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 ÐBßCÑ−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 asŸ¡ß£ , , or .

Definition 2.1 A relation V\ on is called:

transitive ifÀ a +ß ,ß - − \ Ð+V, and ,V-Ñ Ê +V-Þ reflexive ifÀa+−\+V+ antisymmetric ifÀ a +ß , − \ Ð+V, and ,V+ Ñ Ê Ð+ œ ,Ñ symmetric ifÀ a +ß , − \ +V, Í ,V+ (that is, the set V is “symmetric” with respect to thediagonal ? œÖÐBßBÑÀB−\ש\‚\ ).

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 ? œÖÐBßBÑÀB−\×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 Ðllœ"ÑÞunless V

319 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\ߟ ) 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 irreflexiveÐa+ − \ , 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 ŸE on E©\ by restricting the order ŸEÞ to More formally, ŸœŸ∩E‚EE ( ). 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 ŸŸE for the order E on .

If Ð\ß Ÿ Ñ is a poset (or chain), and E © \ , then ÐEß Ÿ Ñ is also a poset (or chain). For example, every subset of Ðߟё is a chain.

2) For Dß A −‚ Ð œ the set of complex numbers Ñ , define D £ A iff lDl Ÿ lAlß where Ÿ is the usual order in ‘‚ . Ðᜥ is not a poset. (Why? )

3) Let Ð\ßg Ñ be a topological space. For 0ß 1 − GÐ\Ñ , define

0Ÿ1 iff aB−\ , 0ÐBÑŸ1ÐBÑ .

As a set, Ÿ œ ÖÐ0ß 1Ñ − GÐ\Ñ ‚ GÐ\Ñ À aB −‘ ß 0ÐBÑ Ÿ 1ÐBÑ× .

Notice that, in contrast to part 2), we are allowing ourselves an ambiguity in the notation here because we are using “ŸŸGÐ\Ñ ” with two different meanings: we are defining an order “ ” in , but the comparison “0ÐBÑ 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 0ß 1 are given by 0ÐBÑ œ B and 1ÐBÑ œ B# , then 0 ŸÎ1 and 1ŸÎ 0 . When is ÐGÐ\Ñß Ÿ Ñ a chain? ÐThe answer is not “ iff l\l Ÿ "Þ” Ñ

320 4) The following two diagrams represent posets, each with 5 elements. Line segments upward from B to C indicate that BŸC . 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 VV is a collection of sets. We can define ŸEŸFE©F on by iff . Then ÐߟÑVV is a poset. In this case, we say that has been ordered by inclusion . In particular, for any set \, we can order cc Ð\Ñ by inclusion. What conditions on \ will guarantee Ð Ð\Ñß Ÿ ) is a chain ?

6) Suppose VV is a collection of sets. We can define ŸEŸFEªF on by iff . Then ÐߟÑVV is a poset. In this case, we say that has been ordered by reverse inclusion . In particular, for any set \Ð\Ñ , we can order c by reverse inclusion. What conditions on \ will guarantee that ÐÐ\Ñߟc ) 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 3ÀVV Ä 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 ŸÐߟÑ\ . The “order structure” of the poset aB 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×−RBB and RŸÖB× for every R−R . So the poset ÐߟÑaB has a largest element. ÐIs the converse true? Ñ

ii) If \ is first countable and UB"#5 œ ÖR ß R ß ÞÞÞß R ß ÞÞÞ× is a countable neighborhood base at BR−5 , then for every aB there is a such that RŸR5B . Thus the poset Ða ßŸÑ contains a countable subset ÖR"# ß R ß ÞÞÞß R 5 ß ÞÞÞ× whose members become “arbitrarily large” in the poset.

iii) The poset ÐߟÑaB is usually not a chain. But it does have interesting

321 order property: for any RßR−"#aa B ßbR− $ B such that RŸR " $ and RŸR # $ (let RœR∩R$"#ÑÞ

It will turn out (in Chapter 9) that this poset ÐߟÑaB 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) the largest (or last ) element in \CŸB if for all C−\ ii) the smallest (or first ) element in \BŸC if for all C−\ iii) a maximal element in \ if ÐC B ÊCœBÑ for all C−\ iv) a minimal element in \ if ÐCŸB ÊCœBÑ for all C−\ .

It is clear that a largest element in Ð\ß Ÿ Ñ , if it exists, is unique. (If DD"# and were both largest, then DŸD"# and DŸD #""# so DœD .)

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 +ŸB for all +−EàB is called aleast upper bound (sup) for E if B is the smallest upper bound for EÞ The set E\E might have many upper bounds, one upper bound, or no upper bounds in . If has more than one upper bound, E\\ might or might not have a least upper bound in . But if has a least upper boundB , then the least upper bound is unique (why? ).

vi) An element B−\ is called a lowerbound for E if BŸ+ for all +−EàB is called a greatest lower bound (inf) for EB if is the largest lower bound for EÞ The set E might have many lower bounds, one lower bound, or no lower bounds in \E . If 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 bD−\Î with BDC , then B is called an immediate predecessor of CC and 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 sup Ö.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\Ÿ©\‚\ÞŸŸ\"""# is an order on , then So if and are orders on , it makes sense to ask whether Ÿ©Ÿ"# , or vice-versa. If we look at the set c œÖŸÀŸ is a partial

322 order on \× , then c is partially ordered by inclusion. A linear order Ÿ is a maximal element in Ðß©Ñc (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−\ÀB+× or ÖB−\À,B× ( +ß,−\ ) are a subbase . (As usual, we write BC as shorthand for “BŸC and BÁCÞ ” )

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,לÐ+ß,Ñ ÖB−\À+ŸB,לÒ+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"לÖ5− À5#× œÐ∞ß#Ñ ; and for 8" , Ö8לÐ8"ß8"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 \E is a chain with the order topology. There are two possible topologies on :

a) The order Ÿ\E gives an order topology gŸ on and we can give the subspace topology ÐÑÞgŸ E

b) EŸ has an ordering E (inherited from the order Ÿ\ÑE on and we can use it to give 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 g‘Ÿ on is the usual topology on ‘ , and this topology produces a subspace topology for which #EÞ is isolated in

But in the order topology on E , each basic open set containing 2 must have the form

ÖB−EÀB+לÐ+ß#Ó where +" . So # is not isolated in ÐEßggŸŸEE Ñ . In fact, the space ÐEß Ñ is homeomorphic to Ð!ß "Ó (why? ).

Is there any necessary inclusion ©ª or between ggŸŸE and ÐÑlE ? Can you state any hypotheses on \E or that will guarantee that ggŸŸE œ ( ÑlE ?

323 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 \ œÖ+ß,ß-× with the partial order represented by the following diagram:

\+ is the only open set containing (why? ), so the order topology is not XÞÐ" Can you find a poset for which the order topology is not even X! ?ÑÐ\ßŸÑ By contrast, the order topology for any chain has good separation properties. For example, it is easy to show that the order topology on a chain must be XÀ#

Suppose +Á,−\ ; we can assume +,Þ If + is the immediate predecessor of , , we can let Y œÖB−\ÀB,× and Z œÖB−\ÀB+× . But if there exists a point - satisfying +-, , then we can define Y œÖB−\ÀB-× and Z œÖB−\ÀB-×Þ Either way, we have a pair of disjoint open sets YZ and with B−YC−ZÞ and

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 \Ð\ßŸÑ , rather than , 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 ŸŸ"# , , ... .

Definition 3.5 A function 0À\Ä] between ordered sets is called an order isomorphism if 0 is bijection and both 0+Ÿ,0Ð+ÑŸ0Ð,ÑÐ and order preservin gthat is , if and only if . Since 0 is one- to-one, it follows that BC if and only if 0ÐBÑ0ÐCÑÞÑIf such an 0\] exists, we say that and are order isomorphic and write \¶] . If 0\¶0Ò\Ó©]0\] is not onto, then and is an order isomorphism of into .

Between ordered sets, “¶ ” will always refer to order isomorphism.

324 Theorem 3.6 Let \ß] ß and ^ be chains Ðparts i-iv) are also true for posets ).

i) \¶\ ii) \¶] iff ] ¶\ iii) if \¶] and ] ¶^ , then \¶^ iv) \¶] implies l\lœl]lÞ v) if \] and are finite chains, then \¶]l\lœl]lÞ iff

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 0ÐBÑ œ B and 0ÐBÑ 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, 0Ð8Ñ 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 0À Ä is any bijection, then we can use 0 to create a chain Ђ‚‘ ß ŸÑ : simply define D"# ŸD iff 0ÐDÑŸ0ÐDÑ " # . Then Ð ß ŸÑ¶Ð ß ŸÑÞ

Of course, this chain Ðߟт 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 DŸD"# and DŸD $% , then DDŸDDÞ "$#%

In the same way, a bijection 0À 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 88 À 8 − × ∪ Ö!× and ] œ Ö!× ∪ Ö À 8 − ×Þ The order topologies on \] and are the usual topologies, and the reflection 0ÐBÑœB is a homeomorphism (in fact, an isometry) between themÞ " The largest element in ]"ß is and it has an immediate predecessor, # Þ But ! is the largest element in \\Þ and it has no immediate predecessor in Since an order isomorphism preserves largest elements and immediate predecessors, there is no order isomorphism between \] and .

325 The next theorem tells when an ordered set is order isomorphic to the set of rational numbers.

Theorem 3.7 (Cantor) Suppose that (PŸ , ) 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 6 8 ß ÞÞÞ×, and let 0Ð; " Ñ œ 6 " . Then inductively define 0Ð; 8 Ñ œ some 6 − P chosen so that 0Ð;8" Ñ Á 0Ð; Ñß ÞÞÞß 0Ð; 8" Ñ and so that 0Ð; 8 Ñ has the same order relations to 0Ð; " Ñß ÞÞÞß 0Ð; 8" Ñ as ;8"8" has to ; ß ÞÞÞß ; . 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 1À ÄP . 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 1À ÄP .

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© ßF©P where EßF are finite. Let 1 be an order isomorphism from EF onto .

a) If ;− E , there exists an order isomorphism 2 that extends 1 and for which domÐ2Ñ œ E ∪ Ö;× and ran Ð2Ñ © 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;"8 ß ÞÞÞß ; × and that 1Ð; 33 Ñ œ 6 . Pick an 6 − P  F that has the same order relations to 6"8 ß ÞÞÞß 6 as ; does to ; " ß ÞÞÞß ; 8 Þ ( For example: if ; is greater that all of ;"8 ß ÞÞÞß ; ß pick 6 greater than all of 6 "8 ß ÞÞÞß 6 ; if 3 and 4 are the largest and smallest subscripts for which ;;;34 , then choose 6 between 6 34 and 6Þ) 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 PPÁgPP is a countable chain. Since and has no last element, must be infinite. Without loss of generality, we may assume that P∩ œgÞ

We define an order isomorphism 1P between  and 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 7 8 ß ÞÞÞ×Þ Each element of and P appears exactly once in this list.

326 Define1œg! , and continue by induction. Suppose 8! and that an order isomorphism 1ÀEÄF8" 8" 8" has been defined, where E© 8" and F©PÞ 8"

If 7−888" and 7ÂE , use the Lemma to get an order isomorphism 1 8 that extends 18" and for which dom Ð1ÑœE 8 8" ∪Ö7לE 8 8 Þ Let F 8 œ ran Ð1Ñ©P 8 .

If 7−P888" and 7ÂF ß use the Lemma to get an order isomorphism 1 8 that extends 18" and for which ran Ð1ÑœF 8 8" ∪Ö7לF 8 8 Þ Let E 8 œ dom Ð1Ñ© 8  .

If 78 − dom Ð1 8" Ñ ∪ ran Ð1 8" Ñ , let 1 8 œ 1 8" ß E 8 œ E 8" and F 8 œ F 8" Þ

By induction, 1811888" is defined for all , and since extends and we can define an order isomorphism ∞ The construction guarantees that dom and ran 1 œ8œ! 18 Þ Ð1Ñ œ Ð1Ñ 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 \l\l"ß\ is a countable order-dense and then is order isomorphic to exactly one of the following chains À

a)  ∩ Ð!ß "Ñ ¶ b)  ∩ Ò!ß "Ñ c)  ∩ Ð!ß "Ó d)  ∩ 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 1À\Ä ∩Ò!ß"Ñ by setting 1Ð+Ñœ! and 1ÐBÑœ2ÐBÑ for BÁ+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 0ÐBÑ 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 ∩ .

327 For short, write Ð!ß"Ñ∩ œÐ 0 ß"Ñ and Ð!ß"Ó∩ œÐ!ß" ] .

In Ð!ß "Ñ , choose a strictly increasing sequence of irrationals Ð:8! Ñ Ä " −‘ . Let : œ ! . 11 For 8 !À let S888"8 œÐ:ß: Ñ∩ , Y œÐ22 : 88" ß : Ñ∩ , 11 and ZœÐ":ß":Ñ∩88"822 . Each of SßYß 888 and Z is clopen in Ðß"Ñ 0 and in Ð!ß "] .

∞∞∞ 1 We have Ð!ß "Ó œ S888 ∪ Ö"× and Ð!ß "Ñ œ Y ∪ Ö2 × ∪ Z 8œ! 8œ! 8œ! ∞∞ 1 Define a map 9À Ð!ß "Ó Ä Y88 ∪ Ö2 × ∪ Z by 8œ! 8œ!

9 ±SYS#88#8 = an increasing linear map from onto

9 ±SZS#8"8#8  " = a decreasing linear map from onto

1 9Ð"Ñ œ 2

Then 9 is a homeomorphism. Ð Since the sets S888 ßY ßZ are clopen, it is clear that 9 is continuous at all points except perhaps 1 and that 9" is continuous at all points except 1 perhaps 2 . 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\H be a countable dense set in the order topology on . Then satisfies the hypotheses of Cantor's Theorem ÐÑwhy? so there exists an (onto) order isomorphism 0ÀÄH . For each irrational :−‘, let : œÖ;−  À;:× and extend 0 to an order isomorphism 1À ‘ Ä\ by defining 0Ð:Ñ œsup 0Ò: 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 ÐPßŸÑ is order isomorphic to a subset of Ð ߟÑÞ ()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 \G has a subset 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 88 totally unordered subsets (see E5 ) and the 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 “order- isomorphic” 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 ./7=ßßß  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) !g is the order type of " is the order type of Ö!× # is the order type of Ö!ß "× . . . 8 is the order. type of Ö 0 ß"ßÞÞÞß8"× . . . =! 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 QR and . We say that ./ Ÿ if there exists an order isomorphism 0Q of into R . We write ./  if ./ Ÿ but ./ ÁÐ that is, Q¶ÎRÑÞ(Check that the definition is independent of the chains QR and chosen to represent .

331 and /. )

Example 4.3 Let . be the order type of a chain Qg . Since 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 BŸ C iff CŸB . We write .= for the order type of Ð\ߟ ). For example, ! is the order ‡‡ type of the chain Ö× ... , 2, 1 of negative integers. Since ==!! ŸÎŸ!! and ==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 ThereforeŸ is not even a partial ordering among order types.

Definition 4.4 For α −E , let ÐQßαα Ÿ Ñ be pairwise disjoint chains and suppose that the index set E is also a chain. We define the ordered sum as the chain , , where we define α−E QÐQŸÑααα−E Bß C − Qand B Ÿ C , or BŸC if αα  α"−EßB−Q α" 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 ÐQß ŸQR Ñ and ÐRß Ÿ Ñ , the ordered sum ÐQ  Rß Ÿ Ñ is formed by putting RÐÑQ to the right of “larger than” and using the old orders inside each of Q andR .

Definition 4.5 Suppose E−E is a chain and that for each α. , we have an order type α . Let the 's be represented by pairwise disjoint chains . Then as order type of the chain ..αααQ  α−E , . In particular, if and are order types represented by disjoint chains and , then ÐQŸÑα−E α ./ QR ./ÐQRßŸÑ is the order type of the ordered sum . (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œ./ 7 . ÐÑ /7 . 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 7Á8−=, !! Á = 8Á = ! 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 +83!! is larger than every . Less abstracting, ==  could also be represented by 11 the chain Ö 188 : 8−×∪Ö 2 : 8−ש .

‡‡‡ ==!!!!!! 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.)

‡  ™0==has order type œ! ! . For each sequence + œ Ð+ "#$ ß + ß + ÞÞÞÑ − Ö!ß "× , we can define an order type

0000+œ "+ "#$ 1 +"+ . . .

It is not hard to show that the map +È0+ 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 GG is a chain representing 0+ and let the finite components of (listed in order of increasing size) be J"# ß J ß ÞÞÞ ß J 8 ß ÞÞÞ where J 8 has order type "  + 8 .

 w Let + Á , − Ö!ß "× and suppose G is a chain that represents 0, . Call the finite wwwww components of G (listed in order of increasing size) J"# ßJ ßÞÞÞßJ 8 ßÞÞÞ where J 8 has order type ",8 .

ww An order isomorphism between GG and would necessarily carry JJ8 to 8 for every 88"+",". But this is impossible since, for some , one of 88 and 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 ÐQß ŸQR ) and ÐRß Ÿ Ñ be chains representing the order types ./ and . We define the product ./ to be the order type of ÐR ‚ Qß Ÿ ), where Ð8"" , 7 ) Ÿ Ð8 ## , 7 ) iff 8 "R#  8 or (8œ8"# and 7Ÿ "Q# 7 ). This ordering ŸR‚Q on 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†œÐ""цÁ††œ 2 ===== !!!!! 1 1 . 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: # where each . A subset of that represents # ===! œ!!   ÞÞÞ  = !  ÞÞÞ œ 8−8 . .= 8 œ !  =! " 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>Ñœ / ./  .7 .

Note: Other books may define ./ “in reverse” as the order type of the lexicographically ordered set Q‚R. Under that definition, =!†œ2 = ! 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"ÓÞ# " 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. Suppose +œw

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 DßA−‚ , exactly one of BœAßBA , or DA holds ii) for all ?ß Aß D −‚ À if D  A , then D  ?  A  ? iii) ifBß 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 EF©EFEF is a set of type ( and , then either or 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 EF and 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 EF , 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 0À‘ 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 \+Á,−\Ö+ß,× is automatically a chain: if , then set 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 Ðß"Ñ0 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 Ÿ and that applies to ordinals.

Theorem 5.3 If α" and are ordinals, so are α"α"† and .

Proof Let α" and be represented by disjoint well-ordered sets EF and . 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 GEF of must contain a smallest element.

F is well-ordered so, if G©F , then G has a smallest element. Otherwise G∩EÁg 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 GÀ more precisely, let ,Ö,−FÀ+−EßÐ,ß+Ñ−G×Þ+! be the smallest element in for some Then let ! be the smallest element in Ö+−EÀ 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" 88 À 8 − × ∪ Ö#  À 8 − × ∪ Ö#×ßneither " 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 Q0ÀQÄQß is well-ordered. If is a one-to-one order-preserving map of Q into Q0Ð7Ñ 77−Q , then for all .

The theorem says that 0 cannot move an element “to the left.” Notice that Theorem 5.5 is false for 1 chains in general: for example, consider Qœ‘ and 0ÐBÑœ2 B . On the other hand, if you try to construct a counterexample using Qœ , you will probably see how the proof of the theorem should go.

338 Proof Suppose not. Then EœÖ7−QÀ0Ð7Ñ7×Ág . Let 7! be the smallest element in E . We get a contradiction by asking “what is 0Ð0Ð7! ÑÑ ” ?

Since 7!!! − E , 0Ð7 Ñ  7 . Since 0 preserves order and is one-to-one, 0Ð0Ð7 !! ÑÑ 0Ð7 Ñ which means that 0Ð7!! Ñ − E . But that is impossible because 7 is the smallest element of EÞ ì

Corollary 5.6 If Q0QQ is well-ordered, then the only order isomorphism from onto is the identity map 0Ð7Ñ œ 7.

(Note that the theorem is false for chains in general: if Qœ‘ , then 0ÐBÑœB$ is an onto order isomorphism. C orollaries 5.6 and 5.7 indicate that well-ordered sets have a very “rigid” structure.)

Proof Let 0À QÄQ be an order isomorphism. By Theorem 5.5, 0Ð7Ñ 7 for all 7 . If 0 is not the identity, then EœÖ7À0Ð7Ñ7×Ág . Let 7! be the least element of E .

339

Then 7Ð0Ñì! cannot be in ran (why? ).

Corollary 5.7 Suppose QR and are well-ordered. If 0ÀQÄR1ÀQÄR and are order isomorphisms from QR onto , then 0œ1 . (So M and N can be order isomorphic “in only one way.”Ñ

Proof If 0Á1 , 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 !Q is the smallest element .) If we need to carefully distinguish between the first elements in two well-ordered sets QßR 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 œÖB−QÀB7×. We can write this set using the “interval notation” Ò!QQ ß7Ñ . If a discussion involves only a single well-ordered set QÒ!ß7Ñ , we may simply write Q or even just Ò!ß 7ÑÞ

340 Notice that:

i) For every 7−Q , QÁÒ!ß7Ñ , 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 0ÀQÄR is an order isomorphism of Q onto R , then 0ÒÒ!ß 7ÑÓ œ Ò!ß 0Ð7ÑÑ

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 QR©QQ is well-ordered and . is not order isomorphic to an initial segment of RÑQ . In particular (when RœQ , is not order isomorphic to an initial segment of itself.

Proof Suppose 8−R©Q . If 0ÀQÄR©Q is one-to-one and order preserving, then Theorem 5.5 gives us that 0Ð8Ñ 8 for each 8 − RÞ Therefore ß for each 8ß ran Ð0ÑÁ Ò!R ß8ÑÞ ì

Corollary 5.11 No two initial segments of QQ are order isomorphic (so each initial segment of , 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 QR and . We say that ./Q if is order isomorphic to an initial segment of RQ¶Ò!ß8Ñ that is for some 8−R . If QRœŸœ is order isomorphic to we write ./ . We write ././ if or ./ . (Check that the definition is independent of the choice of well-ordered sets QR and 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 QR and :

QRR is order isomorphic to a proper subset of but not to itself Ô Ð‡Ñ QR is order isomorphic to an initial segment of .

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 ÐÑ*, we will be using the new definition 5.12 of ./Ÿ for ordinals .

341 The relationŸ among ordinals is clearly reflexive and transitive. The next theorem implies that Ÿ is antisymmetricŸ and therefore any set of ordinals is partially ordered by .

Theorem 5.13 If . and / are ordinals, then at most one of the relations ././ß œ , and ./  can be true.

Proof Let QR and 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 ./Q , then is isomorphic to an initial segment of R . If /. were also true, then R would, in turn, be isomorphic to an initial segment of QQ . By composing these isomorphisms, we would have order isomorphic to an initial segment of itselfì which is impossible.

Notation For an ordinal ..αα , let ordÐÑœÖÀ is an ordinal and α..  × . If œ! , then ord ÐÑœg . and if ...! , then !− ord Ð Ñß so ord( ) Ág . Like any set of ordinals, we know that ord Ð . 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 ordÐÑ. 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 QÐѶQ , then ord . Therefore ordÐÑ... is a well-ordered set of ordinals and ord ÐÑ represents .

Proof For each ordinal α.−ÐÑ ord , we have α.α  , so can be represented by some initial segment Ò!ß 7αα Ñ of QÞ Define 0 À ord Ð.α Ñ Ä Q by 0Ð Ñ œ 7 . This function 0 is one-to-one since different ordinals cannot be represented by the same initial segment of Q0 , and clearly preserves order.

If 7−Q , then the initial segment Ò!ß7Ñ 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 0Ðα Ñ so 0 is onto. Therefore ordÐѶQ. . ì

We will often write ordÐÑ. in “interval” notation:

For an ordinal .. , Ò!ß Ñ œ Ö αα À is an ordinal and α.  × œ ord Ð . 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"Ñœ ord Ð"ÑœÖ!×

2 is represented by Ò!ß #Ñ œ ord( #Ñ œ Ö!ß "×

ã

===!!! is represented by Ò!ß Ñ œ ord Ð 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, “ à8 ” indicates that =! comes “after” all the natural numbers )

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 œg 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"α" iff . This seems to accomplish what we want. For example:

"$ because "−$

!"#$ÞÞÞ==!!  " because !−"−#−$−ÞÞÞ− == !! − "−ÞÞÞ

==!!  "(because == !! −  "(

343 etc.

If \Ÿ is any set well-ordered by , we can then define its ordinal number Ðœ “the ordinal number associated with \Ñ ” as follows: from the axioms ZFC one can prove the existence of a function (set) with domain \ that is defined “recursively” by :

aC−\ 0ÐCÑœran ( 0lÖD−\ÀDC×Ñ

Then “the ordinal number of \Ð0Ñ ” is defined to be the set ran .

For example, for the well-ordered set \ œ Ö"ß $ß &× , what is the function 0 and what is the ordinal number of \ ?

0Ð"Ñ œ ran Ð 0lÖD − \ À D  "×Ñ œ ran Ð0lg Ñ œ g

0Ð$Ñ œ ran Ð 0lÖD − \ À D  $×Ñ œ ran Ð0lÖ"×Ñ œ Ög×

0Ð&Ñ œ ran Ð 0lÖD − \ À D  &×Ñ œ ran Ð0lÖ"ß $×Ñ œ Ögß Ög××

The ordinal number of \ is ran Ð0Ñ 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 88Þ for any 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 œß  (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 œß  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.αα Ñ œ Ö À is an ordinal and α.  × (Theorem 5.14).

The theorem is certainly true if ./œ! or œ! , so we assume both . ! and / ! .

Let HœÒ!ß. Ñ∩Ò!ß / ÑœÖ αα À 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 HÁg Ð since !−H ). Therefore H represents some ordinal $$.$/! . We claim that Ÿ and Ÿ .

1) To show that $.ŸÁ , we assume $. 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 HœÒ!ß##$ Ñ . If that is true , then H represents ; but H represents and therefore $#.œÞ

Ò!ß#α#α.α# Ñ © H À If − Ò!ß Ñ , then − Ò!ß Ñ and  œ smallest element in Ò!ß.α Ñ  H , so − H .

H©Ò!ß#α ÑÀ If −H , then α# and are comparable since both are in the well-ordered set Ò!ß. Ñ .

We examine the possibilities:

i) #αœ−HÂH : impossible, since α and #

ii) #αÀ impossible, since that would mean #α. and #α/  , forcing # −Ò!ßÑ∩Ò!ßÑ . / œH which is false.

Therefore α#−Ò!ßÑÞ , so α #

2) A similar argument (interchanging “./ ” and “ ” throughout) shows that if $/Á , then $#/œ.

345 Since $.ŸŸ and $/ , there are only four possibilities:

a) $. and $/  , in which case $ −Ò!ß . Ñ∩Ò!ß / ÑœHœÒ!ß $ Ñ which is impossible.

Therefore one of the remaining three cases must be true:

b) $.œœ and $/ , in which case ./ œ c) $.œ and $/ , 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 Àα − E× is a family of pairwise disjoint nonempty sets, there is a set such that, for all , | . F© Eααα E ∩F± œ" (This is clearly equivalent to the statement that . If is in the product, let ran( ); on EÁgα 0 Fœ 0 the other hand, if such a set F00ÐÑœE∩F exists, define by α the unique element of α . An element is a function that “chooses” one element from each ) 0−E0ÐÑEÞα α α 2) (Zermelo's Theorem ) Every set can be well-ordered, i.e., for every set \ there is a subset Ÿ\‚\ of such that Ð\ßŸÑ 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 78 and are cardinal numbers, at least one (and thus, by Theorem I.13.2 , exactly one) of the relations 78ß7œ8ß78 holds. (Therefore any set of cardinals is a chain.)

Proof According to Zermelo's Theorem, we may assume that QR and are well-ordered sets representing the cardinals 78 and , so that Q and R also represent ordinals ./ and .

346 By Theorem 5.15, either ./.œ8ß , or ./ ; therefore either QR and are order isomorphic (in which case 7œ8 ) 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 QR and are well-ordered sets representing ./ and . If Q is order isomorphic to a subset of RQ (so ./Ÿ in the sense of Definition 4.2 ), then is order isomorphic to RR or to an initial segment of (so ./Ÿ in the sense of Definition 5.12 ).

Proof Without loss of generality, we may assume Q©RÞ If Q is not order isomorphic to R , then ./ÁQ. If is also not isomorphic to an initial segment of R , then ./  is also false Þ Therefore the Trichotomy Theorem 5.15 gives ./R . Then 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 [Ÿ is linearly ordered by . We need to show that if E is a nonempty subset of [E , then contains a smallest element. Pick αα −E . If is itself the smallest in EÖ−EÀ×, we are done. If not, then ""α is nonempty and well-ordered because it is a subset of Ò!ßα"" Ñ so it contains a smallest element !! , and is the smallest element in [ . ì

Corollary 5.21 Ò AC] 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 orderŸH (among cardinals) is a linear order. Let be a nonempty subset of G7−HQ7Q and, for each cardinal , let represent . By Zermelo's Theorem, each set can be well- ordered, after which it represents some ordinal .. . By Theorem 5.20, the set of all such 's contains a smallest element .!!!! represented by QÞ Then 7 œlQl is clearly the smallest cardinal in H . ì

Example 5.22

1) The set GœÖ7À7 is a cardinal and i! 7Ÿ-× has a smallest element. It is called the  immediate successor of iii-œi!"" and is denoted by ! or . The statement 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 77 we can consider the smallest element  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 iœiiœi"##$ , , and so on.

The Generalized Continuum Hypothesis is the statement

GCHÀ77œ# “for every infinite cardinal , 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\ßg Ñ by

AÐ\Ñ œ i!  min ÖlUU l À 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 [œÖ"À−[ׇ .. , and represent each ordinal . " in [‡ by a well-ordered set QQ.."1. We may assume the 's are pairwise disjoint. (If not, replace each Q." with , ordered in the obvious way.) Form the ordered sum: that is, let Q‚Ö"×." . Wœ ÖQ." À . −[×, and order W by

Bß C − Q, and B Ÿ C in Q BŸC if .." "  B−Q./" ßC−Q " and ./ 

Clearly, ŸWÐWßŸÑ well-orders , so represents an ordinal 5 . Since each Q©W." , we have .5"Ÿ by Theorem 5.19. Since ..5  "Ÿ 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 55 larger than every ordinal in [Ò!ß"Ñ[Ág . Then (it contains 55 ) and the smallest element in Ò!ß  "Ñ  [ is sup [ . ì

Example 5.25

1) supÖ!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) supÖÀαα 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 iÒ!ßÑ"""""" (why? ). Since == represents , so there are exactly ii ordinals = , that is, exactly countable ordinals. Each countable ordinal can be represented by a subset of  ÐÑsee Example 4.7 , so there are exactly i" nonisomorphic well-ordered subsets of  .

Since =α="" is the smallest uncountable ordinal, each  is a countable ordinal that is represented by Ò!ßαα Ñ. Therefore 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.

#8 The ordinals ==!! , ... , 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 =! ==!!ÐÑ= ! the end of this chapter.) Then it turns out that ==!! , " , ... , = ! , ... , are still countable ordinals. =! =!!= =!!== ! If you accept that, then you should also believe that %====!!œÖ sup , !! , , ! , ... ל ( “ = ! 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.

%! %!!ÐÑ% 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 Ö5αα Àαα −E } and Ö7 À −E× are sets of cardinals and 5 αα 7 for each α −E , then . (An infinite sum of cardinals is defined in the obvious way: if the ' s are 5 αα 7 Oα pairwise disjoint sets with cardinality , then 55œlOlÞÑααα Proof Exercise

Theorem 5.27 If a set of cardinals GœÖ5Àα −E× contains no largest element, then 5 5 ααα ! for each α! −E .

7œ5αα!! Proof For any particular α! −E , let Þ  7œ!α for αα Á ! Then the lemma gives 5 7œ5 . ααα! If 5œ5 for some α , 5 would be the largest element in G which, by hypothesis, does not  αα!!! α exist. Therefore 55 for every α −Eì .  αα! !

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 G7G is a set of cardinals, then there is a cardinal larger than every member of (and therefore there is no “set of all cardinals”).

Proof If G57œ# has a largest element , then let 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 G every cardinal in .

Proof Without loss of generality, we may assume that if :−G , then all cardinals smaller than : are also in GG55 (why? and where is this used in what follows? ). If has a largest element , then is the least upper bound. Otherwise, pick a cardinal 7G greater than all the cardinals in and let WœÖ:À: is a cardinal and :Ÿ7× . Then WGÁg (it contains 7 ) and the smallest cardinal in this set is the least upper bound for Gì .

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55 that is, indexed in such a way that α" 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 ( œi ), i Ðœi Ñ , ... , i , ... , i , i , ... , i , ... , i# , ... , i !"!" # 8==!! " == !! =! % ! ß

ißi%=!""ß ... , ... .

Thus, iœ==α=!"" sup ÖiÀ8×i8!=α= and = sup ÖiÀ×i " . is the first cardinal with uncountably many (i" ) cardinal predecessors.

iα In this notation, GCH states that for every ordinal α , #œiα" .

By definition, -œ |‘ |. So where is - is this list of cardinals? The continuum hypothesis states “-œiÞ" ” However, it is a fact that for each ordinal α"α , there exists an ordinal  such that the assumption “iœ-" ” 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 i! “calculated” in ZFC: is #œiœi""(" ? ? œi=! ?

On the other hand, one cannot consistently assume “-œiα ” for an arbitrary choice of α : even in ZFC, certain iiαα 's are provably excluded. In fact, if is the least upper bound of a strictly increasing i! sequence of smaller cardinals, we will prove iiα αα (and therefore iÁ- ). It is also true, but we i! will not prove it, that these are the only excluded iα 's it is consistent to assume 2 œiα i! for any cardinal iiiαα for which α œÞ (Solovay, 1965)

350 At the heart of what we need is a classical theorem about cardinal arithmetic.

Theorem 6.1 (Konig¨ Ñ Suppose that for each α −E , 7αα and 8 are cardinals with 7 αα 8 . Then . αα7αα 8 Proof Proving “Ÿ ” is straightforward; proving “ ” takes a little more work.

Let sets QRαα and represent 78 αα and . We may assume the R α 's are pairwise disjoint and, since 78αα, that each Q α is a proper subset of R α . For each α , pick and fix an element 8−RQ α α α and define by: 0À Qαα Ä R

Bœif αα! for B−Qαα! , 0ÐBÑœD− R , where DÐα Ñœ   8Áα if αα!

The Q0α 's are disjoint so is well defined, and clearly 0 is one-to-one, so we conclude that . αα7Ÿαα 8 We now show that is impossible. We do this by showing that if is 7œαα 8 2ÀQÄ αα  R one-to-one, then 2 cannot be onto.

Let T œ ran Ð2Ñ œ 2Ò Qαα Ó œ 2ÒQ Ó . Let 2ÒQ αα Ó œ T . Since 2 is one-to-one,  th lTαα l œ 7 so, for each αα , there are at most 7 α different coordinates of points in T α  that is, |ÖDαααα ÀD−T ×lŸ7 8 . Then we can pick a point A αααα −R with A ÁD for every D−Tαα, i.e., A is not the α -th coordinate of any point in T α .

Define by . Then for all , so ran . A− Rααα AÐαα ÑœA AÂT AÂT œ Ð2Ñ ì

Example 6.2 Suppose we have a strictly increasing sequence of cardinals

! Á 7!"  7  ÞÞÞ  7 5  ÞÞÞ .

For each , let , so . By Konig's¨ Theorem, ∞∞∞ 58œ77855"555œ! 78œ7 5 5œ! 5 5œ" 5 ∞ . Ÿ75œ! 5 In particular: if 7œi , then ∞∞ i i . Since i Ÿ ∞ i (why? ), we have 555œ! 5 5œ! 5=!  5œ! 5 ∞ i iŸi i! . =! 5œ! 5 =!

i! Since -œ- , we conclude that -Ái=! .

Note: A similar argument shows that if a cardinal k is the least upper bound of a sequence of strictly increasing cardinals, then 555Á-i! , so .

Exercise 6.3 For a cardinal 5Ÿ5 , there are how many ordinals with cardinality ?

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!ßαα ÑÞ Therefore \ and Ò!ß Ñ are homeomorphic in their order topologies. Therefore to think about “spaces of ordinals” we only need look at initial segments of ordinals Ò!ßα Ñ . 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 \œÒ!ßα Ñ consists of all sets

ÖB − Ò!ßα###α Ñ À B  × œ Ò!ß Ñ ,  and ÖB − Ò!ßα""α" Ñ À B  × œ Ð ß Ñ , 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##α Ñß where Ÿ Ð #α œ corresponds to the empty intersection Ñ ÐßÑ"# , where " ! and # Ÿ α

If αααœ!ß then \œÒ!ß Ñœg ; so suppose !Þ What does an efficient neighborhood base at a point 7α−Ò!ß Ñ look like? .

If 7αœ ! À Ö!× œ Ò!ß "Ñ is open in Ò!ß Ñ . Therefore ! is an isolated point in Ò!ß α Ñ and ÖÖ!×× is an open neighborhood base at ! .

If !7α  , then any basic open set containing 7 must contain a set of the form Ð 57 ß Ó :

if 7#− Ò!ß Ñ , then 77 − Ð!ß Ó © Ò!ß # Ñ if 7−Ð "# ß Ñ , then 7 −Ð "7 ß Ó©Ð "# ß Ñ

Each set ÐßÓœÐß"Ñ57 57 is open; and each set ÐßÓ 57 is also closed because its complement Ò!ß57α  "Ñ ∪ Ð ß Ñis open. Therefore ÖÐ 5757 ß Ó À ! Ÿ  × is a neighborhood base of clopen sets at 7.

Putting together these open neighborhood bases, we get that

U5757αœ ÖÖ!×× ∪ ÖÐ ß Ó À ! Ÿ   ×

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!ßα Ñ is Hausdorff Þ Since there is a neighborhood base of closed (in fact, clopen) neighborhoods at each point 7α−!ßÑ [ , we know even more: Theorem VII.2.7 tells us that Ò!ßÑ α is a X$-space. But still more is true.

Theorem 7.1 For any ordinal αα , Ò!ß Ñ 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!ßαα Ñ is X" , so need to prove that Ò!ß Ñ is normal. Suppose E and F are disjoint closed sets in Ò!ßα ÑÞ

If 7577− Eß let Y7 œ a basic open set of form Ð ß Ó disjoint from F (or, Y7 œ Ö!× if œ ! ) If 7577−F , let Zœ7 a basic open set of form ÐßÓ disjoint from E (or, ZœÖ!×7 if œ! )

We claim that if 77"#−E and −F , then Y77"# ∩Z œg :

The statement is clearly true if 77"# or œ! so suppose both are ! . Then Y7" œÐ 57 "" ß Ó

and ZœÐßÓÞ7# 57## We can assume without loss of generality that 7 "  7 # .

If ÐßÓ∩ÐßÓÁg57"" 57 ## , then we have 7 " −ÐßÓ 57 ## which would mean that Z∩EÁgÞ7#

If and , then and are disjoint open sets with and Yœ77−E Y77 Zœ −F Z Y Z YªE ZªFÞñ ()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 "α5"5""" is a limit ordinal in Ò!ß Ñß then for all  , Ð ß Ó Á Ö × . Therefore is not isolated in the order topology. If ""# is a nonlimit ordinal, thenÖ × œ Ö!× or, for some , ÖלÐß"#" Ó. Either way, Ö× " is open so " is isolated. Therefore the isolated points in Ò!ßÑ α 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 αα 's is Ò!ß Ñ connected?

Theorem 7.4 Suppose ααα" ! . The ordinal space Ò!ß Ñ is compact iff Ò!ß Ñ œ Ò!ß Ó for some ordinal "α Ò!ßÑ that is, iff contains a largest element " .

Proof Suppose Ò!ßαh##αα Ñ has no largest element. Then œ ÖÒ!ß Ñ À  × is an open cover of Ò!ß Ñ . The sets in h# are nested so, if there were a finite subcover, there would be a single set Ò!ß Ñ covering Ò!ßα#α# Ñ. That is impossible since − Ò!ß Ñ  Ò!ß ÑÞ

Conversely, suppose Ò!ßα"" ÑœÒ!ß ÓÞ If œ! , then Ò!ß " ÓœÖ!× is compact, so we assume " ! .

Let h"h be an open cover of Ò!ß Ó . We can assume consists of basic open sets, that is, sets of the form Ö!× or Ð57 ß Ó . In that case, necessarily, Ö!× − h Þ

353 Let """""""""œ "  ! . For some 5  " ß we have a set Ð 5" ß Ó − h Þ If 5 œ !ß then ÖÖ!×ß Ð 5" ß Ó× is a finite subcover.

If 5"55"h55"5""  ! , then for some#" ß − ( ## ß Ó − Þ If # œ ! , then ÖÖ!×ß Ð ##"" ß Óß Ð ß Ó× is a finite subcover.

We proceed inductively. Having chosen Ðß5"55 Ó , if 5 5 ! we can choose ( 5 5"5" ß " Ó− h so that 55"55"5"−Ð ß 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 555"#$ÞÞÞ ... 5 8 . This is impossible because then the well-ordered set Ö×55"# , , ..., 5 8 , ... would have no smallest element.

When we reach 5h5"5"88""œ ! , we have a finite subcover from À ÖÖ!×ß ( ß8 Óß ÞÞÞß Ð ß Ó× . ñ

Example 7.5 Ò!ß====!!!! Ñis not compact, but Ò!ß  "Ñ œ Ò!ß Ó is compact. In fact, Ò!ß Ó is " homeomorphic to Ö"× ∪ Ö" 8 À 8 − ×Þ

8. The Spaces Ò!ß=="" Ñ and Ò!ß Ó

Theorem 8.1 For each 8=α=αα==!8" ß suppose  . Then œ sup Ö 8!" À8 ×  that is, the sup of a countable set of countable ordinals is countable.

Proof Ò!ßαα=88" Ñ is countable so Ò!ß Ñ is countable. Since has uncountably many 8=! predecessors, there is an ordinal #=−Ò!ß"88 Ñ Ò!ß α Ñ . Then #α  for each 8 , so 8=! αα=#=œÖÀ8ןÞñsup 8! "

Corollary 8.2 Ò!ß=="" Ó and Ò!ß Ñ are not separable.

Proof If H is a countable subset of Ò!ß=α""" Ó , then œ sup ÐH  Ö == ×Ñ  Þ Then cl H © Ò!ßα= Ó ∪ 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 ÐÑα=αα=α8"8"8 is a sequence in Ò!ßÑ . Let œ sup Ö À8−×Þ The 's are all in the closed set Ò!ßαα= Ó , so Ð8" Ñ ÄÎñ .

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 Ð α= 8" Ñ in Ò!ß , ) can converge to = " Þ

This implies that Ò!ß=5757" Ó is not first countable. Of course, ÖÐ ß Ó À  × 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 Ê\ is countably compact Ê\ is pseudocompact.  \ is sequentially compact  We asserted that, in general, no other implications are valid. The following corollary shows that “sequentially compact” ÊÎÊ “compact” (and therefore “countably compact”Î “compact” and “pseudocompact”ÊÎ “compact” ).

Corollary 8.5 Ò!ß=" Ñ is sequentially compact.

Proof Suppose ÐÑα=8" is a sequence in Ò!ßÑ . We need to show that ÐÑ α 8 has a convergent subsequence in Ò!ß=α" ÑÞ Without loss of generality, we may assume that all the 8 's are distinct (why? )

The sequence Ðαααα8888 Ñ has either an increasing subsequence "#   ÞÞÞ 5  ÞÞÞ or a decreasing subsequence αα88"#  ÞÞÞ  α 85  ÞÞÞ

The argument is completely parallel to the one in Lemma IV.2.10 showing that a sequence in ‘ααα has a monotone subsequence. Call 885 a peak point of the sequence if 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 αα88"#  ÞÞÞ  α 85  ÞÞÞ

If ÐÑα8 has infinitely many peak points, then we can choose a subsequence of peak points

αα88""#ß ÞÞÞ5 ß ÞÞÞ and for these, αα 88 ÞÞÞ α 85 ÞÞÞ Þ Since the α 8 's are distinct, we

have αα88"#  ÞÞÞ  α 85  ÞÞÞ .

However, a strictly decreasing sequence of ordinals αα88"#  ÞÞÞ  α 85  ÞÞÞ is impossible.

Therefore Ðαααα8888 Ñ has a subsequence of the form "#   ÞÞÞ 5  ÞÞÞ . Setting

ααœsup Ö8"8"5 À 5 œ "ß #ß ÞÞÞ×  = , we see that then Ð αα= Ñ Ä − 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 =α=="""ÐßÓÖ× that is, the intersection must contain a “tail” . Therefore is not a KÒ!ßÓÞ$-set (and therefore not a zero set) in ="

Proof Let ÖR5"5 À 5 œ "ß #ß ÞÞÞ× be a collection of neighborhoods of =α . For each 5 , there is a for which =α="5"−Ð ß Ó© int R 55 ©R Þ Let α œ sup Ö α 5 À5− = × " Þ Then ∞∞int . In particular, ∞ 5œ"Rª55" 5œ" RªÐßα= Ó  5œ" RÁÖ×Þñ 5" =

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=α=α=""" ÓÑ , then 0 is constant on the “tail” Ò ß Ó for some  .

Proof Suppose 0Ð= Ñ œ < . By Corollary 8.6, 0" ÒÖ<×Ó œ∞ 0 " ÒÐ< "" ß <  ÑÓ contains a tail " 8œ" 88 Ðßα="" ÓÞ Therefore 0lÒßÓœ<Þñ α=

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: <œ0Ð=" 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=α=α=""" ÑÑ , then 0 is constant on the “tail” Ð ß Ñ for some  .

th Proof Let XœÒßα α="" Ñœ “the α tail” Þ By Corollary 8.5, Ò!ßÑ = 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 0ÒXα Ó is a countably compact subset of ‘ . Since countable compactness and compactness are equivalent for subsets of ‘ (Theorem IV.8.17), each 0ÒXα Ó is a nonempty compact set: 0ÒXα Ó © 0ÒX! Ó ©‘ Þ

The Xαα" 's are nested À 0ÒX Ó © 0ÒX Ó © 0ÒX! Ó if "α  Þ Therefore 0ÒX α Ó's have the finite intersection property, and by compactness 0ÒXα Ó Á g (see Theorem IV.8.4 ). In fact, we claim the intersection α= " contains a single numberÀ 0ÒXα Ó œ Ö<×Þ α= "

If <ß = − 0ÒXα Ó , then 0 assumes each of the values <ß = for arbitrarily large values of α . α= " Therefore we can pick an increasing sequence α"α"""##    ÞÞÞ  α" 88   ÞÞÞ such that 0ÐÑœ<α"#α"=α#88 and 0ÐÑœ=Þ Let œ sup Öß 88"8 À8−×Þ Then ÐÑÄ and Ð"#888 Ñ Ä. By continuity Ð0Ð α ÑÑœ Ð<Ñ Ä # and Ð0Ð " ÑÑ œ Ð=ÑÄ # . and we conclude # œ<œ=Þ

356 We claim that 0´< on some tail of Ò!ß="=#"""8 ÑÞ First notice that if  , then b  for which "" 0ÒX#8 Ó © Ð<88 ß<  ÑÞ

"" If not, then Ö 0ÒX# Ó À "#=  "" × ∪ Ö 0ÒÒ!ß = ÑÓ  Ð< 88 ß <  Ñ À 8 −  × would be a family of closed subsets of 0ÒX! Ó with the finite intersection property, so this family would have a nonempty intersection. That is impossible since0ÒX# Ó œ Ö<× and "#=" < Â∞ 0ÒÒ!ß= ÑÓ  Ð< "" ß <  Ñ. 8œ" " 88

"" Pick ###"#" so that 0ÒX##" Ó ©Ð<"ß<"ÑÞ Pick  so that 0ÒX# Ó©Ð<## ß< Ñ and continue "" inductively to pick ##8" 8 so that 0ÒX#8" Ó © Ð< 8" ß <  8" ÑÞ

Let 7#=œsup Ö À8− × Þ Then 0ÒXÓ©0Ò X Ó© 0ÒX Ó©∞ Ð<"" ß< ÑœÖ<×ß 8"7#88 # 8œ" 88 so 0lX7 œ <Þ ñ

(Since 0 is bounded on the compact set Ò!ß7= Ó , we see in a different way that Ò!ß" Ñ is pseudocompact.)

Corollary 8.9 Every continuous function 0ÀÒ!ß=‘" ÑÄ can be extended in a unique way to a continuous function JÀÒ!ß=" ÓÄÒ!ß=" Ó.

Proof For some <−‘α== ß0œ< on a tail Ò ß"" ÑÞ Let JlÒ!ß Ñœ0 and define JÐ = " Ñœ<Þ Any continuous extension K0 of must agree with J since J and K agree on the dense set Ò!ßÑÞñ="

Note: Ò!ß=" Ó is a compact XÒ!ßÑ#" space which contains = as a dense subspace. We call Ò!ß=" Ó a compactification of Ò!ß=" Ñ. The property stated in Corollary 8.9 is a @/

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 0ÐBÑ œ sin ÐB 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 7 plane \XÒ!ß"Ó7\ is not normal however it can be embedded in the % -space for some . Any space that is XX" but not % 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 ‡ XœX ÖÐ=="! ß) × We claim that X is not normal. Let

EœÖÐ=="! ß8Ñ−XÀ8 ל “the right edge of X ” and FœÖÐßα=!" Ñ−XÀ α  = ל “the top edge of X ”

EFand are disjoint sets and closed in X (although not, of course, in X‡ ).

357 Suppose YX is an open set in containing the “right edge” EÐß8Ñ−E©Y . For each point =" , we can choose a basic open set Ðα=8" ß Ó ‚ Ö8× © Y . Let α œ sup Ö α 8 À 8 œ !ß "ß #ÞÞÞ×  = " Þ Then Ðα= ß" Ó‚Ö8שY for all 8 that is, E is contained in a “vertical strip” Ð α= ß"! Ó‚Ò!ß = Ñ inside Y .

Suppose ZX is an open set in containing the “top edge” FÐ"ßÑ−F . Since α=! , there is a basic open set Öα= "ׂÐ8ß! Ó©Z . But then Ð α "ß8"Ñ−Y ∩Z , so Y ∩Z Ág .

Exercise 8.11 Show that every continuous function 0ÀXÄ‘ can be continuously extended to a function JÀX‡ Ä‘ Þ Hint: 0lÒ!ß=="! Ñ ‚ Ö × has constant value < on some tail, and for each 8  =!" ß 0l Ò!ß = Ó ‚ Ö8× has a constant value <8"!8 on a tail. Define J ÐÐ== ß ÑÑ œ <Þ Prove that Ð< Ñ Ä < and then show that the extension JÐßÑÞ is continuous at =="!

As in the remark following Corollary 8.9, XX‡ is a compactification of and the functional extension property in the exercise characterizes XXX‡ as the so-called Stone-Cech compactification of . Since ‡ is compact, J0 must be bounded so, in retrospect, must have been bounded in the first place. Therefore XX is pseudocompact. But is not countably compact  because the “right edge” E is an infinite set that has no limit point in XX . 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 α"#ßß À

a) if "α"α! , then   b) if α"œ , then there exists a unique # such that α "# 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 =α"! œœ" , . Problems arise because ordinal addition is not commutative. c) "αα œ iff α= ! .

E16. Let F©E , where ÐEßŸÑ be a chain. F is called inductive if

for all >−E, Ö+−EÀ+>שFÊ>−F .

Prove that if EEE is the only inductive subset of , then is well-ordered.

E17. Let \J\ be a first countable space. Suppose that for each α=" ,α is a closed subset of and that J©J whenever ααŸ = . Prove that ÖJÀ× α= is closed in \Þ αα"# "# " α "

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 EœÒ!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 PÞP homeomorphic to ‘ is normal but not metrizable.

E19. a) Let EF and be disjoint closed sets in Ò!ßÓ=" . Prove that at least one of EF and is compact and bounded away from =" . (A set GG©Ò!ßÓ is bounded away from =αα="" if for some .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 ÐB88 Ñ and ÐC Ñ are sequences in \ œ Ò!ß= " ) such that, for all 8 , B 888" Ÿ C Ÿ B . Show that both sequences converge and have the same limit. b) Show that if 0 À \ Ä \ is such that 0ÐBÑ 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 Ág . To do this: if any point ÐBß" Ñ is in U, we're done. So suppose this is false and define 0ÐBÑ to be the least ordinal  B such that ÐBß 0ÐBÑÑ 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=5 " Ñ is not -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 [!ßα5 Ñ is -compact iff ...

()You might begin by thinking about the spaces Ò!ß=="# Ñ , Ò!ß Ñ , and Ò!ß ==! Ñ .

E22. A space \0À\Ä is called functionally countable if every continuous ‘ has a countable range.

a) Show that \œÒ!ß=" Ñ is functionally countable. b) Let ]œH∪Ö:× where H is uncountable and :ÂH . 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 ÖÐααα ß 1Ð Ñ À is isolated in \× © \ ‚ ] )

E23. A space \\ is called ] -compact if is homeomorphic to a closed subspace of the product ]7 for some cardinal 7\Ò!ß"Ó\X . For example, is -compact iff compact and # . 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 \\Ò!ß"Ó is Tychonoff and that however is embedded in 7 , its projection in every direction is compact. (More precisely, suppose that for all possible embeddings h À \ Ä Ò!ß "Ó7 , we have that 11ααÒ2Ò\ÓÓ is compact for all projections . This statement is certainly true, for example, if \\ is compact.) Prove or disprove that must be compact.

E24. An infinite cardinal is called sequential if ∞ for some sequence of cardinals . 55œ5558œ" 88

a) Prove that if 555 is sequential, then i! . b) Assume GCH. Prove that if 5555œ# is infinite and ii5!! , then . c) Assume GCH. Prove that an infinite cardinal 555 is sequential iff i! .

360 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 8œ!ß"ß#ßÞÞÞ We would like to show that all the T8 's are true. (The initial value 8œ! is not important. We might want to prove the propositions T"$Ÿ88! for, say, = .) For example, we might have in mind the propositions

8Ð8"Ñ T8 À !"#ÞÞÞ8œ #

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 TTÊTÑT!5"58 is true and if ( is true is true , then is true for all 8=! Þ

2) (Principle of Complete Induction ) If TÐT45ÊT!4 is true and if is true for all 5 is trueÑT , then 8! 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=!8 Ó À T is false × Á g , then it would contain a smallest element 5  ! . This is impossible: since TT5" is true, 5 must be true.

You might expect a principle analogous to 1) could be used in every well-ordered sets Ò!ßα Ñ , 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 αα be an ordinal and X©Ò!ßÑÞ If

1) !−T and 2) a"α − Ò!ß Ñ [ Ò!ß " Ñ ©TT Ê " − ] , then T œÒ!ßÑα .

Proof If X Á Ò!ßα"α"" Ñ , then there is a smallest − Ò!ß Ñ  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=#α! ÑÞ For each  , we have a proposition TT## and we want to show that all the '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 XœÖ#α  ÀT# is true שÒ!ßÑÞ α If we show that T! is true, and if ß 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.

361 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!Ñœg©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œ!# 's with #" . 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À for example, we might say

let 0Ð!Ñ œ 13 and, for each 8  ! , let 0Ð8Ñ œ #0Ð8  "Ñ (**)

We than say that “0 is defined for all 8œ!ß"ß#ßÞÞÞ .” 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 0Ð5"Ñ is defined, and therefore (by ** ) so is 0Ð5Ñ .

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 TT"# in terms of objects already defined (that is, in terms of TT # 's with #" ). Then " is defined for all "αÐ . The 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 TT#" is defined ... , then 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 TT"" ” to mean choosing 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"αœ GÐ\Ñ and want to define continuous functions T" − GÐ\Ñ for each  Þ

The sets T# already chosen (“defined”) in X#" for can be described efficiently by a Ò!ß" Ñ function

To define the set TT" in terms of the preceding #'s means that we need to define T"" using < . We need a function (“rule”)VVÐÑ""" so that

362 The conclusion that we have “completed” the process and that T" is defined for all "α Ò!ßα Ñ means that there is a function J−X"α" where, for each  ßJÐÑœ the T" selected at the earlier stage by V""  that is, JÐ"Ñœ V ÐJlÒ!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 αX is an ordinal. Let be a set and Ò!ß" Ñ suppose that, for each "αÀÄ , we have a function V" X X . Then there exists a unique function Ò!ßα Ñ J −X"α"" such that, for each  , 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 8œ!ß"ßÞÞÞ . Informally we might say:

# # Let E!8!œ 1 and for 8  ! , let I 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:

Ò!ß8Ñ Let Ò!ßα=X Ñ œ Ò!ß!!8 Ñ and œ . For each 8  = , define V À  Ä  as follows:

Ò!ß!Ñ g For 8 œ ! À œ œ Ög× and define V! ÐgÑ œ " Ò!ß8Ñ # # For 8  ! À if < − , 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

JÐ!Ñœ V!! (1 JlÒ!ß!ÑÑ œ V ÐgÑœ ## JÐ"Ñœ V" (1 JlÒ!ß"ÑÑ œ JÐ!Ñ œ œ 1 ## JÐ#Ñœ V# (2F lÒ!ß#ÑÑ œ JÐ!Ñ  JÐ"Ñ œ

ÞÞÞ etc. ...

which is just what we wanted: domJœÒ!ß=!8 Ñ , so I œJÐ8Ñ is defined for all 8=! Þ

363 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 αα"" can be written uniquely in the form œ8 where is either 0 or a limit ordinal, and 8=! .

Proof If αα is finite, then œ!8 where 8 =! .

Suppose αα is infinite. If is a limit ordinal, we can write αααœ!Þ If is not a limit ordinal, then ααα has an immediate predecessor which, for short, we denote here as “" .” If " 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 αα  "   #  ÞÞÞ  α  8  ÞÞÞ would be an infinite decreasing sequence of ordinals. If "αœ8 is a limit ordinal, then α" œ8Þ

To prove uniqueness, suppose α"œ8œ8 "ww where each of "" ß w is ! or a limit ordinal and 8ß8ww are finite. If "" or œ!À say " œ! . Then " 8 is finite, so " ww œ! and 8œ8 . So suppose "" and w are both limit ordinals. We have an order isomorphism 0Ò!ß8Ñ from " onto Ò!ß""ww  8 Ñ. Since Ò!ß  8Ñ contains 8  " ordinals after its largest limit ordinal " , the same must be true in the range Ò!ß""""ww  8 Ñ , and therefore 8 œ 8 w . Let 1 œ 0lÒ!ß ÑÞ Then 1 À Ò!ß Ñ Ä Ò!ß w Ñ is an order isomorphism, so ""œÞñw

Definition 10.2 Suppose α"œ8 , where " is a limit ordinal or !ß8 and is finite. We say that α is even if 88 is even, and that α is odd if is odd.

For example, every limit ordinal ααœ! is even.

Definition 10.3 Suppose Ð\ß .Ñ is a metric space. Let Zg!. œ ß the collection of open sets. For each !α= ", and suppose that Z" has been defined for all "  α Þ Then let

ZZ"ααœÖKÀK is a countable intersection of sets from Ö À  × Ð if is odd) α  " is a countable union of sets from (if is even)  ZZ"αα"œÖKÀK Ö À  × α is the family of Borel sets in UZα=œÖ α À" × Ð\ß.ÑÞ

364 The sets in ZZ"# are the KK$$ sets; the sets in are countable unions of sets and are traditionally called KK$5 sets; the sets in Z$ are called $5$ -sets, etc.

Theorem 10.4 Suppose Ð\ß .Ñ is a metric space.

1) If α"=" , then Zα" © Z , so

ZZZ!"#© © © ÞÞÞ © Zα" © ÞÞÞ © Z © ÞÞÞ Ð α"=   " Ñ

2) ZαZαα is closed under countable unions if is even, and is closed under countable intersections if α is odd.

3) U is closed under countable unions, intersections and complements. Also, if FF"# and are in U , so is FFÞ"#

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"α ÐÑ . In particular, any one set from ZZα" is in .

2) Suppose αZ is even and K"# ß K ß ÞÞÞß K 8 ß ÞÞÞ −α . Each K 8 is a countable union of sets from . Therefore ∞ is also a countable union of sets from , so ÖÀ×Z"α" 8œ" K8  ÖÀ× Z"α" ∞ The proof is similar if is odd. 8œ"K−8 Zαα Þ

3) Suppose F"# , F ,..., F 8 ß ÞÞÞ −UZα= . For each 8 , F 8 −α8 for some 8  " .

If ααœ sup Ö8" À8œ"ß#ßÞÞÞ× =ZZZ , then ααα8 © © " for every 8 . By part 2), one of the collections ZZαα or " is closed under countable intersections and the other under countable unions. Therefore ∞∞ and are both in 8œ"FF88 8œ" ZUα " ©Þ To show that U is closed under complements, we first prove, using transfinite induction, that if K−ZZαα ßthen \K−" Þ

αZœ!ÀIf K−! , then K is open so \K is closed. But a closed set in a metric space is

a K\K−Þ$ set, so Z"

Suppose the conclusion holds for all "α=" . We must show it holds also for Zα .

Let K−Zα Þ If is odd, then Kœ∞ K ß where K − Z"α Ð  ÑÞ α"8œ" 8888 By the induction hypothesis, \K8"" ©ZZ"α8 © for all 8 . Since α " is even, we have that ∞∞ (The 8œ"Ð\K88" Ñœ\ 8œ" K œ\K−Zα Þ case when α is even is similar).

If F−##UZα=ZUU , then F−αα for some  "#"" , so \F− © Þ So if F− , then F∩Ð\FÑœFF−"#"#U Þñ

Part 3) of the preceding theorem shows why the definition of the Borel sets only uses ordinals α= " . Once we get to , the process “closes off” that is, continuing with additional UZα=œÖ α À" ×  countable unions and intersections produces no new sets.

365 Definition 10.3 presents the construction of the Borel sets “from the bottom up.” It has the advantage of exhibiting how the sets in UU are constructed step by step. However, it is also possible to define “from the top down.” This approach is neater, but it gives less insight into which sets are Borel.

Definition 10.5 A family ÆÆÆ of subsets of \\− is called a 5-algebra if and is closed under complements, countable intersections, and countable unions.

Suppose Tc5T is a collection of subsets of \Ð\Ñ . Then is (the largest) -algebra containing . It is also clear that the intersection of a collection of 55 -algebras is a -algebra. Therefore the smallest 5T-algebra containing 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 U5 of Borel sets in Ð\ß .Ñ is the smallest -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 U5 is a -algebra containing the open sets. We must show U is the smallest© that is, if U5w is a -algebra containing the open sets, then UUw . We show this by transfinite induction.

w We are given that ZU! ©Þ

ww Suppose that ZU"α©Þ©Þ for all "α=" We must show ZU

∞ Assume αZ is odd. If K−α" then Kœ8œ" K88 where K − Z"α8 for some 8  Þ ww By hypothesis each ZU"8 − and U is closed under countable intersections so ww K−UZU. Therefore α © Þ (The case when α is even is entirely similar .)

ww Since ZUαα© for all α=" , we get that U œ©Þñ ZU α= "

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=ZßllŸ-Þ" α

αVœ!À A separable metric space has a countable base for the open sets Þ Since every open set is the i! union of a subfamily of VgZcV , we have l.! lœl lŸl Ð ÑlŸ# œ-Þ

Assume that llŸ-Z"α=α" for all " . Since has only countably many predecessors, Since each set in is a countable intersection or union of a sequence of sets from llŸ-Þ"α ZZ"α , we have  i! "αZZZ"α"llŸl "α lŸ-œ-Þ

Therefore |UZlœlα lŸi" †-œ-Þ ñ α= "

366 ()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: Uc œ X. œ Ð\ÑÞ

" 2) If \ œ Ö!× ∪ Ö8 À 8 − × , then every subset is a K$ , but \ is not discrete. Therefore

ZZZ!"#© œ œ ÞÞÞ œ ZαU œ ÞÞÞ œ Þ Á

3) The following facts are true but harder to prove:

a) For each α=Ð\ß.Ñ" , there exists a metric space for which

ZZ!"© © ÞÞÞ © ZZαα œ " œ ÞÞÞ œ Z " œ ÞÞÞ œ U ÁÁÁ In other words, the Borel construction continually adds fresh sets until the αth stage but not thereafter.

b) In ‘Z , α"ÁÑ Z for all α " =" 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 \aB−\B is called point-finite if , is in only finitely many sets from Y .

Definition 10.11 An open cover hαœÖYα À −E× of \ is called shrinkable if there exists an open cover iαœÖZααα À −E× of \ such that, for each α , cl Z ©Y . 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 \EF is shrinkable and let and be disjoint closed sets in \ . The open cover hi œÖ\Eß\F× has a shrinkage œÖZßZ"# × and the sets Y œ\ cl ZZœ\Z"# and cl are disjoint open sets containing EF and . Therefore \ is normal.

367 Conversely, suppose \œÖYÀ−E×\ is normal and let hαα be a point-finite open cover of . Without loss of generality, we may assume that the index set EÒ!ßÑ is a segment of ordinals # (why? ) so that hα#œÖYα À  ×Þ

Let Since , we can use normality to choose an open set such that Jœ\!!!!α! YÞα J©Y Z J©Z©!!cl Z©Y ! ! and ÖZ×∪ÖYÀ ! α α !× covers \ .

Suppose !α#  and that for all "α  we have defined an open sets Z""" such that cl Z ©Y and such that covers . Letting , ÖZ"" À"α  × ∪ ÖY À "α × \ J α œ \  Ð"α Z ""α ∪ "α Y Ñ © Y we can use normality to choose an open set Zααααα with J ©Z © cl Z ©Y Þ

Clearly, ÖZ"" À"α   "× ∪ ÖY À "α  "× covers \.

By transfinite recursion, the ZœÖZÀ×α 's are defined for all α# , and we claim that iα 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=œÖÐ∞ß8ÑÀ8! × . If we go through the procedure described above, we get so we might have chosen : that would give J! œ \ 8! Ð  ∞ß 8Ñ œ g Z! œ g J©Z©!!cl Z©Y ! ! and ÖZ×∪ÖYÀ8!× ! 8 would still be a cover. Continuing, we can see that at every stage we could choose Z8!"85 œ g and that ÖZ ß Z ß ÞÞÞZ × ∪ ÖY À 5  8× is still a cover. But when we're done, the collection i=œÖZ8! À8 × 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 hα  say Yαα"8 ß ÞÞÞß Y . Let be the largest of these indices so that . If is in one of the 's with , we're done. Otherwise BÂ"α Y"" B Z"α  . Either way is in a set in , so is a cover B−\Ð"α Z""αα ∪ "α Y ÑœJ ©Z Bii Þ 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 \B−\ is called locally finite if every point has a neighborhood that intersects at most finitely many of the sets in the cover.

Corollary 10.14 Suppose hαœÖYα À −E× is a locally finite open cover of a normal space \ . Then there exist continuous functions 0À\ÄÒ!ß"Óα such that

i) 0±\Yœ!αα for every α −E ii) for every . Ö0ÐBÑÀα α −Eל " 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 iαœÖZα À −E× . For each α , we can use Urysohn's Lemma (VII.5.2) to pick a continuous function 1À\ÄÒ!ß"Óααααα such that 1lZœ" cl and 1l\Yœ!Þ

368 Each point BY0ÐBÑœ! is in only finitely many αα 's, so for all but finitely many α 's and therefore . Each point has a neighborhood which intersects only finitely many 's 1ÐBÑ œ 1α ÐBÑ −‘ B RB Yα α−E so 1lRB is essentially a finite sum of continuous functions and 1 is continuous (see Exercise III.E21).

Since each is in some set cl , so . Therefore is never so each 1ÐBÑα B Zαα!! 1 ÐBÑ œ " 1ÐBÑ ! 0 α ÐBÑ œ 1ÐBÑ is continuous. The 0ñα 's clearly satisfy both i) and ii).

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 Eww œÖB−\ÀB is a limit point of E× . E is called the derived set of EE . If is closed, it is easy to see that E©Eww and that E is also closed.

We will use the derived set operation ÐÑw repeatedly in a definition by transfinite recursion.

Let Eœ\!" and, for each α=  , define

EœEw if α" œ " α " if is a limit ordinal  Eœα" ÖEÀ"α × α

Eααis closed for all α= "!" and E ª E ª ÞÞÞ ª E ª ÞÞÞ . This sequence is called the derived sequence of \ .

For some α=EœE"" , we have αα because otherwise, for each α , we could choose a point B−EEααα" and ÖBÀ αα= " × would be a subset of \ with cardinal i " . Let α ! be the smallest α"α for which EœEαα!!" . It follows that EœE "α ! for all  ! .

Since \EEœE is compact metric, the closed set ααα!!! is complete, and because " , every point in

EEœgαα!! is a limit point. Therefore , since a nonempty complete metric space with no isolated points contains at least - points (Theorem IV.3.6).

369 We know αα! œ! is impossible (because that would mean Eα! œE!! œ\œgÑ . If were a limit ordinal, then EœÖEÀ×Ág"α (because ÖEÀ "α × is a family of nonempty compact α"!  !! " sets with the finite intersection property). Therefore α"!! must have an immediate predecessor .

The definition of α! implies that EÁg""!! . In fact, E must be finite (if it were an infinite set in the compact space , would have a limit point and then w ) Let \E"α"! E"! œEÁgÞ!! 8œlEl=! Þ

In this way, we arrive at a pair Ðß8Ñ"! , where

E\ÐÑ"! is the last nonempty derived set of "=!" , and

E8Ð!8ÑÞ"! contains points =!

Since homeomorphisms preserve limit points and intersections, it is clear that the construction in any space homeomorphic to \Ðß8Ñ will produce the same pair "! .

Theorem 10.16 (Sierpinski-Mazurkiewicz) Let "=!"8 and = ! . Two nonempty compact, countable metric spaces \] and are homeomorphic iff they are associated with the same pair Ðß8Ñ"! . (Therefore Ðß8Ñ"! is a “topological invariant” that characterizes nonempty compact countable metric spaces.) For any such pair Ðß8Ñ"! , 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 Ðß8Ñ"!"!", namely i†iœiÞñ

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!× ∪ Ö8 À 8 − ×Þ

370 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\l\lŸ-Þ satisfies

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 Ð Pol, ŠapirovskiÑ If \l\lŸ-Þ is first countable, Hausdorff and Lindelof,¨ then

Proof For each :−\ , choose a countable open neighborhood base i: at : and, for each E©\ , let the collection of all the basic neighborhoods of all the points in . iiE œÖ : À:−Eל E For each countable family of sets for which , pick a point ii©\Ág;−\ÞE  ii i Define TÐEÑœ cl ( E∪Ö all such ;i 's chosen for E×ÑÞ

Notice that if lElœ- , then lTÐEÑlœ- .

371 i! Since lEl œ - , we have liElœ-. There are at most - œ - countable families ii © E , 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 ∪ Ö all such ;i 's chosen for E× namely -i! œ -Þ

Now fix , once and for all, a set E©\ with lElœ-Þ (If no such E exists, we are done! ) By recursion, we now “build up” some new sets EEαα from . The idea is that the new E 's always have cardinalityŸ- and that the new sets eventually include all the points of \ .

Let EœE! . For each ordinal α= " , define

TÐE Ñif α" œ " Eœ " . α { : } if is a limit ordinal   E" "α α

For each α=lElœ-À" , α

lE! l œ -Þ Suppose lE" l œ - for all "α  Þ

If α"œ" , then EœTÐEÑlEœ-α"α , so | . If is a limit ordinal, then { α"αlElœlα" E À  ×lœ-†i! œ-Þ Let Then F œ ÖEαα Àα= """ ×Þ - Ÿ lFl œ l ÖE À α=  ×l Ÿ - † i œ -Þ We claim that Fœ\ and, if so, we are done.

F\B−F is closed in : if cl , then (using first countability) there is a sequence ÐBÑF8 in with

ÐB88 Ñ Ä BÞ If B − Eαα8 and we let αα= œ sup Ö 8"8 ×  , then every B is in E . Therefore B−cl Eαα œE" ©FÞ

Since F\F is a closed subspace of the Lindelof¨¨ space , is Lindelof.

If FÁ\ß then we can pick a point ;−\F . For each :−F , choose an open neighborhoodZ−::i such that ;ÂZ : . The Z : 's form an open cover of F , so a countable collection of these sets, say i œÖZ: À:−G , for some countable G©F× covers F . Thus and . But this is impossible : F©ii ;Â

Since GG©E is countable, we have that α for some α=" and i is a countable subfamily of ii for which \ ÁgÞ By definition of TÐEÑœE Ñ , a point Eα  αα" was put into the set . This contradicts the fact that covers ;−\iαii E" ©F Fñ.

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.

372 Theorem 10.17 (Mazurkiewicz) There exists a set E©‘# such that lE∩Plœ# for every straight line P.

Proof Let$$$ be the first ordinal of cardinal -Ò!ßÑ . Since the well-ordered segment represents , there are -Þ ordinals α$

In fact, there are - limit ordinals Þ$$ There are certainly infinitely many (say 7 ) limit ordinals  (why?)every and, by Lemma 10.1, ordinal α$ can be written uniquely in the form α"œ8, where " is a limit ordinal (or ! ) and 8 is finite. Therefore there are 7†iœ7! ordinals -$$. But has predecessors. Therefore 7œ-Þ

Since ‘‘# has exactly -- points and exactly straight lines, we can index both # and the set of straight # lines using the ordinals less than $‘ : œ Ö:00 À 0$  × and ÖP À 0$  ×Þ

We will define points +αα for each α$ , and the set E œÖ+À× α$ will be the set we want.

# Let +P!! be the first point of ‘ (as indexed above) not on .

Suppose that we have defined points +0α for all 0α$ . We need to define +Þ Let

EEα0œÖ+ À0α  × Note ll-α XœÖPÀPαα is a straight line containing 2 points of E } Note lXα l  -

"α"αœPÂX least ordinal so that α

(that is, PXÑ"αα is the first line listed that is not in

WœÖ:À:α"α is a point of intersection of Pα with a line in X× Note lWα l  -

Since lEαα∪W l-, there are points on P "α not in E αα ∪W : let + α be the first

:P∪W0"αα listed that is on α but not in E .

By recursion, we have now defined +œÖ+À×Þαα for all α$ . Let E α$

We claim that for each straight line Pl , E ∩PlŸ# .

If l∩Pl#EE , then we could pick 3 points +ß+ß+−P∩"#α , where, say ß "#α  Þ

Then +ß+−"#E α , so that P−X α . Since + α −P "α (by definition) and + α −P , we have

+−P∩Pα"α . Therefore +−W αα which contradicts the definition of +Þ α

To complete the proof, we will show that for each straight line Pl , E ∩Pl # .

We begin with a series of observations:

a) If αα$Ÿß then " Ÿ " . Ò Clearly X©X . But P is the first line not "# αα"# αα "# "α" in XP and is the first line not in X , so "" ŸÓ α""#"#α2 αααÞ

b) If ααα$"#ß3 then "αα" ÞÒ "3 Otherwise, by a), " ααα "#$ œ " œ " . Then +ß+ß+ are on P œP œP . Then P contains 3 points of E which is ααα""""#$ ααα"#$ " α $ impossible.Ó In particular, if αα"$ , are distinct limit ordinals, there is a third ordinal α #

between them so ""αα"$and must be distinct.

373

c) For any #$ , there is α$ such that "α ÞÒ- # Since there are limit ordinals  $ , there are -Ÿ- distinct values for "##α . They cannot all be , since has fewer than predecessors.Ó

Finally, if Pœ some P#α"α , pick an α"# so that  . Since Pα is the first line ÂX ß we get that PœP#α −X Þ Therefore P contains 2 points of E α ©EÞñ

More generally, it can be shown more that if for each line P7 we are given a cardinal number P with # #Ÿ7P Ÿ-, then there exists a set E©‘ such that for each P , lE∩Plœ7ÞP

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 ZLÊ AC

Proof 1 Let ÖEα Àα − E× be a collection of pairwise disjoint nonempty sets. Consider the poset set for all , , ordered by inclusion. is a nonempty poset because ccœÖW©α−EEÀα α lW∩Eα lŸ"× g−c.

Suppose ÖW À"c − M× is a chain in and let F œ ÖW À − M× . We claim that F − c Þ "" " Suppose lF ∩ E l # for some α Þ Consider two points B Á C − F ∩ E Then B − W and α! ! α! "" C−W for some "" , −M . Since the W 's are a chain, either W ©W or W ©W . "# "# """"""#"2

Therefore one of these sets, say Wß""α""! contains both BßCÞ Therefore lW∩ A l" . But this is

impossible since W−"" cc . Therefore F contains at most one point from each Eα so F−Þ

Since F is an upper bound for the chain in c , Zorn's Lemma says that there is a maximal element Q−c Þ

Since Q−cα ßlQ∩ElŸ"αα for every . If Q∩E! œg for some α! , then we could choose an w +−Eαα! form the set Q œQ∪Ö+תQÞ Then lQ∩E lŸ" would still be true for every α , so we Á w would have Q−c , which is impossible because Q is maximal. Therefore lQ∩Elœ"α for every α. ñ

Notice that the function given by 0ÀEÄ Eαα 0œÖÐα ßCÑ−E‚α−EEÀα C−Q∩E × is in the product {A , so we see that {A . ααÀ−E×αα À−E×Ág

Proof 2 Let ÖEα Àα − E× be a collection of nonempty sets. Let

374 and for each c""œÖ1ÀFÄ Eα" ÀF©E 0Ð Ñ−E −F×

If 1ß1−"#cc , then the function 1ß1 "# are sets of ordered pairs. So we can order by inclusion: 1Ÿ1"# iff 1©1Þ "# (This relation is just “functional extension”: 1"# Ÿ1 iff dom Ð1Ñ© " dom Ð1Ñ # and 1l#""dom Ð1Ñœ1Þ) Ðcc ßŸÑ is a nonempty poset because g− .

Suppose is a chain in . Define . Since the 's form a chain, their Ö1333 À 3 − M×c 1 œ Ö1 À 3 − M× 1 union is a function where dom Moreover, if , then dom 1ÀFÄ Eα ß Fœ3−M Ð1ÑÞ33"" −F − Ð1Ñ for some 3 , so 1ÐÑœ1ÐÑ−E""3 " . 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 0E is if not, we could extend the definition of 0 by adding to its domain a point αα from E dom Ð0Ñ and defining 0Ð Ñ to be a point in the nonempty set Eα Þ Therefore so . 0− ÖEÀαααα −E×ß ÖEÀ −E×Ág 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 \œg , so we assume \ÁgÞ Let c%œÖE©\À E is -discrete × and partially order c by inclusion © . If +−\ , then Ö+×−cc ß so Ág .

Suppose is a chain in . We claim is -discrete. ÖGαα Àαc − E× Ð ß Ÿ Ñα−E G % If BßC− G , then G and C−G for some αα , Þ Since the G 's form a chain, so α−E αα"# α"# α Gαααα"##" ©G or G ©G Þ Without loss of generality, G αα "# ©G . Then BßC−G α # , and this set is %% -discrete. Therefore .ÐBß CÑ Þ

Therefore . Clearly, is an upper bound for the chain . By Zorn's αα−EG−ααcα −E G ÖGÀ−E× α Lemma ÐߟÑc has a maximal element. ñ

Notice that when we use Zorn's Lemma, an upper bound that we produce for a chain in ПÑc , for example, in the preceding paragraphmight not itself be a maximal element in . For α−EGα c 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 is . Then is an upper bound for the chain in . However, ccÖEß G× ÖEß G× œ G 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 ÖEßGßH× , 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 BÁC−HßFÐBÑ∩FÐCÑœg8 "", so, by CCC, the family ÖFÐBÑÀB−H× " 8 must be countable. #8 #8 #8 Therefore is countable, and we claim that the countable set ∞ is dense. HHœH8 8œ" 8

" Suppose D−\H and, for %% ! , choose 8 so that 8  Þ Since H8 is a maximal "" 88-discrete set, H∪ÖD×88 is not -discrete, so there is a point B−H©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 IfÖ\Àα αα −E× is a nonempty collection of sets, then b −E such that l\lŸl\lαα __ for every αα−E  in other words, for each −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], Let for all , is one-to-one , ordered by inclusion. α−E\ÁgÞαααcα1 œÖF©\À −E lF × Ðß©Ñcc is a nonempty poset since g− .

If is any chain in , we claim that . Otherwise there would be points ÖF33 À 3 − M×cc F œ F − BÁC−F and an α11 for which αα ÐBÑœ ÐCÑ . Since the F3 's form a chain, we would have both Bß C − F33 for some 3 , and this would imply that 1α lF is not one-to-one.

FÖFÀ3−M× is an upper bound for the chain 3 in cc , so by Zorn's Lemma contains a maximal element, QÞ

_ __ We claim that for some α1−E , αα lQÀQÄ\ is onto .

Otherwise we would have \αα1α ÒQÓÁg for every . Using [AC] again, we could choose a point Since , would be C œ ÐCαααααα Ñ −α−E Ð\ 111 ÒQÓÑ Á gÞ C  ÒQÓ lÐQ ∪ ÖC×Ñ one-to-one for all αc so that Q∪ÖC×− Þ Since CÂQ , Q∪ÖC× is strictly larger than Q and that is impossible because Q is maximal.

__ ___" Therefore 1911 ααlQÀQÄ\ is a bijection and the map αααααα œ ‰Ð lQÑ À\ Ä\ is "" for 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 OœGÐ\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) MÁO 2) +ß,−M Ê +,−M 3) +−M and 5−O Ê 5+−M

In other words, an ideal in OO is a proper subset of 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©Nß then MœNÞ A maximal ideal is a maximal element in the poset of all ideals of O , ordered by inclusion.

For example, two ideals in OœG (‘ ) are:

M œÖ1À1œ03 where 0ß3−GБ ) and 3 is the identity function 3ÐBÑœB× For example, 1ÐBÑ œ B/BB − M (using 0ÐBÑ œ / ÑÞ

377 MœÖ0À0Ð!Ñœ!× In fact, QGÐÑ is a maximal ideal in ‘ (This take a small amount of work to verify; see Exercise E33.)

Maximal ideals, QQO , are important in ring theory for example, if is a maximal ideal in , then the quotient ring OÎQ is actually a field.

Using Zorn's Lemma, we can prove

Theorem 11.8 Let OMO be a commutative ring with unit. Every ideal in is contained in a maximal idealQÞ ( Q might not be unique.)

Proof Let cccœ ÖNÀN is an ideal and M©N×. Ð ß ©Ñ is a nonempty poset since I − . We want to show that c contains a maximal element.

Suppose JJJ is a chain in . Let . Since we only need to check ÖÀ−E×α αcαœÖÀ α −E× M©Nß that NN−Þ is an ideal to show that c

If +ß , −JJ , then +,− −#" and J for some #" ß − EÞ Since the J α '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 kJJJ−O , 5+− " © . Therefore is closed under addition and superclosed under multiplication.

Finally, JJkkÁO : If N œO , then 1 −N so 1 −α for some α . Then, for all −O , œ kJK†−N1αα . So œ , which is impossible since J α is an ideal.

By Zorn's Lemma, we conclude that c contains a maximal element QÞ ñ

Basis for a Vector Space It is assumed here that you know the definition of a vector space Z over a field OO . ( is the set of “scalars” which can multiply the vectors in i . If “field” is unfamiliar, then you may just assume Oœ‘ , , 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 ZOÞF©Z is a vector space over a field A collection of vectors is called a basis for Z@−Z if each nonzero can be written in a unique way as a finite linear combination of elements of FOF@−Z using nonzero coefficients from . More formally, is a basis if for each , there exist unique nonzero , ..., and unique bb , ..., such that 8 b . @Á!ßαα"8 −O "8 −F @œ 3œ" α 33 ZF is called finite dimensional if a finite basis exists.

Definition 11.10 A set of vectors G©Z is called linearly independent if, whenever αα"8 , ..., −O and c , ..., c and 8 c , then 0 (in other words, the only linear "8− G3œ"ααα33 œ ! " œ ÞÞÞ œ 8 œ combination of elements of G! adding to is the trivial combination).

378 Theorem 11.11 Every vector space ZÁÖ! } 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 œÖG©Z ÀG is linearly independent × , ordered by © . For any !Á@−Z , we have Ö@× −cc ßso Ð ß © Ñ is a nonempty poset.

Let be a chain in . We claim is linearly independent, so that . ÖGαα Àαc − M× G œ G G − c If αα , ...,−O and --,..., −G and 8 α - œ! , then each - is in some G . "8"8 3œ" 33 3 α3 ‡ The G©GGααα 's form a chain with respect to , so one of the 3 's call it contains 8 all the others. Then is a linear combination of elements from ‡‡ , and !œ3œ"α33 - Gαα G is linearly independent. So all the α3 's must be ! .

Therefore, G−cαc and G is an upper bound for the chain ÖGÀα −M× . By Zorn's Lemma, has a maximal element FÞ

We claim that FZ is a basis for .

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 @ÂF , 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@×Ð necessarily involving @!ÀÑ with sum

, ..., and , ..., , with and 8 + . b,"8, bαα""8 −O " Á!3œ" α" 33 , @œ!

Since " Á! , we can solve the equation and write @œ8 α3 , . 3œ" " 3

We complete the proof by showing that such a representation for @ is unique .

Suppose that we have 87ww with , w and , w . 3œ"αααα33,œ@œ 3œ" 33 , 3 3 −O,,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 55ww ww Then 5 ( )ww and is linearly 3œ""#33,œ@œ33 3œ" ,Þ  3œ" "# 33  ,3 œ! F independent, so "#33œ for 3 œ "ß ÞÞÞ , 5 .

Example 11.12 In Example II.5.14 we showed that the only continuous functions 0À‘‘ Ä satisfying the functional equation

0ÐBCÑ œ 0ÐBÑ0ÐCÑ for all BßC −‘ Ð‡Ñ are the linear functions 0ÐBÑ œ -B for some - − ‘ . Now we can see that there are other functions, necessarily discontinuous, that satisfy (*).

379 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" ,  ; 33 , 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 0ÐBÑ œ ;B . Clearly 0ÐBCÑ œ 0ÐBÑ0ÐCÑÞ

Although the definition of 0 looks complicated, perhaps it could happen, for a cleverly chosen - −‘ ßthat 0ÐBÑ œ -B for all B ? No: we show that 0ÐBÑ œ -B is not possible (and therefore 0 is not continuous).

1) If 0ÐBÑ œ -B for some constant - , we would have:

0Ð #Ñ œ - † # −  , and 0Ð"Ñœ-†"œ-− . But -Á! , or else 0ÐBÑœ! for every B , whereas 0Ð,Ñœ"†,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 0Ð,Ñœ! . Therefore the equation 0ÐBÑœ! has infinitely many solutions so 0ÐBÑ 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: œÖB− ∩Ò!ß"ÓÀB:× for each irrational : − Ò!ß "ÓÞ There are uncountably many M: 's and each one has measure ! . In this case, , so also has measure . M©:: M ! 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 Zαα (  =" ) and the collection of Borel sets = { : }. UZα= α  "

a) Let Yα=!" be the family of closed sets in \ . For , define families

Ö−×∞ JJ: Y" , α if α is even Y œ 8œ" 88"8 8 α ÖJÀJ−×∞ Y", α if αis odd  8œ" 88"8 8

YY" is the collection of countable unions of closed sets (called J5 -sets) and # is the family of countable intersections of JJ55$ sets (called -sets).

Prove that YZα"©© and ZY α " for all α"= " .

It follows that {} : . We can build the Borel sets “from the bottom up” beginning UYα=œ α " 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 0À\Ä] is called Borel- measurable (or B-measurable , for short) if 0ÒFÓ" is a Borel set in \ whenever F is a Borel set in ] . Prove that 00ÒSÓ\S] is B-measurable iff " is Borel in whenever 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 ( are closed sets in the normal space with 8 . Prove J3 3 œ "ß ÞÞÞß 8Ñ_ \3œ" J3 œ g that there exist open sets such that and 8 = . Z3333 Ð3 œ "ß ÞÞÞß 8Ñ J © Z3œ" Z g Hint: Use the characterization of normality in Theorem 10.12.

# E26. Prove that the continuum hypothesis (-œi" ) 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-iE : If CH is false, then " . Suppose meets every horizontal line only countably often and that E∪Fœ‘# . Show that F meets some vertical line uncountably often by letting U be the union of any iÒU∩EÓ"\ horizontal lines and examining 1 . )

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 hT be any open cover and let 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 . Then consider , where Tα#œÖYÀ×α""α ÖZÀ "# × Zœ ÖYÀ×Þ α"

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‚TÑ ” Hint: Suppose ŸTŸ is a linear ordering on . Can be enlarged?

E30. Let 7\ be a cardinal. A space has caliber 7 if, whenever h is a family of open sets with , there is a family such that and . ±hihii ±œ7 © ± ±œ7 ÖZÀZ− ×Ág

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 \i\ has caliber " , then satisfies the countable chain condition (see Definition 11.4).

d) Let \i be a set of cardinal " with the cocountable topology. Is \\ separable? Does satisfy the countable chain condition? Does \i have caliber " ? (For notational convenience, you can assume, without loss of generality, that \œÒ!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 HœÖ=XI ÀI− × be a set of distinct points such that H∩  œg . Let ^œ  ∪H , with the following topology:

i) points of  are isolated ii) a basic neighborhood of ==II is any set containing 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" À 8 −Y × ∪ Ö!×. P/> E œ Ò!ß "Ó  . E is relatively discrete and therefore countable. For each 8  B − E, choose a different GBBB −Y and replace G by G ∪ ÖB×Þ)

E33. Suppose \ is Tychonoff. For :−\ , let Q:: œÖ0−GÐ\ÑÀ0Ð:Ñœ!× . Clearly, Q is an ideal in GÐ\Ñ.

a) Prove that QGÐ\Ñ: is a maximal ideal in .

b) Prove that is \GÐ\ÑQ is compact iff every maximal ideal in is of the form : for some :−\.

E34. Prove that there exists a subset E of ‘ that has only countably many distinct “translates” EœÖ+<À+−E×< that is only countably many of the sets EœÖ+<À+−E×<−< , ‘ are distinct).

Hint: Consider ‘ as a vector space over the field , and pick a basis F,−F . Pick a point 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 countabl e ordinals. (See Example 5.25, part 3)

If .α and are ordinals, let

..0ααœÖ0−Ò!ß ÑÒ!ß Ñ À0ÐÑœ 0 for all but finitely many 0α  }.

We can think of a point 0Ò!ßÑ0 in ..α as a “transfinite sequence” in , where has well-ordered domain Ò!ßα Ñ rather than . Using “sequence-like” notation, we can write:

α .0α.œÖÐaa000 ÑÀ 0 Ÿ  , !Ÿ+  ß and œ! for all but finitely many 0 ×

We put an ordering on .α by:

α Given ÐÑÁÐÑaab00 b in ./ , let be the largest index for which // Á Þ We write Ð+00// Ñ  Ð, Ñ if ab  .

Example For α.œ œ 2, .α consists of 4 pairs ordered as follows: ÐÑÐÑÐÑÐÑÞ 0,0 1,0 0,1 1,1

We also use ...ααα to denote the order type associated with ÐŸÑ , . It turns out that ÐŸÑ , is well- ordered, so this order type is actually an ordinal number.

=! Example =! is represented by the set of all sequences in Ö× 0, 1, 2, ... which are eventually 0 Þ (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 0α! 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,

ÐÑÐÑÐÑ 5,0,0,1,2,0,0,0,0,... 107,12,1,3,5,0,0,0,0,... 103,7,0,0,6,0,0,0,0...,

=! 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 !!! , , , ... are all countable ordinals. ÐÑÞHere, αα""## is understood, as usual, to mean () = ! =!!= =!!!== 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 .." œÐŸÑ . Prove that for ordinals .α. and , αα is also an ordinal, i.e., . , is a well-ordered set (not trivial!). For ordinals .α" ,ß! , 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 \œÒ!ß=ααα" Ñ . For each −\ , let µ " . In the quotient space \ε , exactly one point is isolated.

2. i! can be written as a sum of countably many smaller cardinals. i=!

8 3. Let 8−=‘= Þ A continuous function 0ÀÒ!ß"" ÑÄ 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 0À\ Ä‘ such that B − coz Ð0Ñ © S .

6. Let GW0ÀWÄ be a non-Borel subset of the unit circle "" and let ‘ 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 9 ÐBÑ œ 0ÐBß +Ñ and < Ð+ß CÑ œ 0Ð+ß CÑ are Borel measurable.

7. Let GG be a dense subset of ‘‘ . Then is not well-ordered (in the usual order on ).

8. Let HœÖ0− À0Ð8Ñœ! for all but finitely many 8 }, with the lexicographic (“dictionary") ordering ŸÐHßŸÑ . Then is order isomorphic to  .

9. Suppose ŸŸ" and # are linear orders on a set \Ÿ©ŸŸœŸ . If "# , then "# .

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\ßg 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 iαα limit ordinals = .

16. If ααα denotes the order type of the irrationals, then # œ .

17. If α=\œÒ!ßÓ\" , then the ordinal space α is not metrizable because is not normal.

386 18. Suppose c is a nonempty collection containing all subsets of \ with a certain property (*). Suppose is a chain of subsets of and that . By Zorn's Lemma, is a ÖTααα À − E×cc T − T maximal subset of \© (with respect to ) having property (*).

19. For any infinite cardinal 55 , there are exactly  ordinals with cardinality 5 .

20. Suppose ααα0 is an infinite order type. If "œ" , then there is an order type (possibly ! ) ‡ such that α=œ! 0=! .

21. If Ÿ\ a linear order on , then the “reversed relation” ŸBŸCCŸB‡‡ defined by iff is also a linear order on \ .

22. For any infinite cardinal 5#œ5 , 55 (without GCH).

8Ð8  " Ñ 23. If ŸlWlœ8Ÿ is a linear ordering on a finite set S and , then | | = 2 .

24. If EF and are disjoint closed sets in Ò!ßÑ=" , 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 ii!! have the lexicographic order ŸŸ . ( , ) 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 Ö×∪ÖB− 0‘‘ ÀlBl"× is homeomorphic to .

30. If hαœÖUα À −E× is a point-finite open cover of the Sorgenfrey plane W‚W , then there must be an open cover iαœÖVVUααα À −E× of W‚W such that, for each α −E , cl Ð Ñ© .

31. There is a countably infinite compact connected metric space.

32. If Ÿ\ is a linear ordering on , then there can be linear ordering Ÿw for which Ÿ©Ÿ©\‚\w Á

387