nonempty open sets YZ\ÞHααα and in Let be a countable dense set in \HœY∩H and let αα for each αα" . If Á−Eß there is a point :− Yα" ßZ ∩H because H is dense. Then :−H α but :ÂH " œ Y " ∩H since BÂYÀ"" therefore HÁH α" . Therefore the map 9c ÀEÄÐHÑ given by 9α ÐÑœHα is one-to- one, so lElŸlc ÐHÑlœ#i! 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.
Theorem 3.9 Suppose . is connected iff each is connected. \œα−E \αα Ág \ \
Proof \À\Ä\\\ is nonempty so each map 1αα 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 \Þααα is connected For each α , pick a point :−\ . For each finite set , let . is homeomorphic to the finite product , J©E \JJ œαα−J \αα ‚ −EJ Ö:× \ α −J \ α so each is connected. Let is a finite subset of . Each contains the point \HœÖ\ÀJE×\JJ J :œÐ:Ñα , so Corollary V.2.10 tells us that H is connected. Unfortunately, HÁ\ (except in trivial cases; why?).
But we claim that H is dense in \ . If Y œ Yαα"8 ß ÞÞÞß Y is any nonempty basic open set in \ , we need to show that H ∩ Y Á g . Choose a point Bαα33 − Y for each 3 œ "ß ÞÞÞß 8 , and define B − \ having coordinates Bœif αα BÐα Ñ œ α3 3 :Áα if αααα"#, ,..., 8
If we let J œ Ö+"# ßαα ß ÞÞÞß 8 × , then B − \ J ∩ Y © H ∩ Y , so H ∩ Y Á gÞ
Therefore \œ cl H is 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 . Then is compact iff \œα−E \α Ág \ each \α is compact.
The proof “Ê ” in Tychonoff's Theorem is quite easy (why?). And the easy theorem that a finite product of compact spaces is compact was in Exercise IV.E.26. .
259 Exercises
E10. Let \œÒ!ß"ÓÒ!ß"Ó have the product topology.
a) Prove that the set of all functions in \ with finite range (sometimes called step functions ) is dense in \ .
b) By Theorem 3.5, \ is separable. Describe a countable set of step functions which is dense in \.
c) Let E œ ÖB − \ À 0 is the characteristic function of a singleton set Ö<× ×Þ Prove that EE , with the subspace topology, is discrete and not separable. Is closed?
d) Prove that ED\RD has exactly one limit point, , in and that if is a neighborhood of , then ER is finite.
E11. “Boxes” of the form , where is open in , are a base for the box topology ÖYααα Àα − E× Y \ on { . Throughout this problem, we assume that products have the box α−E\Þαα Ö\Àα −E× topology rather than the usual product topology.
a) Show that the “diagonal map” 0À‘‘ Äi! given by 0ÐBÑ œ ÐBß Bß Bß ÞÞÞÑ is not continuous, but that its composition with each projection map is continuous.
b) Show that Ò!ß"Ói! is not compact.
Hint: let E!" œÒ!ß"Ñ and E œÐ!ß"Ó . Consider the collection h of all sets of the form i! E%%"# ‚ E ‚... ‚ E % 8 ‚ ÞÞÞ , where Ð%%"# ß ß ÞÞÞß % 8 ß ÞÞÞÑ − Ö!ß "× .
Note: [ ith the product topology, in contrast,Ò!ß"Ó7 is compact for any cardinal 7 by the Tychonoff Product Theorem (3.10) which we will prove later.
c) Show that ‘‘i! is not connected by showing that the set EœÖB−i! ÀB is an unbounded sequence in ‘× is clopen.
d) Suppose Ð\ß .Ñ and Ð\αα ß . Ñ Ðα − EÑ are metric spaces. Prove that a function (with the box topology) is continuous iff each coordinate function is 0À\Ä \α 0αα œ1 ‰0 continuous and each B−\ has a neighborhood on which all but a finite number of the 0α 's are constant.
E12. State and prove a theorem that gives a necessary and sufficient condition for a product of spaces to be path connected.
E13. Prove the following more general version of Theorem 3.8:
Suppose , and that, for each , there exist disjoint nonempty open sets \œα−E \α Ágα −E YZ\\ααα and in . If is separable, then lElŸ-Þ
260 4. Embedding Spaces in Products
It is often possible to embed a space in a product Such embeddings will give us some nice \\Þ α 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 0ÀÒ!ß"ÓÄ"#‘‘ and 0ÀÒ!ß"ÓÄ #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‘11# defined by the family Ö cos # Bß sin # B× is not a homeomorphism between Ò!ß "Ó and ranÐ/Ñ © ‘# (why? what is ranÐ/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 \\−EÑ and ααα (αα are topological spaces and that 0À\Ä\ for each Þ The evaluation map defined by the family is the function given by Ö0αα Àα − E× / À \ Äα−E \
/ÐBÑÐα Ñ œ 0α ÐBÑ
Then is the point in the product whoseth coordinate function is :\. In more informal /ÐBÑ \ααα 0 ÐBÑ coordinate notation, /ÐBÑ œ Ð0α ÐBÑÑÞ
Exercise 4.2 Suppose \œ \Þ For each α1 there is a projection map À\Ä\ . What is α−E α αα the evaluation map defined by the family ÖÀ−E×1αα ?
261 Definition 4.3 Suppose \\−EÑ and ααα (α are topological spaces and that 0À\Ä\Þ We say that the family Ö0α Àαα − E× separates points if, for each pair of points B Á C − \ , there exists an − E for which 0ÐBÑÁ0ÐCÑαα .
Clearly, the evaluation map is one-to-one for all /À\Äα−E \α Í BÁC−\ß/ÐBÑÁ/ÐCÑ Ífor all BÁC−\ there is an αα −E for which /ÐBÑÐ Ñœ0αα ÐBÑÁ0ÐCÑœ/ÐCÑÐ α Ñ ÍÖ0À−E×the family α α separates points.
Theorem 4.4 Suppose \0À\ÄÐ\ßÑ has the weak topology generated by the maps αααg and that the family Ö0α Àα − E× separates points. Then / is an embedding that is, / a homeomorphism between and . \ /Ò\Ó ©\ α
Proof Since Ö0α × separates points, / is one-to-one and / is continuous because each composition 1αα‰/œ0 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 \Yœ0ÒZÓZ has the weak topology, a subbasic open set has the form α , where is open in " " \α. But then /ÒY Ó œ /Ò0αα ÒZ ÓÓ œ1 ÒZ Ó ∩ /Ò\Ó is an open set in /Ò\ÓÞ ñ (Note: might not be open in , but that is irrelevant. See the earlier example where /ÒY Ó \α /ÐBÑ œ ÐB#B 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"8 ß ÞÞÞß Y ß ÞÞÞ× for the open sets. For each 8 , let 0888 ÐBÑ œ .ÐBß \ Y ÑÞ Then 0 À \ Ä Ò!ß "Ó is continuous and, since \ Y 8 is closed, we have 088 ÐBÑ! iff B−Y Þ If BÁC−\ , there is an 8 such that B−Y 88 and CÂY Þ Then 0888 ÐCÑœ!Á0 ÐBÑ, so 0 's separate points.
We claim that the topology gg.A8 on \0 is actually just the weak topology generated by the 's. Because the functions 0\8..Aare continuous if has the topology ggg , we know that ªÞ To show gg.A©, suppose B−Y − g . Þ For some 8 , we have B−Y 8 ©Y and therefore 0 8 ÐBÑœ-!Þ But " - Z8 œ08 ÒÐß"ÓÓ# is a subbasic open set in the weak topology and B−Z88 ©Y ©YÞ Therefore Y−gA.
By Theorem 4.4, /À\ ÄÒ!ß"Óii! is an embedding, so \ ¶/Ò\Ó©Ò!ß"Ó! Þ We sometimes write this top as \ © Ò!ß "Óii! . 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 "Ói!
ii) separable metric spaces can get quite complicated, so the subspaces of a cube Ò!ß "Ói! are more complicated than we imagined.
∞ Since Ò!ß "Ói! ¶ Ò!ß" Ó œ L (the “Hilbert cube”) © j , we can also say that topologically 8œ" 8 # the separable metrizable spaces are precisely the subspaces of j# . This is particularly amusing because of the almost “Euclidean” metric .j on # : ∞ # " .ÐBß CÑ œ Ð ÐB88 C Ñ Ñ# 8œ" 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 Lj and # is a “universal separable metric space” (But they are not homeomorphic: Lj is compact but # is not.)
Example 4.6
Suppose Ð\ß .Ñ is a metric space, not necessarily separable, and that ÖYα Àα − E× is a base for the topology g. , where 7 œ lElÞ An argument completely analogous to the reasoning in Example 4.5 top (just replace “8©Ò!ß"Ó ” everywhere with “α ”) shows that \ 7 . Therefore topologically 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.
The weight AÐ\Ñ of a topological space Ð\ßgUUg Ñ is defined as min Öl l À is a base for × 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 finitei in which case the “! ” guarantees that AÐ\Ñ i! Ð convenient for purely technical reasons that don't matter in these notes).
The density $gÐ\Ñ is defined as min ÖlHl À H is dense in Ð\ß Ñ× i! Þ For a metrizable space, it is not hard to prove that AÐ\Ñ œ$ 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 \ © Ò!ß "ÓAÐ\Ñ œ Ò!ß "Ó$ Ð\Ñ (*)
Notice that, for a given space Ð\ß .Ñ , the exponents in this statement are not necessarily the smallest top possible. For example, (*) says that ‘ © Ò!ß "ÓAБ Ñ œ Ò!ß "Ó i! , 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 L77 with weight such that every metric space \7 with weight can be ii!! embedded in H77ÞL In other words, 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 .
263 Without going into all the details, you can think of L77 as a “star” with 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 0À\Ä\αα” is sometimes not so easy to check. The following definition and theorem can sometimes help.
Definition 4.7 Suppose \\−EÑ and ααα (αα are topological spaces and that, for each , 0À\Ä\Þ We say that the collection YαœÖ0α À −E×separates points from closed sets if whenever J is a closed set in \ and B  J , there is an α such that 0αα ÐBÑ Â cl 0 ÒJ ÓÞ
Example 4.8 Let \œ‘Y and œGÐ ‘ Ñ . Suppose J is a closed set in ‘ and <ÂJ . There is an open interval Ð+ß,Ñ for which <−Ð+ß,Ñ©‘‘ J . Define 0−GÐ Ñ with a graph like the one shown in the figure:
Then ! œ 0Ð<Ñ Â cl 0ÒJ ÓÞ Therefore GБ Ñ separates points and closed sets.
The same notation continues in the following lemma.
" Lemma 4.9 The family UαœÖ0α ÒZÓÀ −EßZ open in \α × is a base for the topology in \ iff
i) the 0 's are continuous, and α ii) Ö0α Àα − E× separates points and closed sets.
In particular, i)\\ ii) imply that U is a sub base for the topology on so that has the weak topology generated by the 0α 's.
264 Note: the more open (closed) sets there are in \Ö0À−E× , the harder it is for a given family α α to succeed in separating points and closed sets. In factß the lemma shows that a family of continuous functions Ö0α Àα − E× succeeds in separating points and closed sets only if g is the smallest topology that makes the 0α 's continuous.
" Proof Suppose UU is a base. Then the sets 0ÒZÓα in are open, so the 0α 's are continuous and i) holds. To prove ii), suppose J\BÂJZ\ is closed in and . For some α and some open in α , we " " have B−0ÒZÓ©\JÞα Then 0ÐBÑ−Zαα and Z∩0ÒJÓœgÐ since 0ÒZÓ∩JœgÑα . Therefore 0ÐBÑÂαα cl 0ÒJÓ so ii) also holds.
Conversely, suppose i) and ii) hold. If B−S and S is open in \ , we need to find a set " " 0ÒZÓ−ααUαsuch that B−0ÒZÓ©SÞ Since BÂJœ\S , condition ii) gives us an for which " " 0αα ÐBÑ Âcl 0 ÒJ Ó . Then 0 α ÐBÑ − Z œ \ αα cl 0 ÒJ ÓÞ Then B − 0αα ÒZ Ó and we claim 0 ÒZ Ó © S : " Suppose D − 0α ÒZ Ó so 0ααα ÐDÑ − Z œ \ cl 0 ÒJ ÓÞ But if D−J, then 0ÐDÑ−0ÒJÓœ0Òαα α cl JÓ© cl 0ÒJÓÞ α Therefore D−\JœSÞñ
Theorem 4.10 Suppose 0À\Ä\αα is continuous for each αα −E . If the collection Ö0À−E× α
i) separates points and closed sets, and
ii) separates points then the evaluation map is an embedding. /À\Ä \α
Proof Since the 0\α 's are continuous, Lemma 4.9 gives us that has the weak topology. Then Theorem 4.4 implies that /ñ is an embedding.
If the space \X we are trying to embed is a " -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 0À\Ä\αα is continuous for each α −E . If \ is a X" -space and separates points and closed sets, then evaluation map is an embedding. Ö0α Àα − E× / À \ Ä \α
Proof By Theorem 4.10, it is sufficient to show that the 0BÁC−\α 's separate points, so suppose . Since B Â the closed set ÖC× , there is an α for which 0αα ÐBÑ Â cl 0 ÒÖC×Ó . Therefore 0 αα ÐBÑ Á 0 ÐCÑß so Ö0α Àα − E× separates points. ñ
265 Exercises
E14. A space Ð\ßgaa Ñ is called a X!BC space if whenever B Á C − \ , then Á (equivalently, either there is an open set YBC containing but not , or vice-versa). Notice that the X! condition is weaker than XXX"!! (see example III.2.6.4). Clearly, a subspace of a -space is .
a) Prove that a nonempty product is iff each is . \œ Ö\Àααα −E× X!! \ 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 \X is ! iff \ is homeomorphic to a subspace of W for some cardinal 7 . (Nearly all interesting spaces are X! , so nearly all interesting spaces are (topologically) just subspaces of W77 for some cardinal .) HintÊY\ : for each open set in , let ;Y be the characteristic function of YÞ Use an embedding theorem.
E15. a) L et Ð\ß .Ñ be a metric space. Prove that GÐ\Ñ ( œ the collection of continuous real-valued functions on \Þ ) separates points from closed sets (Since \XGÐ\Ñ is " , therefore also separates points).
b) Suppose \X is any ! 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 is separable for each . Prove that : satisfies CCC. \αααα−E \œ Ö\ −E× (||)There isn't much to prove when AŸ c; why?
Hint: Let ÖÀ>−X× U> be any such collection. We can assume all the U> 's are basic open sets (why?). Prove that if W© T and ||WŸ c, then || WŸi! and hence X must be countable.)
E17. There are several ways to define dimension for topological spaces. One classical method is the following inductive definition.
Define dimgœ 1. For :−\ , we say that \ has dimension Ÿ 0 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, \Ÿ:: has dimension 0 at iff has a neighborhood base consisting of clopen sets. We say \Ÿ8:: has dimension at if has a neighborhood base consisting of sets with n 1 dimensional frontiers. We say \Ÿ has dimension n, and write dim Ð\ÑŸ\Ÿ8 n, if has dimension at p for each p−\Ÿ8Ð\ÑŸ X and that dim(X)œÐ\Ñœ∞ n if dim( ) but dimÎ8" . We say dim if dim Ð\Ñ Ÿ 8 is false for every 8 − Þ While this definition of dimÐ\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 “dimÐ\Ñ œ 8 ” is a topological property of \ . There is a theorem stating that dim(‘‘‘887 )œ8 , from which it follows that is not homeomorphic to if m Á n. In proving the theorem, showing dim(ÐÑŸ‘‘88 n is easy; the hard part is showing that dim ÐÑŸ Î n 1.
a) Prove that dim(‘ )œ" . b) Let Gœ!Þ be the Cantor set. Prove that dim(G ) 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 \]X and are ! -spaces. Part a) outlines a sufficient condition enabling ] to be top embedded in a product of \] 's, i.e., that ©\ 7 for some cardinal m. Parts b) and c) look at some corollaries.
7 a) Theorem Let \]X and be ! spaces. Then ] can be topologically embedded in \ for some cardinal 7JC if, for every closed set ©] and every point ÂJ , there exists a continuous function 0À]Ä\8 such that fy ( )Â cl fF [ ] (here, n is finite and depends on f ).
Proof Let XœÖ>œÐCßJÑÀJ is closed in ] and CÂJ× and, for each such pair 8 >œÐCß0Ñ, let 0>> be the function given in the hypothesis. Let \ œ\ (the space containing the values of ). Then clearly is homeomorphic to 7 for some 0Ö\À>−X×\>> , so it suffices to show can be embedded in . Let 7]Ö\À>−X× > be defined as follows: for , has for its -th coordinate 2À] Ä Ö\À>−X×> C−] 2Ð>Ñ > 0>> ÐCÑß i.e., 2ÐCÑÐ>Ñ œ 0 ÐCÑÞ
i) Show 2 is continuous.
ii) Show 2 is one-to-one.
iii) Show that 22Ò]Ó2 is a closed mapping onto its range to complete the proof that is homeomorphism between and . (Note: the converse of the +]2Ò]Ó©Ö\À>−X× > theorem is also true. Both the theorem and converse are due to S. Mrowka.)
b) Let F denote Sierpinski space {{0,1},{g ,{0},{0,1}}. Use the theorem to show that every T! space YFm can be embedded in 7 for some .
c) Let D denote the discrete space Ö× 0,1 . Use the theorem to show that every X" -space ] satisfying dimÐ] Ñ œ ! (see Exercise E14) can be embedded in Dm7 for some . Parts b) and c) are due to Alexandroff.
Mrowka also proved that there is no T"" -space \] such that every T -space can be embedded in X7 for some m.
E19Þ Let \ œ Ö!ß "× and consider the product space \7 , where 7 is an infinite cardinal.
267 a) Show that \77 contains a discrete subspace of cardinality . b) Show that AÐ\7 Ñ œ 7 Ðsee Example 4.6 ).
E20. A space \ElElŸ"Þ is called totally disconnected every connected subset satisfies Prove that a totally disconnected compact Hausdorff space is homeomorphic to a closed subspace of Ö!ß "×7 for 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!Ñ in the space P , Example III.9.8. ∞ Let ii! for and ! . Prove that no point in the EœÖB−\43 ÀBœ+ 34× Eœ4œ! E©\ 4 (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 −E we have a map 1αα À\ Ä] , where \ α is a topological space and ] is a set. Certainly there is a topology for ]1 that will make all the α 's continuous: for example, the trivial topology on ]] . But what is the largest topology on that will do this? Let
" gαœÖS©] À for all −E , 1α ÒSÓ is open in \α ×
" It is easy to check that gα is a topology on ]1ÒSÓ\ . For each , by definition, each set α is open in α 1so ggα makes all the α 's continuous. Moreover, if F©]FÂ and , then for at least one , " 1ÒFÓα is not open in \α so adding F to g would “destroy” the continuity of at least one 1Þα Therefore g is the largest possible topology on ]1 making all the α 's continuous.
Definition 5.1 Suppose 1À\Ä]αα for each αg −E . The strong topology on ] generated by the " maps 1]1ßS−1ÒSÓαα is the largest topology on making all the maps continuous and g iff α is open in \α for every α.
The strong topology generated by a collection of maps Ö1α Àα − E× is “dual” to the weak topology in 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‰1α is continuous; but a map 0 into a space with the weak topology generated by mappings 00‰0ααis continuous iff all the compositions are continuous (see Theorem 2.6)
Theorem 5.2 Suppose ]Ö1À−E×Þ has the strong topology generated by a collection of maps α α If ^ is a topological space and 0À] Ä^ , then 0 is continuous iff 0‰1αα À\ Ä^ is continuous for each α −E.
10α Proof For each α −E , we have \αα Ä] Ä^ , and the 1 's are continuous since ] has the strong topology.
If 00‰1Þ is continuous, so is each composition α
" Suppose each 0‰1α is continuous and that Y is 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 \αα Þ But 1α Ò0 ÒYÓÓœ Ð0‰1 Ñ ÒYÓ which is open because 0 ‰1 α 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 1œ1α and 1 is onto.
Definition 5.3 Suppose Ð\ßg Ñ is a topological space and 1 À \ Ä ] is onto . The strong topology on ]1 generated by is also called the quotient topology on ]] . If has the quotient topology from 1 , we
269 say that 1À\Ä] is a quotient mapping and we say ] is a quotient of \ . We also say the ] is a quotient space of \]œ\Î1 and sometimes as .
From our earlier discussion we know that if \1À\Ä] is a topological space and , then the quotient topology on ] is gg œ ÖY À 1" ÒY Ó is open in \× . Thus Y − 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 ]1 making 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 #11 Bß sin # BÑ accomplishes. It turns out (see below ) that the usual topology on W11" is the same as quotient topology generated by the map . Therefore we can say that 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 1 to “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 11 (see below ). Therefore we can say that is a quotient mapping and the the “figure-eight” is a quotient of W" .
Suppose we are given an onto map 1ÀÐ\ßgg ÑÄÐ]ßw Ñ . How can we tell whether 1 is a quotient map that is, how can we tell whether g w is the quotient topology? By definition, we must check that Y−ggw" iff 1 ÒYÓ− Þ Sometimes it is fairly straightforward to do this. But the following theorem can sometimes make things much easier.
Theorem 5.4 Suppose 1ÀÐ\ßgg ÑÄÐ]ßw Ñ is continuous and onto. If 1 is open (or closed), then 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 ggww , but if makes 1]1 continuous, then so would any smaller topology on . The theorem tells us that if is both continuous and open (or closed), then ggww is completely determined by 1 : is the largest topology making 1 continuous.
Proof Suppose Y © ] Þ W / must show Y −ggggw" iff 1 ÒY Ó − Þ If Y − w , then 1 " ÒY Ó − because 1 is continuous. On the other hand, suppose 1" ÒY Ó −g . Since 1 is onto, 1Ò1 " ÒY ÓÓ œ Y and so, if 1 is open, Y œ 1Ò1" ÒY ÓÓ −g w 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 1ÒJÓ" is closed in \Þ With that observation, the proof that a continuous, closed, onto map 11 is a quotient map is exactly parallel to the proof when is open. If \]X11 is compact and is # , then must be closed, so is a quotient mapping. ñ
Note: Theorem 5.4 implies that the map 1ÐBÑ œ Ð cos #11 Bß sin # BÑ from Ò!ß "Ó to W" is a quotient map, but 11 is not open. The same formula can be used to define a quotient map 1ÀÄW‘ " which is not closed (why?). Exercise E25 gives examples of quotient maps 1 that are neither
270 open nor closed.
Suppose µ\B−\B is an equivalence relation on a space . For each , the equivalence class of 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 relationBµDBD namely, iff and 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™™ Ä Îµ by 1Ð+ÑœÒ+Ó and give ™ ε the quotient topology. A set Y is open in ™™Îµ iff 1" ÒYÓ is open in , and that is true for every Y © ™™ ε because is discrete. Therefore the quotient ™Îµ 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 topology Þ Points at distance !\ in have been “identified with each other” to become one point (the equivalence class) in ]Þ For ÒBÓß ÒDÓ − ] , define .ww 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:
If ÒBÓœÒBÓww and ÒDÓœÒDÓ , then .ÐBßBÑœ! w and .ÐDßD w Ñœ!Þ Therefore .ÐBß DÑ Ÿ .ÐBß Bwwww Ñ .ÐB ß D Ñ .ÐD ß DÑ œ .ÐB ww ß D Ñ , and similarly .ÐBww ß D Ñ Ÿ .ÐBß DÑ . Thus .ÐB ww ß D Ñ œ .ÐBß DÑ so . w ÐÒB w Óß ÒD w ÓÑ œ . w ÐÒBÓß ÒDÓÑÞ
It is easy to check that .]..ÐÒBÓßÒDÓÑœ!www is a pseudometric on . In fact, is a metric : if , 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 ggg..ww. We claim that œÞ To see this, first notice that
.ww . ÒCÓ − F%% ÐÒBÓÑ iff . ÐÒCÓß ÒBÓÑ %% iff .ÐCß BÑ iff C − F ÐBÑ
" .ww . . . Therefore 1 ÒF%% ÐÒBÓÑÓ œ F ÐBÑ and 1 ÒF %% ÐBÑÓ œ F ÐÒBÓÑ . But then we have
w Y−g. w iff Y is a union of . -balls iff 1ÒYÓ" is a union of .-balls iff 1ÒYÓ" is open in \ iff Y−g Þ
271 The metric space Ð] ß .w 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 11"". would be a homeomorphism if only 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 #11 Bß sin # BÞ ) 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 W1" has its usual topology and is continuous. However, since \W is compact and " is Hausdorff, Theorem 5.4 gives that the usual topology on W1" is the quotient topology and is 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 WW"" but then found that the quotient topology on the set 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 1 is (clearly) continuous and onto. Again, 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"Ñ and 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 Mo¨¨bius 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 ‘%>2 using color as a 4 dimension. To each point ÐBßCßDÑ on the “bottle” pictured above, add a 4th coordinate to get ÐBßCßDß<ÑÞ Now color the points on the bottle in varying shades of red and let < be a number measuring the “intensity of red coloration at a point.” Do the coloring in such a way that the surface “blushes” as it intersects itself so that the points of “intersection” seen above in ‘$ will be different (in their 4th coordinates).
Alternately, you can think of the Klein bottle as a parametrized surface traced out by a moving point T œÐBßCßDß>Ñ where BßCßD depend on time > and > is recorded as a 4th -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, TµT for each T−W" . Convince yourself that the quotient Wε"" is W .
2) Let HÖT−ÀlTlŸ"×## be the unit disk ‘ . Identify antipodal points on the boundary of HTµTT−W©HÞHε#"#: that is, if The quotient 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"Ó and define ÐBß"ѵÐCß"Ñ and ÐBß"ѵÐCß"Ñ for 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 αg−E , let Ð\ßαα Ñ be a topological space, and assume that the sets \ α are pairwise disjoint. The topological sum (or “free sum”) of the 's is the space where \Ð\ßÑααα−E g is open in for every }. We denote the topological sum by gαœÖS©α−E \αα ÀS∩\ \ α −E . In the case , we use the simpler notation \lElœ#\\Þα "# α−E
In , each is a clopen subspace. Any set open (or closed) in is open (or closed) in the sum. \\αα \ α α−E The topological sum can be pictured as a union of the disjoint pieces , all “far apart” from \\αα α−E 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 EE"# is the same as E∪E "# with subspace topology Let F"Ð!ß!ÑF" be an open disk with radius centered at and let # 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 α−E
1) If the 's are nonempty and separable, then is separable iff . \\lElŸiαα ! α−E 2) If the 's are nonempty and second countable, then is second countable iff . \\lElŸiαα ! α−E 3) A function is continuous iff each is continuous. 0À \αα Ä^ 0l\ α−E
277 4) If 1 is a metric for , then the topology on is the same as where .Ÿαα \ \ αg. α−E . ÐBß CÑif Bß C − \ .ÐBß CÑ œ αα " otherwise so is metrizable if all the 's are metrizable. You should be able to find similar statements for \\αα α−E 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 0ÀEÄ]E©\ , where . In the sum \] , define BµC iff Cœ0ÐBÑÞ If 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 \F and 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 EE"" and clockwise onto the points of FF "# and respectively.. # Then W]0 is a “sphere with a handle .”
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 WQ# . Let be a Mobius¨ strip and let F be the curve that bounds it. Of course, F¶W" . Let 0ÀEÄF be 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 \B−\Y is a 2-manifold if each has an open neighborhood that is homeomorphic to ‘# . Thus, a 2-manifold looks “locally” just like the Euclidean plane. A surface is a Hausdorff 2-manifold.
Theorem 5.19 Let \\W be a compact, connected surface. The is homeomorphic to a sphere # or to W# with a finite number of handles and crosscaps attached.
You can read more about this theorem and its proof in Algebraic Topology: An Introduction (William Massey).
278 Exercises
# E22. a) Letµ 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 \]Ð\εт] and each equivalence class is homeomorphic to the same space , then 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 (.\εт¶‚¶\‘‘‘ 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 (BßCѵÐBßCÑ"" ## iff B " C"# œB # C . Prove that ‘# /µ is homeomorphic to some familiar space.
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 1 is a quotient map that is neither open nor closed.
b) Let ‘g# have the topology 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 0ÐBßCÑ œ 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 E and F of‘# , EF¶E∪F iff ... ”
E27. Let ‘g# have the topology 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 =−W>−X homeomorphic to Ð\=> ‚ ] ÑÞ =−Wß>−X
E31. This problem outlines a proof (due to Ira Rosenholtz) that every nonempty compact metric space \G\ is a continuous image of the Cantor set . From Example 4.5, we know that is homeomorphic to a subspace of Ò!ß "Ói! Þ
∞ +4 a) Prove that the Cantor set G©‘ consists of all reals of the form $4 where 4œ! each +œ!4 or 2.
∞∞ ++44 b) Prove that Ò!ß "Ó is a continuous image of G . Hint: Define 1Ð $#44 Ñ œ Þ 4œ! 4œ!
c) Prove that the cube Ò!ß "Ói! is a continuous image of G . Hint: By Corollary 2.21, G ¶ Ö!ß #×ii!! ¶ G. Use 1 from part b) to define 0 À G i ! Ä Ò!ß "Ói! by 0ÐB"# ß B ß ÞÞÞß ÞÞÞÑ œ Ð1ÐB " Ñß 1ÐB # Ñß ÞÞÞß ÞÞÞÑ
d) Prove that a closed set O©G is a continuous image of G . (Hint: GG is homeomorphic to the “middle two-thirds" set w consisting of all reals of the form ∞ ,4 wwwwwBC '#4 . G has the property that if BßC−G , then ÂG Þ If O is closed in G , we can map 4œ! GÄOww by sending each point B to the point in O nearest to B .)
e) Prove that every nonempty compact metric space \G is a continuous image of .
.
280 Chapter VI Review
Explain why each statement is true, or provide a counterexample.
1. Ð!ß "Ñii!! is open in Ò!ß "Ó Þ
2. Suppose J is a closed set in Ò!ß "Ó ‚‘1 . Then # ÒJ Ó is a closed set in ‘ Þ
3. i! is discrete
4. If GG is the Cantor set, then there is a complete metric on i! which produces the product topology.
‘‘ 5. Let ‘‘ have the box topology. A sequence Ð088 Ñ Ä 0 − iff Ð0 Ñ Ä 0 uniformly.
8 Ò!ß"Ó 6. Let 0888 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 1ÐBÑœB for all B− . Give an example of a sequence Ð0Ñ8 of distinct functions in ‘‘ that converges to 1 .
8. Let GG be the Cantor set. Then is homeomorphic to the topological sum GG .
# 9. The projection maps 11BC and from ‘‘Ä separate points from closed sets.
10. The letter NM is a quotient of the letter .
11. Suppose µ\B−\ßÒBÓ is an equivalence relation on and that for represents its equivalence class. If B\ÒBÓ is a cut point of , then is a cut point of the quotient space \ε .
12. If 1À \ Ä ] is a quotient map and ] is compact T## , then ] is compact T .
13. Suppose E\−E is infinite and, that in each space α (α ) there is a nonempty proper open subset . Then is not a basic open set of the product , and, in fact, cannot even be SSαα \œ\S α α open in the product.
14. If ∞ , where the 's are disjoint clopen sets in , then ∞ ( the \œ8œ" E88 E \ \z8œ" E 8 œ topological sum of the E8 's).
∞∞ 15. Let \88 œ Ö!ß "× and ] œ with their usual topologies. Then \ 8 is homeomorphic to ] 8 Þ 8œ" 8œ"
∞ 16. Let \88 œ Ö!ß "× and ] œ with their usual topologies. Then 8œ" \ 8 is homeomorphic to ∞ 8œ"]Þ8 17. Let E œ ÖÐBß Cß DÑ −‘‘‘$# À B C #CD D # l sin ÐBCDÑl × , and let 0À $ Ä be given by 0ÐBßCßDÑ œ B#. Then 0 ÒEÓ is open but not closed in ‘ .
281 18. Suppose is a connected subset of and that ∞∞ is dense in Then each E\ÐÁgÑE\Þ88888œ" 8œ" \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 .
" 55 21. Let EœÖ8 À8−‘ ש , and let 7 be any cardinal number. E is a closed set in ‘ Þ
22. If G is the Cantor set, then G‚G‚G‚G is homeomorphic to G‚G .
33. W‚W"" is homeomorphic to the “infinity symbol”: ∞
34. 31. Let T be the set of all real polynomials in one variable, with domain restricted to Ò!ß "Ó , for which ranÐT Ñ © Ò!ß "ÓÞ Then T is dense in Ò!ß "ÓÒ!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 { } be a nonempty product space. Each is a quotient of the product space \Àααα −E \ Ö\α Àα − E× 39. Suppose \\ does not have the trivial topology. Then #- cannot be separable.
40. Every countable space \ is a quotient of
41. ‚i is homeomorphic to the sum of ! disjoint copies of .
42. Suppose int , where . Then for every , int . BœÐBÑ−αααα E E© \α1 B − ÒEÓ 43. The unit circle, " , is homeomorphic to a product , where each W\\©Ò!ß"Óα−E αα (i.e., W" can be “factored” into a product of subspaces of Ò!ß "Ó ).
44.‘i! is homeomorphic to .
282