Chapter VIII
Ordered Sets, Ordinals and Transfinite Methods
1. Introduction
In this chapter, we will look at certain kinds of ordered sets. If a set \ is ordered in a reasonable way,
then there is a natural way to define an “order topology” on \ . Most interesting (for our purposes)
will be ordered sets that satisfy a very strong ordering condition: that every nonempty subset contains
a smallest element. Such sets are called well-ordered. The most familiar example of a well-ordered set
is  and it is the well-ordering property that lets us do mathematical induction in 
In this chapter we will see “longer” well ordered sets and these will give us a new proof method called
“transfinite induction.” But we begin with something simpler.
2. Partially Ordered Sets
Recall that a relation V on a set \ is a subset of \ ‚ \ (see Definition I.5.2). If ÐBß CÑ − V , we write
BVCÞ An “order” on a set \ is refers to a relation on \ that satisfies some additional conditions.
Order relations are usually denoted by symbols such as Ÿ ,  , ¡ ß or £ .
Definition 2.1 A relation V on \ is called:
transitive if À
reflexive if À
antisymmetric if À
symmetric if À
a +ß ,ß - − \
a+ − \
a +ß , − \
a +ß , − \
Ð+V, and ,V-Ñ Ê +V-Þ
+V+
Ð+V, and ,V+ Ñ Ê Ð+ œ ,Ñ
+V, Í ,V+ (that is, the set V is “symmetric”
with respect to the diagonal
? œ ÖÐBß BÑ À B − \× © \ ‚ \ ).
Example 2.2
1) The relation “ œ ” on a set \ is transitive, reflexive, symmetric, and
antisymmetric. Viewed as a subset of \ ‚ \ , the relation “ œ ” is the diagonal set
? œ ÖÐBß BÑ À B − \×Þ
2) In ‘, the usual order relation  is transitive and antisymmetric, but not
reflexive or symmetric.
3) In ‘, the usual order Ÿ is transitive, reflexive and antisymmetric. It is not
symmetric.
4) On any set of cardinal numbers V we have a relation Ÿ . It is transitive, reflexive
and antisymmetric (by the Cantor-Schroeder-Bernstein Theorem I.10.2), but not
symmetric Ðunless lVl œ "ÑÞ
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Definition 2.3 A relation Ÿ on a set \ is called a partial order if Ÿ is transitive, reflexive and
antisymmetric. The pair Ð\ß Ÿ Ñ is called a partially ordered set (or, for short, poset).
A relation Ÿ on a set \ is called a linear order if Ÿ is a partial order and, in addition, any two
elements in \ are comparable: a +ß , − \ either + Ÿ , or , Ÿ +. In this case, the pair Ð\ß Ÿ ) is
called a linearly ordered set. For short, a linearly ordered set is also called a chain.
We write +  , to mean that + Ÿ , and + Á ,Þ For any sort of order relation Ÿ on \ , we can invert
the order notation and write , + Ð,  +Ñ to mean the same thing as + Ÿ , Ð+  ,ÑÞ
In some books, a partial order is defined as a “strict” relation  which is transitive and
irreflexive Ða+ − \ , a 
Î +ÑÞ In that case, we can define + Ÿ , to mean “+  , or + œ , ” to
get a partial order in the sense defined above. This variation in terminology creates no real
mathematical problems: the difference is completely analogous to worrying about whether
“ Ÿ ” or “  ” should be called the“usual order” on ‘ .
Example 2.4
1) Suppose E © \Þ For any kind of order Ÿ on \ß we can get an order Ÿ E on E © \ by
restricting the order Ÿ to EÞ More formally, Ÿ E œ Ÿ ∩ (E ‚ E). We always assume that a subset
of an ordered set has this natural “inherited” ordering unless something else is explicitly stated. With
that understanding, we usually omit the subscript and also write Ÿ for the order Ÿ E on E.
If Ð\ß Ÿ Ñ is a poset (or chain), and E © \ , then ÐEß Ÿ Ñ is also a poset (or chain).
every subset of Ð‘ß Ÿ Ñ is a chain.
For example,
2) For Dß A − ‚ Ð œ the set of complex numbersÑ, define D £ A iff lDl Ÿ lAlß where Ÿ is
the usual order in ‘. Ð‚ß £ Ñ is not a poset. (Why?)
3) Let Ð\ß g Ñ be a topological space. For 0 ß 1 − GÐ\Ñ, define
0 Ÿ 1 iff aB − \ , 0 ÐBÑ Ÿ 1ÐBÑ.
As a set,
Ÿ œ ÖÐ0 ß 1Ñ − GÐ\Ñ ‚ GÐ\Ñ À aB − ‘ß 0 ÐBÑ Ÿ 1ÐBÑ×.
Notice that, in contrast to part 2), we are allowing ourselves an ambiguity in the notation here because
we are using “ Ÿ ” with two different meanings: we are defining an order “ Ÿ ” in GÐ\Ñ, but the
comparison “0 ÐBÑ Ÿ 1ÐBÑ” refers to the usual order a different ordered set, ‘. Of course, we could be
more careful and write 0 £ 1 for the new order on GÐ\Ñ, but usually we won't be that fussy when the
context makes clear which meaning of “ Ÿ ” we have in mind.
ÐGÐ\Ñß Ÿ Ñ is a poset but usually not a chain: for example, if \ œ ‘ and 0 ß 1 are given by 0 ÐBÑ œ B
and 1ÐBÑ œ B# , then 0 Ÿ
Î 1 and 1 Ÿ
Î 0 . When is ÐGÐ\Ñß Ÿ Ñ a chain? ÐThe answer is not “ iff
l\l Ÿ "Þ”Ñ
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4) The following two diagrams represent posets, each with 5 elements. Line segments upward
from B to C indicate that B Ÿ C. In Figure (i), for example, . Ÿ - and - Ÿ + (so . Ÿ +); in Figure (i),
+ and , are not comparable.
Figure (ii) shows a chain: + Ÿ , Ÿ - Ÿ . Ÿ /Þ
5) Suppose V is a collection of sets. We can define Ÿ on V by E Ÿ F iff E © F . Then
ÐVß Ÿ Ñ is a poset. In this case, we say that V has been ordered by inclusion. In particular, for any set
\ , we can order c Ð\Ñ by inclusion. What conditions on \ will guarantee Ðc Ð\Ñß Ÿ ) is a chain ?
6) Suppose V is a collection of sets. We can define Ÿ on V by E Ÿ F iff E ª F . Then
ÐVß Ÿ Ñ is a poset. In this case, we say that V has been ordered by reverse inclusion. In particular, for
any set \ , we can order c Ð\Ñ by reverse inclusion. What conditions on \ will guarantee that
Ðc Ð\Ñß Ÿ ) is a chain ?
For a given collection V, Examples 5) and 6) are quite similar: one is a “mirror image” of the other.
The identity map 3 À V Ä V is an “order-reversing isomorphism” between the posets.
For our purposes, the “reverse inclusion” ordering on a collection of sets will turn out to be more
useful. For a point B in a topological space \ , we can order the neighborhood system aB by reverse
inclusion Ÿ . The “order structure” of the poset ÐaB ß Ÿ Ñ reflects some topological properties of \
and indicates just how complicated the “neighborhood structure” at B is. For example,
i) If B is isolated, then ÖB× − RB and R Ÿ ÖB× for every R − RB . So the poset
ÐaB ß Ÿ Ñ has a largest element. ÐIs the converse true?Ñ
ii) If \ is first countable and UB œ ÖR" ß R# ß ÞÞÞß R5 ß ÞÞÞ× is a countable
neighborhood base at B, then for every R − aB there is a 5 such that
R Ÿ R5 . Thus the poset ÐaB ß Ÿ Ñ contains a countable subset
ÖR" ß R# ß ÞÞÞß R5 ß ÞÞÞ× whose members become “arbitrarily large” in the poset.
iii) The poset ÐaB ß Ÿ Ñ is usually not a chain. But it does have interesting
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order property: for any R" ß R# − aB ß bR$ − aB such that R" Ÿ R$ and R# Ÿ R$
(let R$ œ R" ∩ R# ÑÞ
It will turn out (in Chapter 9) that this poset ÐaB ß Ÿ Ñ is the inspiration for defining the notion of
“convergent nets,” a kind of convergence that is more powerful than “convergent sequences.” Unlike
sequences, convergent nets will be able to determine closures (and therefore the topology) in any
topological space.
Definition 2.5 Suppose Ð\ß Ÿ Ñ is a poset and let E © \Þ An element B − \ is called
i)
ii)
iii)
iv)
the largest (or last) element in \ if C Ÿ B for all C − \
the smallest (or first) element in \ if B Ÿ C for all C − \
a maximal element in \ if ÐC B Ê C œ BÑ for all C − \
a minimal element in \ if ÐC Ÿ B Ê C œ BÑ for all C − \ .
It is clear that a largest element in Ð\ß Ÿ Ñ, if it exists, is unique. (If D" and D# were both largest, then
D" Ÿ D# and D# Ÿ D" so D" œ D# .)
In Figure (i): both +ß , are maximal elements and .ß / are minimal elements. This poset has no largest
or smallest element. Suppose a poset Ð\ß Ÿ Ñ has a unique maximal element D . Must D also be the
largest element in Ð\ß Ÿ Ñ ?
In Figure (ii): + is the smallest (and also a minimal element); / is the largest (and also a maximal)
element.
v) Suppose E © \ and B − \Þ We say that B − \ is an upper bound for E if + Ÿ B for all
+ − Eà B is called a least upper bound (sup) for E if B is the smallest upper bound for EÞ The
set E might have many upper bounds, one upper bound, or no upper bounds in \ . If E has
more than one upper bound, E might or might not have a least upper bound in \ . But if \ has
a least upper bound B, then the least upper bound is unique (why?).
vi) An element B − \ is called a lower bound for E if B Ÿ + for all + − Eà B is called a
greatest lower bound (inf) for E if B is the largest lower bound for EÞ The set E might have
many lower bounds, one lower bound, or no lower bounds in \ . If E has more than one
lower bounds, E might or might not have a greatest lower bound in \ , but if \ has a greatest
lower bound B, then the greatest lower bound is unique (why?).
vii) If B  C − \ and if bÎ D − \ with B  D  C, then B is called an immediate predecessor
of C and C is called an immediate successor of B. In a poset, an immediate predecessor or
successor might not be unique; but if \ is a chain, then an immediate predecessor or
successor, if it exists, must be unique. (Why?)
In Figure (i), the upper bounds on Ö.ß /× are +ß ,ß -, and - œ supÖ.ß /×; the set Ö.ß /× has
no lower bounds. Both +, , are immediate successors of - . The elements .ß / have no
immediate predecessor (in fact, no “predecessors” at all). In Figure (ii), the immediate
predecessor of - is , and . is the immediate successor of -Þ
Example 2.6 If Ÿ " is an order on \ , then Ÿ " © \ ‚ \Þ So if Ÿ " and Ÿ # are orders on \ , it
makes sense to ask whether Ÿ " © Ÿ # , or vice-versa. If we look at the set c œ Ö Ÿ À Ÿ is a partial
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order on \×, then c is partially ordered by inclusion. A linear order Ÿ is a maximal element in
Ðc ß © Ñ (Why? Is the converse true?)
3. Chains
Definition 3.1 Let Ð\ß Ÿ Ñ be a chain. The order topology on \ is the topology for which all sets of
the form ÖB − \ À B  +× or ÖB − \ À ,  B× (+ß , − \ ) are a subbase. (As usual, we write B  C as
shorthand for “B Ÿ C and B Á CÞ” )
It's handy to use standard notation when working with chains: but we need to be careful not to read too
much into the notation. For example, if +ß , − \ :
ÖB − \ À +  B  ,× œ Ð+ß ,Ñ
ÖB − \ À + Ÿ B  ,× œ Ò+ß ,Ñ
ÖB − \ À B Ÿ +× œ Ð  ∞ß +Ó
For the chain in Figure (ii), above, we see how the interval notation can be misleading if not used
thoughtfully: Ð+ß ,Ñ œ gß Ð.ß ∞Ñ œ Ö/×ß Ð  ∞ß ,Ñ œ Ö+×, Ð+ß -Ñ œ Ö,×, and Ð+ß /Ñ œ Ò,ß .ÓÞ
Example 3.2
1) The order topology on the chain in Figure (ii) is the discrete topology.
2) The order topology on  is the usual (discrete) topology: Ö"× œ Ö5 −  À 5  #×
œ Ð  ∞ß #Ñ; and for 8  ", Ö8× œ Ð8  "ß 8  "ÑÞ
Example 3.3  and  each have an order inherited from ‘, and their order topologies are the same as
the usual subspace topologies. But, in general, we have to be careful about the topology on E © \
when \ is a chain with the order topology. There are two possible topologies on E:
a) The order Ÿ gives an order topology gŸ on \ and we can give E the subspace
topology ÐgŸ ÑE Þ
b) E has an ordering Ÿ E (inherited from the order Ÿ on \Ñ and we can use it to give E an
order topology. More formally, we could write this topology as gŸE .
Unfortunately, these two topologies might not be the same. Let E œ Ð!ß "Ñ ∪ Ö#× © ‘. The order
topology gŸ on ‘ is the usual topology on ‘, and this topology produces a subspace topology for
which # is isolated in EÞ
But in the order topology on E, each basic open set containing 2 must have the form
ÖB − E À B  +× œ Ð+ß #Ó where +  ". So # is not isolated in ÐEß gŸE Ñ. In fact, the space ÐEß gŸE Ñ is
homeomorphic to Ð!ß "Ó (why? ).
Is there any necessary inclusion © or ª between gŸE and ÐgŸ ÑlE ? Can you state any
hypotheses on \ or E that will guarantee that gŸE œ (gŸ ÑlE ?
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Example 3.4 We defined the order topology only for chains, but the same definition could be used in
any ordered set Ð\ß Ÿ Ñ. We usually restrict our attention to chains because otherwise the order
topology may not be very nice. For example, let \ œ Ö+ß ,ß -× with the partial order represented by
the following diagram:
\ is the only open set containing + (why?), so the order topology is not X" Þ ÐCan you find a poset for
which the order topology is not even X! ?Ñ By contrast, the order topology for any chain Ð\ß Ÿ Ñ has
good separation properties. For example, it is easy to show that the order topology on a chain must
be X# À
Suppose + Á , − \ ; we can assume +  ,Þ If + is the immediate predecessor of ,, we can let
Y œ ÖB − \ À B  ,× and Z œ ÖB − \ À B  +×. But if there exists a point - satisfying
+  -  ,, then we can define Y œ ÖB − \ À B  -× and Z œ ÖB − \ À B  -×Þ Either way,
we have a pair of disjoint open sets Y and Z with B − Y and C − Z Þ
In fact, the order topology on a chain is always X% , but this is much messier and we will not prove it
here. (Most of our interest in this Chapter will be with ordered sets that are much “better” than chains
(well-ordered sets) where it is relatively easy to prove normality.)
From now on, when the context is clear, we will simply write \ , rather than Ð\ß Ÿ Ñ, for an ordered
set. We will denote the orders in many different sets all with the same symbol “ Ÿ ”, letting the
context settle which order is being referred to. If it is necessary to distinguish carefully between
orders, then we will occasionally add subscripts such as Ÿ " , Ÿ # , ... .
Definition 3.5 A function 0 À \ Ä ] between ordered sets is called an order isomorphism if 0 is
bijection and both 0 and order preserving  that is, + Ÿ , if and only if 0 Ð+Ñ Ÿ 0 Ð,Ñ. ÐSince 0 is oneto-one, it follows that B  C if and only if 0 ÐBÑ  0 ÐCÑÞÑ If such an 0 exists, we say that \ and ] are
order isomorphic and write \ ¶ ] .
If 0 is not onto, then \ ¶ 0 Ò\Ó © ] and 0 is an order isomorphism of \ into ] .
Between ordered sets, “ ¶ ” will always refer to order isomorphism.
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Theorem 3.6 Let \ß ] ß and ^ be chains Ðparts i-iv) are also true for posets).
i)
ii)
iii)
iv)
v)
\¶\
\ ¶ ] iff ] ¶ \
if \ ¶ ] and ] ¶ ^ , then \ ¶ ^
\ ¶ ] implies l\l œ l] lÞ
if \ and ] are finite chains, then \ ¶ ] iff l\l œ l] lÞ
Proof The proof is very easy and is left as an exercise. ñ
Even though the proof is easy, there are some interesting observations to make.
1) To show that \ ¶ \ , the identity map 3 might not be the only order isomorphism possible.
For example, when \ œ ‘ß the functions 0 ÐBÑ œ B and 0 ÐBÑ œ B$ are both order isomorphisms
between ‘ and ‘ . How many order isomorphisms exist between  and ?
2) A chain can be order isomorphic to a proper subset of itself. For example, 0 Ð8Ñ œ #8 is an
order isomorphism between  and „ (the set of even natural numbers). Both  and „ are order
isomorphic to the set of all prime numbers. Must every two countable infinite chains be order
isomorphic?
3) An order isomorphism between \ and ] preserves largest, smallest, maximal and minimal
elements (if they exist). Therefore Ð!ß "Ñ and Ò!ß "Ó are not order isomorphic: for example, Ò!ß "Ó has a
smallest element and Ð!ß "Ñ doesn't. Similarly,  is not order isomorphic to the set of integers ™.
An order isomorphism preserves “betweenness,” so ™ is not order isomorphic to : in ,
there is a third element between any two elements, but this is false in ™.
4) Let ‚ be the set of complex numbers. If 0 À ‚ Ä ‘ is any bijection, then we can use 0 to
create a chain Ð‚ß Ÿ Ñ: simply define D" Ÿ D# iff 0 ÐD" Ñ Ÿ 0 ÐD# Ñ. Then Ð‚ß Ÿ Ñ ¶ Ð‘ß Ÿ ÑÞ
Of course, this chain Ð‚ß Ÿ Ñ is not be very interesting from the point of view of algebra or analysis:
we imposed an arbitrary ordering on ‚ that has nothing to do with the algebraic structure of ‚. For
example, there is no reason to think that D" Ÿ D# and D$ Ÿ D% , then D"  D$ Ÿ D#  D% Þ
In the same way, a bijection 0 À  Ä  can be used to give  a (new) order Ÿ so that the two chains
are order isomorphic. In this case, 0 is just a sequence that enumerates : ;" ß ;# ß ÞÞÞß ;8 ß ÞÞÞ and the
new order on  is simply defined as ;"  ;#  ÞÞÞ  ;8  ÞÞÞ . More generally, if 0 À \ Ä ] is a
bijection and one of \ or ] is ordered, we can use 0 to “transfer” the order to the other set in such a
way that Ð\ß Ÿ Ñ ¶ Ð] ß Ÿ ÑÞ
5) Order isomorphic chains are clearly homeomorphic in their order topologies. But the
converse is false. Suppose \ œ Ö  8" À 8 − × ∪ Ö!× and ] œ Ö!× ∪ Ö 8" À 8 − ×Þ The order
topologies on \ and ] are the usual topologies, and the reflection 0 ÐBÑ œ  B is a homeomorphism
(in fact, an isometry) between themÞ
The largest element in ] is "ß and it has an immediate predecessor, #" Þ But ! is the largest
element in \ and it has no immediate predecessor in \Þ Since an order isomorphism preserves largest
elements and immediate predecessors, there is no order isomorphism between \ and ] .
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The next theorem tells when an ordered set is order isomorphic to the set of rational numbers.
Theorem 3.7 (Cantor) Suppose that (P, Ÿ ) is a nonempty countable chain such that
a) a+ − P, b, − P with +  , (“no last element”)
b) a+ − Pß b, − P with ,  + (“no first element”)
c) a+ß , − P, if +  , then b- − P such that +  -  ,.
Then ÐPß Ÿ Ñ is order isomorphic to .
When a chain that satisfies the third condition  that between any two elements there must
exist a third element  we say that \ is order-dense. Then Theorem 3.7 can be restated as: a
nonempty countable order-dense chain with no first or last element is order isomorphic to .
We might attempt a proof in the following way: enumerate  œ Ö;" ß ;# ß ÞÞÞß ;8 ß ÞÞÞ× and
P œ Ö6" ß 6# ß ÞÞß 68 ß ÞÞÞ×, and let 0 Ð;" Ñ œ 6" . Then inductively define 0 Ð;8 Ñ œ some 6 − P chosen so that
0 Ð;8 Ñ Á 0 Ð;" Ñß ÞÞÞß 0 Ð;8" Ñ and so that 0 Ð;8 Ñ has the same order relations to 0 Ð;" Ñß ÞÞÞß 0 Ð;8" Ñ as
;8 has to ;" ß ÞÞÞß ;8" . Such a choice is always be possible since P satisfies a), b), c). In this way we end
up with one-to-one order preserving map 1 À  Ä P. However, 1 would not necessarily be onto. The
“back-and-forth” construction used in the following proof is designed to be sure we end up with an
onto order preserving map 1 À  Ä P.
First we prove a lemma: it states that an order isomorphism between finite subsets of  and P can
always be extended to include one more point in its domain (or, one more point in its range).
Lemma: Suppose E © ß F © P where Eß F are finite. Let 1 be an order isomorphism from
E onto F .
a) If ; −   E, there exists an order isomorphism 2 that extends 1 and for
which domÐ2Ñ œ E ∪ Ö;× and ranÐ2Ñ © PÞ
b) If 6 − P  F , there exists an order isomorphism 2 that extends 1 and for
which domÐ2Ñ ©  and ran Ð2Ñ œ F ∪ Ö6×Þ
Proof of a) Suppose E œ Ö;" ß ÞÞÞß ;8 × and that 1Ð;3 Ñ œ 63 . Pick an 6 − P  F that has the same
order relations to 6" ß ÞÞÞß 68 as ; does to ;" ß ÞÞÞß ;8 Þ ( For example: if ; is greater that all of
;" ß ÞÞÞß ;8 ß pick 6 greater than all of 6" ß ÞÞÞß 68 ; if 3 and 4 are the largest and smallest subscripts
for which ;3  ;  ;4 , then choose 6 between 63 and 64 Þ ) This choice is always possible since
P satisfies a), b), and c). Define 2 by 2Ð;Ñ œ 6 and 2ÐBÑ œ 1ÐBÑ for B − E.
Proof for b) The proof is almost identical. ñ
Proof of Theorem 3.7 P is a countable chain. Since P Á g and P has no last element, P must be
infinite. Without loss of generality, we may assume that P ∩  œ gÞ
We define an order isomorphism 1 between  and P in stages. At each stage, we enlarge the function
we have by adding a new point to its domain or range.
Let Q œ  ∪ P œ Ö7" ß 7# ß ÞÞÞß 78 ß ÞÞÞ×Þ Each element of  and P appears exactly once in this list.
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Define 1! œ g, and continue by induction. Suppose 8  ! and that an order isomorphism
18" À E8" Ä F8" has been defined, where E 8" ©  and F8" © PÞ
If 78 −  and 78 Â E8" , use the Lemma to get an order isomorphism 18 that extends
18" and for which domÐ18 Ñ œ E8" ∪ Ö78 × œ E8 Þ Let F8 œ ranÐ18 Ñ © P.
If 78 − P and 78 Â F8" ß use the Lemma to get an order isomorphism 18 that extends
18" and for which ranÐ18 Ñ œ F8" ∪ Ö78 × œ F8 Þ Let E8 œ domÐ18 Ñ © .
If 78 − domÐ18" Ñ ∪ ranÐ18" Ñ, let 18 œ 18" ß E8 œ E8" and F8 œ F8" Þ
By induction, 18 is defined for all 8, and since 18 extends 18" and we can define an order
isomorphism 1 œ ∞
8œ! 18 Þ The construction guarantees that domÐ1Ñ œ  and ranÐ1Ñ œ PÞ ñ
“Being order isomorphic” is an equivalence relation among ordered sets so any two chains having the
properties in Cantor's theorem are order isomorphic to each other. Since order isomorphic chains have
homeomorphic order topologies, we have a topological characterization of  in terms of order.
Corollary 3.8 A nonempty countable order-dense chain with no largest or smallest element is
homeomorphic to .
The following corollary gives a characterization of all order-dense countable chains.
Corollary 3.9 If \ is a countable order-dense and l\l  "ß then \ is order isomorphic to exactly one
of the following chains À
a)
b)
c)
d)
 ∩ Ð!ß "Ñ ¶ 
 ∩ Ò!ß "Ñ
 ∩ Ð!ß "Ó
 ∩ Ò!ß "Ó
Proof By looking at largest and smallest elements, we see that no two of these chains are order
isomorphic. Therefore no chain is order isomorphic to more than one of them.
If \ has no largest or smallest element then, Cantor's theorem gives \ ¶  and a second
application of Cantor's Theorem gives that  ¶  ∩ Ð!ß "Ñ.
If \ has a smallest element, +, but no largest element, then \  Ö+× is nonempty and has no
smallest element (why?). \  Ö+× clearly satisfies the other hypotheses of the Cantor's theorem so
there is an order isomorphism 2 À \  Ö+× Ä  ∩ Ð!ß "Ñ. Then define an order isomorphism
1 À \ Ä  ∩ Ò!ß "Ñ by setting 1Ð+Ñ œ ! and 1ÐBÑ œ 2ÐBÑ for B Á +Þ.
The proofs of the other cases are similar. ì
No two of the chains a)-d) mentioned in Corollary 3.9 are order isomorphic, but they are, in fact, all
homeomorphic topological spaces. We can see that b) and c) are homeomorphic by using the map
0 ÐBÑ œ "  B , but the other homeomorphisms are not so obvious. Here is a sketch of a proof,
contributed by Edward N. Wilson, that Ð!ß "Ñ ∩  is homeomorphic to Ð!ß "Ó ∩ .
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For short, write Ð!ß "Ñ ∩  œ Ð0ß "Ñ and Ð!ß "Ó ∩  œ Ð!ß "] .
In Ð!ß "Ñ, choose a strictly increasing sequence of irrationals Ð:8 Ñ Ä " − ‘. Let :! œ !.
For 8 ! À let S8 œ Ð:8 ß :8" Ñ ∩ , Y8 œ Ð 12 :8 ß 12 :8" Ñ ∩ ,
and Z8 œ Ð"  12 :8" ß "  12 :8 Ñ ∩ . Each of S8 ß Y8 ß and Z8 is clopen in Ð0ß "Ñ and
in Ð!ß "] .
We have Ð!ß "Ó œ  S8 ∪ Ö"× and Ð!ß "Ñ œ  Y8 ∪ Ö 12 × ∪  Z8
∞
∞
Define a map 9À Ð!ß "Ó Ä  Y8 ∪ Ö 12 × ∪  Z8 by
8œ!
∞
∞
8œ!
8œ!
8œ!
∞
8œ!
9 ± S#8 = an increasing linear map from S#8 onto Y8
9 ± S#8  " = a decreasing linear map from S#8  " onto Z8
9Ð"Ñ œ
1
2
Then 9 is a homeomorphism. ÐSince the sets S8 ß Y8 ß Z8 are clopen, it is clear that 9 is
continuous at all points except perhaps 1 and that 9" is continuous at all points except
perhaps 12 . These special cases are easy to check separately.Ñ
There is, in fact, a more general theorem that states that every infinite countable metric space with no
isolated points is homeomorphic to . This theorem is due to Sierpinski (1920).
We can also characterize the real numbers as an ordered set.
Theorem 3.10 Suppose \ is a nonempty chain which
i) has no largest or smallest element
ii) is order-dense
iii) is separable in the order topology
iv) is Dedekind complete (that is, every nonempty subset of \ which has an upper
bound in \ has a least upper bound in \ ).
Then \ is order isomorphic to ‘ (and therefore \ , with its order topology, is homeomorphic to ‘).
Proof We will not give all the details of a proof, However, the ideas are completely straightforward
and the details are easy to fill in.
Let H be a countable dense set in the order topology on \ . Then H satisfies the hypotheses of
Cantor's Theorem Ðwhy?Ñ so there exists an (onto) order isomorphism 0 À  Ä H. For each irrational
: − ‘, let : œ Ö; −  À ;  :× and extend 0 to an order isomorphism 1 À ‘ Ä \ by defining
0 Ð:Ñ œ sup 0 Ò : Ó. ì
Remark In a separable space, any family of disjoint open sets must be countable (why?). Therefore
we could ask whether condition iii) in Theorem 3.10 can be replaced by
328
iii)w
every collection of disjoint open intervals in \ is countable.
In other words, can we say that
(**)
a nonempty chain satisfying i), ii), iii)w , and iv), must \ be order isomorphic to ‘ ?
The Souslin Hypothesis (SH) states that the answer to (**) is “yes.” The status of SH was famously
unknown for many years. Work of Jech, Tennenbaum and Solovay in the 1970's showed that SH is
consistent with and independent of the axioms ZFC for set theory  that is, SH is undecidable in ZFC.
We could add either SH or its negation as an additional axiom in ZFC without introducing an
inconsistency. If one assumes that SH is false, then there is a nonempty chain satisfying i), ii), iii)w ,
and iv) but not order isomorphic to ‘: such a chain is called a Souslin line.
SH was of special interest for a while in connection with the question “if \ is a T% -space, is \ ‚ Ò!ß "Ó
necessarily X% ? ” In the 1960's, Mary Ellen Rudin showed that if she had a Souslin line to work
with  that is, if SH is false  then the answer to the question was “no.” ÐSee the remarks following
Example III.5.7Ñ
There are lots of equivalent ways of formulating SH—for example, in terms of graph theory. There is
a very nice expository article by Mary Ellen Rudin on the Souslin problem in the American
Mathematical Monthly, 76(1969), 1113-1119. The article was written before the consistency and
independence results of the 1970's and deals with aspects of SH in a “naive” way.
329
Exercises
E1. Prove or disprove: if V is a relation on \ which is both symmetric and antisymmetric, then V
must be the equality relation “ œ ”.
E2. State and prove a theorem of the form:
A power set cÐ\Ñ, ordered by inclusion, is a chain iff . . .
E3. Suppose Ð\ß Ÿ Ñ is a poset in which every nonempty subset contains a largest and smallest
element. Prove that Ð\ß Ÿ Ñ is a finite chain.
E4. Prove that any countable chain ÐPß Ÿ Ñ is order isomorphic to a subset of Ðß Ÿ ÑÞ
(Hint: See the “Caution” in the proof of Cantor's Theorem 3.7.)
E5. Let Ð\ß Ÿ Ñ be an infinite poset. A subset G of \ is called totally unordered if no two distinct
elements of G are comparable, that is:
a+ − G a, − G Ð+ Ÿ ,Ñ Í Ð+ œ ,Ñ
Prove that either \ has a subset G which is an infinite chain or \ has a totally unordered infinite
subset G .
E6. Let Ð\ß Ÿ Ñ be a poset in which the longest chain has length 8 (8 − ). Prove that \ can be
written as the union of 8 totally unordered subsets (see E5) and the 8 is the smallest natural number
for which this is true.
.
330
4. Order Types
In Chapter I, we assumed that we can somehow assign a cardinal number l\l to each set \ß in such a
way that l\l œ l] l iff there exists a bijection 0 À \ Ä ] Þ Similarly, we now assume that we can
assign to each chain an “object” called its order type and that this is done in such a way that two chains
have the same order type iff the chains are order isomorphic. Just as with cardinal numbers, an exact
description of how this can be done is not important here. In axiomatic set theory, all the details can
be made precise. Of course, then, the order type of a chain turns out itself to be a certain set (since
“everything is a set” in ZFC). For our purposes, it is enough just to take the naive view that “orderisomorphic” is an equivalence relation among chains, and that each equivalence class is an order type.
We will usually denote order types by lower case Greek letters such as .ß / ß 7 ß =  with a few
traditional exceptions mentioned in the next example. If . is the order type of a chain Q , we say that
Q represents ..
Example 4.1
1) Two chains with the same order type are order isomorphic. Since the order isomorphism is
a bijection, chains with the same order type also have the same cardinal number. But the converse is
false:  and  have the same cardinal number but they they have different order types because the sets
are not order isomorphic.
However, two finite chains have the same cardinality iff they are order isomorphic.
Therefore, for finite chains, we will use the same symbol for both the cardinal number and the order
type. ÐIn the precise definitions of axiomatic set theory, the cardinal number and the order type of
a finite chain do turn out to be the same set! Ñ
2)
! is the order type of g
" is the order type of Ö!×
# is the order type of Ö!ß "×
.
.
.
8 is the order
.
type of Ö0ß "ß ÞÞÞ ß 8  "×
.
.
.
=! is the order type of 
(The subscript “!” hints at bigger things to come.)
=! is also the order type of „ œ Ö#ß %ß 'ß ÞÞÞ× since this chain is order isomorphic to .
Notice that each order type in this example is represented by “the set
of preceding order types.”
Definition 4.2 Let . and / be order types represented by chains Q and R . We say that . Ÿ / if
there exists an order isomorphism 0 of Q into R . We write .  / if . Ÿ / but . Á / Ðthat is,
Q¶
Î R ÑÞ (Check that the definition is independent of the chains Q and R chosen to represent .
331
and / . )
Example 4.3 Let . be the order type of a chain Q . Since g is order isomorphic to a subset of Q
we have ! Ÿ .. More generally, !  "  #  ÞÞÞ  =! .
Suppose Ð\ß Ÿ Ñ has order type .. With a little reflection, we can create a new chain Ð\ß Ÿ ‡ ) by
defining B Ÿ ‡ C iff C Ÿ B. We write .‡ for the order type of Ð\ß Ÿ ‡ ). For example, =‡! is the order
type of the chain Ö... ,  2,  1× of negative integers. Since =! Ÿ
Î =‡! and =‡! Ÿ
Î =! (why? ), we see
that two order types may not be comparable.
The relation Ÿ between order types is reflexive and transitive but it is not antisymmetric  for
example, let . and / be order types of the intervals Ð!ß "Ñ and Ò!ß "Ó: then . Ÿ / and / Ÿ . but . Á / Þ
Therefore Ÿ is not even a partial ordering among order types.
Definition 4.4 For α − E, let ÐQα ß Ÿ α Ñ be pairwise disjoint chains and suppose that the index set E
is also a chain. We define the ordered sum  α−E Qα as the chain Ðα−E Qα , Ÿ Ñ, where we define
Bß C − Qα and B Ÿ α C, or
B Ÿ C if 
α  " − Eß B − Qα and C − Q"
We can “picture” the ordered sum as laying the chains Qα “end-to-end” with larger α's further to the
right. In particular, for two disjoint chains ÐQ ß Ÿ Q Ñ and ÐR ß Ÿ R Ñ, the ordered sum ÐQ  R ß Ÿ Ñ
is formed by putting R to the right of Гlarger than”Ñ Q and using the old orders inside each of Q
and R .
Definition 4.5 Suppose E is a chain and that for each α − E, we have an order type .α . Let the
.α 's be represented by pairwise disjoint chains Qα . Then  α−E .α as order type of the chain
Ðα−E Qα , Ÿ Ñ. In particular, if . and / are order types represented by disjoint chains Q and R , then
.  / is the order type of the ordered sum ÐQ  R ß Ÿ Ñ. (Check that sum of order types is
independent of the disjoint chains used to represent the order types. )
Example 4.6 Addition of order types is clearly associative: Ð.  / Ñ  7 œ .  Ð/  7 Ñ. It is not
commutative. For example =!  " Á "  =! , since a chain representing the left side has a largest
element but a chain representing the right side does not. In general, for 8 − , 8  =! œ =! while, if
7 Á 8 − , =! Á =!  8 Á =!  7. Of course, chains representing these different order types all
have cardinality i! .
The order type =!  =! is represented by the ordered set Ö!ß "ß #ß $ß ÞÞÞ à +" , +# , +$ , ÞÞÞ× where “ ; ”
indicates that each +3 is larger than every 8. Less abstracting, =!  =! could also be represented by
the chain Ö1  81 : 8 − × ∪ Ö2  81 : 8 − × © .
=‡!  =! is the order type of the set of integers ™. Is =‡!  =! œ =!  =!‡ ? (Why or why not? Give an
example of a subset of  that represents =!  =‡! ÞÑ
Example 4.7 It is easy to prove that every countable chain G is order isomorphic to a subset of :
just list the elements of G and inductively define a one-to-one mapping into  that preserves order at
each step (see the “attempted” argument that precedes that actual proof of Cantor's Theorem 3.7)
332
Theorem 3.7). But if every countable order type can be represented by some subset of , then there
can be at most - different countable order types.
As a matter of fact, we can prove that there are exactly - different countable order types. A
sketch of the argument follows: the details are easy to fill in (or see W. Sierpinski, Cardinal
and Ordinal Numbers  and old “Bible” on the subject with much more information than
anybody would want to know.)
™ has order type 0 œ =‡!  =! . For each sequence + œ Ð+" ß +# ß +$ ÞÞÞÑ − Ö!ß "× , we can define
an order type
0+ œ 0  "  +"  0  1  +#  0  "  +$  . . .
It is not hard to show that the map + È 0+ is one-to-one. Here is a sketch of the argument:
We say that two elements of a chain to be in the same component if there are only
finitely many elements between them. (This use of the word “component” has nothing
to do with connectedness.) It is clear all elements in the same component are smaller
(or larger) than all elements in a different component; this observation lets us order the
components of the chain.
Suppose G is a chain representing 0+ and let the finite components of G (listed in
order of increasing size) be J" ß J# ß ÞÞÞ ß J8 ß ÞÞÞ where J8 has order type "  +8 .
Let + Á , − Ö!ß "× and suppose G w is a chain that represents 0, . Call the finite
components of G w (listed in order of increasing size) J"w ß J#w ß ÞÞÞ ß J8w ß ÞÞÞ where J8w has
order type "  ,8 .
An order isomorphism between G and G w would necessarily carry J8 to J8w for every
8. But this is impossible since, for some 8, one of "  +8 and "  ,8 is " and the
other is #.
Thus, there are at least as many different countable order types as there are sequences
+ − Ö!ß "× , namely - .
Definition 4.8 Let ÐQ ß Ÿ Q ) and ÐR ß Ÿ R Ñ be chains representing the order types . and / . We
define the product ./ to be the order type of ÐR ‚ Q ß Ÿ ), where Ð8" , 7" ) Ÿ Ð8# , 7# ) iff 8"  R 8#
or (8" œ 8# and 7" Ÿ Q 7# ).
This ordering Ÿ on R ‚ Q is called the lexicographic (or dictionary) order since the
pairs are ordered “alphabetically.”
Example 4.9
a) # † =! œ =! . To see this, we can represent =! by  and 2 by Ö!ß "×. Then the chain
representing 2 † =! , listed in increasing (lexicographic) order, is  ‚ Ö!ß "× œ ÖÐ"ß !Ñß Ð"ß "Ñß Ð#ß !Ñß
333
Ð#ß "Ñß Ð$ß !Ñß Ð$ß "Ñß ... , Ð8ß !Ñ, Ð8ß "Ñ, ÞÞÞ ×, and this chain is order isomorphic to . More generally,
8 † =! œ =! for each 8 − Þ
However, =! † 2 Á =! . The product =! † 2 is represented by Ö!ß "× ‚  ordered as:
ÖÐ!ß "Ñß Ð!ß #Ñß ÞÞÞß Ð!ß 8Ñß ÞÞÞ ; Ð"ß "Ñß Ð"ß #Ñß ÞÞÞß Ð"ß 8Ñß ÞÞÞ×
This chain is not order isomorphic to  À  has only one element with no “immediate predecessor,”
while this set has two such elements. In fact, =! † 2 = =!  =! . Thus, multiplication of order types is
not commutative.
b) =! œ 2 † =! œ Ð"  "Ñ † =! Á 1 † =!  1 † =! œ =!  =! . Thus the “right distributive” law
fails.
c) =#! can be represented by the lexicographically ordered chain  ‚ . In order of increasing
size, this chain is:
ÖÐ"ß "Ñß Ð"ß #Ñß ÞÞÞ à Ð"ß 8Ñß ÞÞÞ à Ð#ß "Ñß Ð#ß #Ñß ÞÞÞß Ð#ß 8Ñß ÞÞÞ à ÞÞÞ à ÞÞÞ à Ð8ß "Ñß Ð8ß #Ñß ÞÞÞ ß à ÞÞÞ ×
The chain looks like countably many copies of  placed end-to-end, so we can also write:
=#! œ =!  =!  ÞÞÞ  =!  ÞÞÞ œ  8− .8 where each .8 œ =! . A subset of  that represents =#!
is Ö5 
"
8
À 5ß 8 œ "ß #ß $ß ÞÞÞ ×.
Exercise Prove that
1) multiplication of order types is associative
2) the left distributive law holds for order types: .Ð/  >Ñ œ ./  .7 .
Note: Other books may define ./ “in reverse” as the order type of the lexicographically ordered set
Q ‚ R . Under that definition, =! † 2 œ =! and #=! œ =!  =! Á =! . Also, under the “reversed”
definition, the right distributive law holds but the left distributive law fails.
Which way the definition is made is not important mathematically. The arithmetic of order types
under one definition is just a “mirror image” of the arithmetic under the other definition. You just
need to be aware of which convention a writer is using.
334
Exercises
E7. Give Ò!ß "Ó# the lexicographic order Ÿ , and let Ð+ß ,Ñ represent an open interval in Ò!ß "Ó# Þ
Describe what a “small” open interval around each of the following points looks like: Ð!ß "# Ñß Ð!ß "Ñß
Ð "# ß !Ñß Ð"ß !Ñ.
E8. For each + − , we can write + uniquely in the form a œ 2< (2=  "Ñ for integers <ß = 0.
w
Suppose + w œ 2< (2= w  ") − . Define + Ÿ + w if + œ + w , or <  < w , or < œ < w and =  = w . What
is the order type of (, Ÿ )? Does a nonempty subset of (, Ÿ ) necessarily contain a smallest
element?
E9. Show that it is impossible to define an order Ÿ on the set ‚ of complex numbers in such a way
that all three of the following are true:
i) for all Dß A − ‚, exactly one of B œ Aß B  A, or D  A holds
ii) for all ?ß Aß D − ‚ À if D  A, then D  ?  A  ?
iii) if B ß C  !, then BC  !
(Hint: Begin by showing that if such an order exists, then  "  !. But notice that this, in itself, is not
a contradiction.)
E10. Give explicit examples of subsets of  which represent the order types:
a)
=!  "  =!
b)
=#!  =!
c)
=#!  # † =!  $
d)
=#!  =! † 2  $
e)
=#!  =#!
f)
=$!  =!
E11. Let ( be the order type of .
335
a) Give an example of a set F ©  such that neither F nor   F has order type ( .
b) Prove that if E is a set of type ( and F © E, then either F or E  F contains a subset
of type (.
c) Prove or disprove:
(  "  ( œ (.
Hint: Cantor's characterization of  as an ordered set may be helpful.
E12. A chain Ð\ß Ÿ Ñ is called an (" -set if the following condition holds in \ À
(*) Whenever E and F are countable subsets of \ such that +  , for every choice of
+ − E and , − F , then b- − \ such that +  -  , for all + − E and all , − F
More informally, we could paraphrase condition (*) as: for countable subsets E,F of \ß
“E  F ” Ê b- − \ such that “E  -  F ”
a) Show that ‘ is not an (" -set.
b) Prove that every (" -set is uncountable.
Hint: there is a one line argument; note that g is countable.
c) By b), an (" -set Ð\ß Ÿ Ñ satisfies ± \ ± i" , and so l\l - if we assume the continuum
hypothesis, CH. Prove that l\l - without assuming CH.
Hint: show how to define a one-to-one function 0 À ‘ Ä \ ; begin by defining 0 on .
More generally, a chain Ð\ß Ÿ Ñ is called an (α -set if, whenever E and F are subsets of \ , both of
cardinality  iα , then there is a B − \ such that “E  B  F ”. So, for example, an (! -set is
simply an order dense chain.
336
5. Well-Ordered Sets and Ordinal Numbers
We now look at a much stronger kind of order Ÿ on a set.
Definition 5.1 A poset (\ß Ÿ Ñ is called well-ordered if every nonempty subset of \ contains a
smallest element.
The definition implies that a well-ordered set \ is automatically a chain: if + Á , − \ , then set Ö+ß ,×
has a smallest element, so either + Ÿ , or , Ÿ +.
 and all its subsets are well-ordered. The set of integers, ™, is not well-ordered since, for example, ™
itself contains no smallest element. ‘ is not well-ordered since, for example, the nonempty interval
Ð0ß "Ñ contains no smallest element.
Since a well-ordered set \ is a chain, it has an order type. These special order types are very nicely
behaved and have a special name.
Definition 5.2 An ordinal number (or simply ordinal) is the order type of a well-ordered set.
Since we know how to add and multiply order types, we already know how to add and multiply
ordinals and get new ordinals. We also have a Definition 4.2 for  and Ÿ that applies to ordinals.
Theorem 5.3 If α and " are ordinals, so are α  " and α † " .
Proof Let α and " be represented by disjoint well-ordered sets E and F . Then α  " is represented
by the ordered sum E  F . We must show this set is well-ordered. Since E  F is a chain, we only
need to check that a nonempty subset G of E  F must contain a smallest element.
F is well-ordered so, if G © F , then G has a smallest element. Otherwise G ∩ E Á g and, since E is
well-ordered, there is a smallest element - − G ∩ E. In that case - is the smallest element of G .
Similarly, we need to show that the lexicographically ordered product F ‚ E is well-ordered. If G is a
nonempty subset of F ‚ E, let ,! be the smallest first coordinate of a point in G À more precisely, let
,! be the smallest element in Ö, − F À for some + − Eß Ð,ß +Ñ − G×Þ Then let +! be the smallest
element in Ö+ − E À Ð,! ß +Ñ − G×. Then Ð,! ß +! Ñ is the smallest element in G . (Intuitively, Ð,! ß +! Ñ is
the point at the “lower left corner” of GÞ The fact that F and E are well-ordered guarantees that
such a point exists.) ñ
Some examples of ordinals (increasing in size) are
0, 1, 2, ... , =! , =!  ", =!  #, ... , =!  8, ... , =! † #, =! † #  ", ... , =! † #  8, ... ,
ÞÞÞ =! † 3, ... , =! † 8, ..., =#! , =#!  ", ... , =#!  =! , =#!  =!  ", ..., =#!  =! † #, ÞÞÞ
337
All these ordinals can be represented by countable well-ordered sets (in fact, by subsets of ) so we
refer to them as “countable ordinals.” We will see later (assuming AC), there are well-ordered sets of
arbitrarily large cardinality  so that this list of ordinals barely scratches the surface.
Exercise 5.4 Find a subset of  that represents the ordinal =#!  =! † $  #.
Here are a few very simple properties of well-ordered sets. Missing details should be checked as
exercises.
1) In a well-ordered set \ , each element + except the largest (if there is one) has an
“immediate successor”— namely, the smallest element of the nonempty set ÖB − \ À B  +×.
However, an element in a well-ordered set might not have an immediate predecessor: for example in
Ö"  8" À 8 − × ∪ Ö#  8" À 8 − × ∪ Ö#×ß neither " nor # has an immediate predecessor. This set
represents the ordinal =!  =!  "Þ
2) A subset of a well-ordered set, with the inherited order, is well-ordered.
3) Order isomorphisms preserve well-ordering: if a poset is well-ordered, so is any order
isomorphic poset. An order isomorphism preserves the smallest element in any nonempty subset.
The following theorems indicate how order isomorphisms between well-ordered sets are much less
“flexible” than isomorphisms between chains in general. .
Theorem 5.5 Suppose Q is well-ordered. If 0 À Q Ä Q is a one-to-oneß order-preserving map of Q
into Q , then 0 Ð7Ñ 7 for all 7 − Q .
The theorem says that 0 cannot move an element “to the left.” Notice that Theorem 5.5 is false for
chains in general: for example, consider Q œ ‘ and 0 ÐBÑ œ 12 B. On the other hand, if you try to
construct a counterexample using Q œ , you will probably see how the proof of the theorem should
go.
338
Proof Suppose not. Then E œ Ö7 − Q À 0 Ð7Ñ  7× Á g. Let 7! be the smallest element in E.
We get a contradiction by asking “what is 0 Ð0 Ð7! ÑÑ” ?
Since 7! − E , 0 Ð7! Ñ  7! . Since 0 preserves order and is one-to-one, 0 Ð0 Ð7! ÑÑ  0 Ð7! Ñ which
means that 0 Ð7! Ñ − E. But that is impossible because 7! is the smallest element of EÞ ì
Corollary 5.6 If Q is well-ordered, then the only order isomorphism 0 from Q onto Q is the identity
map 0 Ð7Ñ œ 7.
(Note that the theorem is false for chains in general: if Q œ ‘, then 0 ÐBÑ œ B$ is an onto order
isomorphism. Corollaries 5.6 and 5.7 indicate that well-ordered sets have a very “rigid” structure.)
Proof Let 0 À Q Ä Q be an order isomorphism. By Theorem 5.5, 0 Ð7Ñ 7 for all 7. If 0 is
not the identity, then E œ Ö7 À 0 Ð7Ñ  7× Á g. Let 7! be the least element of E.
339
Then 7! cannot be in ranÐ0 Ñ (why? ). ì
Corollary 5.7 Suppose Q and R are well-ordered. If 0 À Q Ä R and 1 À Q Ä R are order
isomorphisms from Q onto R , then 0 œ 1. (So M and N can be order isomorphic “in only one way.”Ñ
Proof If 0 Á 1, then 0 " 0 and 1" 0 are two different order isomorphisms from Q onto Q , and that
is impossible by the preceding corollary. ì
Exercise 5.8 Find two different order isomorphisms between ‘ and the set of positive reals ‘ .
We have already seen that order types, in general, are not very nicely behaved. Therefore, during this
the following discussion about well-ordered sets and ordinal numbers, there is a certain amount of
fussiness in the notation  to make sure we do not jump to any false conclusions. Much of this
fussiness will drop by the wayside as things become clearer.
In a nonempty well-ordered set Q , we will often refer to the smallest element as !. (In fact, without
loss of generality, we can literally assume ! is the smallest element Q .) If we need to carefully
distinguish between the first elements in two well-ordered sets Q ß R we may write them as !Q and
!R . (This might be necessary if, say, R © Q and the smallest elements of R and Q are different.)
But usually this degree of care is not necessary.
Definition 5.9 Suppose 7 − Q , where Q is well-ordered. The initial segment of Q determined
by 7 œ ÖB − Q À B  7×. We can write this set using the “interval notation” Ò!Q ß 7ÑQ .
If a discussion involves only a single well-ordered set Q , we may simply write Ò!ß 7ÑQ or even
just Ò!ß 7ÑÞ
340
Notice that:
i) For every 7 − Q , Q Á Ò!ß 7Ñ , so Q is not an initial segment of itself, and we will see in
Theorem 5.10 that much more is true: Q cannot even be order isomorphic to an initial segment of
itself.
ii) Order isomorphisms preserve initial segments: if 0 À Q Ä R is an order isomorphism of
Q onto R , then 0 ÒÒ!ß 7ÑÓ œ Ò!ß 0 Ð7ÑÑ
iii) Given any two initial segments in a well-ordered set Q , one of them is an initial segment
of the other. More precisely, if 7  8 − Q , then “the initial segment in Q determined by
7” œ ÖB − Q À B  7× œ ÖB − Ò!ß 8Ñ À B  7× œ “the initial segment in Ò!ß 8Ñ determined by 7.”
Theorem 5.10 Suppose Q is well-ordered and R © Q . Q is not order isomorphic to an initial
segment of R . In particular (when R œ Q Ñ, Q is not order isomorphic to an initial segment of itself.
Proof Suppose 8 − R © Q . If 0 À Q Ä R © Q is one-to-one and order preserving, then Theorem
5.5 gives us that 0 Ð8Ñ 8 for each 8 − R Þ Thereforeß for each 8ß ranÐ0 Ñ Á Ò!R ß 8ÑÞ ì
Corollary 5.11 No two initial segments of Q are order isomorphic (so each initial segment of Q , as
well as Q itself, represents a different ordinal).
Proof One of the two segments is an initial segment of the other, so by Theorem 5.10 the segments
cannot be order isomorphic. ì
Definition 5.12 Suppose . and / are ordinals represented by the well-ordered sets Q and R . We say
that .  / if Q is order isomorphic to an initial segment of R  that is Q ¶ Ò!ß 8Ñ for some 8 − R .
If Q is order isomorphic to R we write . œ / . We write . Ÿ / if .  / or . œ / . (Check that the
definition is independent of the choice of well-ordered sets Q and R representing . +8. /.)
Note: We already have a different definition (4.2) for . Ÿ / when we think of . and / as
arbitrary order types. It will turn out  for ordinals  that the two definitions are equivalent,
that is:
for well-ordered sets Q and R :
Q is order isomorphic to a proper subset of R but not to R itself
Ô Ð‡Ñ
Q is order isomorphic to an initial segment of R .
The equivalence Ð*Ñ is not true for chains in general: for example, each of Ò!ß "Ó and Ð!ß "Ñ is
order isomorphic to a subset of the other, but neither is order isomorphic to an initial segment
of the other (why?)
Until Corollary 5.19, where we prove the equivalence Ð*Ñ, we will be using the new definition
5.12 of . Ÿ / for ordinals.
341
The relation Ÿ among ordinals is clearly reflexive and transitive. The next theorem implies that Ÿ is
antisymmetric  and therefore any set of ordinals is partially ordered by Ÿ .
Theorem 5.13 If . and / are ordinals, then at most one of the relations .  / ß . œ / , and .  / can
be true.
Proof Let Q and R represent . and / . If . œ / , then Q ¶ R . In this case, .  / and /  . are
impossible since a well-ordered set cannot be order isomorphic to an initial segment of itself (Theorem
5.10).
If .  / , then Q is isomorphic to an initial segment of R . If /  . were also true, then R would, in
turn, be isomorphic to an initial segment of Q . By composing these isomorphisms, we would have Q
order isomorphic to an initial segment of itself  which is impossible. ì
Notation For an ordinal ., let ord Ð.Ñ œ Öα À α is an ordinal and α  .×. If . œ !, then ord Ð.Ñ œ g
and if .  !, then ! − ordÐ.Ñß so ord(.) Á g. Like any set of ordinals, we know that ordÐ.Ñ is
partially ordered by Ÿ .
It turns out that much more is true: every set of ordinals is actually well-ordered by Ÿ , but to see that
takes a few more theorems. However, ord(.) is a very special set of ordinals and, for starters, Theorem
5.14 tells us that ordÐ.Ñ is well-ordered by Ÿ . Theorem 5.14 also gives us a very nice “standard” way
to pick a well-order that represents an ordinal ..
Theorem 5.14 If . is an ordinal represented by the well-ordered set Q , then ordÐ.Ñ ¶ Q . Therefore
ordÐ.Ñ is a well-ordered set of ordinals and ordÐ.Ñ represents ..
Proof For each ordinal α − ordÐ.Ñ, we have α  ., so α can be represented by some initial segment
Ò!ß 7α Ñ of Q Þ Define 0 À ordÐ.Ñ Ä Q by 0 ÐαÑ œ 7α . This function 0 is one-to-one since different
ordinals cannot be represented by the same initial segment of Q , and 0 clearly preserves order.
If 7 − Q , then the initial segment Ò!ß 7Ñ in Q represents some ordinal α  .. But α is represented
by Ò!ß 7α Ñ. Since different initial segments of Q are not isomorphic, we get 7 œ 7α œ 0 ÐαÑ so 0 is
onto. Therefore ordÐ.Ñ ¶ Q . ì
We will often write ordÐ.Ñ in “interval” notation:
For an ordinal ., Ò!ß .Ñ œ Öα À α is an ordinal and α  .× œ ordÐ.Ñ
By 5.14, Ò!ß .Ñ is well-ordered and represents the ordinal .; therefore any ordinal . can be represented
by the set of preceding ordinals.
For example,
0 is represented by the set of preceding ordinals, namely Ò!ß !Ñ œ ord(!Ñ œ g
342
1 is represented by Ò!ß "Ñ œ ordÐ"Ñ œ Ö!×
2 is represented by Ò!ß #Ñ œ ord(#Ñ œ Ö!ß "×
ã
=! is represented by Ò!ß =! Ñ œ ordÐ=! Ñ œ Ö!ß "ß #ß ÞÞÞß 8ß ÞÞÞ×
=!  " is represented by Ò!ß =!  "Ñ œ Ö!ß "ß #ß ÞÞÞß 8ß ÞÞÞ à =! ×
(here, “ à ” indicates that =! comes “after” all the natural numbers 8)
ã
α is represented by the set of previously defined ordinals
etc.
ã
Some comments about axiomatics
The informal definition of ordinals is good enough for our purposes, However, the preceding list
roughly illustrates how one can define ordinals in axiomatic set theory ZFC. For example, in ZFC the
ordinal # is defined by # œ Ö!ß "× (rather than saying that the set Ö!ß "× represents the ordinal #).
Definition
0 œ g
1 œ Ö!× œ Ög×
2 œ Ö!ß "× œ Ögß Ög××
ã
=! œ Ö!ß "ß #ß ÞÞÞß 8ß ÞÞÞ ×
=!  " œ Ö!ß "ß #ß ÞÞÞß 8ß ÞÞÞà =! ×
ã
etc.
and in general, an ordinal α œ the set of previously defined ordinals. Of course, this
presentation is still a little vague: in particular, some sort of “induction” in ZFC is needed
to justify the “etc.” where an ordinal is defined in terms of ordinals already defined.
Once we have defined ordinals (as sets) in ZFC, we need to say how they are compared, that is, how to
define Ÿ . We do this for ordinals α and " by writing α  " iff α − " . This seems to
accomplish what we want. For example:
"  $ because " − $
!  "  #  $  ÞÞÞ  =!  =!  " because ! − " − # − $ − ÞÞÞ − =! − =!  " − ÞÞÞ
=!  =!  "( because =! − =!  "(
343
etc.
If \ is any set well-ordered by Ÿ , we can then define its ordinal number Ð œ “the ordinal number
associated with \ ӄ as follows: from the axioms ZFC one can prove the existence of a function (set)
with domain \ that is defined “recursively” by :
a C − \ 0 ÐCÑ œ ran ( 0 lÖD − \ À D  C×Ñ
Then “the ordinal number of \ ” is defined to be the set ranÐ0 Ñ.
For example, for the well-ordered set \ œ Ö"ß $ß &×, what is the function 0 and what is the
ordinal number of \ ?
0 Ð"Ñ œ ran Ð 0 lÖD − \ À D  "×Ñ œ ran Ð0 lg Ñ œ g
0 Ð$Ñ œ ran Ð 0 lÖD − \ À D  $×Ñ œ ran Ð0 lÖ"×Ñ œ Ög×
0 Ð&Ñ œ ran Ð 0 lÖD − \ À D  &×Ñ œ ran Ð0 lÖ"ß $×Ñ œ Ögß Ög××
The ordinal number of \ is ran Ð0 Ñ œ Ö!ß Ög×ß Ögß Ög××× œ Ö!ß "ß #× œ $
In axiomatic set theory, cardinals are viewed as certain special ordinals: an ordinal α is called a
cardinal if for all ordinals "  α there is no bijection between " and α  that is a cardinal is an “initial
ordinal”  meaning that it's “the first ordinal with a given size.” From that point of view =!
is a cardinal because there is a bijection between =! and 8 for any 8  =! Þ Earlier, we gave this
cardinal the name i! . But =!  " is not a cardinal because there is a bijection between =! and =!  "Þ
In Theorem I.13.2, we proved that at most one of the relations  ß œ ß  (as defined for cardinal
numbers) can hold between two cardinals. ÐContext will make clear whether “ Ÿ ” refers to the
ordering of cardinals or ordinals.Ñ We also stated in Chapter I that (assuming AC) at least one of the
relations  ß œ ß  must hold between any two cardinals  that is, for any two sets one must be
equivalent to a subset of the other. We are almost ready to prove that statement  in fact, this statement
about cardinal numbers follows easily (assuming AC) from the corresponding result about ordinal
numbers which we now prove.
Theorem 5.15 (Ordinal Trichotomy Theorem) If . and / are ordinals, then at least one of the
relations .  / , . œ / , .  / must hold (and so, by Theorem 5.13, exactly one of these relations
holds).
Proof We already know that certain special sets of ordinals are well-ordered: for example
Ò!ß .Ñ œ Öα À α is an ordinal and α  .× (Theorem 5.14).
The theorem is certainly true if . œ ! or / œ !, so we assume both .  ! and /  !.
Let H œ Ò!ß .Ñ ∩ Ò!ß / Ñ œ Öα À α is an ordinal for which α  . and α  / }. Because we do not yet
know that . and / are comparable, the situation might look something like the following:
344
H is well-ordered because H © Ò!ß . Ñ, and H Á g Ðsince ! − H). Therefore H represents some
ordinal $  !. We claim that $ Ÿ . and $ Ÿ / .
1) To show that $ Ÿ ., we assume $ Á . and prove $  ..
Ò!ß .Ñ represents .. Since H © Ò!ß .Ñ and H represents $ Á ., we conclude H Á Ò!ß .Ñ,
so Ò!ß .Ñ  H Á g . Let # be the smallest element in Ò!ß .Ñ  H. Ò!ß # Ñ is an initial
segment of Ò!ß .Ñ.
We claim H œ Ò!ß # Ñ . If that is true, then H represents # ; but H represents $ and
therefore $ œ #  .Þ
Ò!ß # Ñ © H À If α − Ò!ß # Ñ , then α − Ò!ß .Ñ and α  # œ smallest element in
Ò!ß .Ñ  H, so α − H.
H © Ò!ß # Ñ À If α − H, then α and # are comparable since both are in the
well-ordered set Ò!ß .Ñ.
We examine the possibilities:
i) # œ α: impossible, since α − H and #  H
ii) #  α À impossible, since that would mean
#  α  . and #  α  / , forcing # − Ò!ß .Ñ ∩ Ò!ß / Ñ
œ H  which is false.
Therefore α  # , so α − Ò!ß # ÑÞ
2) A similar argument (interchanging “.” and “/ ” throughout) shows that if $ Á / ,
then $ œ #  / .
345
Since $ Ÿ . and $ Ÿ / , there are only four possibilities:
a) $  . and $  / , in which case $ − Ò!ß .Ñ ∩ Ò!ß / Ñ œ H œ Ò!ß $ Ñ  which is impossible.
Therefore one of the remaining three cases must be true:
b) $ œ . and $ œ / , in which case . œ /
c) $  . and $ œ / , in which case .  /
d) $ œ . and $  / , in which case .  /
ñ
Corollary 5.16 Any set of ordinals is linearly ordered with respect to the ordinal ordering Ÿ .
(We shall see in Theorem 5.20 that even more is true: every set of ordinals is well-ordered.)
We simply state the following theorem. A proof of the equivalences can be found, for example, in Set
Theory and Metric Spaces (Kaplansky) or Topology (Dugundji).
Theorem 5.17 The following statements are equivalent. (Moreover, each is consistent with and
independent of the axioms ZF for set theory:)
1) (Axiom of Choice) If ÖEα À α − E× is a family of pairwise disjoint nonempty sets, there is
a set F © Eα such that, for all α, |Eα ∩ F ± œ ".
(This is clearly equivalent to the statement that Eα Á g. If 0 is in the product, let F œ ran(0 ); on
the other hand, if such a set F exists, define 0 by 0 ÐαÑ œ the unique element of Eα ∩ F . An element
0 − Eα is a function that “chooses” one element 0 ÐαÑ from each Eα Þ)
2) (Zermelo's Theorem) Every set can be well-ordered, i.e., for every set \ there is a subset
Ÿ of \ ‚ \ such that Ð\ß Ÿ Ñ is well-ordered.
3) (Zorn's Lemma) Suppose Ð\ß Ÿ Ñ is a nonempty poset. If every chain in \ has an upper
bound in \ , then \ contains a maximal element.
We will look at some powerful uses of Zorn's Lemma later. For now, we are mainly interested in
Zermelo's Theorem.
It is tradition to call 2) Zermelo's Theorem and 3) Zorn's Lemma. They appear as “proven” results in
the early literature, but the “proofs” used some form of the Axiom of Choice (AC). Generally, we have
been casual about mentioning when AC is being used. However in the following theorems, for
emphasis, ÒEGÓ indicates that the Axiom of Choice is used in one of these equivalent forms.
Corollary 5.18 [AC, Cardinal Trichotomy] If 7 and 8 are cardinal numbers, at least one (and thus,
by Theorem I.13.2 , exactly one) of the relations 7  8ß 7 œ 8ß 7  8 holds. (Therefore any set of
cardinals is a chain.)
Proof According to Zermelo's Theorem, we may assume that Q and R are well-ordered sets
representing the cardinals 7 and 8, so that Q and R also represent ordinals . and / .
346
By Theorem 5.15, either . œ / , .  8ß or .  / ; therefore either Q and R are order isomorphic (in
which case 7 œ 8) or one is order isomorphic to an initial segment of the other (so 7  8 or 8  7).
ì
In fact, the Cardinal Trichotomy Corollary, is equivalent to the Axiom of Choice. (See Gillman, Two
Classical Surprises Concerning the Axiom of Choice and the Continuum Hypothesis, Am. Math.
Monthly 109(6), 2002, pp. 544-553 for this and other interesting results that do not depend on
techniques of axiomatic set theory.) Over 200 equivalents to the Axiom of Choice are given in
Equivalents of the Axioms of Choice ÐRubin & Rubin, North-Holland Publishing, 1963).
The following corollary tells us that, among ordinals, the two definitions of “ Ÿ ” (Definition 4.2 and
Definition 5.12) are equivalent.
Corollary 5.19 Suppose Q and R are well-ordered sets representing . and / . If Q is order
isomorphic to a subset of R (so . Ÿ / in the sense of Definition 4.2), then Q is order isomorphic to
R or to an initial segment of R (so . Ÿ / in the sense of Definition 5.12).
Proof Without loss of generality, we may assume Q © R Þ If Q is not order isomorphic to R , then
. Á / . If Q is also not isomorphic to an initial segment of R , then .  / is also falseÞ Therefore the
Trichotomy Theorem 5.15 gives .  / . Then R is order isomorphic to an initial segment of a subset
Q of itself  which violates Theorem 5.10. ì
Theorem 5.20 Every set [ of ordinals is well-ordered. In particular, every nonempty set of ordinals
contains a smallest element.
Proof Theorem 5.15 implies that [ is linearly ordered by Ÿ . We need to show that if E is a
nonempty subset of [ , then E contains a smallest element. Pick α − E. If α is itself the smallest in
E, we are done. If not, then Ö" − E À "  α× is nonempty and well-ordered  because it is a subset of
Ò!ß αÑ  so it contains a smallest element "! , and "! is the smallest element in [ . ì
Corollary 5.21 ÒAC] Every set G of cardinal numbers is well-ordered. In particular, every nonempty
set of cardinal numbers contains a smallest element.
Proof We know that the order Ÿ (among cardinals) is a linear order. Let H be a nonempty subset of
G and, for each cardinal 7 − H, let Q represent 7. By Zermelo's Theorem, each set Q can be wellordered, after which it represents some ordinal .. By Theorem 5.20, the set of all such .'s contains a
smallest element .! represented by Q! Þ Then 7! œ lQ! l is clearly the smallest cardinal in H. ì
Example 5.22
1) The set G œ Ö7 À 7 is a cardinal and i!  7 Ÿ -× has a smallest element. It is called the
immediate successor of i! and is denoted by i
! or i" . The statement - œ i" is the Continuum
Hypothesis which, we recall, is independent of the axioms ZFC. If CH is assumed as an additional
axiom in set theory, then G œ Öi" × œ Ö-×.
347
2) More generally, for any cardinal 7 we can consider the smallest element 7 in the set
Ö5 À 5 is a cardinal and 7  5 Ÿ #7 ×. We call 7 the immediate successor of 7. In particular, we

write i
" œ i# , i# œ i$ , and so on.
The Generalized Continuum Hypothesis is the statement
GCH À “for every infinite cardinal 7, 7 œ #7 .”
GCH and µ GCH are equally consistent with the axioms ZFC. (Curiously, ZF  GCH implies AC.
This is discussed in the Gillman article cited after Corollary 5.18.)
3) In Example VI.4.6, we (provisionally) defined the weight of a topological space Ð\ß g Ñ by
AÐ\Ñ œ i!  min ÖlU l À U is a base for g ×.
We now see that the definition makes sense  because there must exist a base of smallest cardinality.
Theorem 5.23 If [ is a set of ordinals, then there exists an ordinal greater than any ordinal in [ .
(Therefore there is no “set of all ordinals.”)
Proof Let [ ‡ œ Ö.  " À . − [ ×, and represent each ordinal .  " in [ ‡ by a well-ordered set
Q.1 . We may assume the Q." 's are pairwise disjoint. (If not, replace each Q." with
Q." ‚ Ö.  "×, ordered in the obvious way.) Form the ordered sum: that is, let W œ ÖQ." À
. − [ ×, and order W by
Bß C − Q." , and B Ÿ C in Q."
B Ÿ C if 
B − Q." ß C − Q/ " and .  /
Clearly, Ÿ well-orders W , so ÐWß Ÿ Ñ represents an ordinal 5 . Since each Q." © W , we have
.  " Ÿ 5 by Theorem 5.19. Since .  .  " Ÿ 5 for each . − [ , 5 is larger than any ordinal in
[. ì
Corollary 5.24 Every set [ of ordinals has a least upper bound, denoted sup [  that is, there is a
smallest ordinal every ordinal in [ .
Proof Without loss of generality we may assume that if α  . − [ , then α − [ (why? and where
is this assumption used in what follows?). If [ contains a largest element, then it is the least upper
bound. Otherwise, pick an ordinal 5 larger than every ordinal in [ . Then Ò!ß 5  "Ñ  [ Á g (it
contains 5 ) and the smallest element in Ò!ß 5  "Ñ  [ is sup [ . ì
Example 5.25
1) sup Ö!ß "ß #ß ÞÞÞß × œ =! , and sup{!ß "ß #ß ÞÞÞ, =! × œ =!
2) We say that an ordinal “has cardinal 7” if it is represented by a well-ordered set with
cardinality 7. In particular, countable ordinals are those represented by countable well-ordered sets.
348
3) sup Öα À α is a countable ordinal × is called =" . Since there is no largest countable ordinal
(why?), we see that =" is the smallest uncountable ordinal. A set representing =" must have cardinal
i" (why?). Since Ò!ß =" Ñ represents =" , so there are exactly i" ordinals  =" , that is, exactly i"
countable ordinals. Each countable ordinal can be represented by a subset of  Ðsee Example 4.7Ñ, so
there are exactly i" nonisomorphic well-ordered subsets of .
Since =" is the smallest uncountable ordinal, each α  =" is a countable ordinal that is represented by
Ò!ß αÑ. Therefore α has only countably many predecessors and =" is the first ordinal with uncountably
many (i" ) predecessors.
The spaces Ò!ß =" Ñ and Ò!ß ="  "Ñ œ Ò!ß =" Ó, with the order topology, have some interesting properties
that we will look at later. These properties hinge on the fact that =" is the smallest uncountable
ordinal.
The ordinals =! , ... , =!  8, ... , =! † #, ... , =#! , ... , =8! , ... are all mere countable ordinals. For ordinals
α, " it is possible to define “ordinal exponentiation” α" . (The definition is sketched in the appendix at
=
Ð= ! ! Ñ
the end of this chapter.) Then it turns out that ==! ! , ==! !  ", ... , =!
, ... , are still countable ordinals.
=
= !
supÖ=! , ==! ! ,=! ! ,
=
= !
= !
=! !
If you accept that, then you should also believe that %! œ
, ... × ( œ “=! to the
=! power =! times” ) is still countable. Roughly, each element in the set has only countably many
predecessors and the set has only countable many elements, so the least upper bound of the set still has
only countably many predecessors—namely, all the predecessors of its predecessors.
%
Ð% ! Ñ
But once you get up to %! , you can then form %!%! , %! ! , ..., take the least upper bound again, and still
have only a countable ordinal %" . And so on. So =" , the first ordinal with uncountably many
predecessors is way beyond all these: “the longer you look at =" , the farther away it gets” (Robert
McDowell).
We now look at some similar results for cardinals.
Lemma 5.26 If Ö5α À α − E} and Ö7α À α − E× are sets of cardinals and 5α 7α for each α − E,
then 5α 7α . (An infinite sum of cardinals is defined in the obvious way: if the Oα ' s are
pairwise disjoint sets with cardinality 5α , then 5α œ lOα lÞ Ñ
Proof Exercise
Theorem 5.27 If a set of cardinals G œ Ö5α À α − E× contains no largest element, then 5α  5α!
for each α! − E.
Proof For any particular α! − E, let 
7α! œ 5α!
7α œ ! for α Á α!
Then the lemma gives 5α 7α œ 5α! .
Þ
If 5α œ 5α! for some α! , 5α! would be the largest element in G which, by hypothesis, does not
exist. Therefore 5α  5α! for every α! − E. ì
349
The conclusion may be true even if G has a largest element: for example, suppose G œ Ö"ß #×.
Can you give an example involving an infinite set G of cardinals?
Corollary 5.28 If G is a set of cardinals, then there is a cardinal 7 larger than every member of G
(and therefore there is no “set of all cardinals”).
Proof If G has a largest element 5 , then let 7 œ #5 . Otherwise, use the preceding theorem and let
7 œ Ö5 À 5 − G×Þ ì
Corollary 5.29 Every set G of cardinal numbers has a least upper bound, that is, there is a smallest
cardinal every cardinal in G .
Proof Without loss of generality, we may assume that if : − G , then all cardinals smaller than : are
also in G (why? and where is this used in what follows?). If G has a largest element 5 , then 5 is the
least upper bound. Otherwise, pick a cardinal 7 greater than all the cardinals in G and let
W œ Ö: À : is a cardinal and : Ÿ 7×. Then W  G Á g (it contains 7) and the smallest cardinal in this
set is the least upper bound for G . ì
6. Indexing the Infinite Cardinals
By Corollary 5.21, the set of infinite cardinals less than a given cardinal 5 is well-ordered, so this set is
order isomorphic to an initial segment of ordinals. Therefore this set of cardinals can be “faithfully
indexed” by that segment of ordinals  that is, indexed in such a way that 5α  5" iff α  " . When
the infinite cardinals are listed in order of increasing size and indexed by ordinals, they are denoted by
i's with ordinal subscripts. In this notation, the first few infinite cardinals are

i ! , i " ( œ i
! ), i# Ð œ i" Ñ , ... , i8 , ... , i=! , i=! " , ... , i=! =! , ... , i=#! , ... , i%! ß
i%! "ß ... ß i=" , ... .
Thus, i=! œ sup Öi8 À 8  =! × and i=" = sup Öiα À α  =" ×. i=" is the first cardinal with uncountably
many (i" ) cardinal predecessors.
In this notation, GCH states that for every ordinal α, #iα œ iα" .
By definition, - œ |‘|. So where is - is this list of cardinals? The continuum hypothesis states
“- œ i" Þ” However, it is a fact that for each ordinal α, there exists an ordinal "  α such that the
assumption “i" œ - ” is consistent with ZFC. ÐPerhaps ‘ is more mysterious than you thought.Ñ
These results imply that not even the simplest exponentiation #i! involving an infinite cardinal can be
“calculated” in ZFC: is #i! œ i" ? œ i"( ? œ i=! " ?
On the other hand, one cannot consistently assume “- œ iα ” for an arbitrary choice of α: even in
ZFC, certain iα 's are provably excluded. In fact, if iα is the least upper bound of a strictly increasing
sequence of smaller cardinals, we will prove iiα!  iα (and therefore iα Á - ). It is also true, but we
will not prove it, that these are the only excluded iα 's  it is consistent to assume 2i! œ iα
for any cardinal iα for which iiα! œ iα (Solovay, 1965)Þ
350
At the heart of what we need is a classical theorem about cardinal arithmetic.
¨ Ñ Suppose that for each α − E, 7α and 8α are cardinals with 7α  8α . Then
Theorem 6.1 (Konig
α 7α  α 8α .
Proof Proving “ Ÿ ” is straightforward; proving “  ” takes a little more work.
Let sets Qα and Rα represent 7α and 8α . We may assume the Rα 's are pairwise disjoint and, since
7α  8α , that each Qα is a proper subset of Rα . For each α, pick and fix an element 8α − Rα  Qα
and define 0 À Qα Ä Rα by:
for B − Qα! , 0 ÐBÑ œ D − Rα , where DÐαÑ œ 
B
8α
if α œ α!
if α Á α!
The Qα 's are disjoint so 0 is well defined, and clearly 0 is one-to-one, so we conclude that
α 7α Ÿ α 8α .
We now show that 7α œ 8α is impossible. We do this by showing that if 2 À Qα Ä Rα is
one-to-one, then 2 cannot be onto.
Let T œ ranÐ2Ñ œ 2ÒQα Ó œ 2ÒQα Ó. Let 2ÒQα Ó œ Tα . Since 2 is one-to-one,
lTα l œ 7α so, for each α, there are at most 7α different αth coordinates of points in Tα  that
is, |ÖDα À D − Tα ×l Ÿ 7α  8α . Then we can pick a point Aα − Rα with Aα Á Dα for every
D − Tα , i.e., Aα is not the α-th coordinate of any point in Tα .
Define A − Rα by AÐαÑ œ Aα . Then A  Tα for all α, so A  T œ ranÐ2Ñ. ì
Example 6.2 Suppose we have a strictly increasing sequence of cardinals
! Á 7!  7"  ÞÞÞ  75  ÞÞÞ .
∞
∞
For each 5 , let 85 œ 75" , so 75  85 . By Konig's
Theorem, ∞
¨
5œ! 75 
5œ! 85 œ
5œ" 75
.
Ÿ ∞
7
5œ! 5
∞
∞
In particular: if 75 œ i5 , then ∞
5œ! i5 
5œ! i5 . Since i=! Ÿ
5œ! i5 (why?), we have
i!
i= !   ∞
5œ! i5 Ÿ i=! .
Since - œ - i! , we conclude that - Á i=! .
Note: A similar argument shows that if a cardinal k is the least upper bound of a sequence of strictly
increasing cardinals, then 5 i!  5 , so 5 Á - .
Exercise 6.3 For a cardinal 5 , there are how many ordinals with cardinality Ÿ 5 ?
351
7. Spaces of Ordinals
Let \ be a set of ordinals with the order topology. Since \ is well-ordered, \ is order isomorphic to
some initial segment of ordinals Ò!ß αÑÞ Therefore \ and Ò!ß αÑ are homeomorphic in their order
topologies. Therefore to think about “spaces of ordinals” we only need look at initial segments of
ordinals Ò!ß αÑ. We will look briefly at some general facts about these spaces. In Section 8, we will
consider the spaces Ò!ß =" Ñ and Ò!ß ="  "Ñ œ Ò!ß =" Ó in more detail. These particular spaces have some
interesting properties that arise from the fact that =" is the first uncountable ordinal.
According to Definition 3.1, a subbase for the order topology on \ œ Ò!ß αÑ consists of all sets
ÖB − Ò!ß αÑ À B  # × œ Ò!ß # Ñ , #  α
and
ÖB − Ò!ß αÑ À B  " × œ Ð" ß αÑ, " 0
The set of finite intersections of such sets is a base, so (check this!) a basic open set has one of the
following forms:
Ò!ß # Ñß where # Ÿ α
Ð" ß # Ñ, where " ! and # Ÿ α
Ð# œ α corresponds to the empty intersectionÑ
If α œ !ß then \ œ Ò!ß αÑ œ g; so suppose α  !Þ What does an efficient neighborhood base at a point
7 − Ò!ß αÑ look like? .
If 7 œ ! À Ö!× œ Ò!ß "Ñ is open in Ò!ß αÑ. Therefore ! is an isolated point in Ò!ß αÑ and
ÖÖ!×× is an open neighborhood base at !.
If !  7  α, then any basic open set containing 7 must contain a set of the form Ð5 ß 7 Ó:
if 7 − Ò!ß # Ñ, then 7 − Ð!ß 7 Ó © Ò!ß # Ñ
if 7 − Ð" ß # Ñ, then 7 − Ð" ß 7 Ó © Ð" ß # Ñ
Each set Ð5 ß 7 Ó œ Ð5 ß 7  "Ñ is open; and each set Ð5 ß 7 Ó is also closed because its complement
Ò!ß 5  "Ñ ∪ Ð7 ß αÑ is open. Therefore ÖÐ5 ß 7 Ó À ! Ÿ 5  7 × is a neighborhood base of clopen
sets at 7 .
Putting together these open neighborhood bases, we get that
U œ ÖÖ!×× ∪ ÖÐ5 ß 7 Ó À ! Ÿ 5  7  α×
is a clopen base for the topology.
We noted in Example 3.4 that any chain with the order topology is HausdorffÞ Therefore every ordinal
space Ò!ß αÑ is HausdorffÞ Since there is a neighborhood base of closed (in fact, clopen)
neighborhoods at each point 7 − [!ß αÑ, we know even more: Theorem VII.2.7 tells us that Ò!ß αÑ is a
X$ -space. But still more is true.
Theorem 7.1 For any ordinal α, Ò!ß αÑ is X% .
352
As remarked earlier, every chain with the order topology is X%  but the proof is much simpler for
well-ordered sets.
Proof We know that Ò!ß αÑ is X" , so need to prove that Ò!ß αÑ is normal. Suppose E and F are
disjoint closed sets in Ò!ß αÑÞ
If 7 − Eß let Y7 œ a basic open set of form Ð5 ß 7 Ó disjoint from F (or, Y7 œ Ö!× if 7 œ !)
If 7 − F , let Z7 œ a basic open set of form Ð5 ß 7 Ó disjoint from E (or, Z7 œ Ö!× if 7 œ !)
We claim that if 7" − E and 7# − F , then Y7" ∩ Z7# œ g:
The statement is clearly true if 7" or 7# œ ! so suppose both are  !. Then Y7" œ Ð5" ß 7" Ó
and Z7# œ Ð5# ß 7# ÓÞ We can assume without loss of generality that 7"  7# .
If Ð5" ß 7" Ó ∩ Ð5# ß 7# Ó Á g, then we have 7" − Ð5# ß 7# Ó which would mean that Z7# ∩ E Á gÞ
If Y œ 7 −E Y7 and Z œ 7 −F Z7 , then Y and Z are disjoint open sets with Y ª E and Z ª FÞ ñ
(Why doesn't the same proof work for chains with the order topology? )
Definition 7.2 An ordinal " is called a limit ordinal if "  ! and " has no immediate predecessor; "
is called a nonlimit ordinal if " œ ! or " has an immediate predecessor (that is, " œ #  " for some
ordinal # ).
Example 7.3
1) If " is a limit ordinal in Ò!ß αÑß then for all 5  " , Ð5 ß " Ó Á Ö" ×. Therefore " is not
isolated in the order topology. If " is a nonlimit ordinal, then Ö" × œ Ö!× or, for some # ,
Ö" × œ Ð# ß " Ó. Either way, Ö" × is open so " is isolated. Therefore the isolated points in Ò!ß αÑ are the
exactly the points that are not limit ordinals.
2) Ò!ß =! Ñ is discrete: it is homeomorphic to .
3) Ò!ß =!  "Ñ is homeomorphic to Ö" 
"
8
À 8 − × ∪ Ö"×Þ
4) For what α's is Ò!ß αÑ connected?
Theorem 7.4 Suppose α  !. The ordinal space Ò!ß αÑ is compact iff Ò!ß αÑ œ Ò!ß " Ó for some ordinal
"  that is, iff Ò!ß αÑ contains a largest element " .
Proof Suppose Ò!ß αÑ has no largest element. Then h œ ÖÒ!ß # Ñ À #  α× is an open cover of Ò!ß αÑ.
The sets in h are nested so, if there were a finite subcover, there would be a single set Ò!ß # Ñ covering
Ò!ß αÑ. That is impossible since # − Ò!ß αÑ  Ò!ß # ÑÞ
Conversely, suppose Ò!ß αÑ œ Ò!ß " ÓÞ If " œ !, then Ò!ß " Ó œ Ö!× is compact, so we assume "  !.
Let h be an open cover of Ò!ß " Ó. We can assume h consists of basic open sets, that is, sets of the form
Ö!× or Ð5 ß 7 Ó. In that case, necessarily, Ö!× − h Þ
353
Let "" œ "  !. For some 5"  "" ß we have a set Ð5" ß "" Ó − h Þ If 5" œ !ß then ÖÖ!×ß Ð5" ß "" Ó× is a
finite subcover.
If 5"  !, then for some "# ß 5" − (5# ß "# Ó − h Þ If 5# œ !, then ÖÖ!×ß Ð5# ß "# Óß Ð5" ß "" Ó× is a finite
subcover.
We proceed inductively. Having chosen Ð55 ß "5 Ó, if 55  ! we can choose (55" ß "5" Ó − h so that
55 − Ð55" ß "5" ÓÞ
We continue until 58 œ ! occurs  and this must happen in a finite number of steps because otherwise,
we would generate an infinite descending sequence of ordinals 5"  5#  5$  ...  58  ÞÞÞ . This is
impossible because then the well-ordered set Ö5" , 5# , ..., 58 , ... × would have no smallest element.
When we reach 58 œ !, we have a finite subcover from h À ÖÖ!×ß (58 ß "8 Óß ÞÞÞß Ð5" ß "" Ó×.
Example 7.5 Ò!ß =! Ñ is not compact, but Ò!ß =!  "Ñ œ Ò!ß =! Ó is compact.
homeomorphic to Ö"× ∪ Ö"  8" À 8 − ×Þ
ñ
In fact, Ò!ß =! Ó is
8. The Spaces Ò!ß =" Ñ and Ò!ß =" Ó
Theorem 8.1 For each 8  =! ß suppose α8  =" . Then α œ sup Öα8 À 8  =! ×  ="  that is, the
sup of a countable set of countable ordinals is countable.
Proof Ò!ß α8 Ñ is countable so 8  =! Ò!ß α8 Ñ is countable. Since =" has uncountably many
predecessors, there is an ordinal # − Ò!ß =" Ñ  8  =! Ò!ß α8 Ñ. Then #  α8 for each 8, so
α œ sup Öα8 À 8  =! × Ÿ #  =" Þ ñ
Corollary 8.2 Ò!ß =" Ó and Ò!ß =" Ñ are not separable.
Proof If H is a countable subset of Ò!ß =" Ó, then α œ sup ÐH  Ö=" ×Ñ  =" Þ Then
cl H © Ò!ß αÓ ∪ Ö=" × Á Ò!ß =" ÓÞ
A dense set in Ò!ß =" Ñ is also dense in Ò!ß =" Ó, so Ò!ß =" Ñ is not separable.
ñ
Corollary 8.3 In Ò!ß =" Óß no sequence from Ò!ß =" Ñ can converge to =" .
Proof Suppose Ðα8 Ñ is a sequence in Ò!ß =" Ñ. Let α œ sup Öα8 À 8 − ×  =" Þ The α8 's are all in
the closed set Ò!ß αÓ, so Ðα8 Ñ Ä
Î =" . ñ
354
Example 8.4 In Example III.9.8, we saw a rather complicated space P in which sequences are not
sufficient to describe the topology. Corollary 8.3 gives an example that may be easier to “see” À
=" − cl Ò!ß =" Ñ, but no sequence Ðα8 Ñ in Ò!ß ,=" ) can converge to =" Þ
This implies that Ò!ß =" Ó is not first countable. Of course, ÖÐ5 ß 7 Ó À 5  7 × is a countable neighborhood
base at each point 7  ="  so the “problem” point is =" . The neighborhood poset a=" (ordered by
reverse inclusion) is very nicely ordered (in fact, well-ordered) but the chain of neighborhoods is just
“too long” and we cannot “thin it out” enough to get a countable neighborhood base at =" .
In contrast, the basic neighborhoods of Ð!ß !Ñ in the space P were very badly “entangled”Þ The
neighborhood system aÐ!ß!Ñ had a very complicated order structure  too complicated for us to find a
countable subset of aÐ!ß!Ñ that “goes arbitrarily far out” in the poset. (See discussion in Example
2.4.6).
In Theorem IV.8.11 we proved that certain implications hold between various “compactness-like”
properties in a topological space \ .
 \ is compact
or
(*) 
 \ is sequentially compact
Ê \ is countably compact Ê \ is pseudocompact.
We asserted that, in general, no other implications are valid. The following corollary shows that
“sequentially compact” Ê
Î “compact” (and therefore “countably compact” Ê
Î “compact” and
“pseudocompact” Ê
Î “compact” ).
Corollary 8.5 Ò!ß =" Ñ is sequentially compact.
Proof Suppose Ðα8 Ñ is a sequence in Ò!ß =" Ñ. We need to show that Ðα8 Ñ has a convergent
subsequence in Ò!ß =" ÑÞ Without loss of generality, we may assume that all the α8 's are distinct (why?)
The sequence Ðα8 Ñ has either an increasing subsequence α8"  α8#  ÞÞÞ  α85  ÞÞÞ or a
decreasing subsequence α8"  α8#  ÞÞÞ  α85  ÞÞÞ
The argument is completely parallel to the one in Lemma IV.2.10 showing that a sequence in
‘ has a monotone subsequence. Call α8 a peak point of the sequence if α8 α5 for all
5 8Þ
If the sequence has only finitely many peak points, then after some α8" there are no peak
points and we can choose an increasing subsequence α8"  α8#  ÞÞÞ  α85  ÞÞÞ
If Ðα8 Ñ has infinitely many peak points, then we can choose a subsequence of peak points
α8" ß ÞÞÞα85 ß ÞÞÞ and for these, α8" α8# ÞÞÞ α85 ÞÞÞ Þ Since the α8 's are distinct, we
have α8"  α8#  ÞÞÞ  α85  ÞÞÞ .
However, a strictly decreasing sequence of ordinals α8"  α8#  ÞÞÞ  α85  ÞÞÞ is impossible.
Therefore Ðα8 Ñ has a subsequence of the form α8"  α8#  ÞÞÞ  α85  ÞÞÞ . Setting
α œ supÖα85 À 5 œ "ß #ß ÞÞÞ×  =" , we see that then Ðα8 Ñ Ä α − Ò!ß =" ÑÞ ñ
355
An example of a pseudocompact space that is not countably compact is given in Exercise E32.
In Chapter X, we will discuss a space " that is compact (therefore countably compact and
pseudocompact ) but not sequentially compact (see Example X.6.5). That will complete the set
of examples showing that “no other implications exist” other than those stated in (*).
Corollary 8.6 In Ò!ß =" Ó, the intersection of a countable collection of neighborhoods of =" is again a
neighborhood of ="  that is, the intersection must contain a “tail” Ðαß =" Ó. Therefore Ö=" × is not a
K$ -set (and therefore not a zero set) in Ò!ß =" ÓÞ
Proof Let ÖR5 À 5 œ "ß #ß ÞÞÞ× be a collection of neighborhoods of =" . For each 5 , there is a α5 for
which =" − Ðα5 ß =" Ó © int R5 © R5 Þ Let α œ supÖα5 À 5 − ×  =" Þ Then
∞
∞
∞
5œ" R5 ª
5œ" int R5 ª Ðαß =" Ó. In particular,
5œ" R5 Á Ö=" ×Þ ñ
Since Ò!ß =" Ó is compact, we know that each 0 − GÐÒ!ß =" ÓÑ is bounded. In fact, something more is
true. (Why does the corollary state something “more” ?)
Corollary 8.7 If 0 − GÐÒ!ß =" ÓÑ, then 0 is constant on the “tail” Òαß =" Ó for some α  =" .
"
Proof Suppose 0 Ð=" Ñ œ <. By Corollary 8.6, 0 " ÒÖ<×Ó œ ∞
ÒÐ<  8" ß <  8" ÑÓ contains a tail
8œ" 0
Ðαß =" ÓÞ Therefore 0 lÒαß =" Ó œ <Þ ñ
Proving Corollary 8.7 was relatively easy because we can see immediately what the constant value
would have to be for the theorem to be true: < œ 0 Ð=" ÑÞ A more remarkable thing is that the same
result holds for Ò!ß =" ÑÞ But to prove that fact, we have no “initial guess” about what constant value 0
might have on a tail, so we have to work harder.
Theorem 8.8 If 0 − GÐÒ!ß =" ÑÑ, then 0 is constant on the “tail” Ðαß =" Ñ for some α  =" .
Proof Let Xα œ Òαß =" Ñ œ “the αth tail”Þ By Corollary 8.5, Ò!ß =" Ñ is countably compact so the
closed set Xα is also countably compact. It is easy to see that a continuous image of a countably
compact space is countably compact, so 0 ÒXα Ó is a countably compact subset of ‘. Since countable
compactness and compactness are equivalent for subsets of ‘ (Theorem IV.8.17), each 0 ÒXα Ó is a
nonempty compact set: 0 ÒXα Ó © 0 ÒX! Ó © ‘Þ
The Xα 's are nested À 0 ÒXα Ó © 0 ÒX" Ó © 0 ÒX! Ó if "  αÞ Therefore 0 ÒXα Ó's have the finite intersection
property, and by compactness α=" 0 ÒXα Ó Á g (see Theorem IV.8.4). In fact, we claim the intersection
contains a single number À α=" 0 ÒXα Ó œ Ö<×Þ
If <ß = − α=" 0 ÒXα Ó, then 0 assumes each of the values <ß = for arbitrarily large values of α.
Therefore we can pick an increasing sequence α"  ""  α#  "#  ÞÞÞ  α8  "8  ÞÞÞ such
that 0 Ðα8 Ñ œ < and 0 Ð"8 Ñ œ =Þ Let # œ sup Öα8 ß "8 À 8 − ×  =" Þ Then Ðα8 Ñ Ä # and
Ð"8 Ñ Ä # . By continuity Ð0 Ðα8 ÑÑ œ Ð<Ñ Ä # and Ð0 Ð"8 ÑÑ œ Ð=Ñ Ä # . and we conclude
# œ < œ =Þ
356
We claim that 0 ´ < on some tail of Ò!ß =" ÑÞ First notice that if "  =" , then b #8  " for which
0 ÒX#8 Ó © Ð<  8" ß <  8" ÑÞ
If not, then Ö 0 ÒX# Ó À "  #  =" × ∪ Ö 0 ÒÒ!ß =" ÑÓ  Ð<  8" ß <  8" Ñ À 8 −  × would be a
family of closed subsets of 0 ÒX! Ó with the finite intersection property, so this family would
have a nonempty intersection. That is impossible since " #=" 0 ÒX# Ó œ Ö<× and
"
"
< Â ∞
8œ" 0 ÒÒ!ß =" ÑÓ  Ð<  8 ß <  8 Ñ.
Pick #" so that 0 ÒX#" Ó © Ð<  "ß <  "ÑÞ Pick ##  #" so that 0 ÒX## Ó © Ð<  "# ß <  "# Ñ and continue
"
"
inductively to pick #8"  #8 so that 0 ÒX#8" Ó © Ð<  8"
ß <  8"
ÑÞ
"
"
Let 7 œ supÖ#8 À 8 − ×  =" Þ Then 0 ÒX7 Ó © 0 Ò X#8 Ó © 0 ÒX#8 Ó © ∞
8œ" Ð<  8 ß <  8 Ñ œ Ö<×ß
so 0 lX7 œ <Þ ñ
(Since 0 is bounded on the compact set Ò!ß 7 Ó, we see in a different way that Ò!ß =" Ñ is pseudocompact.)
Corollary 8.9 Every continuous function 0 À Ò!ß =" Ñ Ä ‘ can be extended in a unique way to a
continuous function J À Ò!ß =" Ó Ä Ò!ß =" Ó.
Proof For some < − ‘ß 0 œ < on a tail Òαß =" ÑÞ Let J lÒ!ß =" Ñ œ 0 and define J Ð=" Ñ œ <Þ
Any continuous extension K of 0 must agree with J since J and K agree on the dense set Ò!ß =" ÑÞñ
Note: Ò!ß =" Ó is a compact X# space which contains Ò!ß =" Ñ as a dense subspace. We call Ò!ß =" Ó
a compactification of Ò!ß =" Ñ. The property stated in Corollary 8.9 is a @/<C special property for a
compactification to have À in fact, it characterizes Ò!ß =" Ó as the so-called Stone-Cech compactification
of Ò!ß =" ÑÞ We will discuss compactifications in Chapter 10.
By way of contrast, notice that Ò  "ß !Ó is compactification of Ò  "ß !Ñ which, just as above, is
obtained by adding a single point to the original space. However, the continuous function
0 À Ò  "ß !Ñ Ä ‘ defined by 0 ÐBÑ œ sin Ð B" Ñ cannot be continuously extended to the point !.
We saw in Example VII.5.10 that a subspace of a normal space need not be normal: the Sorgenfrey
plane \ is not normal however it can be embedded in the X% -space Ò!ß "Ó7 for some 7. Any space \
that is X$ " but not X% works just as well. However, these examples are not very explicit  it is hard to
#
“picture why” the normality of Ò!ß "Ó7 isn't inherited by the subspace \ . The “picturing” may be easier
in the following example.
Example 8.10 Let X ‡ œ Ò!ß =" ] ‚ Ò!ß =! ], sometimes called the “Tychonoff plank.” X ‡ is compact X#
and therefore X% . Discarding the “upper right corner point,” we are left with the (open) subspace
X œ X ‡  ÖÐ=" ß =! )× We claim that X is not normal. Let
E œ ÖÐ=" ß 8Ñ − X À 8  =! × œ “the right edge of X ” and
F œ ÖÐαß =! Ñ − X À α  =" × œ “the top edge of X ”
E and F are disjoint sets and closed in X (although not, of course, in X ‡ ).
357
Suppose Y is an open set in X containing the “right edge” E. For each point Ð=" ß 8Ñ − E © Y , we can
choose a basic open set Ðα8 ß =" Ó ‚ Ö8× © Y . Let α œ sup Öα8 À 8 œ !ß "ß #ÞÞÞ×  =" Þ Then
Ðαß =" Ó ‚ Ö8× © Y for all 8  that is, E is contained in a “vertical strip” Ðαß =" Ó ‚ Ò!ß =! Ñ inside Y .
Suppose Z is an open set in X containing the “top edge” F. Since Ðα  "ß =! Ñ − F, there is a basic
open set Öα  "× ‚ Ð8ß =! Ó © Z . But then Ðα  "ß 8  "Ñ − Y ∩ Z , so Y ∩ Z Á g.
Exercise 8.11 Show that every continuous function 0 À X Ä ‘ can be continuously extended to a
function J À X ‡ Ä ‘Þ
Hint: 0 lÒ!ß =" Ñ ‚ Ö=! × has constant value < on some tail, and for each 8  =! ß 0 l Ò!ß =" Ó ‚ Ö8×
has a constant value <8 on a tail. Define J ÐÐ=" ß =! ÑÑ œ <Þ Prove that Ð<8 Ñ Ä < and then show
that the extension J is continuous at Ð=" ß =! ÑÞ
As in the remark following Corollary 8.9, X ‡ is a compactification of X and the functional extension
property in the exercise characterizes X ‡ as the so-called Stone-Cech compactification of X . Since X ‡
is compact, J must be bounded  so, in retrospect, 0 must have been bounded in the first place.
Therefore X is pseudocompact. But X is not countably compact  because the “right edge” E is an
infinite set that has no limit point in X . X is an example of a pseudocompact space that is not
countably compactÞ
358
Exercises
E13. Suppose G © ‘ and that G is well-ordered (in the usual order on ‘). Prove that G is countable.
E14. Let =‡! denote the order type of the set of nonpositive integers, with its usual ordering.
a) Prove that a chain Ð\ß Ÿ Ñ is well-ordered iff \ contains no subset of order type =‡! .
b) Prove that if Ð\ß Ÿ Ñ is a chain in which every countable subset is well-ordered, then \ is
well-ordered.
c) Prove that every infinite chain either has a subset of order type =‡! or one of order type =! .
E15. Prove the following facts about ordinal numbers αß " ß # À
a) if "  !, then α  "  α
b) if α  " , then there exists a unique # such that α œ "  #
We might try using b) to define subtraction of ordinals: α  # œ " if α œ "  # .
However this is perhaps not such a good idea. (Consider =! œ α, " œ "). Problems arise
because ordinal addition is not commutative.
c) "  α œ α iff α = ! .
E16. Let F © E, where ÐEß Ÿ Ñ be a chain. F is called inductive if
for all > − E,
Ö+ − E À +  >× © F Ê > − F .
Prove that if E is the only inductive subset of E, then E is well-ordered.
E17. Let \ be a first countable space. Suppose that for each α  =" , Jα is a closed subset of \ and
that Jα" © Jα# whenever α" Ÿ α#  =" . Prove that ÖJα À α  =" × is closed in \Þ
E18. Let E be well-ordered. Order P œ E ‚ Ò!ß "Ñ lexicographically and give the set the order
topology.
a) What does a “nice” neighborhood base look like at each point in P? Discuss some other
properties of this space.
b) If E œ Ò!ß =" Ñ, then P  Ö(!ß !Ñ× is called is the “long line.” Show that P is path connected
and locally homeomorphic to ‘ but it cannot be embedded in ‘. (See Topology, J. Munkres, 2nd
edition, p. 159 for an outline of a proof.)
c) Each point in P homeomorphic to ‘Þ P is normal but not metrizable.
E19. a) Let E and F be disjoint closed sets in Ò!ß =" Ó. Prove that at least one of E and F is compact
and bounded away from =" . (A set G is bounded away from =" if G © Ò!ß αÓ for some α  =" .Ñ
b) Characterize the closed sets in Ò!ß = " ) that are zero sets.
c) Prove that Ò!ß = " Ñ and Ò!ß = " Ó are not metrizable.
359
E20. a) Suppose ÐB8 Ñ and ÐC8 Ñ are sequences in \ œ Ò!ß =" ) such that, for all 8, B8 Ÿ C8 Ÿ B8" .
Show that both sequences converge and have the same limit.
b) Show that if 0 À \ Ä \ is such that 0 ÐBÑ B for every B, then there is an B such that ÐBß BÑ
is a limit point of the graph of 0 in \ ‚ \ .
c) Prove that \ ‚ Ò!ß =" ] is not normal.
(Hint: Let ? be the diagonal of \ ‚ \ and B = X ‚ Ö=" ×. Show that if U and V are open with
? © Y and F © Z , then Y ∩ Z Á g. To do this: if any point ÐBß =" Ñ is in U, we're done. So suppose
this is false and define 0 ÐBÑ to be the least ordinal  B such that ÐBß 0 ÐBÑÑ Â Y . Use part b). )
E21. A space \ is called 5 -compact if \ can be written as a countable union of compact sets.
a) Prove Ò!ß = " Ñ is not 5 -compact.
b) Using part a) (or otherwise), prove Ò!ß = " Ñ is not Lindelof.
¨
c) State and prove a theorem of the form: an ordinal space [!ß αÑ is 5 -compact iff ...
(You might begin by thinking about the spaces Ò!ß =" Ñ, Ò!ß =# Ñ, and Ò!ß ==! Ñ.)
E22. A space \ is called functionally countable if every continuous 0 À \ Ä ‘ has a countable range.
a) Show that \ œ Ò!ß =" Ñ is functionally countable.
b) Let ] œ H ∪ Ö:× where H is uncountable and :  H. Give ] the topology for which all the
points of H are isolated and for which the basic neighborhoods of : are those cocountable sets
containing :. Show ] is functionally countable.
c) Prove that \ ‚ ] is not functionally countable.
(Hint: Let 1À \ Ä H be one-to-one. Consider the set L œ ÖÐαß 1ÐαÑ À α is isolated in \× © \ ‚ ] )
E23. A space \ is called ] -compact if \ is homeomorphic to a closed subspace of the product ] 7
for some cardinal 7. For example, \ is Ò!ß "Ó-compact iff \ compact and X# . An ‘-compact space is
called realcompact.
a) Prove that if \ is both realcompact and pseudocompact, then \ is compact.
(Note: the converse is clear)
b) Suppose that \ is Tychonoff and that however \ is embedded in Ò!ß "Ó7 , its projection in
every direction is compact. (More precisely, suppose that for all possible embeddings h À \ Ä Ò!ß "Ó7 ,
we have that 1α Ò2Ò\ÓÓ is compact for all projections 1α . This statement is certainly true, for example,
if \ is compact.) Prove or disprove that \ must be compact.
E24. An infinite cardinal 5 is called sequential if 5 œ ∞
8œ" 58 for some sequence of cardinals 58  5 .
a) Prove that if 5 is sequential, then 5 i!  5 .
b) Assume GCH. Prove that if 5 is infinite and 5 i!  5 , then 5 i! œ #5 .
c) Assume GCH. Prove that an infinite cardinal 5 is sequential iff 5 i!  5 .
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9. Transfinite Induction and Transfinite Recursion
“...to understand recursion, you have to understand recursion....”
Suppose we have a sequence of propositions T8 that depend on 8 œ !ß "ß #ß ÞÞÞ We would like to show
that all the T8 's are true. (The initial value 8 œ ! is not important. We might want to prove the
propositions T8 for, say, "$ Ÿ 8  =! .) For example, we might have in mind the propositions
T8 À
!  "  #  ÞÞÞ  8 œ
8Ð8"Ñ
#
As you should already know, the proof can be done by induction. Induction in elementary courses
takes one of two forms (stated here using T! as the initial proposition):
1) (Principle of Induction) If T! is true and if (T5" is true Ê T5 is trueÑ, then T8 is true for
all 8  =! Þ
2) (Principle of Complete Induction) If T! is true and if ÐT4 is true for all 4  5 Ê T5 is
trueÑ, then T8 is true for all 8  =! Þ
Formally, 2) looks weaker than 1), because it has a stronger hypothesis. But in fact the versions 1) and
2) are equivalent statements about Ò!ß =! ÑÞ (Why?) Sometimes form 2) is more convenient to use. For
example, try using both versions of induction to prove that every natural number greater than " has a
factorization into primes.
The Principle of Induction works because Ò!ß =! Ñ is well-ordered:
If Ö8 − Ð!ß =! Ó À T8 is false× Á g, then it would contain a smallest element 5  !. This is
impossible: since T5" is true, T5 must be true.
You might expect a principle analogous to 1) could be used in every well-ordered sets Ò!ß αÑ, not just
in Ò!ß =! ÑÞ But an ordinal " might not have an immediate predecessor, so version 1) might not make
sense. So we work instead with 2): we can generalize “ordinary induction” if we state it in the form of
complete induction.
Theorem 9.1 (Principle of Transfinite Induction) Let α be an ordinal and X © Ò!ß αÑÞ If
1) ! − T and
2) a" − Ò!ß αÑ [ Ò!ß " Ñ © T Ê " − T ] ,
then T œ Ò!ß αÑ.
Proof If X Á Ò!ß αÑ, then there is a smallest " − Ò!ß αÑ  X Þ By definition of " , Ò!ß " Ñ © X . But then
2) implies " − X , contrary to the choice of " .
Using Theorem 9.1 is completely analogous to using complete induction in Ò!ß =! ÑÞ For each #  α,
we have a proposition T# and we want to show that all the T# 's are all true. (For example, we might
have a set O# © \ somehow defined for each #  α, and T# might be the proposition “O# is
compact.”) Let X œ Ö#  α À T# is true× © Ò!ß αÑÞ If we show that T! is true, and ifß assuming that
assume T# is true for all #  " , we can then prove that T" must be true, then Theorem 9.1 implies that
T#. is true for all #  α. We will look at several examples in Section 10.
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Note: In the statement of Theorem 9.1, part 1) is included only for emphasis. In fact, 1) is
automatically true if we know 2) is true: for if we let " œ ! in 2)ß then Ò!ß !Ñ œ g © X is true, so
! − X  that is, T! is true. But in actually using Theorem 9.1 (as described in the preceding
paragraph) and trying to prove that 2) is true, the first value " œ ! requires us to show T! is true with
“no induction assumptions” since there are no T# 's with #  " œ !. Doing that is just
verifying that 1) is true.
We can also define objects in a similiar way  by transfinite recursion. Elementary definitions by
recursion should be familiar  for example, we might say À
let 0 Ð!Ñ œ 13 and,
for each 8  ! , let 0 Ð8Ñ œ #0 Ð8  "Ñ (**)
We than say that “0 is defined for all 8 œ !ß "ß #ß ÞÞÞ.” We draw that conclusion by arguing in the
following informal way: if not, then there is a smallest k − Ð!ß =! Ñ for which 0 is not defined: this is
impossible because then 0 Ð5  "Ñ is defined, and therefore (by **) so is 0 Ð5Ñ.
This argument depends only on the fact that Ò!ß =! Ñ is well-ordered, so it generalizes to the following
principle.
Informal Principle 9.2 (Transfinite Recursion) For each "  α, suppose a rule is given that defines
an object T" in terms of objects T# already defined (that is, in terms of T# 's with #  " ). Then T" is
defined for all "  α. ÐThe principle implies that P! is defined “absolutely”  that is, without any
previous T# 's to work with  since there are no T# 's with  !×.
This “informal” statement is a reasonably accurate paraphrase of a precise theorem in axiomatic set
theory, and the informal proof is virtually identical to the one given above for simple recursion on
Ò!ß =! ÑÞ As stated in 9.2, this principle is strong and clear enough for everyday use, and we will
consider it “proven” and useable.
Principle 9.2 is “informal” because it does seem a little vague in spots  “some object”, “a rule is
given”, “if T# is defined ... , then T" is defined ...” Without spending a lot of time on the set theoretic
issues, we digress to show how the statement can be made a little more precise.
In axiomatic set theory, the “objects” T" will, of course, be sets (since everything is a set). We
can think of “defining T" ” to mean choosing T" from some specified set X . In the context of
some problem, X is the “universal set” in which the objects T" will all live. For example, we
might have X œ GÐ\Ñ and want to define continuous functions T" − GÐ\Ñ for each "  αÞ
The sets T# already chosen (“defined”) in X for #  " can be described efficiently by a
function <" − X Ò!ß" Ñ À for #  " , <" Ð# Ñ œ T# − X Þ
To define the set T" in terms of the preceding T# 's means that we need to define T" using <" .
We need a function (“rule”) V" so that V" Ð<" Ñ gives the new set T" − X . In other words, we
want V" À X Ò!ß"Ñ Ä X Þ
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The conclusion that we have “completed” the process and that T" is defined for all "  α
means that there is a function J − X Ò!ßαÑ where, for each "  αß J Ð" Ñ œ the T" selected at the
earlier stage by V"  that is, J Ð"Ñ œ V" ÐJ l Ò!ß " ÑÑ .
This leads us to the following formulation. The “full” formulation of the standard theorem
about transfinite recursion in axiomatic set theory needs to be a little stronger still, so we call
this version  which we state without proof  the “weak” version. It is more than adequate for
our purposes here.
Theorem 9.3 (Transfinite Recursion, Weak Form) Suppose α is an ordinal. Let X be a set and
suppose that, for each "  α, we have a function V" À X Ò!ß" Ñ Ä X . Then there exists a unique function
J − X Ò!ßαÑ such that, for each "  α, J Ð" Ñ œ V" ÐJ l Ò!ß " ÑÑ.
Proof See, for example, Topology (J. Dugundji)
The following example illustrates what the Recursion Theorem 9.3 in a concrete example. When all is
said and done, it looks just the way an informal, simple definition by recursion (Principle 9.2) should
look.
Example 9.4 We want to define numbers I8 for every 8 œ !ß "ß ÞÞÞ . Informally we might say:
#
Let E! œ 1 and for 8  !, let I8 œ I! #  ÞÞÞ  I8"
.
The informal Principle 9.2 lets us conclude that I8 is defined for all 8  =! .
In terms of the more formal Theorem 9.3, we can describe what is “really” happening as follows:
Let Ò!ß αÑ œ Ò!ß =! Ñ and X œ . For each 8  =! , define V8 À  Ò!ß8Ñ Ä  as follows:
For 8 œ ! À  Ò!ß!Ñ œ  g œ Ög× and define V ! ÐgÑ œ "
For 8  ! À if < −  Ò!ß8Ñ , define V8 Ð<Ñ œ <# Ð!Ñ  ÞÞÞ  <# Ð8  "Ñ
(Note that the R8 's are defined explicitly for each n, not recursively.).
The theorem states that there is a unique function J − Ò!ß=! Ñ such that
J Ð!Ñ œ V! ( J lÒ!ß !Ñ Ñ œ V! ÐgÑ œ 1
J Ð"Ñ œ V" (J lÒ!ß "Ñ Ñ œ J Ð!Ñ# œ 1# œ 1
J Ð#Ñ œ V# (F lÒ!ß #Ñ Ñ œ J Ð!Ñ#  J Ð"Ñ# œ 2
ÞÞÞ etc. ...
which is just what we wanted: dom J œ Ò!ß =! Ñ, so I8 œ J Ð8Ñ is defined for all
8  =! Þ
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10. Using Transfinite Induction and Recursion
This section presents a number of examples using recursion and induction in an essential way. Taken
together, they are a miscellaneous collection, but each example has some interest in itself.
Borel Sets in Metric Spaces
The classical theory of Borel sets is developed in metric spaces (\ß .ÑÞ The collection of Borel sets in
a metric space is important in analysis and also in set theory. Roughly, Borel sets are the sets that can
be generated from open sets by the operations of countable union and countable intersection “applied
countably many times.” Therefore a Borel set is only a “small” number of operations “beyond” the
open sets, and Borel sets are fairly well behaved. We will use transfinite recursion (the informal
version) to define the Borel sets and prove a few simple theorems. When objects are defined by
recursion, proofs about them often involve induction.
We begin with a simple lemma about ordinals.
Lemma 10.1 Every ordinal α can be written uniquely in the form α œ "  8 where " is either 0 or a
limit ordinal, and 8  =! .
Proof If α is finite, then α œ !  8 where 8  =! .
Suppose α is infinite. If α is a limit ordinal, we can write α œ α  !Þ If α is not a limit
ordinal, then α has an immediate predecessor which, for short, we denote here as “α  ".” If α  " is
not a limit ordinal, then it has an immediate predecessor α  #. Continuing in this way, we get to a
limit ordinal after a finite number of steps, 8  for otherwise α  "  α  #  ÞÞÞ  α  8  ÞÞÞ
would be an infinite decreasing sequence of ordinals. If " œ α  8 is a limit ordinal, then α œ "  8Þ
To prove uniqueness, suppose α œ "  8 œ " w  8 w where each of " ß " w is ! or a limit ordinal
and 8ß 8 w are finite. If " or " w œ ! À say " œ !. Then "  8 is finite, so " w œ ! and 8 œ 8 w . So
suppose " and " w are both limit ordinals. We have an order isomorphism 0 from Ò!ß "  8Ñ onto
Ò!ß " w  8 w Ñ. Since Ò!ß "  8Ñ contains 8  " ordinals after its largest limit ordinal " , the same must be
true in the range Ò!ß " w  8 w Ñ, and therefore 8 œ 8 w . Let 1 œ 0 lÒ!ß " ÑÞ Then 1 À Ò!ß " Ñ Ä Ò!ß " w Ñ is an
order isomorphism, so " œ " w Þ ñ
Definition 10.2 Suppose α œ "  8, where " is a limit ordinal or !ß and 8 is finite. We say that α is
even if 8 is even, and that α is odd if 8 is odd.
For example, every limit ordinal α œ α  ! is even.
Definition 10.3 Suppose Ð\ß .Ñ is a metric space. Let Z! œ g. ß the collection of open sets. For each
!  α  =" , and suppose that Z" has been defined for all "  αÞ Then let
Zα œ ÖK À K is a countable intersection of sets from ÖZ" À "  α×
 Z œ ÖK À K is a countable union of sets from ÖZ À "  α×
α
"
U œ ÖZα À α  =" × is the family of Borel sets in Ð\ß .ÑÞ
364
Ðif α is odd)
(if α is even)
The sets in Z" are the K$ sets; the sets in Z# are countable unions of K$ sets and are traditionally called
K$5 sets; the sets in Z$ are called K$5$ -sets, etc.
Theorem 10.4 Suppose Ð\ß .Ñ is a metric space.
1) If α  "  =" , then Zα © Z" , so
Z! © Z" © Z# © ÞÞÞ © Zα © ÞÞÞ © Z" © ÞÞÞ
Ðα  "  =" Ñ
2) Zα is closed under countable unions if α is even, and Zα is closed under countable
intersections if α is odd.
3) U is closed under countable unions, intersections and complements. Also, if F" and F# are
in U , so is F"  F# Þ
Proof 1) Suppose α  "  =" . Z" is defined as the collection of all countable unions (if " is even)
or intersections (if is odd) of sets in the preceding families Zα Ðα  " Ñ. In particular, any one set
from Zα is in Z" .
2) Suppose α is even and K" ß K# ß ÞÞÞß K8 ß ÞÞÞ − Zα . Each K8 is a countable union of sets
ÖZ" À "  α×, so
from ÖZ" À "  α×. Therefore ∞
8œ" K8 is also a countable union of sets from
∞
8œ" K8 − Zα Þ The proof is similar if α is odd.
3) Suppose F" , F# ,..., F8 ß ÞÞÞ − U. For each 8, F8 − Zα8 for some α8  =" .
If α œ sup Öα8 À 8 œ "ß #ß ÞÞÞ ×  =" , then Zα8 © Zα © Zα" for every 8. By part 2), one of the
collections Zα or Zα" is closed under countable intersections and the other under countable unions.
∞
Therefore ∞
8œ" F8 and
8œ" F8 are both in Zα" © U Þ
To show that U is closed under complements, we first prove, using transfinite induction, that
if K − Zα ß then \  K − Zα" Þ
α œ ! À If K − Z! , then K is open so \  K is closed. But a closed set in a metric space is
a K$ set, so \  K − Z" Þ
Suppose the conclusion holds for all "  α  =" . We must show it holds also for Zα .
Let K − Zα Þ If α is odd, then K œ ∞
8œ" K8 ß where K8 − Z"8 Ð"8  αÑÞ
By the induction hypothesis, \  K8 © Z"8 " © Zα" for all 8. Since α  " is
∞
even, we have that ∞
8œ" Ð\  K8 Ñ œ \ 
8œ" K8 œ \  K − Zα" Þ (The
case when α is even is similar).
If F# − U , then F# − Zα for some α  =" , so \  F# − Zα" © U Þ So if F" − U , then
F" ∩ Ð\  F# Ñ œ F"  F# − U Þ ñ
Part 3) of the preceding theorem shows why the definition of the Borel sets only uses ordinals α  =" .
Once we get to U œ ÖZα À α  =" ×, the process “closes off”  that is, continuing with additional
countable unions and intersections produces no new sets.
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Definition 10.3 presents the construction of the Borel sets “from the bottom up.” It has the advantage
of exhibiting how the sets in U are constructed step by step. However, it is also possible to define U
“from the top down.” This approach is neater, but it gives less insight into which sets are Borel.
Definition 10.5 A family Æ of subsets of \ is called a 5-algebra if \ − Æ and Æ is closed under
complements, countable intersections, and countable unions.
Suppose T is a collection of subsets of \ . Then c Ð\Ñ is (the largest) 5 -algebra containing T . It is
also clear that the intersection of a collection of 5 -algebras is a 5 -algebra. Therefore the smallest
5 -algebra containing T exists: it is the intersection of all 5 -algebras containing T .
The following theorem could be taken as the definition of the family of Borel setsÞ
Theorem 10.6 The family U of Borel sets in Ð\ß .Ñ is the smallest 5 -algebra containing all the open
sets of \ .
Proof The rough idea is that our previous construction puts into U all the sets that need to be there to
form a 5 -algebra, but no others.
We have already proven that U is a 5 -algebra containing the open sets. We must show U is the
smallest  that is, if U w is a 5 -algebra containing the open sets, then U © U w . We show this by
transfinite induction.
We are given that Z! © U w Þ
Suppose that Z" © U w for all "  α  =" Þ We must show Zα © U w Þ
Assume α is odd. If K − Zα then K œ ∞
8œ" K8 where K8 − Z"8 for some "8  αÞ
By hypothesis each Z"8 − U w and U w is closed under countable intersections so
K − U w . Therefore Zα © U w Þ (The case when α is even is entirely similar.)
Since Zα © U w for all α  =" , we get that U œ α=" Zα © U w Þ ñ
The next theorem gives us an upper bound for the number of Borel sets in a separable metric space.
Theorem 10.7 If Ð\ß .Ñ is separable metric space, then lU l Ÿ - .
Proof We prove first that for each α  =" ß lZα l Ÿ -Þ
α œ ! À A separable metric space has a countable base V for the open setsÞ Since every open set is the
union of a subfamily of V, we have lg. l œ lZ! l Ÿ lc ÐVÑl Ÿ #i! œ -Þ
Assume that lZ" l Ÿ - for all "  α  =" . Since α has only countably many predecessors,
l" α Z" l Ÿ -Þ Since each set in Zα is a countable intersection or union of a sequence of sets from
" α Z" , we have lZα l Ÿ l" α Z" l  Ÿ - i! œ -Þ
Therefore |U l œ lα=" Zα l Ÿ i" † - œ -Þ ñ
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(Can you see a generalization to arbitrary metric spaces?)
Corollary 10.8 There are non-Borel sets in ‘.
For those who know a bit of measure theory: every Borel set in ‘ is Lebesgue measurable. Since a
subset of a set of measure ! is measurable, all #- subsets of the Cantor set G are measurable.
Therefore there are Lebesgue measurable subsets of G that are not Borel sets.
Example 10.9
1) If Ð\ß .Ñ is discrete, then every subset is open, so every subset is Borel: U œ X. œ c Ð\ÑÞ
2) If \ œ Ö!× ∪ Ö 8" À 8 − ×, then every subset is a K$ , but \ is not discrete. Therefore
Z! © Z" œ Z# œ ÞÞÞ œ Zα œ ÞÞÞ œ U Þ
Á
3) The following facts are true but harder to prove:
a) For each α  =" , there exists a metric space Ð\ß .Ñ for which
Z! © Z" © ÞÞÞ © Zα œ Zα" œ ÞÞÞ œ Z" œ ÞÞÞ œ U
Á
Á Á
In other words, the Borel construction continually adds fresh sets until the αth stage but not
thereafter.
b) In ‘, Zα Á Z" for all α  "  =" Ñ  that is, new Borel sets appear at every stage
in the construction.
A New Characterization of Normality
Definition 10.10 A family Y of subsets of a space \ is called point-finite if aB − \ , B is in only
finitely many sets from Y .
Definition 10.11 An open cover h œ ÖYα À α − E× of \ is called shrinkable if there exists an open
cover i œ ÖZα À α − E× of \ such that, for each α, cl Z α © Yα . i is called a shrinkage of h .
Theorem 10.12 \ is normal iff every point-finite open cover of \ is shrinkable. (In particular, every
finite open cover of a normal space is shrinkable: see Exercise VII.E18.)
Proof Suppose every point-finite open cover of \ is shrinkable and let E and F be disjoint closed sets
in \ . The open cover h œ Ö\  Eß \  F× has a shrinkage i œ ÖZ" ß Z# × and the sets Y œ \  cl
Z" and Z œ \  cl Z# are disjoint open sets containing E and F . Therefore \ is normal.
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Conversely, suppose \ is normal and let h œ ÖYα À α − E× be a point-finite open cover of \ .
Without loss of generality, we may assume that the index set E is a segment of ordinals Ò!ß # Ñ (why?)
so that h œ ÖYα À α  # ×Þ
Let J! œ \  α! Yα Þ Since J! © Y! , we can use normality to choose an open set Z! such that
J! © Z! © cl Z! © Y! and ÖZ! × ∪ ÖYα À α  !× covers \ .
Suppose !  α  # and that for all "  α we have defined an open sets Z" such that cl Z" © Y" and
such that ÖZ" À "  α× ∪ ÖY" À " α× covers \ . Letting Jα œ \  Ð" α Z" ∪ " α Y" Ñ © Yα ,
we can use normality to choose an open set Zα with Jα © Zα © cl Zα © YαÞ
Clearly, ÖZ" À "  α  "× ∪ ÖY" À " α  "× covers \ .
By transfinite recursion, the Zα 's are defined for all α  # , and we claim that i œ ÖZα À α  # × is a
cover of \ .
Notice that there is something here that needs to be checked: we know that we have a cover
ÖZ" À "  α× ∪ ÖY" À " α× at each step in the process, but do we still have a cover when
we're finished? To see explicitly that there is an issue, consider the following example.
Let \ be the set of reals with the “left-ray” topology (a normal space) and consider the open
cover h œ ÖÐ  ∞ß 8Ñ À 8  =! ×. If we go through the procedure described above, we get
J! œ \  8! Ð  ∞ß 8Ñ œ g so we might have chosen Z! œ g: that would give
J! © Z! © cl Z! © Y! and ÖZ! × ∪ ÖY8 À 8  !× would still be a cover. Continuing, we can
see that at every stage we could choose Z8 œ g and that ÖZ! ß Z" ß ÞÞÞZ8 × ∪ ÖY5 À 5  8× is still
a cover. But when we're done, the collection i œ ÖZ8 À 8  =! × is not a cover! Of course,
the cover h is not point-finite.
Suppose B − \Þ Then B is in only finitely many sets of h  say Yα" ß ÞÞÞß Yα8 . Let α be the largest of
these indices so that B Â " α Y" . If B is in one of the Z" 's with "  α, we're done. Otherwise
B − \  Ð" α Z" ∪ " α Y" Ñ œ Jα © Zα . Either way B is in a set in i , so i is a coverÞ ñ
Question: what happens if “point-finite” is changed to “point-countable” in the hypothesis? Could
the “max” in the argument be replaced by a “sup” ?
Definition 10.13 A cover of \ is called locally finite if every point B − \ has a neighborhood that
intersects at most finitely many of the sets in the cover.
Corollary 10.14 Suppose h œ ÖYα À α − E× is a locally finite open cover of a normal space \ . Then
there exist continuous functions 0α À \ Ä Ò!ß "Ó such that
i) 0α ± \  Yα œ ! for every α − E
ii) Ö0α ÐBÑ À α − E× œ " for every B − \ .
The collection of functions Ö0α × is called a partition of unity subordinate to h .
Proof It is easy to see that a locally finite open cover of \ is point-finite. By Theorem 10.12, h has a
shrinkage i œ ÖZα À α − E×. For each α, we can use Urysohn's Lemma (VII.5.2) to pick a continuous
function 1α À \ Ä Ò!ß "Ó such that 1α lcl Zα œ " and 1α l\  Yα œ !Þ
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Each point B is in only finitely many Yα 's, so 0α ÐBÑ œ ! for all but finitely many α's and therefore
1ÐBÑ œ  1α ÐBÑ − ‘. Each point B has a neighborhood RB which intersects only finitely many Yα 's
α−E
so 1lRB is essentially a finite sum of continuous functions and 1 is continuous (see Exercise III.E21).
Since each B is in some set cl Zα! , so 1α! ÐBÑ œ ". Therefore 1ÐBÑ is never ! so each 0α ÐBÑ œ
1α ÐBÑ
1ÐBÑ
is
continuous. The 0α 's clearly satisfy both i) and ii). ñ
A Characterization of Countable Compact Metric Spaces
In 1920, the journal Fundamenta Mathematicae was founded by Zygmund Janiszewski, Stefan
Mazurkiewicz and Waclaw Sierpinski. It was a conscious attempt to raise the profile of Polish
mathematics thorough a journal devoted primarily to the exciting new field of topology. To reach the
international community, it was agreed that published articles would be in one of the most popular
scientific languages of the day: French, German or English. Fundamenta Mathematicae continues
today as a leading mathematical journal with a scope broadened somewhat to cover set theory,
mathematical logic and foundations of mathematics, topology and its interactions with algebra, and
dynamical systems.
An article by Sierpinski and Mazurkiewicz appeared in the very first volume of this journal
characterizing compact, countable metric spaces in a rather vivid way. We will prove only part of this
result: our primary purpose here is just to illustrate the use of transfinite recursion and induction and
the omitted details are messy.
Suppose Ð\ß .Ñ is a nonempty compact, countable metric space.
Definition 10.15 For E © \ , define Ew œ ÖB − \ À B is a limit point of E×. Ew is called the derived
set of E. If E is closed, it is easy to see that Ew © E and that Ew is also closed.
We will use the derived set operation Ð w Ñ repeatedly in a definition by transfinite recursion.
Let E! œ \ and, for each α  =" , define
Eα œ E"

Eα œ ÖE" À "  α×
w
if α œ "  "
if α is a limit ordinal
Eα is closed for all α  =" and E! ª E" ª ÞÞÞ ª Eα ª ÞÞÞ . This sequence is called the derived
sequence of \ .
For some α  =" , we have Eα" œ Eα  because otherwise, for each α, we could choose a point
Bα − Eα  Eα" and ÖBα À α  =" × would be a subset of \ with cardinal i" . Let α! be the
smallest α for which Eα! œ Eα! " . It follows that E" œ Eα! for all "  α! .
Since \ is compact metric, the closed set Eα! is complete, and because Eα! œ Eα! " , every point in
Eα! is a limit point. Therefore Eα! œ g, since a nonempty complete metric space with no isolated
points contains at least - points (Theorem IV.3.6).
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We know α! œ ! is impossible (because that would mean Eα! œ E! œ \ œ gÑ. If α! were a limit
ordinal, then Eα! œ ÖE" À "  α! × Á g (because ÖE" À "  α! × is a family of nonempty compact
sets with the finite intersection property). Therefore α! must have an immediate predecessor "! .
The definition of α! implies that E"! Á g. In fact, E"! must be finite (if it were an infinite set in the
compact space \ , E"! would have a limit point and then E"w ! œ Eα! Á g)Þ Let 8 œ lE"! l  =! Þ
In this way, we arrive at a pair Ð"! ß 8Ñ, where
E"! is the last nonempty derived set of \ Ð"!  =" Ñ, and
E"! contains 8 points Ð!  8  =! ÑÞ
Since homeomorphisms preserve limit points and intersections, it is clear that the construction in any
space homeomorphic to \ will produce the same pair Ð"! ß 8Ñ.
Theorem 10.16 (Sierpinski-Mazurkiewicz) Let "!  =" and 8  =! . Two nonempty compact,
countable metric spaces \ and ] are homeomorphic iff they are associated with the same pair Ð"! ß 8Ñ.
(Therefore Ð"! ß 8Ñ is a “topological invariant” that characterizes nonempty compact countable
metric spaces.) For any such pair Ð"! ß 8Ñ, there exists a nonempty compact countable metric space
associated with this pair.
Corollary 10.17 There are exactly i" nonhomeomorphic compact countable metric spaces.
Proof The number of different compact countable metric spaces is the same as the number of pairs
Ð"! ß 8Ñ, namely i" † i! œ i" Þ ñ
Example 10.18 In the following figure, each column contains a compact countable subspace of ‘"
with the
invariant pair listed.
Each figure is built up using sets order-isomorphic to
"
Ö!× ∪ Ö 8 À 8 − ×Þ
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Alexandroff's Problem
A compact metric space Ð\ß .Ñ is second countable and therefore satisfies l\l Ÿ - (Theorem II.5.21).
In 1923, Alexandroff and Urysohn conjectured that a stronger result is true:
A first countable compact Hausdorff space \ satisfies l\l Ÿ -Þ
This conjecture was not settled until 1969, in a rather famously complicated proof by Arhangel'skii.
Here is a proof of an even stronger result  “compact” is replaced by “Lindelof”
¨  that comes from a
few years after Arhangel'skii's work.
Theorem 10.16 ÐPol, ŠapirovskiÑ If \ is first countable, Hausdorff and Lindelof,
¨ then l\l Ÿ -Þ
Proof For each : − \ , choose a countable open neighborhood base i: at : and, for each E © \ , let
iE œ Öi: À : − E× œ the collection of all the basic neighborhoods of all the points in E.
For each countable family of sets i © iE for which \  i Á g, pick a point ;i − \  i Þ
Define T ÐEÑ œ cl (E ∪ Öall such ;i 's chosen for E×ÑÞ
Notice that if lEl œ - , then lT ÐEÑl œ - .
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Since lEl œ - , we have liE l œ -. There are at most - i! œ - countable families i © iE , so
lE ∪ Öall such ;i 's chosen for E×l œ - . Since \ is first countable, sequences suffice to
describe the topology, and since \ is Hausdorff, sequential limits are unique. Therefore
lT ÐEÑl is no larger than the number of sequences in E ∪ Öall such ;i 's chosen for
E×  namely - i! œ -Þ
Now fix, once and for all, a set E © \ with lEl œ -Þ (If no such E exists, we are done!) By
recursion, we now “build up” some new sets Eα from E. The idea is that the new Eα 's always have
cardinality Ÿ - and that the new sets eventually include all the points of \ .
Let E! œ E.
For each ordinal α  =" , define
T ÐE Ñ
Eα œ   "
{E" : "  α}
if α œ "  "
.
if α is a limit ordinal
For each α  =" , lEα l œ - À
lE! l œ -Þ
Suppose lE" l œ - for all "  αÞ
If α œ "  ", then Eα œ T ÐE" Ñ, so lEα | œ - .
If α is a limit ordinal, then lEα l œ l{E" À "  α×l œ - † i! œ -Þ
Let F œ ÖEα À α  =" ×Þ Then - Ÿ lFl œ lÖEα À α  =" ×l Ÿ - † i" œ -Þ
We claim that F œ \ and, if so, we are done.
F is closed in \ : if B − cl F , then (using first countability) there is a sequence ÐB8 Ñ in F with
ÐB8 Ñ Ä BÞ If B8 − Eα8 and we let α œ supÖα8 ×  =" , then every B8 is in Eα . Therefore
B − cl Eα œ Eα" © FÞ
Since F is a closed subspace of the Lindelof
¨ space \ , F is Lindelof.
¨
If F Á \ß then we can pick a point ; − \  F . For each : − F , choose an open
neighborhood Z: − i: such that ; Â Z: . The Z: 's form an open cover of F , so a countable
collection of these sets, say i œ ÖZ: À : − G , for some countable G © F× covers F . Thus
F © i and ; Â i . But this is impossible:
Since G is countable, we have that G © Eα for some α  =" and i is a countable
subfamily of iEα for which \  i Á gÞ By definition of T ÐEα Ñ œ Eα  " Ñ, a point
;i − \  i was put into the set Eα" © F . This contradicts the fact that i covers
F. ñ
The Mazurkiewicz 2-set
The circle W " is a subset of the plane which intersects every straight line in at most two points. We use
recursion to construct something more bizarre.
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Theorem 10.17 (Mazurkiewicz) There exists a set E © ‘# such that lE ∩ Pl œ # for every straight
line P.
Proof Let $ be the first ordinal of cardinal - . Since the well-ordered segment Ò!ß $ Ñ represents $ , there
are - ordinals α  $ Þ
In fact, there are - limit ordinals  $ Þ There are certainly infinitely many (say 7) limit ordinals  $
(why?) and, by Lemma 10.1, every ordinal α  $ can be written uniquely in the form
α œ "  8, where " is a limit ordinal (or !) and 8 is finite. Therefore there are 7 † i! œ 7 ordinals
 $ . But $ has - predecessors. Therefore 7 œ -Þ
Since ‘# has exactly - points and exactly - straight lines, we can index both ‘# and the set of straight
lines using the ordinals less than $ : ‘# œ Ö:0 À 0  $ × and ÖP0 À 0  $ ×Þ
We will define points +α for each α  $ , and the set E œ Ö+α À α  $ × will be the set we want.
Let +! be the first point of ‘# (as indexed above) not on P! .
Suppose that we have defined points +0 for all 0  α  $ . We need to define +α Þ Let
Eα œ Ö+0 À 0  α×
Xα œ ÖP À P is a straight line containing 2 points of Eα }
"α œ least ordinal so that P"α  Xα
(that is, P"α is the first line listed that is not in Xα Ñ
Wα œ Ö: À : is a point of intersection of P"α with a line in Xα ×
Note lEα l  Note lXα l  -
Note lWα l  -
Since lEα ∪ Wα l  - , there are points on P"α not in Eα ∪ Wα : let +α be the first
:0 listed that is on P"α but not in Eα ∪ Wα .
By recursion, we have now defined +α for all α  $ . Let E œ Ö+α À α  $ ×Þ
We claim that for each straight line P, lE ∩ Pl Ÿ #.
If lE ∩ Pl  #, then we could pick 3 points +" ß +# ß +α − P ∩ E, where, sayß "  #  αÞ
Then +" ß +# − Eα , so that P − Xα . Since +α − P"α (by definition) and +α − P, we have
+α − P ∩ P"α . Therefore +α − Wα  which contradicts the definition of +α Þ
To complete the proof, we will show that for each straight line P , lE ∩ Pl #.
We begin with a series of observations:
a) If α" Ÿ α#  $ ß then "α" Ÿ "α# . Ò Clearly Xα" © Xα# . But P"α" is the first line not
in Xα" and P"α2 is the first line not in Xα# , so "α" Ÿ "α# Þ Ó
b) If α"  α#  α3  $ ß then "α"  "α3 Þ Ò Otherwise, by a), "α" œ "α# œ "α$ . Then
+α" ß +α# ß +α$ are on P"α" œ P"α# œ P"α$ . Then P"α$ contains 3 points of E which is
impossible.Ó In particular, if α" , α$ are distinct limit ordinals, there is a third ordinal α#
between them so "α" and "α$ must be distinct.
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c) For any #  $ , there is α  $ such that "α  # Þ Ò Since there are - limit ordinals  $ ,
there are - distinct values for "α . They cannot all be Ÿ # , since # has fewer than predecessors.Ó
Finally, if P œ some P# , pick an α so that "α  # . Since P"α is the first line  Xα ß we
get that P œ P# − Xα Þ Therefore P contains 2 points of Eα © EÞ ñ
More generally, it can be shown more that if for each line P we are given a cardinal number 7P with
# Ÿ 7P Ÿ - , then there exists a set E © ‘# such that for each P, lE ∩ Pl œ 7P Þ
11. Zorn's Lemma
Zorn's Lemma (ZL) states that if every chain in a nonempty poset Ð\ß Ÿ Ñ has an upper bound in \ ,
then \ contains a maximal element. We remarked in Theorem 5.17 that the Axiom of Choice (AC)
and Zermelo's Theorem are equivalent to Zorn's Lemma.
As a first example using Zorn's Lemma, we prove part of Theorem 5.17, in two different ways.
Theorem 11.1 ZL Ê AC
Proof 1 Let ÖEα À α − E× be a collection of pairwise disjoint nonempty sets. Consider the poset set
c œ ÖW © α−E Eα À for all α, lW ∩ Eα l Ÿ "×, ordered by inclusion. c is a nonempty poset because
g − c.
Suppose ÖW" À " − M× is a chain in c and let F œ ÖW" À " − M×. We claim that F − c Þ
Suppose lF ∩ Eα! l # for some α! Þ Consider two points B Á C − F ∩ Eα! Then B − W" " and
C − W" # for some "" , "# − M . Since the W" 's are a chain, either W"" © W"2 or W"# © W"" .
Therefore one of these sets, say W"" ß contains both Bß CÞ Therefore lW"" ∩ Aα! l  ". But this is
impossible since W"" − c . Therefore F contains at most one point from each Eα so F − c Þ
Since F is an upper bound for the chain in c , Zorn's Lemma says that there is a maximal element
Q − cÞ
Since Q − c ß lQ ∩ Eα l Ÿ " for every α. If Q ∩ Eα! œ g for some α! , then we could choose an
+ − Eα! form the set Q w œ Q ∪ Ö+× ª Q Þ Then lQ ∩ Eα l Ÿ " would still be true for every α, so we
Á
w
would have Q − c , which is impossible because Q is maximal. Therefore lQ ∩ Eα l œ " for every
α. ñ
Notice that the function 0 À E Ä Eα given by 0 œ ÖÐαß CÑ − E ‚ α−E Eα À C − Q ∩ Eα ×
is in the product {Aα À α − E×, so we see that {Aα À α − E× Á g.
Proof 2 Let ÖEα À α − E× be a collection of nonempty sets. Let
374
c œ Ö1 À F Ä Eα À F © E and 0 Ð" Ñ − E" for each " − F×
If 1" ß 1# − c , then the function 1" ß 1# are sets of ordered pairs. So we can order c by inclusion:
1" Ÿ 1# iff 1" © 1# Þ (This relation is just “functional extension”: 1" Ÿ 1# iff domÐ1" Ñ © domÐ1# Ñ
and 1# ldomÐ1" Ñ œ 1" Þ) Ðc ß Ÿ Ñ is a nonempty poset because g − c .
Suppose Ö13 À 3 − M× is a chain in c . Define 1 œ Ö13 À 3 − M×. Since the 13 's form a chain, their
union is a function 1 À F Ä Eα ß where F œ 3−M domÐ13 ÑÞ Moreover, if " − F , then " − domÐ13 Ñ
for some 3, so 1Ð" Ñ œ 13 Ð" Ñ − E" . Therefore 1 − c , and 1 is an upper bound on the chain. By Zorn's
Lemma, c has a maximal element, 0 .
By maximality, the domain of 0 is E  if not, we could extend the definition of 0 by adding to its
domain a point α from E  domÐ0 Ñ and defining 0 ÐαÑ to be a point in the nonempty set Eα Þ
Therefore 0 − ÖEα À α − E×ß so ÖEα À α − E× Á g. ñ
In principle, it should be possible to rework any proof using transfinite induction into a proof that uses
Zorn's Lemma and vice-versa. However, sometimes one is much more natural to use than the other.
We now present several miscellaneous examples that further illustrate how Zorn's Lemma is used.
The Countable Chain Condition and %-discrete sets in Ð\ß .Ñ
Definition 11.2 Suppose Ð\ß .Ñ is a metric space and %  !. A set E © \ is called %-discrete if
.ÐBß CÑ % for every pair B Á C − EÞ
Theorem 11.3 For every %  !, Ð\ß .Ñ has a maximal %-discrete set.
For example, ™ is a maximal "-discrete set in ‘.
Proof Let %  !. The theorem is clearly true if \ œ g, so we assume \ Á gÞ
Let c œ ÖE © \ À E is %-discrete× and partially order c by inclusion © .
If + − \ , then Ö+× − c ß so c Á g.
Suppose ÖGα À α − E× is a chain in Ðc ß Ÿ Ñ. We claim α−E Gα is %-discrete.
If Bß C − α−E Gα , then Gα" and C − Gα# for some α" , α# Þ Since the Gα 's form a chain, so
Gα" © Gα# or Gα# © Gα" Þ Without loss of generality, Gα" © Gα# . Then Bß C − Gα# , and this set
is %-discrete. Therefore .ÐBß CÑ %Þ
Therefore α−E Gα − c . Clearly, α−E Gα is an upper bound for the chain ÖGα À α − E×. By Zorn's
Lemma Ðc ß Ÿ Ñ has a maximal element. ñ
Notice that when we use Zorn's Lemma, an upper bound that we produce for a chain in Ðc , Ÿ Ñ  for
example, α−E Gα in the preceding paragraph  might not itself be a maximal element in c . For
example, suppose c is a poset that contains exactly four sets Eß Fß Gß H ordered by inclusion. A
particular c is indicated in the following diagram:
375
One chain in c is ÖEß G×. Then ÖEß G× œ G is an upper bound for the chain in c . However,
G © H so G is not a maximal element in c .
It is always true that an upper bound for a maximal chain in c (such as ÖEß Gß H×, above) is
a maximal element in c (why?). The statement that “in any poset, every chain is contained in
a maximal chain” is called the Hausdorff Maximal Principle  and it is yet another equivalent
to the Axiom of Choice.
Definition 11.4 A space \ satisfies the countable chain condition (CCC) if every family of disjoint
open sets in \ is countable.
We already know that every separable space satisfies CCC. The following theorem shows that CCC is
equivalent to separability among metric spaces.
Theorem 11.5 Suppose Ð\ß .Ñ is a metric space satisfying CCC. Then Ð\ß .Ñ is separable.
Proof For each 8 − , we can use Theorem 11.3 to get a maximal 8" -discrete subset H8 . For
B Á C − H8 ß F " ÐBÑ ∩ F " ÐCÑ œ g, so, by CCC, the family ÖF " ÐBÑ À B − H8 × must be countable.
#8
#8
#8
Therefore H8 is countable, and we claim that the countable set H œ ∞
8œ" H8 is dense.
Suppose D − \  H and, for %  !, choose 8 so that 8"  %Þ Since H8 is a maximal
"
"
8 -discrete set, H8 ∪ ÖD× is not 8 -discrete, so there is a point B − H8 © H
"
with .ÐBß DÑ  8  %Þ ñ
Sets of cardinals are well-ordered
We proved this result earlier (Corollary 5.21) using ordinals. The following proof, due to Metelli and
Salce, avoids any mention of ordinals  but it makes heavy use of Zorn's Lemma and the Axiom of
376
Choice. The statement that any set of cardinals is well-ordered is clearly equivalent to the following
theorem.
_
Theorem 11.6 If Ö\α À α − E× is a nonempty collection of sets, then b α − E such that l\α_ l Ÿ l\α l
_
for every α − E  in other words, for each α − E there exists a one-to-one map 9αα
À \α_ Ä \α Þ
Proof Assume all the \α 's are nonempty (otherwise the theorem is obviously true). Then by [AC],
α−E \α Á gÞ Let c œ ÖF © \α À for all α − E, 1α lF is one-to-one×, ordered by inclusion.
Ðc ß © Ñ is a nonempty poset since g − c .
If ÖF3 À 3 − M× is any chain in c , we claim that F œ F3 − c . Otherwise there would be points
B Á C − F and an α for which 1α ÐBÑ œ 1α ÐCÑ. Since the F 3 's form a chain, we would have both
Bß C − F3 for some 3, and this would imply that 1α lF3 is not one-to-one.
F is an upper bound for the chain ÖF3 À 3 − M× in c , so by Zorn's Lemma c contains a maximal
element, Q Þ
_
We claim that for some α − E, 1 α_ lQ À Q Ä \ α_ is onto.
Otherwise we would have \α  1α ÒQ Ó Á g for every α . Using [AC] again, we could choose
a point C œ ÐCα Ñ − α−E Ð\α  1α ÒQ ÓÑ Á gÞ Since Cα  1α ÒQ Ó, 1α lÐQ ∪ ÖC×Ñ would be
one-to-one for all α so that Q ∪ ÖC× − c Þ Since C Â Q , Q ∪ ÖC× is strictly larger than Q
and that is impossible because Q is maximal.
_
œ 1α ‰ Ð1α_ lQ Ñ" À \α_ Ä \α is "  " for
Therefore 1 α_ lQ À Q Ä \ α_ is a bijection and the map 9αα
every α − E. ñ
Maximal ideals in a commutative ring with unit 1 Suppose O is a commutative ring with a unit
element. (If “commutative ring with unit” is unfamiliar, then just let O œ GÐ\Ñ throughout the whole
discussion. In that case, the “unit” is the constant function ".).
Definition 11.7 Suppose M © O , where O is a commutative ring with unit. A subset M of O is called
an ideal in O if
1) M Á O
2) +ß , − M Ê +  , − M
3) + − M and 5 − O Ê 5+ − M
In other words, an ideal in O is a proper subset of O which is closed under addition and “superclosed”
under multiplication.
An ideal M in O is called a maximal ideal if: whenever N is an ideal and M © N ß then M œ N Þ
A maximal ideal is a maximal element in the poset of all ideals of O , ordered by inclusion.
For example, two ideals in O œ G (‘) are:
M œ Ö1 À 1 œ 0 3 where 0 ß 3 − GБ) and 3 is the identity function 3ÐBÑ œ B×
For example, 1ÐBÑ œ B/B − M (using 0 ÐBÑ œ /B ÑÞ
377
M œ Ö0 À 0 Ð!Ñ œ !×
In fact, Q is a maximal ideal in GÐ‘Ñ (This take a small amount of work to verify; see
Exercise E33.)
Maximal ideals, Q , are important in ring theory  for example, if Q is a maximal ideal in O , then
the quotient ring OÎQ is actually a field.
Using Zorn's Lemma, we can prove
Theorem 11.8 Let O be a commutative ring with unit. Every ideal M in O is contained in a maximal
ideal Q Þ (Q might not be unique.)
Proof Let c œ ÖN À N is an ideal and M © N ×. Ðc ß © Ñ is a nonempty poset since I − c . We want
to show that c contains a maximal element.
Suppose ÖJα À α − E× is a chain in c . Let J œ ÖJα À α − E×. Since M © N ß we only need to check
that N is an ideal to show that N − c Þ
If +ß , − J, then + − J# and , − J" for some # ß " − EÞ Since the Jα 's form a chain, either
J# © N" or J" © N# : say J# © N" . Then +ß , are both in the ideal N" , so +  , − N" © J.
Moreover, if k − O , 5+ − J" © J. Therefore J is closed under addition and superclosed under
multiplication.
Finally, J Á O : If N œ O , then 1 − N so 1 − Jα for some α. Then, for all k − O , k œ
k † 1 − Nα . So Jα œ K, which is impossible since Jα is an ideal.
By Zorn's Lemma, we conclude that c contains a maximal element Q Þ ñ
Basis for a Vector Space It is assumed here that you know the definition of a vector space Z over a
field O . (O is the set of “scalars” which can multiply the vectors in i . If “field” is unfamiliar, then
you may just assume O œ ‘, , or ‚ in the result.) Beginning linear algebra courses usually only
deal with finite-dimensional vector spaces: the number of elements in a basis for Z is called the
dimension of Z . But some vector spaces are infinite dimensional. Even then, a basis exists, as we
now show.
Definition 11.9 Suppose Z is a vector space over a field OÞ A collection of vectors F © Z is called
a basis for Z if each nonzero @ − Z can be written in a unique way as a finite linear combination of
elements of F using nonzero coefficients from O . More formally, F is a basis if for each @ − Z ,
@ Á !ß there exist unique nonzero α" , ..., α8 − O and unique b" , ..., b8 − F such that @ œ 83œ" α3 b3 .
Z is called finite dimensional if a finite basis F exists.
Definition 11.10 A set of vectors G © Z is called linearly independent if, whenever α" , ..., α8 − O
and c" , ..., c8 − G and 83œ" α3 c3 œ !, then α" œ ÞÞÞ œ α8 œ 0 (in other words, the only linear
combination of elements of G adding to ! is the trivial combination).
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Theorem 11.11 Every vector space Z Á Ö!} over a field O has a basis. (The trivial vector space Ö!×
cannot have a basis under our definition: the only nonempty subset Ö!× is not linearly independent.)
Proof We will use Zorn's Lemma to show that there is a maximal linearly independent subset of i and
that it must be a basis.
Let c œ ÖG © Z À G is linearly independent×, ordered by © .
Ö@× − c ß so Ðc ß © Ñ is a nonempty poset.
For any ! Á @ − Z , we have
Let ÖGα À α − M× be a chain in c . We claim G œ Gα is linearly independent, so that G − c .
If α" , ..., α8 − O and -" ,..., -8 − G and 83œ" α3 -3 œ !, then each -3 is in some Gα3 .
The Gα 's form a chain with respect to © , so one of the Gα3 's  call it Gα‡  contains
all the others. Then ! œ 83œ" α3 -3 is a linear combination of elements from Gα‡ , and Gα‡
is linearly independent. So all the α3 's must be !.
Therefore, G − c and G is an upper bound for the chain ÖGα À α − M ×. By Zorn's Lemma, c has
a maximal element FÞ
We claim that F is a basis for Z .
First we show that if @ − Z , then @ can be written as a finite linear combination of
elements of U .
If @ − F , then @ œ 1 † @. If @  F , then F § F ∪ Ö@× so, by
Á
maximality, F ∪ Ö@× is not linearly independent. That means there is a
nontrivial linear combination of elements of F ∪ Ö@× Ðnecessarily involving
@Ñ with sum ! À
b ," , ..., ,8 and b α" , ..., α8 , " − O with " Á ! and 83œ" α3 ,3 + " @ œ !.
Since " Á !, we can solve the equation and write @ œ 83œ" "α3 ,3 .
We complete the proof by showing that such a representation for @ is unique.
w w
w
w
Suppose that we have 83œ" α3 ,3 œ @ œ 7
3œ" α3 ,3 with α3 , α3 − O and ,3 , ,3 − F .
By allowing additional ,'s with ! coefficients as necessary, we may assume that the
same elements of F are used on both sides of the equation, so that
ww
5
53œ" "3 ,3ww œ @ œ 53œ" #3 ,3ww Þ
3œ"
Then
("3  #3 ),3 œ ! and F is linearly
independent, so "3 œ #3 for 3 œ "ß ÞÞÞ,5 .
Example 11.12 In Example II.5.14 we showed that the only continuous functions 0 À ‘ Ä ‘
satisfying the functional equation
0 ÐB  CÑ œ 0 ÐBÑ  0 ÐCÑ for all Bß C − ‘
are the linear functions 0 ÐBÑ œ -B for some - − ‘.
necessarily discontinuous, that satisfy (*).
379
ЇÑ
Now we can see that there are other functions,
Consider ‘ as a vector space over the field , and let F © ‘ be a basis. Choose any ," − F . Then for
8
each B − ‘, there is a unique expression B œ ;B ,"   ;3 ,3 for some ,# ß ÞÞÞß ,8 − F and
3œ#
;B ß ;# ß ÞÞÞß ;8 − . (We can insist that ," be part of this sum by allowing ;B œ ! when necessary.).
Define 0 À ‘ Ä  © ‘ by 0 ÐBÑ œ ;B . Clearly 0 ÐB  CÑ œ 0 ÐBÑ  0 ÐCÑÞ
Although the definition of 0 looks complicated, perhaps it could happen, for a cleverly chosen
- − ‘ß that 0 ÐBÑ œ -B for all B? No: we show that 0 ÐBÑ œ -B is not possible (and therefore 0 is not
continuous).
1) If 0 ÐBÑ œ -B for some constant - , we would have:
0 Ð# Ñ œ - † # − , and
0 Ð"Ñ œ - † " œ - − . But - Á !, or else 0 ÐBÑ œ ! for every B, whereas
0 Ð,Ñ œ " † , œ "Þ So we can divide by - getting


- # Î- œ # − , which is false.
2) Here is a different argument to the same conclusion: For every , − U where , Á ," , we
have , œ ! † ,"  " † ,ß so 0 Ð,Ñ œ !. Therefore the equation 0 ÐBÑ œ ! has infinitely many
solutions so 0 ÐBÑ is not linear.
As we remarked earlier in Example II.5.14, it can be shown that discontinuous solutions 0 for (*) must
be “not Lebesgue measurable” (a nasty condition that implies that 0 must be “extremely
discontinuous.”)
A “silly” example from measure theory (optional) It is certainly possible for an uncountable union
of sets of measure 0 to have measure 0. For example, let M: œ ÖB −  ∩ Ò!ß "Ó À B  :× for each
irrational : − Ò!ß "ÓÞ There are uncountably many M: 's and each one has measure !. In this case,
M: © , so M: also has measure !.
Might it be true that every union of sets of measure 0 (say, in Ò!ß "Ó) must have measure 0? (It is easy
to answer this question: how?) What follows is an “unnecessarily complicated” answer using Zorn's
Lemma.
If every such union had measure zero, we could apply Zorn's Lemma to the poset consisting of
all measure-! subsets of Ò!ß "Ó and get a maximal subset Q with measure ! in Ò!ß "Ó. Since Ò!ß "Ó
does not have measure !, there is a point : − Ò!ß "Ó  Q . Then Q ∪ Ö:× also has measure !,
contradicting the maximality of Q .
380
Exercises
E24. Let Ð\ß .Ñ be a metric space. In Definition 10.3, we defined collections Zα (α  =" ) and the
collection of Borel sets U = {Zα : α  =" }.
a) Let Y! be the family of closed sets in \ . For α  =" , define families
Ö∞ J8 : J8 − Y"8 , "8  α× if α is even
Yα œ  8œ"
Ö ∞
8œ" J8 À J8 − Y"8 , "8  α× if α is odd
Y" is the collection of countable unions of closed sets (called J5 -sets) and Y# is the family of countable
intersections of J5 sets (called J5$ -sets).
Prove that Yα © Z" and Zα © Y" for all α  "  =" .
It follows that U œ {Yα : α  =" }. We can build the Borel sets “from the bottom up” beginning
with either the open sets or the closed sets. Would this be true if we defined Borel sets the same way in
an arbitrary topological space?)
b) Suppose \ and ] are separable metric spaces. A function 0 À \ Ä ] is called Borelmeasurable (or B-measurable, for short) if 0 " ÒFÓ is a Borel set in \ whenever F is a Borel set in ] .
Prove that 0 is B-measurable iff 0 " ÒSÓ is Borel in \ whenever S is open in ] .
c) Prove that there are Ÿ - B-measurable maps 0 from \ to ] .
E25. a) Is a locally finite cover of a space \ necessarily point finite? Is a point finite cover
necessarily locally finite (see Definitions 10.10, 10.13)? Give an example of a space \ and an open
cover h that satisfies one of these properties but not the other.
b) Suppose J3 (3 œ "ß ÞÞÞß 8Ñ are closed sets in the normal space
\ with 83œ" J3 œ g. Prove
_
8
that there exist open sets Z3 Ð3 œ "ß ÞÞÞß 8Ñ such that J3 © Z3 and 3œ" Z 3 = g.
Hint: Use the characterization of normality in Theorem 10.12.
E26. Prove that the continuum hypothesis (- œ i" ) is true iff ‘# can be written as E ∪ F where E has
countable intersection with every horizontal line and F has countable intersection with every vertical
line.
Hints: Ê : See Exercise I.E44. If CH is true, ‘ can be indexed by the ordinals < =" .
É : If CH is false, then -  i" . Suppose E meets every horizontal line only countably often
and that E ∪ F œ ‘# . Show that F meets some vertical line uncountably often by letting U be the
union of any i" horizontal lines and examining 1\ ÒU ∩ EÓ. )
381
E27. A cover h of the space \ is called irreducible if it has no proper subcover.
a) Give an example of an open cover of a noncompact space which has no irreducible
subcover.
b) Prove that \ is compact iff every open cover has an irreducible subcover.
Hint: Let h be any open cover and let T be a subcover with the smallest possible cardinality 7. Let #
be the least ordinal with cardinality 7. Index T using the ordinals less than # , so that
T œ ÖYα À α  # ×. Then consider ÖZ" À "  # ×, where Z" œ ÖYα À α  " ×Þ
E28. Two set theory students, Ray and Debra, are arguing.
Ray: “There must be a maximal countable set of real numbers. Look: partially
order the countable infinite subsets of ‘ by inclusion. Now every chain of such sets
has an upper bound (remember, a countable union of countable sets is countable), so
Zorn's Lemma gives us a maximal element.”
Debra: “I don't know anything about Zorn's Lemma but it seems to me that you can
always add another real to any countable set of real numbers and still have a countable
set. So how can there be a largest one?”
Ray:
“I didn't say largest! I said maximal!”
Resolve their dispute.
E29. Suppose ÐT ß Ÿ Ñ is a poset. Prove that Ÿ can be “enlarged” to a relation Ÿ ‡ such that
ÐT ß Ÿ ‡ Ñ is a chain. ( “ Ÿ ‡ enlarges Ÿ ” means that “ Ÿ © Ÿ ‡ © T ‚ T ”Ñ
Hint: Suppose Ÿ is a linear ordering on T . Can Ÿ be enlarged?
E30. Let 7 be a cardinal. A space \ has caliber 7 if, whenever h is a family of open sets with
± h ± œ 7, there is a family i © h such that ± i ± œ 7 and ÖZ À Z − i × Á g.
a) Prove that every separable space has caliber i" .
b) Prove that any product of separable spaces has caliber i" .
Hint: Recall that a product of - separable spaces is separable (Theorem VI.3.5).
c) Prove that if \ has caliber i" , then \ satisfies the countable chain condition (see Definition
11.4).
d) Let \ be a set of cardinal i" with the cocountable topology. Is \ separable? Does \
satisfy the countable chain condition? Does \ have caliber i" ? (For notational convenience, you can
assume, without loss of generality, that \ œ Ò!ß =" Ñ. )
382
E31. a) Prove that there exists an infinite maximal family X of infinite subsets of  with the
property that the intersection of any two sets from X is finite.
b) Let H œ Ö=I À I − X × be a set of distinct points such that H ∩  œ g. Let ^ œ  ∪ H,
with the following topology:
i) points of  are isolated
ii) a basic neighborhood of =I is any set containing =I and all but at most finitely
many points of I .
Prove ^ is Tychonoff.
c) Prove that ^ is not countably compact. Hint: Consider the set H
d) Prove that ^ is pseudocompact. Hint: This proof uses the maximality of XÞ
E32. According to Theorem V.5.10, the closed interval Ò!ß "Ó cannot be written nontrivially as a
countable union of pairwise disjoint nonempty closed sets. Of course, Ò!ß "Ó can certainly be written as
the union of - such sets: for example, Ò!ß "Ó œ B−Ò!ß"Ó ÖB×.
Prove that Ò!ß "Ó can be written as the union of uncountably many pairwise disjoint closed sets
each of which is countably infinite.
Hint: Use Zorn's Lemma to choose a maximal family Y of subsets of Ò!ß "Ó each homeomorphic to
Ö 8" À 8 − × ∪ Ö!×. P/> E œ Ò!ß "Ó  Y . E is relatively discrete and therefore countable. For each
B − E, choose a different GB − Y and replace GB by GB ∪ ÖB×Þ)
E33. Suppose \ is Tychonoff. For : − \ , let Q: œ Ö0 − GÐ\ÑÀ 0 Ð:Ñ œ !×. Clearly, Q: is an ideal
in GÐ\Ñ.
a) Prove that Q: is a maximal ideal in GÐ\Ñ.
b) Prove that is \ is compact iff every maximal ideal in GÐ\Ñ is of the form Q: for
some : − \ .
E34. Prove that there exists a subset E of ‘ that has only countably many distinct “translates”
E< œ Ö+  < À + − E×  that is only countably many of the sets E< œ Ö+  < À + − E×, < − ‘ are
distinct).
Hint: Consider ‘ as a vector space over the field , and pick a basis F . Pick a point , − F and
consider all reals whose expression as a finite linear combination of elements of F does not
involve ,Þ
383
Appendix
Exponentiation of Ordinals: A Sketch
The appendix gives a brief sketch about exponentiation of ordinals. Some of the details are omitted.
The main point is to explain why ordinals like ==! ! are still countable ordinals. (See Example 5.25, part
3)
If . and α are ordinals, let
.α œ Ö0 − Ò!ß .ÑÒ!ßαÑ À 0 Ð0Ñ œ 0 for all but finitely many 0  α}.
We can think of a point 0 in .α as a “transfinite sequence” in Ò!ß .Ñ, where 0 has well-ordered domain
Ò!ß αÑ rather than . Using “sequence-like” notation, we can write:
.α œ ÖÐa0 Ñ À 0 Ÿ 0  α , ! Ÿ +0  .ß and a0 œ ! for all but finitely many 0 ×
We put an ordering on .α by:
Given Ða0 Ñ Á Ðb0 Ñ in .α , let / be the largest index for which a/ Á b/ Þ We write
Ð+0 Ñ  Ð,0 Ñ if a/  b/ .
Example For α œ . œ 2, .α consists of 4 pairs ordered as follows: Ð0,0Ñ  Ð1,0Ñ  Ð0,1Ñ  Ð1,1ÑÞ
We also use .α to denote the order type associated with Ð.α , Ÿ Ñ. It turns out that Ð.α , Ÿ Ñ is wellordered, so this order type is actually an ordinal number.
Example ==! ! is represented by the set of all sequences in Ö0, 1, 2, ...× which are eventually 0Þ (The
sequences in the set ==! ! turn out to be those which are “eventually !” because each element of =! has
has only finitely many predecessors. In general, the condition that members of .α be ! for all but
finitely many 0  α is much stronger than merely saying “eventually !.”)
The order relation on ==! ! between two sequences is determined by comparing the largest term at which
the sequences differ. For example,
Ð5,0,0,1,2,0,0,0,0,...Ñ  Ð107,12,1,3,5,0,0,0,0,...Ñ  Ð103,7,0,0,6,0,0,0,0...,Ñ
The initial segment of ==! ! representing =! consists of:
Ð!ß !ß !ß ÞÞÞß ÞÞÞÑ  Ð"ß !ß !ß ÞÞÞß ÞÞÞÑ  Ð#ß !ß !ß ÞÞÞß ÞÞÞÑ  ...  Ð8ß !ß !ß ÞÞÞß ÞÞÞÑ  ...
==! ! is clearly a countable ordinal. A little thought shows that ==! ! œ =!  =#!  =$!  ÞÞÞ
α
Once . is defined, it is easy to see that the ordinals
#
#
ÐHere, α" is understood, as usual, to mean α(" ) ÑÞ
{==! !
=
= !
=! ! ,
==! !
=
,
= !
=! ! ,
=
= !
= !
=! !
, ...
are all countable ordinals.
=
= !
= !
=! !
We can then define %! œ sup
,
, ... }. Roughly, %! is “=! raised to the =! power =!
times”. As noted earlier, the sup of a countable set of countable ordinals is still countable: %! is a
countable ordinal, that is, %!  =" !
384
Exercise Prove that ." œ .. Prove that for ordinals . and α, .α is also an ordinal, i.e., Ð.α , Ÿ Ñ is a
well-ordered set (not trivial!). For ordinals ., αß "  !, and .α" œ .α † ." .
For further information, see Sierpinski, Cardinal and Ordinal Numbers, pp. 309 ff.
385
Chapter VIII Review
Explain why each statement is true, or provide a counterexample.
1. Let \ œ Ò!ß =" Ñ. For each α − \ , let α µ α  ". In the quotient space \Î µ , exactly one point is
isolated.
2. ii=!! can be written as a sum of countably many smaller cardinals.
3. Let 8 − Þ A continuous function 0 À Ò!ß =" Ñ Ä ‘8 is constant on a tail of Ò!ß =" Ñ.
4. Every order-dense chain with more than 1 point contains a subset order isomorphic to .
5. If B − S, where S is an open set in \ œ Ò!ß =" Ó ‚ Ò!ß =! Ó  ÖÐ=!  "ß =! Ñ×, then there must be a
continuous function 0 À \ Ä ‘ such that B − coz Ð0 Ñ © S.
6. Let G be a non-Borel subset of the unit circle W " and let 0 À W " Ä ‘ be the characteristic function of
GÞ Then 0 is not Borel measurable but 0 is Borel measurable in each variable separately (that is, for
each + − Ò  "ß "Ó, the functions 9 and < defined by 9ÐBÑ œ 0 ÐBß +Ñ and <Ð+ß CÑ œ 0 Ð+ß CÑ are Borel
measurable.
7. Let G be a dense subset of ‘. Then G is not well-ordered (in the usual order on ‘).
8. Let H œ Ö0 −  À 0 Ð8Ñ œ ! for all but finitely many 8}, with the lexicographic (“dictionary")
ordering Ÿ . Then ÐHß Ÿ Ñ is order isomorphic to .
9. Suppose Ÿ " and Ÿ # are linear orders on a set \ . If Ÿ " © Ÿ # , then Ÿ " œ Ÿ # .
10. Consider the ordinals 1, =! and =! † #. Considering all possible sums of these ordinals (in the six
different possible orders) produces exactly 3 distinct values.
11. If α and " are nonzero ordinals and +  " œ =! , then α" œ =! .
12. The order topology on Ð"ß $Ñ ∪ Ð$ß &Ñ is the same as the subspace topology from ‘.
13. If Ð\ß g Ñ is a finite topological space, then there exists a partial ordering Ÿ on \ for which g is
the order topology.
14. If S is open and J is closed in X œ Ò!ß =" Ó ‚ Ò!ß =! Ó  ÖÐ=" ß =! Ñ× and J © S, then there must
exist an open set Z such that J © Z © cl Z © S.
15. There are exactly iα limit ordinals  =α .
16. If α denotes the order type of the irrationals, then α# œ α.
17. If α  =" , then the ordinal space \ œ Ò!ß αÓ is not metrizable because \ is not normal.
386
18. Suppose c is a nonempty collection containing all subsets of \ with a certain property (*).
Suppose ÖTα À − E× is a chain of subsets of c and that Tα − c . By Zorn's Lemma, Tα is a
maximal subset of \ (with respect to © ) having property (*).
19. For any infinite cardinal 5 , there are exactly 5  ordinals with cardinality 5 .
20. Suppose α is an infinite order type. If "  α œ α  ", then there is an order type 0 (possibly !)
such that α œ =!  0  =‡! .
21. If Ÿ a linear order on \ , then the “reversed relation” Ÿ ‡ defined by B Ÿ ‡ C iff C Ÿ B is also a
linear order on \ .
22. For any infinite cardinal 5 , #5 œ 5 5 (without GCH).
23. If Ÿ is a linear ordering on a finite set S and lWl œ 8, then | Ÿ | =
8Ð8  " Ñ
2
.
24. If E and F are disjoint closed sets in Ò!ß =" Ñ, then at least one of them is countable.
25. A locally finite open cover (by nonempty sets) of a compact space must be finite.
26. Ò!ß =" Ó is homeomorphic to a subspace of the Cantor set.
27. Let i! have the lexicographic order Ÿ . (i! , Ÿ ) is not well-ordered, because the set
E œ ÖB − i! À b5 a8  5 BÐ8Ñ œ 2× contains no smallest element.
28. Every subset of  is a Borel set in ‘.
29. With the order topology, the set Ö0× ∪ ÖB − ‘ À lBl  "× is homeomorphic to ‘.
30. If h œ ÖUα À α − E× is a point-finite open cover of the Sorgenfrey plane W ‚ W , then there must
be an open cover i œ ÖVα À α − E× of W ‚ W such that, for each α − E, cl ÐVα Ñ © Uα .
31. There is a countably infinite compact connected metric space.
w
32. If Ÿ is a linear ordering on \ , then there can be linear ordering Ÿ for which
w
Ÿ © Ÿ ©\‚\
Á
387