Chapter I The Basics of Set Theory
1. Introduction
Every mathematician needs a working knowledge of set theory. The purpose of this chapter is to provide some of the basic information. Some additional set theory will be discussed in Chapter VIII.
Sets are a useful vocabulary in many areas of mathematics. They provide a language for stating interesting results. For example, in analysis: “a monotone function from ‘‘ to is continuous except, at most, on a countable set of points.” In fact, set theory had its origins in analysis, with work done in the late 19th century by Georg Cantor (1845-1918) on Fourier series. This work played an important role in the development of topology, and all the basics of the subject are cast in the language of set theory. However sets are not just a tool; like many other mathematical ideas, “set theory” has grown into a fruitful research area of its own.
Moreover, on the philosophical side, most mathematicians accept set theory as a foundation for mathematics that is, the notions of “set” and “membership in a set” can be taken as the most primitive notions of mathematics, in terms of which all (or nearly all) others can be defined. From this point of view, “everything in mathematics is a set.” To put it another way, most mathematicians believe that “mathematics can be embedded in set theory.”
So, you ask, what is a set? There are several different ways to try to answer. Intuitively and this is good enough for most of our work in this course a set is a collection of objects, called its elements or members . For example, we may speak of “the set of United States citizens” or “the set of all real numbers.” The idea seems clear enough. However, we have not really given a satisfactory definition of a set: to say “a set is a collection of objects...” seems almost circular. After all, what is a “collection” ?
In the early days of the subject, writers tried to give definitions of “set,” just as Euclid attempted to give definitions for such things as “point” and “straight line” (“a line which lies evenly with the points on itself ”). And, as in Euclid's case, these attempts did not really clarify things very much. For example, according to Cantor
Unter einer Menge verstehen wir jede Zusammenfassung Q von bestimmten wohlunterschiedenen Objekten in unserer Anschauung oder unseres Denkens (welche die Elemente von Q genannt werden) zu einem Ganzen [By a set we are to understand any collection into a whole M of definite and separate objects (called the elements of Q ) of our perception or thought.] (German seems to be a good language for this kind of talk. )
More compactly, Felix Hausdorff, around 1914, stated that a set is “a plurality thought of as a unit.”
At this point, there are several ways we could proceed. One possibility is simply to take our intuitive, informal notion of a set and go on from there, ignoring any more subtle issues just as we would not
1 worry about having definitions for “point” and “line” in beginning to study geometry. Another option might be to try to make a formal definition of “set” in terms of some other mathematical objects (assuming, implicitly, that these objects are “more fundamental” and intuitively understood). As a third approach, we could take the notions of “set” and “set membership” as primitive undefined terms and simply write down a collection of formal axioms that prescribe how “sets” behave.
The first approach is sometimes called naive set theory . (“Naive” refers only to the starting point naive set theory gets quite complicated.) Historically, this is the way set theory began. The third option would take us into the subject of axiomatic set theory . Although an enormous amount of interesting and useful naive set theory exists, almost all research work in set theory nowadays requires the axiomatic approach (as well as some understanding of mathematical logic).
We are going to take the naive approach. For one thing, the axiomatic approach is not worth doing if it isn't done carefully, and that is a whole course in itself. In addition, axiomatic set theory isn't much fun unless one has learned enough naive set theory to appreciate why some sort of axiomatization would be important. It's more interesting to try to make things absolutely precise after you have a good overview. However as we go along, we will add some tangential comments about the axiomatic approach to help keep things in a more modern perspective.
2. Preliminaries and Notation
Informal Definition 2.1 A set is a collection of objects called its elements (ormembers ). If E is a set and BB−EBÂEÞ is an element of A , we write . Otherwise, we write
As the informal definition implies, we may also use the word “collection” (or other similar words such as “family” ) in place of “set.” Strictly speaking, these words will be viewed as synonymous. However, we sometimes interchange these words just for variety. Sometime we switch these words for emphasis: for example, we might refer to a set whose elements are also sets as a “collection of sets” or a “family of sets,” rather than a “set of sets” even though these all mean the same thing.
We can describe sets in at least two different ways:
By listing the elements, most useful when the set is a small finite set or an infinite set whose elements can be referred to using an ellipsis “...”
For example: E œ Ö"ß #× œ Ö"ß #ß $ß ÞÞÞ × , the set of natural numbers ™ œ Ö! , „ "ß „ #ß ÞÞÞ × , the set of integers
By abstraction , that is, by specifying a property that describes exactly what elements are in the set. We do this by writing something like ÖB À B has a certain property × .
For example: ‘ œÖBÀB is a real number × , the set of all real numbers p ™™œÖq : pq − ß;− and Á 0}, the set of rational numbers ‘œÖBÀB − and B  }, the set of irrational numbers
Suppose we write ÖB À B−−‘‘ and B# œ "× or ÖB À B and B Á B×Þ No real number is actually a member of either setgà both sets are empty . The empty set is usually denoted by the symbol it is
2 occasionally also denoted by { }. Sometimes the empty set is also called the “null set,” although nowadays that term is more often used as a technical term for a certain kind of set in measure theory. ().By the way, gœ is a Danish letter, not a Greek phi F9 or
It may seem odd to talk about an empty set and even to give it a special symbol; but otherwise we would need to say to say that Öœ"×BÀ B−‘ and B# , which looks perfectly well-formed, is not sets at all. Even worse: consider WœÖBÀB−αα"and Bœ" , where and are irrational × . Do you know whether or not there are any such rational numbersB ? If you're not sure and if we did not allow an empty set, then you would not be able to decide whether or not W is a set! It's much more convenient simply to agree that ÖB À B −α and B œ" , where α" and are irrational × is a set and allow the possibility that it might be empty.
Members of sets could be any kind of objects. In mathematics, however, we are not likely to be interested in a set whose members are aardvarks. We will only use sets that contain various mathematical objects. For example, a set of functions
GÒ+ß ,Ó œ Ö0 À 0 is a continuous real-valued function with domain Ò+ß ,Ó× or a set of sets such as
Ö Ö"×ß Ö"ß #× × or Ög×or Ög, Ög×× .
Of course, if “everything in mathematics is a set,” then all sets in mathematics can only have other sets as members (because nothing else is available).
Once we start thinking that everything is a set, then an interesting thought comes up. If B is a set, then either Bœg or there is an element B"" −B . Since B is a set, either B " œg or there is a set B #" −B , and so on. Is it possible to find a set for which there is an “infinite descending chain” of members
ÞÞÞ−B88" −B − ... −B " −B ?
We say that two sets are equal , EœF , if E and F have precisely the same elements, that is, if B−EÍB−F. For example, ÖBß B× œ ÖB× and ÖBß C× œ ÖCß B× . Two sets whose descriptions look very different on the surface may turn out, on closer examination, to have exactly the same elements and therefore be equal. For example, ÖB À B−−‘‘ and B# œ "× œ ÖB À B and B Á B× , and a scrupulous reader might verify that
ÖB À B −‘ and B&% &B #*B $ "!*B # )B "%! œ 0} = { 7, 2, 1, 5}.
We say that A is a subset of B , written AB,©F provided each element of A is also a member of , that is, for all BßB−EÊB− B . If E©F but EÁF we say E is a proper subset of F . Clearly, EœF if and only if A©F and F©E are both true. ( Notation: “ if and only if ” is often abbreviated to “iff”. )
Note that E©F and F©G implies that E©G .
You should look carefully at each of the following true statements to be sure the notation is clear:
3 B−E iff ÖBשE
‘‚©©© (the set of complex numbers)
g Á Ög× g © Ög× g − Ög×
g©g gÂg g©E for any set A
g − Ög× − ÖÖg×× , but g  ÖÖg×× (so E − F − G doesn't imply E−GÑ
Note that g Á Ög× . The set on the left is empty, while the set on the right has one member, namely the set gÖ×ÁÖÖ× . This might be clearer with the alternate notation: }. The set on the left is analogous to an empty paper bag, while the set on the right is analogous to a bag with an empty bag inside.
We define the power set of a set EÐEÑ , written as c , to be the set of all subsets of A . In symbols, cÐEÑ œ ÖF À F © E×.
Since gg−ÐEÑE is a subset of every set, we have c for every set .
Since E©E for any set E , we also have E−c ÐEÑ for every set E .
cÐÖ"ß #×Ñ œ Ög ß Ö"×ß Ö#×ß Ö"ß #××
cÐÖ"×Ñ œ Ög ß Ö"××
cÐgÑ œ Ög ×
The last three examples suggest that a set E8 with elements has 28 subsets Ðto define a subset F©EÀ for each element B in Eß there are two choices “yes” or “no” about whether B to put into FBß . For each the choice is independent of the other choices. So there are 28 ways to pick the elements to form a subset FÞÑ
4 Exercises
E1. Use induction to prove that if a set \8 has elements, then c Ð\Ñ# has 8 elements.
E2. Use Exercise 1 to explain the meaning of the identity
888 ÞÞÞ 8 8 œ #8 () 8 œ !ß "ß # ÞÞÞ !"# 8"8
5 3. Paradoxes
The naive approach to sets seems to work fine until someone really starts trying to cause trouble. The first person to do this was Bertrand Russell who, around 1902, created Russell's Paradox:
It makes sense to ask whether a set is one of its own membersE that is, for a given set , to ask whether E−E is true or false. The statement E−E is false for the sets you immediately think of: for example Ö"ß #× Â Ö"ß #×Þ However, is the infinite set
E œ Ögß Ög×ß Ögß Ög××ß Ögß Ög×ß Ögß Ög×××ß Ögß Ög×ß Ögß Ög××ß Ögß Ög×ß Ögß Ög××××ß ÞÞÞ ×
a member of itself? How about an even more complicated infinite set? Could it happen that E−E? Whatever the answer, it makes sense to ask.
According to our naive approach for defining sets, we can define a set ´ by writing ´´œÖEÀEÂE×, so that is the set of all sets which are not members of themselves. Then we can ask, for this new set ´´´ , whether −− is true or false. If ´´´ , then must satisfy the requirement for being a member of ´´´ , that isß . On the other hand, if ´´´  , then does meet the membership requirement for ´´´ , so − . Thus, each of the only two possibilities about the set ´´´´´ (that − or ) leads to a contradiction!
Russell's Paradox illustrates that dilemmas can arise from using the method of abstraction to define sets too casually. A way out is to refuse to call ´ a set. To do that, in practice, we will insist that whenever we define a set by abstraction, we only form subsets of sets that already exist. That is, in defining a set by abstraction, we should always write ÖB ÀB−\ and ÞÞÞ × or, for short, ÖB − \ À ÞÞÞ ×where \ is some set that we already have. The result is a subset of X. Since the preceding definition of ´´ doesn't follow this form, is not guaranteed to be a set.
This is the route taken in axiomatic set theory, and it eliminates Russell's paradox. If \ is any set and we consider the set ´´´œÖE−\ÀEÂE× , the dilemma vanishes. For now,  means either that ´´−Â\ (a contradiction) or that ´ an alternative conclusion that we can live with.
Russell's Paradox has the same flavor as many “self-referential” paradoxes in logic. For example, some books in the library mention themselves in the preface the author might say, “In this book, I will discuss ... ”. Other books make no mention of themselves. Suppose Library\ wishes to make a book listing all books that do not mention themselves. Should the new book list itself? That is Russell's Paradox: if it doesn't mention itself, then it should; and if it does mention itself, then it shouldn't. The resolution of the paradox is that the new book really is intended to list “all books in \ which do not mention themselves”— that is, in forming the new book, one is restricted to examining only those books already in the collection \ . With this additional qualification, the paradox disappears. Think about the same paradox in the dedication at the front of these notes: are the notes dedicated to Freiwald?
In everyday mathematics, we usually don't have to worry about this kind of paradox. Almost always, when we form a new set, we have (at least in the back of our minds) a larger set \ of which it is a subset. Therefore, indulging in a bit of sloppiness, we may sometimes write ÖB À ... } rather than the more correct ÖB − \ À ... } simply because the set \ could be supplied on demand, and the notation is simpler.
6 There is another kind of difficulty we can get into when defining sets by abstraction. It arises from the nature of the description “ ... ” when we write ÖB − \ À ... }. This is illustrated by Richard's Paradox:
Consider EœÖB− À B is definable in English using less than 10000 characters × . There are only finitely many English character strings with length < 10000 (a very large number, but finite). Most of these character strings are gibberish, but some of them define a positive integer. So the set E is finite. Therefore there must be natural numbers not in EÀ , and we can pick the smallest such natural number call it 7 .
But if E7 is a well-defined set, then the preceding paragraph has precisely defined , using fewer than 10000 characters (count them!), so 7E is in !
To resolve this paradox carefully involves developing a little more formal machinery than we want to bother with here, but the idea is easy enough. The idea involves requiring that only a certain precise kind of description can be used for the property “ ... ” when defining a set ÖB − \ À ... }.
Roughly, these are the descriptions we can form using existing sets,−a , the logical quantifiers and b , and the familiar logical connectives œ, • , ”ßÊßÍ and c . When all this is made precise (using first order predicate calculus), the “set” E above is not allowed as a set because the description “ ... ” used above E is not a “legal” description. Fortunately, the sets we want to write down in mathematics can be described by “legal” expressions. For example, we could define the set of positive real numbers by
‘‘#œÖB− ÀcÐBœ!Ñ•bCÐC− ‘ •BœC Ñ×
To summarize: there are dangers in a completely naive, casual formation of “sets.” One of the reasons for doing axiomatic set theory is to avoid these dangers by giving a list of axioms that lay out our precise initial assumptions about sets and how they are formed. But fortunately, with some common sense and a little feeling acquired in practice, such dangerous situations rarely arise in the everyday practice of mathematics.
4. Elementary Operations on Sets
We want to have operations that we can use to combine old sets into new ones. The simplest operations are union and intersection .
Informally, the union of two sets is the set consisting of all elements inEFÞ or in (Note: when a mathematician says “either p or q”, this means “either p or q or both.” This is called the “inclusive” use of the word “or.” )
The intersection of two sets is the set of all elements belonging to bothEF and . In symbols,
the union of EF and is E∪FœÖBÀB−EB−F or }, and the intersection of E and F is E∩F œ { BÀB−E and B−F }.
7 Note: You might expect, after the discussion of paradoxes, that the definition of union should read: E∪FœÖB−\À B−E or B−F× and then ask “given E and F , what is \ ?”
In practice, the sets EF and that we combine are always subsets of some larger set \Þ Then there is no need to worry because E ∪ F , we understand, could be written more properly as ÖB − \ À B − E or B − F× . But to cover all possible situations, axiomatic set theory adds a separate axiom (see A5 on p. 12) to guarantee that unions always exist.
This issue doesn't come up for intersections. For example, given sets EF and , we can always write E∩FœÖB−EÀ B−F× .
Examples 4.1 {1,2} ∪∩ {2,3} = {1,2,3} {1,2} {2,3} = {2}
∪œ ‘ ∩œg
The union and intersection of two sets can be pictured with “Venn diagrams” À
E∪F E∩F
We want to be able to combine more than just two sets, perhaps even infinitely many. To express this, we use the idea of an index set .
Definition 4.2 Suppose that for each -A in some set , a set E- is given. We then say that the collection ´-AAœÖE- À − × is indexed by . We might write ´ more informally as ÖE×- -A− or even merely as ÖE- × if the index set A is clearly understood.
Examples 4.3 1) ´Aœ ÖE"# ß E × is indexed by the set œ Ö"ß #×
2) Let MB denote the interval Ò!ßBÓ of real numbers. Then ´‘ œÖMB ÀB− , B 0 × is indexed by A œ the set of nonnegative real numbers.
3) ´Aœ ÖE"# ß E ß ÞÞÞß E 8 ß ÞÞÞ × is indexed by œ
8 4) ´´œg iff can be indexed by A œg
Definition 4.4 Suppose . The union of the family is { } for ´-AœÖE-- À − × ´ E À -A − œÖBÀ some -A−B−E× , . The union is also written more simply, as just ´ . (So B− ´ iff B is an ! -! element of an element of ´Þ ) The intersection of the family, denoted or, more simply , is ÖE- À-A − × ´ ÖB À B − E- for all -A − }. When the specific set A-A is understood or irrelevant, we may ignore the “− ” and just write or . If , we might also write and as ∞∞ and EE--A œ ´ ´ 8œ" E88 8œ" EÞ
Examples 4.5 1) Suppose ´ œÖEßE×Þ"# We can write the union of this family of sets in several different ways: # ´ œ3œ" E3"# œ E ∪ E œ ÖE 3 À 3 œ "ß #×Þ 2) If , then [0, ) and . MBB œ Ò!ß BÑ ©‘ ÖM À B !× œ ∞ ÖM B À B !× œ g
ÖMB À B !× œ Ö!×Þ 3) If E œ 11 , 1 + ©‘ , then ÖE À 8 − × œ [ "ß # ] and 88 nn
ÖE8 À 8 − × œ Ò!ß "Ó ∞∞ 4) If FœÒ8ß∞Ñ©ß888‘ then FœÒ"ß∞Ñ and Fœg 8œ" 8œ"
5) Suppose each set If , then and E©\Þ--A-A œg ÖEÀ − לg . For the intersection: if , then is in every can ÖE- À-A − × œ \ B−\ B E- Ð you name an that doesn't contain ? so EBÑB−ÖEÀ−×-- -AÞ If this argument bothers you, then you can consider the statement ÖE- À-A − × œ\ to be a just a convention motivated by the idea that “the fewer sets in the family, the larger the intersection should be, so that when A œg , the intersection should be as large as possible.”
Theorem 4.6 1) E∪FœF∪E , and E∩FœF∩E .
2) E∪ÐF∪GÑœÐE∪FÑ∪G , and E∩ÐF∩GÑœÐE∩FÑ∩G . More generally,
-A−E∪Ð-./-./ . −Q E∪ / −R EќР-A − E∪ . −Q EÑ∪ / −R E , which we could also write in an alternate form
-A−E∪Ð-./ . −Q E∪ / −R EÑœ αA − ∪Q∪R E α The same equations hold with “∩∪ ” replacing “ ” everywhere.
3)E∩ÐF∪GÑœÐE∩FÑ∪ÐE∩GÑ , and E∪ÐF∩GÑœÐE∪FÑ∩ÐE∪GÑ . More generally, and ÐEÑ∩ÐFÑœ-A−−Q−ß−Q-. . -A. ÐE∩FÑ -.
( -A− EÑ∪Ð-..-A.−Q FÑœ − ß −Q ÐE∪FÑ -.
9 Proof To prove two sets are equal we show that they have the same elements; the most basic way to do this is to show that if BB is in the set on the left hand side (LHS) of the proposed equation, then must also be in the set on the right hand side (RHS) (thereby proving LHS © RHS ) and vice-versa. All parts of the theorem are easy to prove; we illustrate by proving the last equality:
( -A− EÑ∪Ð-..-A.−Q FÑœ − ß −Q ÐE∪FÑ -.
If LHS, then or . B− B− -A− EF-. B− .−Q
If , then for every , so for every and every B− -A− EE∪F--. B−E- -A − B− -A − . −Q , so B− RHS. If , then for every , so for every and every B− .−Q FE∪F.-. B−F. .-A −Q B− − . −Q, so B−RHS.
Therefore LHS© RHS.
Conversely, if LHS, then and , so there exist indices and such BÂ BÂ -A− EF-. BÂ .−Q -.!! that BÂE-.!! and BÂF . Then BÂE∪F-.!!, so BÂ RHS.
Therefore RHS©œì LHS, so RHS LHS.
Remarks Part 1) of the theorem states the commutative laws for union and intersection. Part 2) states the associative laws. Part 3) states the distributive laws.
Exercise Try to write a generalization of the commutative laws for infinite families.
Definition 4.7 If E and F are sets, then EFœÖB−EÀBÂF× is called the complement of F in E . If the set EEF is clearly understood, we might simply refer to as the complement of F , µ sometimes written as FFF- or or .
Theorem 4.8 (DeMorgan's Laws) For sets EF and - ,
1) , and E ÖF-- À-A − × œ ÖE F À -A − × 2) E ÖF-- À-A − × œ ÖE F À -A − × Proof Exercise (DeMorgan's Laws are very simple but important tools for manipulating sets. )
Definition 4.9 The set E‚FœÖÐ+ß,ÑÀ +−E and ,−F× is called the product of E and F . This definition can be extended in an obvious way to any finite product of sets: for example E‚F‚GœÖÐ+ß,ß-ÑÀ+−Eß,−Fß-−G×Þ
If EœF then we sometimes write E## for E‚E . For example, ‘‘‘ = ‚ . Note that E‚F and F‚E are usually not the same. ÐIn fact, E‚FœF‚E iff ... ? )
10 E‚Fis a set of ordered pairs, and Ð if everything in mathematics is a set) an ordered pair should itself be defined as a set. The characteristic behavior of ordered pairs is that Ð+ß ,Ñ œ Ð-ß .Ñ iff + œ - and ,œ.. So we want to define Ð+ß,Ñ to be a set having this property. Any set that behaves in this way is as good as any other to agree on as an official definition for an ordered pair. The most commonly used definition is due to the Polish topologist Kazimierz KuratowskiÐ 1896-1980): we define
Ð+ß ,Ñ œ ÖÖ+×ß Ö+ß ,××
If we use this definition then we can prove that Ð+ß ,Ñ œ Ð-ß .Ñ iff + œ - and , œ . . (Try it and remember, in your argument, that +ß ,ß -ß . may not be distinct! )
Exercise: There are other ways one could define an ordered pair. For example, a possible alternate definition is due to the American mathematician Norbert Wiener (1894-1964): we could define
Ð+ß ,Ñ œ ÖÖÖ+×ß g×ß ÖÖ,××× .
Using this definition, prove that Ð+ß ,Ñ œ Ð-ß .Ñ iff + œ - and , œ . .
Note In the discussion of paradoxes, we stated that sets should only be defined as subsets of other sets. Therefore, a careful definition of E‚F should read:
E‚FœÖÐ+ß,Ñ−\À+−E and ,−F×.
So, if we were doing axiomatic set theory, we would have to provide, as part of the definition, a set \E‚F\EF which we know exists and define to be a subset of . Given and , how in general would we supply \ ? To give the flavor of how axiomatic set theory proceeds, we'll elaborate on this point.
Axiomatic set theory begins with some axioms: the variables Bß Cß ÞÞÞ refer to sets and − refers to the membership relation among them. In axiomatic set theory, everything is a setßB−C so members of sets are other sets. Therefore we see notation like in the axioms: “the set BC is a member of the set .”
A partial list of these axioms includes:
A1) two sets are equal iff they have exactly the same elements : more formally, aB aC ÐB œ C Í aD ÐD − B Í D − CÑÑ
A2) there is an empty set : more formally, bB ( aC cÐC − BÑÑ ()Notice, by A1, this set Bg is unique, so we can give this set a name:
A3) any two sets BC and can be “paired” into a new set ÖBßC× : more formally, aB aC bDaA ÐA − D Í ÐA œ B ” A œ CÑÑ
A4) every set BC has a power set : more formally, aB bC aD ÐD − C Í ÐaA ÐA − D Ê A − BÑÑ
11 A5)the union u of any set x exists : more formally, aB b? aD ÐD − ? Í ÐbC Ð D − C • C − B ÑÑÑ
The last axiom to be mentioned here, A6, is a little harder to write precisely, so we will simply state an informal version of it:
A6) for any setBCB , there is a subset of whose members are all the elements in B that meet an appropriate “legal” description “...”: Roughly,
aB bC aD Ð D − C Í ÐD − B • “...” Ñ Ñ
A6) is a fancier (but still somewhat vague) version of saying that we are allowed to define a set C by writing: CœÖD−BÀ “ ...” ×
These axioms and three others A7) A9) which we will not state here are called the Zermelo-Fraenkel Axioms, or ZF for short. (One of the additional axioms, for example, implies that no set is an element of itself.) The resulting system (the axioms and all the theorems that can be proved from the axioms) is called ZF set theory .
In ZF, by using these axioms, we can make a complete definition of E‚FÞ Suppose + − E and , − F . By axiom A3), the sets ÖEß F×ß Ö+ß ,× and Ö+ß +× œ Ö+× exist so, by axiom A3) again, the set ÖÖ+×ß Ö+ß ,×× œ Ð+ß ,Ñ also exists.
By axiom A5) E∪Fœ the union of the set ÖEßF× exists, and by axiom 4), the sets cccÐE ∪ FÑ and Ð ÐE ∪ FÑÑ exist. Using these sets, we then notice that
+−E∪F and ,−E∪F , so Ö+× © E ∪ F and Ö+ß ,× © E ∪ F , so Ö+× −ccÐE ∪ FÑ and Ö+ß ,× − ÐE ∪ FÑ , so ÖÖ+×ß Ö+ß ,×× ©c ÐE∪ FÑ , so ÖÖ+×ß Ö+ß ,×× œ Ð+ß ,Ñ −cc ÐE ∪ FÑ
Therefore each pair Ð+ß,Ñ which we want to put in the set to be called E‚F happens to be a member of the set \ œcc ÐE ∪ FÑß whose existence follows from the axioms. Of course, not every element of ccÐE ∪ FÑ is an ordered pair. Putting all this together, we could then make the definition
E‚FœÖD−cc ÐE∪FÑÀ Ðb+−EÑÐb,−FÑDœÐ+ß,Ñ×Þ
In the rest of these notes we will usually not refer to the ZF axioms. However, we will occasionally make comments about the axioms when something interesting is going on.
12 Exercises
E3. Which of the following are true when “− ” is inserted in the blank space? Which are true when “© ” is inserted?
a) Ög× ___ Ög , Ög× × b) Ög× ___ Ög , ÖÖg××× c) ÖÖg××___ Ög , Ög×× d) ÖÖg××___ Ög , ÖÖg××× e) ÖÖg××___ Ög , Ög, Ög×× }
E4. a) Suppose EE− and UU are sets and . Prove, or give a counterexample, to each of the following statements:
i) ccUÐEÑ − Ð Ñ ii) ccUÐEÑ − Ð Ñ iii) ccUÐEÑ © Ð Ñ
b) Let E œcc ÐgÑ , F œ ÐEÑ and G œ c ÐFÑ . What is E ∩ F ∩ G ?
c) Give an example of a set E which has more than one element and such that whenever B−E , then B©E . (Such a set is called transitive .)
E5. Explain why the following statement is or is not true:
If , then = E œ Ö Ög×ß ÖÖg×× × ÖF À F −c ÐEÑ× Ö ÖÖg××ß ÖÖÖg××× ×
E6. a) Prove that ccccÐE FÑ ∪ ÐFÑ œ ÐF EÑ ∪ ÐEÑ if and only if ÐE œ F or E ∩ F œ gÑ .
b) State and prove a theorem of the form:
cccÐE FÑ Ög× œ ÐEÑ ÐFÑ if and only if ...
c) For any sets E and F it is true that cc ÐEÑ ∩ ÐFÑ œ c ÐE ∩ FÑ and that ccÐEÑ ∪ ÐFÑ © c ÐE ∪ FÑ. State and prove a theorem of the form:
ccÐEÑ ∪ ÐFÑ œ c ÐE ∪ FÑ if and only if ...
E7. Suppose Eß Fß Gß are sets with E Á g , F Á g and ÐE ‚ FÑ ∪ ÐF ‚ EÑ œ G ‚ G . Prove that EœFœG.
13 E8. Suppose Eß Fß Gß and H are sets, with E Á g and F Á g . Show that if
ÐE ‚ FÑ ∪ ÐF ‚ EÑ œ ÐG ‚ HÑ ∪ ÐH ‚ GÑ , then either ÐE œ G and F œ HÑ or ÐE œ H and F œ GÑÞ
E9. Let , ,..., ,... be subsets of a set . Define lim ∞∞ . EE"# E 8 E Eœ 88œ" 5œ8 E 5 lim is read “lim inf”. In expanded form, the notation means that
lim E8 œ ( E∩E∩E∩ "#$ ...) ∪ ( E∩E∩E∩ #$% ...) ∪ ( E∩E∩E∩ $%& ...) ∪ ... .
Similarly, define lim ∞∞ . ( lim is read “lim sup”) Eœ858œ" 5œ8 E
a) Prove that lim EœÖBÀB888 is in all but at most finitely many E× and that lim E œÖBÀB is in infinitely many of the E×8 .
b) Prove that ∞∞lim lim 8œ"E©888 E© E © 8œ" E 8
c) Assume that all the E\8 's are subsets of a set . Prove that
lim (\E88 ) œ \lim E
d) Prove that lim E8 œ lim E 8 if either E "#$ ©E ©E © ... or E "#$ ªE ªE ª ... .
E10. SupposeEß Fß Gß and H are nonempty sets satisfying
E Á Fß G Á Hß and ÖEß F× Á ÖGß H×Þ Prove that
ÐE ‚ ÐF EÑÑ ∪ ÐF ‚ ÐE FÑÑ Á ÐG ‚ ÐH GÑÑ ∪ ÐH ‚ ÐG HÑÑ
14 5. Functions
Suppose \] and are sets. Informally, a function (or mapping , or map ) 0\from into ] is a rule that assigns to each element B\ in a unique element C] in . We call C the image of B and call B a preimage of C. We write Cœ0ÐBÑ , and we denote the function by 0À\Ä] . This informal definition is usually good enough for our purposes.
\ is called the domain of0œ dom Ð0ÑÞ The range of 0 œ ran Ð0Ñ is the set
ÖC−]ÀCœ0ÐBÑ for some B−\× .
We can think of giving an “input” B0 to and the corresponding “output” is Cœ0ÐBÑ. Then dom( 0 ) is the set of “all allowed inputs” and ran(0 ) is the “set of all outputs” corresponding to those inputs.
The set ]0Ð0Ñ©]Ð0Ñ is sometimes referred to as the codomain of . Of course, ran , but ran may not be the whole codomain ]Ð0Ñœ] . When ran the codomain , we say that 0 is a function from \ onto ] , or simply that 0 is onto . In some books, an onto function is also called a surjection .
If different inputs always produce different outputs, we say that 0 is a one-to-one (or 1 1) function. More formally: 0 is one-to-one if, whenever +ß , − dom(0 ) and + Á , , then 0Ð+Ñ Á 0Ð,Ñ . In some books, a one-to-one function is also called an injection .
If 00 is both one-to-one and onto, we call a bijection between \ and ] . A bijection sets up a perfect one-to-one correspondence between the elements of the two sets \] and . Intuitively, a bijection can exist iff \] and have the same number of elements.
Examples 5.1 (verify the details where necessary)
1) Suppose \œ] and 0À\Ä\ is given by 0ÐBÑœBÞ This bijection is called the identity map on \, sometimes denoted by 3\ .
2) If 0À\Ä] and E©\ , then we can define a new function 1ÀEÄ] by 1ÐBÑœ0ÐBÑ for B−E; 1 is called the restriction of 0 to E , denoted g œ0lE . If 0 is one-to-one then so is 1à but 1 may not be onto even if 001EÁ\ is onto. Note that and are different functions when because 11 11 dom(0Á ) dom Ð1 ). For example, 1œ sin lÒßÓ## is a bijection between ÒßÓ ## and Ò"ß"ÓÞ
3) Let 0À ‚ Ä by the rule 0Ð7ß8Ñœ#78 $ . Then 0 is one-to-one by the Fundamental Theorem of Arithmetic (which states that each natural number has a unique prime factorization). But 0 is not onto because, for example, 35ÂÐ0ÑÞ range
∞ 1 4) Let \œ the interval (1, ∞ ) ©‘‘ , ] œ and define 0ÐBÑœ " t B .> Þ ( This definition ∞ makes sense since the improper integral converges for B" .) For example, 0Ð 2 Ñ œ 1 .> œ" . " t # Then 0À\Ä] is one-to-one but not onto (why?).
15 5) Let 0 À Ö#ß $ß %ß ÞÞÞ× Ä by the rule 0ÐBÑ œ “the least integer 2 which divides B .” For example, 0Ð#Ñ œ #ß 0Ð$Ñ œ $ß 0Ð%Ñ œ #ß 0Ð&Ñ œ &ß 0Ð'Ñ œ #ß and 0Ð*Ñ œ $Þ The function 0 is not one-to-one because 0Ð#Ñ œ 0Ð%Ñ , and 0 is not onto because ran Ð0Ñ is the set of prime numbers.
6) Define 0À Ä by 0Ð8Ñœ the nth prime number. For example 0ÐÑœ# 1 , 0Ð#Ñœ 3, 0Ð3 Ñ œ 5, and 0Ð 4 Ñ œ 7. The function 0 is clearly one-to-one but not onto.
7) Let ’’ be the set of prime numbers and define 0À Ä using the same rule as in Example 6. Then 00À\Ä] is a bijection between ’ and . This illustrates that whether is onto depends on the codomain ]Ð0Ñ . Although ran is determined in Examples 6) and 7) by the domain and the “rule” 0] , the codomain is not automatically determined. Without knowing , we cannot say whether a function 0 is onto.
8) Define 1‘À Ä ÖD − ™ À D !× by 1 ÐBÑ œ “the number of primes Ÿ B ”. Thus, 1 Ð"Ñ œ ! , 11ÐÑœ2 1, Ð 2.5 Ñœ 1, and 1 Ð 5.6 Ñœ 3. The function 1 is onto (why?) but not one-to-one. This function appears as part of the famous Prime Number Theorem which states that
ln lim1ÐBÑ ÐBÑ œ 1. BÄ∞ B
A proof of the Prime Number Theorem is quite difficult. However, a much simpler fact is that lim1ÐBÑ œ ∞ . Why is this simpler fact true? And does it also follow from the Prime Number BÄ∞ Theorem?
8 9) Define 1À‘‘ Ä by 1ÐBÑœ ∞ B . This function is one-to-one but not onto. (Why?) 8œ! 8!
10) Let E be a set and define 0 À E Äcc Ð ÐEÑÑ by
0Ð+Ñ œ “the set of all subsets of E containing + ” œ ÖF −c ÐEÑ À + − F× .
For example, if E œ Ö"ß #ß $× , then
0Ð#Ñ œ ÖÖ#×ß Ö"ß #×ß Ö 2 ß $×ß Ö"ß #ß $×× .
This function is one-to-one. To see this, suppose that +Á, ; then Ö+×−0Ð+Ñ and Ö+× Â 0Ð,Ñ, so 0Ð+Ñ Á 0Ð,Ñ . Is the function onto?
11) Define 11À\‚] Ä] by ÐBßCÑœCß called the projection of \‚] to ] Þ ” When is this projection function 1 1? Must 1 be onto?
12) Let \ œ ÖG À G is a rectifiable curve in the plane ×Þ (A rectifiable curve is one for which an arc-length is defined.) Let 0À\ÄÒ!ß∞Ñ by 0ÐGÑœ “the length of the curve C .” Is 0 one-to- one? onto?
16
Our informal definition of function is good enough for most purposes. However in the spirit of “everything in mathematics is a set,” we should be able to give a more precise definition of a “function 0\]from into ” as a set. We begin by using sets to define a relation .
Definition 5.2 A relation V between \ and ] is a subset of \ ‚ ] . If ÐBß CÑ − V , we write BVC , meaning that “BC is related to ” (by this relation V ).
Example 5.3 Consider ÖÐBß CÑ −‘‘ ‚ À for some , − ‘ , C œ B ,# × . This set of ordered pairs is a relation from ‘‘ to . We could call the relation VŸÀ , but it actually has a more familiar name,
ŸœÖÐBßCÑ−‚‘‘ À for some ,− ‘, CœB,×#
The relation Ÿ is a set. The statement “ BŸC ” means that the pair ÐBßCÑ−ŸÞ Notice that for each B there are many C 's for which ÐBß CÑ − Ÿ À for example Ð"ß #Ñ − Ÿ ß Ð"ß %Þ&Ñ − Ÿ ß ...
Exercise: The relation Ÿßß is a set. What sets are the familiar relations and on ‘?
A function0\]B0 is a special kind of relation between and : one in which every is related by to only one C]Þ in
Definition 5.4 A function 0 from \ into ] , denoted 0À\Ä] is a set 0© \‚] (so 0 is a relation from \] to ) with the additional properties that:
a) for every B−\ , there is a C−] such that ÐBßCÑ−0 , and
b) if ÐBß CÑ − 0 and ÐBß DÑ − 0 , then C œ D.
\00 is called the domain of and the range of is the set
ÖC−] À there exists an B−\ for which ÐBßCÑ−0×
If 0\]ÐBßCÑ−0 is a function from to and , then we use the standard functional notation and write C œ 0ÐBÑ in preference over the relation notation B0C .
Condition a) states that a function 0\] from into is defined at each point of \ ; condition b) states that 00 is single-valued can't have more than one value for a particular B0ÀÄ . For functions ‘‘ , condition b) just states the familiar condition from precalculus that a vertical line cannot intersect the graph of a function 0 in more than one point.
Example 5.5 Let 0À‘‘ Ä be given by 0ÐBÑœB# . More formally, this function is the set of ordered pairs 0œÖÐBßBÑÀB−##‘‘ ש . A drawing of this set of points in ‘ # is the parabola CœB#. In precalculus, this parabola is called the “graph” of the function. But in our formal definition, the parabola is 00 : that is, is defined to be the set of pairs which, in precalculus, is called its graph.
Example 5.6 For sets \] and , we write ]\ to represent the set of all functions from \] to . For example, ‘‘ is the set of all real-valued functions of a real variable.
17 If \œg, then there is exactly one function in ]À\ the empty function g. (Check: g©g‚] and g satisfies the conditions a) and b) in Definition 5.4 Þ ) So ]\ œ Ög×Þ
If \Ág and ]œg, then there are no functions in ]]œg\\ so . (Since \‚gœg , the only relation from \ to ] is g , so g is the only “candidate” for a function from \] to . But since \Ágg , fails to satisfy part a) in Definition 5.4 Þ )
If \ has 8 elements and ] has 7 elements (8ß 7 − ), then there are 78 functions in ]Þ\ (Why? For each B−\ß how many choices are there for 0ÐBÑ ? )
It is usually not necessary to think of a function as a set of ordered pairs; the informal view of a function as a “rule” usually is good enough. The formal definition is included partly to reiterate the point of view that “everything in mathematics is a set.” However, it can also be very useful notationally. For example, if 01 and are functions then, since 01 and are sets, it makes sense to form the sets 0∪1 and 0∩1 and sometimes these sets are new functions.
1) Suppose 0 À Ð ∞ß !Ó Ä‘‘ and 1 À Ò!ß ∞Ñ Ä are given by 0ÐBÑ œ B and 1ÐBÑœBÞ As sets, 0œÖÐBßBÑÀ BŸ!× and 1œÖÐBßBÑÀB!×Þ Then the set 2œ0∪1 is a function with domain ‘‘ . For B− ß 2ÐBÑœ ?. Is the set 5œ0∩1 a function? If so, what is its domain and what is a formula for the function?
#Ñ In general, if 0 and 1 are functions, must 0 ∪ 1 and 0 ∩ 1 always be functions? If so, what is the domain of each? If not, then what conditions must01 and satisfy so that 0∪1 and 0∩1 are functions?
Example 5.7 As another illustration of how “everything in mathematics is a set,” consider the real number system †it consists of the set ‘ , two operations called addition ( ) and multiplication ( ), and an order relation Ÿ . (Subtraction and division are defined in terms of addition and multiplication.) Notice that the operations † and are functions ÀÄ‘‘## and †ÀÄ ‘‘ .
Therefore, ©‘‘# ‚ and, for example, ÐÐ#ß $Ñß &Ñ − Þ In function notation this would be written Ð#ß $Ñ œ & , but the more standard way, in this case, is to write #$œ&Þ
The functions † and are required to obey certain axioms such as ÐBßCÑ−ÐCßBÑ− iff , that isß BCœCB . The order relation Ÿ is also just a certain subset of ‘# Ð see Example 5.3 Ñ .
Therefore we can think of the real number system , its elements and all its ordering and arithmetic as being “captured” in the 4 sets: ‘ßß†ß and Ÿ , and we can gather all this into a single object: the 4-tuple of sets: Ðß߆ߟё .
But this 4-tuple can be viewed as just a complicated ordered pair: ÐÐБ ß Ñß † Ñß Ÿ Ñ . Since each ordered paired here is just a set, the whole real number system, with all its ordering and operations, can be thought of as one single (complicated) set.
6. More About Functions
18 Suppose E©\ , F©] and that 0À\Ä] . The image of the set E , denoted 0ÒEÓ , is the set of all images of the elements of E . More precisely, 0ÒEÓœÖC−] À Cœ0Ð+Ñ for some +−E× . A little less formally we could also write 0ÒEÓ œ Ö0Ð+Ñ À + − E× .
The inverse image set of F , denoted 0ÒFÓ" , is the subset of \ consisting of all preimages of all the elements from F . More precisely, 0" ÒFÓœÖB−\À0ÐBÑ−F× .
Example 6.1 Suppose 0À‘‘ Ä is the function given by 0ÐBÑœB# . Then
0ÒÒ!ß#ÓÓœÒ!ß%Ó 0" ÒÒ!ß*ÓÓœÒ$ß$Ó
0Ò Ö#×Ó œ Ö%× 0" Ò Ö 4 ×Ó œ Ö 22 ß ×
0Ò Ò $ß #Ó Ó œ Ò%ß 9 Ó 0" Ò Ò &ß %Ó Ó œ g
The function 00ÒÓÁ is not onto since ‘‘ .
When FÖ×0ÒÓ0ÐÑ is a one point set such as 4 , we will often write " 4 or even " 4 rather than the more formal 0ÒÖ×Ó" 4 .
The next theorem gives some frequently used properties of image and inverse image sets.
Theorem 6.2 Suppose 0À\ Ä ] , that E-- © \ and F © ] (-A − )
1) 0ÒgÓœg and 0Ò\Ó©] 1)w" 0 ÒgÓœg and 0 " Ò]] œ\
w"" 2) if E"# ©E , then 0ÒEÓ©0ÒE " # ] 2 ) if F "# ©F , then 0 ÒFÓ©0 " ÒFÓ #
3) 3)w" " 0Ò E-- Ó œ 0ÒE Ó 0 Ò F - Ó œ 0 ÒF - Ó 4) 4)w" " 0Ò E-- Ó © 0ÒE Ó 0 Ò F - Ó œ 0 ÒF - Ó Proof The proof of each part is extremely simple. As an example, we prove 4w ).
w Let LHS and RHS denote the left and right sides of 4 ). Then B− LHS iff 0ÐBÑ− F- iff " 0ÐBÑ−F-- for every -A − iff B−0 ÒFÓ for every -A − iff B− RHS. Thus LHS and RHS have the same members and are equal. ì
Example 6.3 In part 4), “©œ ” cannot be replaced by “ ”. For example, suppose E"# œ ÖB −‘‘ À B "×ß E œ ÖB − À B "× and let 0 be the constant function 0ÐBÑ œ " . Then 0ÒE"# Ó œ 0ÒE Ó œ Ö"×, so 0ÒE "# Ó ∩ 0ÒE Ó œ Ö"× , but 0ÒE "# ∩ E Ó œ 0ÒgÓ œ g .
Definition 6.4 Suppose 0À\Ä] and 1À]Ä^ are functions. Their composition 1‰0 is a function from \^ to formed by first appplying 0 , then 1 . More precisely, the composition (1 ‰ 0Ñ À \ Ä ^ is defined by Ð1 ‰ 0ÑÐBÑ œ 1Ð0ÐBÑÑ .
19 01 \Ä] Ä^ |______Å 1‰0
If we have another function 2À ^ Ä [ , we can form the “triple compositions” 2 ‰ Ð1 ‰ 0Ñ and Ð2 ‰ 1Ñ ‰ 0. It is easy to check that these functions \ Ä [ are the same. In other words, the associative law holds for composition of functions and we may write 2‰1‰0 without worrying about parentheses.
012 \Ä] Ä^Ä[ |______Å 2‰1‰0
Here is useful observation about compositions and inverse image sets: if E©^ , then
Ð1 ‰ 0Ñ" ÒE ] œ 0 " Ò1 " ÒEÓÓ
To check this, note that B − RHS iff 0ÐBÑ − 1" ÒEÓ iff 1Ð0ÐBÑÑ − E iff Ð1 ‰ 0ÑÐBÑ − E iff B − LHS.
If 0À\Ä] is a bijection , then 0 gives a perfect one-to-one correspondence between the members of \] and . In that case, we can define a function 1À]Ä\ as follows.
Suppose C−]Þ Then Cœ0ÐBÑ for some B−\ (because 0 is onto ) and there is only one such B (because 0 is one-to-one ). Define 1ÐCÑ œ BÞ Then 1 À ] Ä \ and 1ÐCÑ œ B if and only if 0ÐBÑ œ CÞ
More formally, 1 œ ÖÐCß BÑ − ] ‚ \ À B − 0" ÐCÑ× . Because 0 is a bijection, the inverse image set 0ÐCÑ" contains exactly one member, B .
\] 0 BCÞ Ò . Ñ1
It follows that 0Ð1ÐCÑÑ œ C and 1Ð0ÐBÑÑ œ B , that is, 0 ‰ 1 œ 3]\ À ] Ä ] and 1 ‰ 0 œ 3 À \ Ä \ . The function g defined above is clearly the only possible function with these two properties.
Of course, 11 is also a bijection and the whole discussion could be carried out starting with and using it to define a function 00‰1œ3À]Ä]Þ with the property that ]
Definition 6.5 Suppose 0À\Ä] and 1À]Ä\ and that 1‰0œ3\] and 0‰1œ3 . Then we say 0 and 1 are inverse functions to each other. We write 1œ0" and 0œ1 " œÐ0 " Ñ " Þ ()The idea is that inverse functions “undo” each other.
The preceding discussion proves the first part of the following theorem.
20 Theorem 6.6 Suppose 0À\ Ä ] . Then
1) If 01À]Ä\1‰0œ3 is a bijection, there is a unique bijection for which \ and 0‰1œ3] .
2) If 010 has an inverse function , then is a bijection.
Proof To prove part 2), we note that if 0Ð+Ñ œ 0Ð,Ñ , then 1Ð0Ð+ÑÑ œ 1Ð0Ð,Ñ ). Since 1 ‰ 0 œ 3\ , this implies that +œ, , so 0 is one-to-one. Also, for any C−] we have 1ÐCÑ−\ , and since 0‰1œ3] , we get that 0Ð1ÐCÑÑ œ C . Therefore C is the image of the point B œ 1ÐCÑ in \ , so 0 is onto. ì
Comment on notation: For any function 0À\Ä] and F©] , we can write the inverse image set 0ÒFÓ" , whether or not 0 is bijective and has an inverse function. In particular, for any function 0 we might write 0ÐCÑ" as a shorthand for 0ÒÖC×Óœ " the inverse image set of the one-point set ÖC× .
If 00ÐCÑ0 happens to be bijective, then the notation " is ambiguous: “ " ” might mean the inverse image operation for which 0ÐCÑœÖB×" , or “ 0 " ” might mean the inverse function, in which case 0ÐCÑœB" .
This double use of the symbol 0 " to denote the inverse image set operation (defined for any function) and also to denote the inverse function (when 0 is bijective) usually causes no confusion in practice the intended meaning is clear from the context.
In a situation where the ambiguity might cause confusion, we simply avoid the shorthand and use the more awkward notation 0" ÒÖC×Ó for the inverse image set..
As a simple exercise in notation, convince yourself that 00ÒÖC×Ó is bijective iff " is a one-element subset of \C−]Þ for every
Examples 6.7 Ð In each case, verify the details.)
1) Let lnÀ Ð!ß ∞Ñ Ä‘ be defined by ln B œB 1 .> ( B !Ñ . This function is " > bijective (why?) and therefore has an inverse function ln"ÀÄÐ!ß∞Ñ‘ . The function ln " is often called exp, so exp‰ ln À Ð!ß ∞Ñ Ä Ð!ß ∞Ñ and exp(ln BÑ œ B for every B !Þ Similarly, ln‰ÀÄ exp‘‘ and ln ÐBÑœB exp for each B−ÞÐ ‘It turns out, of course, that expBœ/B , although that equation requires proof.)
2) The function 0 À„ Ä Ö#ß %ß 'ÞÞÞÞ× œ given by 0Ð8Ñ œ #8 is bijective and has the e inverse function 1À„ Ä given by 1Ð/Ñœ 2 .
3) A linear function 0À‘‘88 Ä can be expressed as multiplication by some 8‚8 matrix Q0ÐBÑœQ†B : . Then 0 is bijective iff det( QÑÁ! and, in that case, 0" À‘‘ 8 Ä 8 is multiplication by the inverse matrix Q " and
21 " ! ÞÞÞ ! ! " ÞÞÞ ! 0Ð0ÐBÑÑœQ†Q†Bœ" " †BœB ã ! ! ÞÞÞ "
These are facts you should know from linear algebra.
Exercise (if you have had an analysis course ) Prove that if 0À‘‘ Ä is continuous and onto, then 00 has an inverse iff is strictly monotone. (Hint: Remember the Intermediate Value Theorem. If 0 is not strictly monotone, then there must exist three points +-, such that 0Ð-Ñ is either Ÿ or both of 0Ð+Ñ and 0Ð,Ñ .)
If we look carefully, we see that part 1) of Theorem 6.6 can be broken into two pieces.
Theorem 6.8 1) 0À\ Ä ] is one-to-one iff there exists a function 1 À ] Ä \ such that 1‰0œ3\ (unless \œg , ] Ág , and 0œg ; why? )
2) 0À\ Ä ] is onto iff there exists a function 1 À ] Ä \ such that 0 ‰1 œ 3] .
Proof The ideas are quite similar to the argument establishing the existence of an inverse for a bijection. The proof of 1) is left as an exercise. We will illustrate by proving part 2), which is of special interest because of a subtle point that comes up. If such a 1C−]0Ð1ÐCÑÑœCC exists, then for each we have , so is the image of the point Bœ1ÐCÑ−\. Therefore 0 is onto. Conversely suppose 0C−]0ÐCÑÁg is onto so that, for each , we know that " . Choose an " arbitrary element B − 0 ÐCÑ and define 1ÐCÑ œ B . Clearly, 0 ‰ 1 œ 3] . (Note that if ] œ g , then we must have 0œg and \œg . The description above then defines 1œg , and 1‰0œ3] œg is still true.) ì
The subtle point of the proof lies in the definition of 1ß which in set notation,
1 œ ÖÐCß BÑ − ] ‚ \ À B is an element arbitrarily chosen from 0" ÐCÑ×
In discussing paradoxes, we stated that a set 1 could only be formed by abstraction as 1œÖÐCßBÑ−] ‚\À ... “some legal description”... × . If one is doing axiomatic set theory, the phrase “an arbitrarily chosen element” is not legal; it is “too vagueÞQ ” ore precisely, it cannot be properly written as a description in the first order predicate calculus and therefore the definition of the function 1 would be invalid.
The axioms for set theory (ZF) tell us that certain sets exist Ðg , for example Ñ and give methods to create new sets from old ones. Roughly, these methods are of two types:
i ) “ from the top down” forming a subset of a given set ii) “from the bottom up” somehow piecing together a new set from old ones (for example, by union, or by pairing)Þ It is understood that the “piecing together” must be done in a finite number of steps for example, we cannot say “apply the Pairing Axiom infinitely many times to get the set D ...”
In the present situation ,
22 i) we tried to define 11 (“from the top down”) by describing as a certain subset of ]‚\ ; the problem is that we can't say precisely how to pick each B and therefore we can't use the Subset Axiom A6.
ii) if we try to define 1 “from the bottom up,” we can begin by using the fact that 0 is onto: this means that
aC − ] bB B − 0" ÐCÑ
Therefore for any particular C−] , we can form an ordered pair ÐCßBÑ with C−0" ÐBÑ . But if there are infinitely many pairs ÐCß BÑ , we have no axiom that allows us to “gather” these pairs into a single set 1
Also, none of the ZF axioms A7) A9) (which we didn't state) is any help here.
In spite of all this, our informal description of 1 seems intuitively sound, so another axiom called the Axiom of Choice (AC, for short) is usually added to the set theory axioms ZF so that our intuitive argument can be justified. In one of its many equivalent forms, the axiom of choice states:
[]AC If T œÖEC ÀC−]× is a family of nonempty sets, then there exists a function such that, for each , 2À] Ä ÖEC À C−]× C 2ÐCÑ − EC.
AC guarantees the existence of a function (set) 22ÐCÑ that “chooses” an element from " each set EEœ0ÐCÑC and, here, that's just what we need. If we use the sets C , then AC gives us the function 22 and we can then use to define
1 œ ÖÐCß BÑ À C − ] • B œ 2ÐCÑ×
Defining the function 1 is not always such a delicate matter. In some special cases , we can avoid the whole problem and don't need AC. For example:
i) if \œ , we could quite specifically define 1 by saying “let B be the smallest element in 1ÐCÑ" .” In other words, we could write a perfectly precise definition of g “from the top down” using the language of first order predicate calculus:
1œÖÐCßBÑ−] ‚\À B−0" ÐCÑ • aDÐD−0 " ÐCÑÊD BÑ× .
ii) if ] were finite, we could index its members so that ] œ ÖC"# ß C ß ÞÞÞß C 8 × for some 8− Þ Then we could proceed “from the bottom up” to write the finite list of statements
" bB"" − 0 ÐC Ñ " bB## − 0 ÐC Ñ . . . " bB88 − 0 ÐC Ñ
Using these B"8 ß ÞÞÞß B we can write a definition of 1 À
1œÖÐCßBÑ−] ‚\ÀÐCœC"" •BœBÑ”ÞÞÞ”ÐCœC 88 •BœB Ñ××
23 (The description is “legal” since it involves only finitely many terms.)
Generally speaking, AC is required when it is necessary to choose an element from each nonempty set in an infinite family and there is no way to describe precisely which element to choose from each set. When the choice can be explicitly stated (as when \œ above), AC is not necessary.
To borrow a non-mathematical example from the philosopher Bertrand Russell: if you have an infinite collection of pairs of socks, you need AC to create a set consisting of one sock from each pair, but if you have an infinite collection of pairs of shoes, you don't need AC to create a set containing one shoe form each pair— because you can precisely describe each choice: “from each pair, pick the left shoe.”
The Axiom of Choice, when added to the other axioms of set theory, makes it possible to prove some very nice results. For example, in real analysis, AC can be used to show the existence of a “nonmeasurable” set of real numbers. AC can also be used to show that every vector space (even one that's not finite dimensional) has a basis. AC is equivalent to a mathematical statement called Zorn's Lemma (see Chapter VIII) which you may have met in another course.
The cost of adding AC to the axioms ZF is that it also make it possible to prove some very counter-intuitive results about infinite sets. Here is a famous example:
The Banach-Tarski paradox (1924) states that it is possible to divide a solid ball into six pieces which can be reassembled by rigid motions to form two balls of the same size as the original. The number of pieces was subsequently reduced to five by R. M. Robinson in 1944, although the pieces are extremely complicated. (Actually, it can be done with just four pieces if the single point at the center of the ball is ignored..)
Most mathematicians are content to include AC with the other along with the other axioms and simply to be amused by some of the strange results it can produce. This is because AC seems intuitively very plausible and because many important mathematical results rely on it. We will adopt this attitude and use AC freely when it's needed (and perhaps even when it isn't!), usually without calling attention to the fact.
The system ZF together with the Axiom of Choice is referred to as ZFC set theory for short.
Definition 6.9 A sequence in a set \0ÀÄ\ is a function . The terms of the sequence are 0Ð"Ñ œ B1, 0Ð#Ñ œ B#8 , ... , 0Ðn Ñ œ B , ... . We often denote a sequence informally by the notation ÐB8 Ñ.
For example, the function 0Ð8Ñ œ #8" defines the sequence whose terms are 3, 5, 7, 9, ... The n>2 term of the sequence is B88 œ 2 8" , and we might refer to the sequence as ÐB ) or ( #8" ).
In the spirit of “everything in mathematics is a set”: since a sequence in \\ is a function from to , a sequence (formally) is just a set‚\ in this case, a special subset of .
24 Exercises
E11. a) Show that if , then and ÐBß CÑ − E B − E C − E b) Show that if is a function, then dom and ran 000©0 c) Show that if 0ÀEÄF , then 0©cc ÐE∪FÑ
E1#0ÀÄ1ÀÄ . Let ‘‘‘‘‘ denote the real numbers and and be given by 0ÐBÑ œ3 B# 2 B 1 and 1ÐBÑ œ |2 B 1|. Find the range of 0 ‰0ß 0 ‰1ß 1‰0 , and 1‰1 .
E1$\œÖ× . a) How many bijections exist from the set 1, 3, 5,..., 99 to the set ]œÖ2, 4, 6,..., 100 × ?
b) How many 1-1 maps are there from \œÖ 1, 3, 5..., 99 × into ] œÖ 2, 4, 6,...., 10 !× ?
c) How many 1-1 maps are there from \œÖ 1, 3, 5..., 99 × into ] œÖ 1, 2, 3, ..., 10 !× ?
E1% . Define 0 Àcc Ð\Ñ ‚ Ð] Ñ Ä c Ð\ ‚ ] Ñ by 0ÐÐEß FÑÑ œ E ‚ F . Prove that 0 is onto if and only if one of the sets \] or contains no more than one point.
E1&GÒ+ß,Ó . Let be the set of all continuous functions 0ÀÒ+ß,ÓÄÞ‘ Define a function +, >‘>>ÀGÒ+ß,ÓÄby Ð0Ñœ0Ð# ÑÞ Is "" ? onto?
E1' . Suppose :Ð>Ñ œ Ð<Ð>Ñß =Ð>ÑÑ is a one-to-one map from ‘‘ into # . Define a new map ;À‘‘Ä3 by qt ( ) = Ð<Ð>Ñß <Ð=Ð>ÑÑß =Ð=Ð>ÑÑÑ . Prove that q is one-to-one.
E17. Let _‘ be the set of straight lines in # which do not pass through the origin Ð!ß !ÑÞ Describe geometrically a bijection 0 À_‘ Ä # ÖÐ!ß !Ñ× .
E18. Let 0À\Ä\ and let 08 À\Ä\ denote the result of composing 0 with itself n times. Suppose that for every BB−\ , there exists an nn − such that 08 ( ) œB (note that may depend on B ). Prove that 0 is a bijection.
E19. Let G (‘‘ ) denote the set of all continuous real-valued functions with domain , that is, GœÖ0À0−(‘‘ )‘ and 0ÀG is continuous}. Define a map I ( ‘‘ ) ÄG ) as follows:
for 0−G (‘ ), MÐ0Ñ is the function given by MÐ0ÑÐBÑ œ B 0Ð>Ñ.>. ! Is I one-to-one? onto? ( Hint: The Fundamental Theorem of Calculus is useful here.)
25 E20. Let 0 be the bijection between the set of nonnegative integers and the set ™ of all integers defined by
8 , if 8 is even 0Ð8Ñ œ 2 8(8"Ñ , if is odd 2
Now define a mapping 1À™ Ä as follows:
For any integer 7" we can factor 7 in a unique way into a product of primes 5 83 , where and each is a positive integer. For each , let 7œ3œ" :3 :33" : 8 3 7"
5 08()3 1Ð7Ñ œ3œ" :3 Define 1Ð"Ñ œ " , 1Ð!Ñ œ ! and, for negative integers 5 , define 1Ð5Ñ œ 1Ð 5Ñ .
Prove that 1 is a bijection between ™ and .
8" 8Ð8"Ñ E21. Let +"8 œ! and for 8œ#ß$ßÞÞÞ define + œ 3œ # . Show that the function 3œ" 0À ‚ Ä given by 0Ð7ß8Ñœ+78" 8 is a bijection.
E22. Let UU infinite subsets of . Define 0À Ä ( !ß"Ó as follows: for F−U , 0ÐFÑœ the “ binary decimal” 0.BB"88#$ B... B 8 ... where B œ" if 8−F and B œ 0 if 8ÂF. For example, if „U„œ Ö#ß %ß 'ß ÞÞÞ × − , then 0Ð Ñ œ !Þ!"!"!"ÞÞÞbase 2
Prove or disprove that 0 is onto.
E23. “Measurable sets” in ‘8 are defined in elementary measure theory as a way to introduce a notion of integration that is more general than the Riemann integral. It is true, but not trivial to prove, that not every subset of ‘8 is measurable that is, there exist nonmeasurable sets. Each measurable subset \ is assigned a number .. Ð\Ñ that “measures” its size. For example, ÐÒ+ß ,ÓÑ œ , +Þ
The following argument claims to show in a very simple way that a nonmeasurable subset of Ò!ß "Ó exists. Find the error in the argument. (Hint: You only need to know that if \©Ò!ß"Ó and \ is measurable, then !Ÿ. Ð\ÑŸ" . Of course, the argument should sound very suspicious because it seems that it doesn't matter whether or not you actually know the definition of measurable set.)
Assume each \©Ò!ß"Ó is measurable and has Lebesgue measure . Ð\ÑÞ Then ..Ð\Ñ − Ò!ß "Ó and the number Ð\Ñ may or may not be in the set \ . Let F œ Ö.. Ð\Ñ À \ © Ò!ß "Ó and Ð\Ñ Â \× . Since F © Ò!ß "Óß F is measurable by assumption.
But ..ÐFÑ − F iff ÐFÑ Â F which is impossible. Therefore not every \©Ò!ß"Ó can be measurable. ñ
26 7. Infinite Sets
We can classify sets as either finite or infinite and, as we will see later, infinite sets can be further classified into different “sizes.” We have already used the words “finite” and “infinite” informally, but now we want to make a more careful definition.
Definition 7.1 The sets EF and are equivalent , written EµF , if there exists a bijection 0ÀEÄF .
Clearly, EµE ; EµF implies FµE ; and if EµF and FµG , then EµG .
Definition 7.2 The set E0ÀÄEE is called infinite if there is a one-to-one map . is called finite if E is not infinite.
If E0ÒÓE is infinite, then is a subset of equivalent to , so we could say that EE is infinite iff contains a “copy” of .
Since we have defined “finite” as “not infiniteß ” we should prove some things about finite sets using that definition. For example:
E is finite iff E œ g or E is equivalent to Ö"ß #ß ÞÞÞß 8× for some 8 − . A subset of a finite set is finite. A union of a finite number of finite sets is finite. The image of a finite set under a map 0 is finite. The power set of a finite set is finite.
These statements are easy to believe and the proofs are not really difficult but they are tedious induction arguments and will be omitted.
Examples 7.3
1) E‚F is equivalent to F‚E . To see this, consider the function given by 0Ð+ß ,Ñ œ Ð,ß +Ñ.
2) The mapping 0 À„ Ä œ Ö#ß %ß 'ß ÞÞÞ × given by 0Ð8Ñ œ #8 is a bijection, so „µ . Thus, an infinite set may be equivalent to a proper subset of itself. (This fact was noticed by Galileo in the 17th century.) In fact the following theorem shows that this property actually characterizes infinite sets.
Theorem 7.4 EE is infinite if and only if is equivalent to a proper subset of itself.
Proof Suppose E0ÀÄE0Ð8Ñœ+E is infinite. Then there is a one-to-one map . Let 8 . Break into two pieces: E œ 0Ò Ó ∪ ÐE 0Ò ÓÑ . Define 1 À E Ä ÐE Ö+" ×Ñ by
+Bœ+−0ÒÓif 1ÐBÑ œ 8" 8 BB−E0ÒÓif Then 1EÖ+×E is a bijection between and A" , so is equivalent to a proper subset of itself.
27 Conversely, suppose there is a bijection 0ÀEÄFß where F©E but FÁE . Then EFÁg so we can pick a point +"" −EF . Starting with + ß apply 0 repeatedly to get a sequence #8 +"" ß 0Ð+ Ñß 0Ð0Ð+ " ÑÑ œ 0 Ð+ " Ñß ÞÞÞß 0 Ð+ " Ñß ÞÞÞ Þ 8 All the terms 0Ð+Ñ" must be different. To see this, suppose that two of these points are 884 equal, say 0Ð+Ñœ0"" Ð+Ñ for some 8ß4 . Then (because 0 is one-to-one) we have 8" 84" 8# 842 4 0 Ð+Ñœ0"""" Ð+Ñ, 0 Ð+Ñœ0 Ð+Ñ , .. Þ , and so on until we get to + "" = 0Ð+Ñ . But this is 4 impossible because 0Ð+Ñ"" is in F and + is not. Therefore the map 1À ÄE given by 8 1Ð8Ñ œ 0 Ð+" Ñ is one-to-one, so E is infinite. ì
Definition 7.5 The set E0ÀEÄ is called countable if there exists a one-to-one map .
Thus, a countable set is one which is equivalent to a subset of (namely, the range of 0 ). A countable set may be finite or infinite. (In some books, the word “countable” is defined to mean “countable and infinite.”)
Examples 7.6
1) If E© , then E is countable (because EµE© ÑÞ
2) Any countable set E is either finite or equivalent to .
Proof: Suppose EµN © , where N is infinite. Let 4" be the least element of N , let 4NÖ4×4œ#"8 be the least element in , ... , and in general, let the least element in the set N Ö4"8" ß ÞÞÞ ß 4 × . Then the map 0 À Ä N given by 0Ð8Ñ œ 4 8 is a bijection, so N µ . Therefore Eµ.
By (2), E1ÀÄE is countable and infinite if and only if there is a bijection . The function 1EÞ is a one-to-one sequence whose range is Therefore a countable infinite set is one whose elements can be listed as a sequence: +"# œ 1Ð"Ñß + œ 1Ð#Ñß ÞÞÞ ß + 8 œ 1Ð8Ñß ÞÞÞ Þ
3) The set œÖB− ÀB!× is countable. This is the first really surprising example. It means that we can list all the positive rationals in a sequence even though it is not clear, at first, how to do this. In the usual order on ‘ , there is a third rational between any two rationals so we can't simply list the rationals, for example, in order of increasing size. But the definition of “countable” doesn't require that our list have any connection to size. One way to create the list is to use a “diagonal argument” invented by Cantor. Begin by imagining all the positive rationals arranged into the following “infinite matrix:”
12345678 11111111ÄÄ . . . . . áß á 12345678 22222222 . . . . . Æ ßá 12345677 33333333 . . . . . á 12345678 44444444 ......
28 Then create a bijection 1À Ä by moving back and forth along the diagonals (skipping over a "# rational from the matrix if it has been listed previously): the sequence begins 1Ð"Ñ œ"" ß 1Ð#Ñ œ ß " "$%$#" " 1Ð$Ñ œ# ß 1Ð%Ñ œ $""#$% ß 1Ð&Ñ œ ß 1Ð'Ñ œ ß 1Ð(Ñ œ ß 1Ð)Ñ œ ß 1Ð*Ñ œ ß 1Ð"!Ñ œ & ß ÞÞÞ Þ
Of course, we can't picture the whole “infinite matrix” and we don't have a formula 1Ð8Ñ œ ÞÞÞ . However we have described a definite computational procedure: you would have no trouble finding 1Ð1 &Ñ , and you could find 1Ð****Ñ with enough time and patience. The function 1 is clearly one-to- $(* one and onto (for example, with some effort you could find the 8 for which 1Ð8Ñ œ#"" ÑÞ
Those who prefer formulas can consider the following alternate approach. Each natural number has a unique representation in base 11, using the numerals 0,1,2,3,...,9, / ( with the symbol / representing 10 ) . For a positive rational :Î; reduced to lowest terms, we may reinterpret the symbol string “ :Î; ” as a natural number by thinking of it as a base 11 numeral. For example, 23/31 would be interpreted 38 ,+=/ "" as 2(11)%$ + 3(11) + 10(11) #"! + 3(11) + 1(11) œ 34519 (in base 10). In this way, we define a one-to-one function 00Ðœ from into . For example, 23/31) 34519. So the set is countable.
We can show that the set of negative rationals, , is countable using a similar diagonal argument. Or, we can notice that the function 0À Ä given by 0ÐBÑœB is a bijection; since is equivalent to a countable set, is also countable.
Of course, introducing the term “countable” would be a waste of words if all sets were countable. Cantor also proved that un countable infinite sets exist. This means that not all infinite sets are equivalent. Some are “bigger” than others!
Theorem 7.7 The interval of real numbers Ð!ß "Ñ is uncountable.
The heart of the proof is another “diagonal” argument. A technical detail in the proof depends on the following fact about decimal representations of real numbers:
Sometimes two different decimal expansions can represent the same real number. For example, 0.10000...œ 0.09999... (why?). However, two different decimal expansions can represent the same real number only if one of the expansions ends in an infinite string of 0's and the other ends in an infinite string of 9's.
Proof Consider any function 0 À Ä Ð!ß "Ñ . We will show 0 cannot be onto and therefore no bijection exists between and Ð!ß "Ñ . To begin, write decimal expansions of all the numbers in ran(0 ):
<" œ0Ð"Ñœ 0. B """#"$ B B ÞÞÞB "8 ÞÞÞ <# œ0Ð#Ñœ 0. B #"###$ B B ÞÞÞB #8 ÞÞÞ <$ œ0Ð$Ñœ!B . $"$#$$ B B ÞÞÞB $8 ÞÞÞ ã <8 œ0Ð8Ñœ!B . 8"8#8$ B B ÞÞÞB 88 ÞÞÞ ã where each B34 is one of the digits !ß "ß ÞÞÞß *Þ Now define a real number C œ !ÞC"#$ C C ÞÞÞ C 8 ÞÞÞ. by choosing Cœ"if B Á 1 888 Cœ#888if B œ"
29 Then C−Ð!ß"Ñ and C is not equal to any of the numbers <88 œ0Ð8ÑÞ To see that CÁ< ß notice that >2 i) by construction, the decimal expression for C<8 differs from 8 in the decimal place; and ii) C is not an alternate decimal representation of Of course, we could start over, adding C to the original list. But then the same construction could be repeated. The list can never be complete, that is, ranÐ0Ñ œ Ð!ß "Ñ is impossible. A different way to try to dodge the technical difficulty about non-uniqueness of representations might be: whenever a number 0Ð8Ñ has two different decimal representations, include both in the list and then define C . Is there any problem with that approach? The next theorem tells us some important properties of countable sets. Theorem 7.8 1) Any subset of a countable set is countable. 2) If is countable for each , then ∞ is countable. E8−E88 8œ" 3) If E"# ß E ß ÞÞÞß E 8 are countable, then E " ‚ E # ‚ ÞÞÞ ‚ E 8 is countable. Proof To show that a set is countable, we need to produce a one-to-one map 1 of the set into . 1) If E0ÀEÄF©E is countable, we have, by definition, a one-to-one map . If , then 1œ0±FÀFÄ is also one-to-one, so F is also countable. 2) If theE8 's are not disjoint, then consider the sets F""##" œEßF œE E , ÞÞÞ , F88 œ E ÐE " ∪ ÞÞÞ ∪ E 8" Ñ, ÞÞÞ Þ The F 8 's are countable and for any 7ß 8 we have that In addition, ∞∞ so it's sufficient to prove that ∞ is F∩FœgÞ878œ" Fœ 8 8œ" E 8 8œ" F 8 countable. Thus we will not lose any generality if we assume at the beginning of the proof that the E8 's are disjoint from each other. >2 For each 80ÀEÄ:8 , there is a one-to-one function 88 . Let 8 denote the prime number and define ∞ as follows: 1À8œ" E8 Ä if ∞ , then is in exactly one set ; B−8œ" E88 B E 0ÐBÑ8 for B−E8, let 1ÐBÑœ:8 . Then 1E1 is well-defined because the 8 's are disjoint, and is one-to-one (why? ). 3) Let :"8 ß ÞÞÞ ß : be the first 8 primes and, as before, let 0 8 be a one-to-one map from E 8 into . Define 1ÀE‚E‚ÞÞÞ‚EÄ"# 8 by 0Ð+Ñ"" 0Ð+Ñ ## 0Ð+Ñ 88 1Ð+"8 ß ÞÞÞß + Ñ œ :"# † : † . . . † :8 The function 1ì function is 1 1 (why? ). 30 Example 7.9 1) A union of a finite number of countable sets EE"5 , ... , is countable: we can let and then use part 2) of Theorem 7.8 to conclude that 5∞ is EœEœÞÞÞœg5" 5# 3œ" EœE33 3œ" countable. We can combine this result with Theorem 7.8.2 to state: if A is a countable index set and E−E is a countable set (for each -A ), then is countable.. Stated more informally: a - -A− - union of countably many countable sets is countable. 2) The set œ ∪ Ö!× ∪ is countable because it is the union of three countable sets. 3) If F©E and F is uncountable, then E is uncountable because if E were countable, then its subset F would be countable by Theorem 7.8.1. For example, the set ‘ of real numbers is uncountable because its subset Ð!ß "Ñ is uncountable.. This means that you cannot index the real numbers using : you cannot write “Let ‘ œÖ<ß<ß<ßÞÞÞ×"#$ .” 4) ‘œ∪ß where is the set of irrational numbers. Since ‘ is uncountable and is countable, the set must be uncountable. Thus there are “more” irrational numbers than rationals. 8 5) ‘‘ (8 "Ñ is uncountable, because it contains the “ B" -axis” ‚ Ö!× ‚ ÞÞÞ ‚ Ö!× , a subset which is clearly equivalent to ‘ . 6) Suppose E œ Ö+"# ß + ß ÞÞÞ , + 8 ß ÞÞÞ × ©‘% and that ! . Let M 8 be the open interval centered at +MœÐ+ß+ÑE©M with length %%% , that is, . Then ∞ , and the 88888###8" 8# 8# 8œ" total length of all the intervals Mœ is ∞ %% % . This shows that any countable subset of ‘ 8 8œ" #8" 2 can be “covered” by a sequence of open intervals whose total length is arbitrarily small. (If this does not seem surprising, suppose if you think that ∞ must be , try to prove that must Eœ‘ à8œ" M8 # be in ∞ ) 8œ"MÞ8 % More informally: take a piece of string with length # . Cut it in half and lay one half on ‘ with its center at ++Þ" . Cut the remaining piece in half again and lay one half on ‘ with center at # Continue in this way, always cutting the remaining piece in half and using one piece to cover the next element from E . (Of course, some of the pieces of string, when laid down, may overlap with earlier pieces.) These pieces from the original length of string cover all the rationals. 31 Exercises E24. Prove or give a counterexample: a) if EµF are equivalent and GµH are equivalent, then E∩GµF ∩H . b) if EµF are equivalent and GµH are equivalent, then E∪GµF∪H . c) if EµF , then EF µ FE . d) if EFµBA , then EµF . e) if EßFß and G are nonempty sets and E‚FµE‚G , then FµG . f) if EF is infinite and is countable, then E∪FµF . g) if EF is infinite and is countable, then E∪FµE . E25. If EF is uncountable and is countable, prove that EµEF . E26. a) Give an explicit formula for a bijection between the intervals (∞ß(Ñ and Ð!ß∞Ñ . b) Give an explicit formula for a bijection between the sets Ò!ß "Ó and Ð!ß "Ñ ∪ Ö#× . E27. Prove that the set of all real numbers in the interval Ð!ß "Ñ that have a decimal expansion using only even digits is uncountable. E28. Is the following statement true? If not, what additional assumptions about EF and will make it true? EµF iff there is a set 0œÖÐ+ß,ÑÀ+−Eß,−F× such that each element of E and each element of FÐ+ß,ÑÞ occur in exactly one pair E29. A subset F\ of is called cocountable if \F is countable and cofinite if \F is finite. ()The name is an abbreviation: cocountable œ co mplement is countable . a) Prove that if FG and are cocountable subsets of \ , then F∪G and F ∩G are also cocountable. Is the analogous result true for cofinite sets? b) Show how to write the set of irrational numbers, , as an intersection of countably many cofinite subsets of ‘Þ E30. A collection TT of sets is called pairwise disjoint if whenever Eß F − and E Á F , then E∩FœgÞ For each statement, provide a proof or a counterexample: a) If TT is a collection of pairwise disjoint circles in the plane, then is countable. b) If TT is a collection of pairwise disjoint circular disks in the plane, then is countable. E31. Prove that there cannot exist an uncountable collection of pairwise disjoint open intervals in ‘. 32 E32. Let jj∩‚ be a straight line in the plane. Prove that the set ( ) is either empty, contains exactly one point, or is countably infinite. In each case, give an equation for a specific line j to illustrate that case. E33. Suppose that 0ÀÒ!ß"ÓÄ‘ has the following property: there is a fixed constant, M, such that for every finite set ÖBßB"# ß ÞÞÞßB 8 שÒ!ß"Óß the following inequality is true: | 0() B"8 ÞÞÞ 0() B | Q a) Give an example of such a function f where ranÐ0Ñ is infinite. b) Prove that for such a function f , ÖB−Ò 0, 1 ÓÀ0ÐBÑÁ 0 × is countable. E34. What is wrong with the following argument? For each irrational number :− , pick an open interval Ð a:: ,b Ñ with rational endpoints +8. centered at :ÐÑ‚ . There are only countably many possible pairs a:: ,b because is countable. Since the interval ÐÑ a:: ,b is centered at p, the function F ÀÄ‚ given by F():œÐa:: ,b Ñ is one-to one, . Therefore is equivalent to a subset of ‚ , so is countable. E35. a) A sequence=b.−=œ=. in is called an arithmetic progression if such that 8" 8 for every 8− Þ Prove that the set of arithmetic progressions in is countable. b) A sequence =b5ß6−=œ68 5Þ in is called eventually constant if such that 8 for all Prove that the set of eventually constant sequences in is countable. c) A sequence =b5ß:−=œ= in is called eventually periodic if such that 88: for all 8 5Þ Prove that the set of all eventually periodic sequences in is countable. E36. For a set EÐEÑœÖ0−EÀ8−×Þ , let jÖ!ß"ßÞÞÞß8 × (We can think of E as an “alphabet” and jÐEÑ as the set of all “finite sequences” or “words” formed from this alphabet.) Prove that if EÐEÑ is countable, then j is countable. E37. Let E be an uncountable subset of ‘ . Prove that there is a subset of distinct elements such that ∞ diverges. Ö+88 À 8 œ "ß #ß ÞÞÞ× © E8œ1 + E38. Let ] be a set and, for each 8−‘ , suppose 0ÀÄ]8 . Let g be a function 1ÀÄ] ‘ such that, for every 8 −‘ , the set ÖB − À 1ÐBÑ œ 08! ÐBÑ× is countable. Prove that there is a point B − ‘ such that for all 81ÐBÑÁ0ÐBÑ , !8! . What property of ‘ makes the proof work? 33 E39. For a set W©‘‘ and >− , let W> denote the “translated set” Ö=>À =−W× . a) Show that if W is countable, then b>−‘ such that W>© Þ b) For sequences st and in ‘‘‘ (that is, for =ß>− ), let =>− be the sequence defined by Ð=> ÑÐ8Ñ œ =Ð8Ñ >Ð8Ñ . Show that if f‘‘ is a countable subset of , then b > − such that =>−f for every =− . E40. Give a proof or a counterexample for the following statement: If b-AœÖS- À − × is a collection of open intervals in ‘ with the property that , then . b©œ b‘ E41. a) Show how to write as the union of infinitely many pairwise disjoint infinite subsets. b) Show how to write as the union of uncountably many sets with the property that, given any two of them, one is a subset of the other. c) Show how to write as the union of uncountably many sets with the property that any two of them have finite intersection. (Such sets are called almost disjoint . ) (Hint: These statements that actually are true for any infinite countable set, not just for . For parts b) and c), you may find it easier to solve the problem for instead, and then use a bijection to“convert” your solution for into a solution for the set .) E42. a) Imagine an infinite rubber stamp which, when applied to the plane, inks over all concentric circles with irrational radii around its center. What is the minimum number of stampings necessary to ink over the whole plane? b) What if the stamp, instead, inks over all concentric circles with rational radii around its center? E43. Suppose 0 À‘‘ Ä . Let E œ Öa − ‘ À lim 0ÐBÑ exists but is not equal to 0Ð+Ñ×Þ Prove BÄ+ that A is countable. [Hint: One way to start is to define, for r− , E = Ö a − E À 0Ð+Ñ < lim 0ÐBÑ× and < BÄ+ E< œ Ö+ − E À lim 0ÐBÑ < 0Ð+Ñ×Þ Then E œ ÐE<< ∪ E Ñ (why?). Prove that BÄ+ r− EE<< and are countable.] E44. Let D be a countable set of points in the plane, ‘# . Prove there exist sets EF and such that HœE∪F, where the set EF has finite intersection with every horizontal line in the plane and has finite intersection with every vertical line in the plane. Notes: 1) This problem is fairly hard. You might get an idea by starting the easy special case of Hœ‘ ‚ 2) The statement that “# can be written as the union of two sets A and B where A has countable intersection with every horizontal line and B has countable intersection with every vertical line” is, in fact, equivalent to the continuum hypothesis(). see p. 40 and Exercise VIII.E.26 34 E45. We say that a function 0À‘‘ Ä has a local maximum at a point B if there exists an open interval Ð+ß ,Ñ containing B such that 0ÐBÑ 0ÐCÑ for all C − Ð+ß ,Ñ in other words, 0ÐBÑ is the (absolute) maximum value of 0Ð+ß,Ñ on the interval . a) Give an example of a nonconstant 0 which has a local maximum at every point. Then modify your example, if necessary, to get an example where ranÐ0Ñ is infinite. b) Suppose 0B−ÞÐ0Ñ has a local maximum at every point ‘ Prove that ran is countable. (Hint: if 0ÐBÑ œ C − ran( 0 ), pick an interval Ð+ß,Ñ with rational endpoints containing B and such that C0Ð+ß,Ñ is the maximum of on . ) 35 8. Two Mathematical Applications The relatively simple facts that we know about countable and uncountable sets are enough to prove an interesting fact about the real numbers. Definition 8.1 A real number << is called algebraic if is a root to a nonconstant polynomial equation 88" +88" B + B ÞÞÞ+ "! B+ œ! (*) where the coefficients +!8 ß ÞÞÞ ß + are all integers. (It is clearly equivalent to require that the coefficients be rational numbers because we could multiply both sides by some integer to “clear the fractions” and get a polynomial equation with integer coefficients and the same roots.) A real number is rational if and only if it is the root of a first degree polynomial equation with integer : coefficients: the rational ; is a root of the equation ;B : œ ! . Therefore rational numbers are algebraic. The algebraic numbers are a natural generalization to include certain irrationals: for # example, # is algebraic, since it is a root of the quadratic equation B "œ! ; and "# # is algebraic because it is a root of BB*B(œ!Þ$# A reasonable question would be: are there any nonalgebraic real numbers? (Complex roots of (*) are called complex algebraic numbers .) Theorem 8.2 The set of all algebraic real numbers is countable. Proof For each 8 " , let c8 be the set of all polynomials with integer coefficients and degree 8 . Each polynomial T−c8 is precisely described by the Ð8"Ñ -tuple containing its coefficients: Ð+88" ß+ ß ÞÞÞß+ ! Ñ . Since T has degree 8 , we know that + 8 Á! and so Ð+ 88" ß+ ß ÞÞÞß+ ! Ñ 88 is in the (countable) set Й™ Ö!×Ñ ‚ . The function 0 À c™™8 Ä Ð Ö!×Ñ ‚ given by 0ÐT Ñ œ Ð+88" ß + ß ÞÞÞ ß + ! Ñ is a bijection, so c 8 is countable. For each polynomial T , let ^ÐT Ñ be the set of real roots of the equation T ÐBÑ œ ! . ^ÐT Ñ is a finite set. Then is a union of a countable collection of finite sets, so is Vœ88 Ö^ÐTÑÀT−c × V8 countable. (For a particular value 8−!8 ßV! is the set of all algebraic numbers that are roots of an equation (*) that has degree 8à!" for example, V œ .) By Theorem 7.8(2), ∞ is countable. œV8œ" 8 ñ Definition 8.3 A real number which is not algebraic is called transcendental . (Euler called these numbers “transcendental” because they “transcend the power of algebraic methods.” To be more politically correct, we might call these numbers “polynomially challenged.” ) Corollary 8.4 Transcendental numbers exist. Proof Let “‘““ be the set of transcendental numbers. Since œ∪ and is countable, cannot be empty. ì In fact, this short proof shows much more: not only is ““ nonempty, but must be uncountable! There are “more” transcendental numbers than algebraic numbers. This is an example of a “pure existence 36 proof” meaning that it does not tell us any parrticular transcendental numbers, nor does it give us a way to construct one. To do that is harder. Transcendental numbers were first shown to exist (using different methods) by Liouville in 1844. The numbers / (Hermite, 1873) and 1 (Lindemann, 1882) are transcendental. One method for producing transcendental numbers is contained in a theorem of Gelfand (1934) from from algebraic number theory; it states that ff αα"α is an algebraic number, Á!ß" , and is an algebraic irrational, then " is transcendental. # 1 For example, the theorem implies that #/ is transcendental. The number is also transcendental. This follows from Gelfand's Theorem (which allows complex algebraic numbers) because: /œ/1113# = Ð/Ñ 3 3 and / 3 1 œ cos11 3 sin œ"Þ So /œÐ"Ñ 1 3 , which is transcendental by Gelfand's TheoremÞ As a second application to a different part of mathematics, we will prove a simple theorem from analysis. Theorem 8.5 A monotone function 0ÀÒ+ß,ÓÄ‘ has at most countably many points of discontinuity. (Since Ò+ß ,Ó is uncountable, this impl9ies that a monotone function on Ò+ß ,Ó must be continuous “at most points.”) Proof Assume BŸC implies 0ÐBÑŸ0ÐCÑ . ( If 0 is decreasing, simply apply the following argument to the increasing function 0 .) Therefore, at each point -−Ð+ß,Ñ we have lim0ÐBÑ Ÿ 0Ð-Ñ Ÿ lim 0ÐBÑ (why must these one-sided limits exist ?) BÄ- BÄ- Let 4Ð-Ñ denote the “jump of 0 at - ” œ lim 0ÐBÑ lim 0ÐBÑ . Since 0 is increasing, 4Ð-Ñ ! , BÄ- BÄ- and 0-4Ð-Ñ! is discontinuous at if and only if . 1 Let EœÖ-−Ð+ß,ÑÀ4Ð-Ñ888 }. The set E is finite because the sum of any set of jumps cannot be 1 more than 0Ð,Ñ 0Ð+Ñ . Furthermore, if 4Ð-Ñ !ß then 4Ð-Ñ 8 for some sufficiently large 8 , so -−E8 for some 8 . Therefore every point - of discontinuity of 0 in ( +ß,Ñ must be in the countable set ∞ . (The function might also be discontinuous at an endpoint or , but the set of discontinuities 8œ"E+,8 would still be countable.) ì Corollary 8.6 A monotone function 0À‘‘ Ä has at most countably many points of discontinuity. Proof For every 5−™ , the set H5 of discontinuities of 0 in the interval Ò5ß5"Ó is countable, by Theorem 8.5. So , the set of all discontinuities, is countable. 5−™ Hì5 As a sort of “converse,” it is possible to prove that if E is any countable subset of ‘ , then there exists a monotone function 0À‘‘ Ä for which E is precisely the set of points of discontinuity. (You can find an argument in A Primer of Real Functions (Boas, p. 129)ß or see Exercise E66 below.) 9. More About Equivalent Sets 37 So far, we have seen two kinds of infinite set, countable and uncountable. In this section, we explore the idea of uncountable sets in more detail. Notice that we have no right to assume two sets are equivalent simply because both are uncountable. There are many equivalent uncountable subsets of ‘ . The following theorem gives some examples. The proof could be made much easier by using the Cantor-Schroeder-Bernstein Theorem, which we will discuss later (p. 42). However, at this stage, it is instructive to give a direct proof. Theorem 9.1 Suppose +, . The following intervals in ‘ are equivalent: ‘ œÐ∞ß∞ѵÐ+ß,ѵÐ!ß"ѵÐ+ß,ÓµÐ!ß"ÓµÒ+ß,ѵÒ!ß"ѵÒ+ß,Ó µÒ!ß"ÓµÒ 0, ß∞ѵÒ+ß∞ѵÐ!ß∞ѵÐ+∞ѵÐ∞ß!ÓµÐ∞ß+Ó µÐ∞ß!ѵÐ∞ß+Ñ . (Note: for technical reasons that we will discuss later, it is convenient to consider g and all one-point sets Ö+× as intervals. These, it turns out, are the only finite intervals. Theorem 9.1 says that all infinite intervals in ‘ are equivalent.) Proof The linear map 0ÐBÑ œ Ð, +ÑB + can be used to show that Ð!ß "Ñ µ Ð+ß ,Ñ , Ð!ß"Óµ Ð+ß,Ó, Ò!ß"ѵÒ+ß,Ñ , and that Ò!ß"ÓµÒ+ß,ÓÞ The bijection tanÀ11ßß Ä‘‘ proves that µ 11 Þ ## ## To show that Ð!ß "Ó µ Ð!ß "Ñ , define functions: 3 11 0ÐBÑœ8 #8 B for #8 BŸ#8" for each 8− Each 00œ0 is a bijection from 11, to 11, , as illustrated above. Let ∞ . 8 ##88" ## 88" 8œ" 8 ÐThe graph of 0 consists of all the separate pieces, shown above, taken together.) Then check! is a function with dom ∞∞ dom( and ran ran( ) Ð Ñ 0 Ð0Ñœ8œ" 088 Ñ œ Ð!ß"Ó Ð0Ñ œ 8œ" 0 38 œ Ð!ß "ÑÞ It is easy to check that 0 is a bijection. (Of course, 0 is not continuous, but that is not required in the definition of set equivalence.) We can extend the definition of 0 to prove that Ò!ß "Ó µ Ò!ß "Ñ . Specifically, use 0 to define a function 1 À Ò!ß "Ó Ä Ò!ß "Ñ by 0ÐBÑif B − Ð!ß"Ó 1ÐBÑ œ !Bœ!if Since 01 is a bijection, so is . It is very simple to show that Ò!ß "Ñ µ ( !ß "Ó , that Ð ∞ß !Ó µ Ò!ß ∞Ñ , and that Ð ∞ß !Ñ µ Ð!ß ∞Ñ: just use the maps 2ÐBÑ œ " B and 5ÐBÑ œ B . The function lnÀ Ð!ß ∞Ñ Ä‘‘ proves that Ð!ß ∞Ñ µ . We can use a “projection” to show that Ò!ß "Ñ is equivalent to Ò!ß ∞Ñ . Imagine a ray of light that emanates from ("ß"Ñ , passes through the point B on the C -axis ( !ŸB" ) and then hits the B-axis at a point that we call :ÐBÑ . (See the following illustration ). Then : À Ò!ß "Ñ Ä Ò!ß ∞Ñ is a bijection. B Those who prefer formulas can check that a formula for :ÀÒ!ß"ÑÄÒ!ß∞Ñ is :ÐBÑœ "B . The few remaining equivalences such as Ð+ß ∞Ñ µ Ð!ß ∞Ñ are left for the reader to prove. ì At this point, we might ask “Are all uncountable sets are equivalen t?” Example 9.2 The set ‘‘‘ is not equivalent to . To see this, we use an argument whose flavor is similar to the “diagonal” argument Cantor used to prove that Ð!ß "Ñ is uncountable. Consider any function 9‘ÀÄ ‘‘ . We will show that 9 cannot be onto, and therefore no bijection can exist between ‘‘ and ‘‘ÞB− If ‘ , then 9 ÐBÑ−Þ ‘ Note that 9 ÐBÑ is not a number: 9‘‘ÐBÑ is a function from Ä ; to emphasize this, we will temporarily write 99 ÐBÑ œB Þ Then for C−‘9‘9‘, it makes sense to evaluate BB at a point C− to get a number ÐCÑ− . 39 Now look at the function 0−‘‘ defined by "ÐBÑœ!if 9 0ÐBÑ œ B !ÐBÑÁ!if 9B We claim that for each B−‘9 , ÐBÑœ 9BB Á0Þ This is because the two functions 9 and 0 have different values at the point BÀ9B ÐBÑÁ0ÐBÑ because of the way 0 was defined. Therefore 0Âran(99 Ñ , so is not onto. Therefore no subset of ‘‘ can be equivalent to ‘ . To see this, suppose E© ‘ and that <‘<ÀEÄ ‘. If were onto, we could use < to create an onto map 9‘‘ from to ‘ (which is impossible!) as follows: pick any function 0−‘‘ and define <ÐBÑif B − E 9ÐBÑ œ 0B−Eif ‘ . Therefore ‘‘‘‘ is an uncountable set which is not equivalent to any subset of . In particular, ‘ is not equivalent to ‘ . Intuitively, ‘ is “bigger” than any subset of ‘ . The fact that not all uncountable sets are equivalent leads to an interesting question. We have already seen (see Example 7.6(2) ) that every subset of is either finite or equivalent to the whole set — there are no “intermediate” sizes for subsets of ‘ . How about for ? Could there be a subset of ‘ which is uncountable but not equivalent to ‘ ? Intuitively, such an infinite set would be “bigger than but smaller than ‘ .” It turns out that this question is “undecidable” ! What does that mean? A conjecture of Cantor called the continuum hypothesis (or CH, for short), says that the answer to the question is “no”— that is, CH states that every subset of ‘‘ is either countable or equivalent to . (“Continuum” is an old word for the real number line. ) Cantor was unable to prove his conjecture. In an address to the International Congress of Mathematicians in 1900, David Hilbert presented a list of research problems he considered most important for mathematicians to solve in the new century; the first problem on the list was CH. Now if one believes (philosophically) that the set of real numbers actually exists “somewhere out there” (in some Platonic sense, or, say, in the mind of God) then any proposition about these real numbers, such as CH, must be either true or false we simply have to figure out which. But to prove such a proposition mathematically requires an argument based on some set of assumptions (axioms) and theorems in terms of which we have made our mathematical definition of ‘‘ . Since is defined mathematically in terms of set theory, a proof of CH (or of its negation) needs to be a proof derived from the axioms of set theory. The usual collection of axioms for set theory we have seen some of these axioms is called the Zermelo-Fraenkel system (ZF)Þ If the Axiom of Choice is added for good measure, the axiom system is referred to as ZFC. The fact is that the standard axioms for set theory (ZFC) are not sufficient to prove either CH or its negation. Since CH cannot be proven from ZFC, we say CH is independent of ZFC. But, since the denial of CH also cannot be proven from these axioms, we say that CH is also consistent with ZFC. Taken together, these statements say that CH is undecidable in ZFC, so one can add either CH or µ CH as an additional axiom to ZFC without fear of introducing a contradiction. These facts were established by Kurt Godel¨ (1906-1978) who showed in 1939 that ZFC could not 40 prove the denial of CH, and Paul J. Cohen (1934-2007) who showed in 1963 that ZFC could not prove CH. “In the early 1960's, a brash, young and extremely brilliant Fourier analyst named Paul J. Cohen (people who knew him in high school assure me he was always brash and brilliant) chatted with a group of colleagues at Stanford about whether he would become more famous by solving a certain Hilbert problem or by proving that CH is independent of ZFC. This (informal) committee decided that the latter problem was the ticket. (To be fair, Cohen had been interested in logic and recursive functions for several years; he may have conducted this seance just for fun.) Cohen went off and learned the necessary logic and, in less than a year, had proved the independence. This is certainly one of the most amazing intellectual achievements of the twentieth century, and Cohen was awarded the Fields Medal for the work. But there is more. Proof in hand, Cohen flew off to Princeton to the Institute for Advanced Study to have his result checked by Kurt Godel.¨¨ Godel was naturally skeptical, as Cohen was not the first person to claim to have solved the problem; and Cohen was not even a logician! Godel¨ was also at this time beginning his phobic period. (Toward the end of his life, Godel¨ became convinced that he was being poisoned, and he ended up starving himself to death.) When Cohen went to Godel's¨ home and knocked on the door, it was opened six inches and a hoary hand snatched the manuscript and slammed the door. Perplexed, Cohen departed. However, two days later, Cohen received an invitation for tea at Godel's¨ home. His proof was correct: the master had certified it.” Mathematical Anecdotes , Steven G. Krantz, Mathematical Intelligencer, v. 12, No. 4, 1990, pp. 35-36) By a curious coincidence, Cohen was an analyst interested in Fourier series, and it was Cantor's work on Fourier series that led to his creation of set theory in the first place. Therefore CH has a status, with respect to ZFC, like that of the parallel postulate in Euclidean geometry: the other axioms are not sufficient to prove or disprove it. To the other axioms for Euclidean geometry, you can either add the parallel postulate (to get Euclidean geometry) or, instead, add an axiom which denies the parallel postulate (to get some kind of non-Euclidean geometry). Likewise, to ZFC, one may consistently add CH as additional axiom, or get a “different set theory” by adding an axiom which denies CH. Unlike the situation with AC, most mathematicians prefer to avoid assuming CH or its negation in a proof whenever possible. The reason is that CH (in contrast to AC) does not seem to command intuitive belief and, moreover, no central mathematical results depend on CH. When the use of CH use seems necessary, it is customary to call attention to the fact that it is being used. 41 To reiterate one last point: if you believe that, in some sense, the real numbers actually exist “out there” beyond ourselves, then you believe that the set ‘ , as defined mathematically within ZFC, is just a model of the “real” real number system. This model may be an inaccurate or incomplete fit to reality. You may therefore continue to believe that for the “real” real numbers, CH is either true or false as a matter of fact. People with this point of view would say that the undecidability of CH within ZFC simply reflects the inadequacy of the axiom system ZFC. Godel¨ himself seemed to feel that ZFC is inadequate, although for perhaps different reasons: “I believe that ... one has good reason for suspecting that the role of the continuum problem in set theory will be to lead to the discovery of new axioms which will make it possible to disprove Cantor's conjecture.” Kurt Go¨del, “What is Cantor's Continuum Problem?”, Philosophy of Mathematics , ed. Benacerraf & Putnam, Prentice-Hall, 1964, p. 268 10. The Cantor-Schroeder-Bernstein Theorem The Cantor-Schroeder-Bernstein Theorem (CSB for short) gives a way to prove that two sets are equivalent without actually constructing a bijection between them. It states that two sets are equivalent if each is equivalent to a subset of the other. Theorem 10.2 (CSB) Suppose there exist one-to-one functions 0ÀEÄF and 1ÀFÄE . Then EµF. Proof We divide the set EB−EB into three subsets in the following way. For a point , we say “ has 1ÐBÑ©F 0 ancestors” if " (this, of course, is always true, so every B has 0 ancestors ). We say “B has 1 ancestor” if 1ÐBÑÁg" , that “ B has 2 ancestors” if 0Ð1ÐBÑÑÁg " " , and so on. In general, “B 8 has ancestors” if the inverse image set resulting from the alternating application of first 1018" , then " , then " , ..., times produces a nonempty set. We say “ B has exactly 8B 8B ancestors” if has ancestors but does not have 8"B ancestors. We say “ has infinitely many ancestors” if B 8 has ancestors for every 8− . Let EœÖB−EÀBI has an even number of ancestors × , EÖB−EÀBS = has an odd number of ancestors × , and EÖB−EÀBM = has an infinite number of ancestors × Clearly, these sets are disjoint and EœEISM ∪E ∪E . Define ancestors for a point C−F in a similar way, beginning the definition with an application of 01F" rather than " and divide into the sets FIS , F , and F M . The maps 0±E I ÀE I ÄF S and 0±E M ÀE M ÄF M are bijections (why?). The function 0±ESSI maps E into F , but it may not be onto (why?). However, the map " 1±FÀFÄEII S is a bijection, so it has an inverse bijection Ð1±FÑÀEÄF I S I . We can now define a bijection 2EF from to by piecing these maps together: " 2 œ Ð0 ± EIMI Ñ ∪ Ð0 ± E Ñ ∪ Ð1 ± F Ñ More explicitly, 42 0ÐBÑif B − EIM ∪E 2ÐBÑ œ " ì Ð1lFIS Ñ ÐBÑif B − E Our previous examples of equivalent sets were simply examples : they were not used in the proof of the CSB Theorem. We can therefore, without circular reasoning, use the CSB Theorem to give easier proofs of many of those equivalences, as the first two examples below illustrate. Examples 10.3 1) If E© , then E is finite or equivalent to . For if E© , we have 3ÀEÄ by 3ÐBÑ œ B, and, if E is also infinite we have, by definition, a one-to-one map 0 À Ä E . Therefore, Eµ. 2) Ð!ß"ѵÒ!ß"Ó since each set is equivalent to a subset of the other: Ð!ß"ѵÐ!ß"Ñ©Ò!ß"Ó and [0,1] is easily seen to be equivalent (using a linear map) to 11, © Ð!ß "Ñ. 42 # 1 3) Ð!ß "Ñ µ Ð!ß "Ñ . We have the one-to-one map 0ÐBÑ œ ÐBß 2 ) mapping the interval into the square, and we can easily define a one-to-one map in the opposite direction as follows: for ÐBß CÑ − Ð!ß "Ñ# , write the binary expansions of B and C B œ !ÞB"#$ B B ÞÞÞB 8 ÞÞÞ >A9 and C œ !ÞC "#$ C C ÞÞÞC 8 ÞÞÞ >A9 choosing, in both cases, a binary expansion that does not end with an infinite string of " 's. Then define 1 À Ð!ß "Ñ# Ä Ð!ß "Ñ by “interlacing” the digits to create a base "! decimal: 1ÐBß CÑ œ !ÞB""##$$ C B C B C ÞÞÞB 88 C ÞÞÞ The function 1 is one-to-one: suppose ÐBß CÑ Á ÐBww ß C Ñ . Then 1ÐBß CÑ and 1ÐB ww ß C Ñ have different decimal expansions. Neither decimal expansion ends in an infinite string of 9's, so 1ÐBß CÑ and 1ÐBww ß C Ñ are different real numbers. 4) A simple exercise, left to the reader, is to show that if EµF and GµH , then E ‚ G µ F ‚ H. Then, because Ð!ß "Ñ µ Ò!ß "Ó µ Ò!ß "Ñ µ ‘ , we can immediately write down such equivalences as ‘‘µÐ!ß"ѵ ( !ß"Ñ## µÒ!ß"Ó µÐ!ß"Ñ‚Ò!ß"ѵ # . Ð We intuitively think of Ð!ß"Ñ as 1-dimensional and (!ß "Ñ# as 2-dimensional, so we see that an equivalence between sets doesn't necessarily preserve our intuitive idea of dimension.)Þ Since ‘‘µ##$87 ß it follows that ‘‘‘‘‘ µ‚µ‚ œ ‘ Þ Similarly, ‘ µ ‘ for any 7ß 8 − . 43 Example 10.4 (A tangential comment) Although ‘‘# µ , there is no continuous bijection between them. In fact, if 0À‘‘# Ä is continuous ß then 0 cannot even be one-to-one. The proof is an exercise in the use of the Intermediate Value Theorem from calculus. Define a function 99 of one variable by holding one variable in 0ÐBÑœ0ÐBß!Ñ constant: . Let 0Ð!ß !Ñ œ99 Ð!Ñ œ + and 0Ð"ß !Ñ œ Ð"Ñ œ , . We can assume + , (if + = , , 0 is not one-to- one and we're done). Since 0 is continuous, so is 99 . By the Intermediate Value Theorem, +, assumes all values between + and , ; in particular, for some - − Ð!ß "Ñß 9 Ð-Ñ œ 2 . For that - , define <<< ÐCÑ œ 0Ð-ß CÑ . Because is continuous at 0, ÐCÑ is close to +, <9Ð!Ñ œ Ð-Ñ œ2 for C close to 0. Specifically, we can force + < ÐCÑ , for all C in some $ sufficiently short interval (ßÑ$$ . In particular, then < Ð2 ) −Ð+ß,Ñ . $ But we know that 9<9 assumes all values between +,ßÐÑœÐDÑ and so 2 for some D in (0,1). $$ $ Therefore 0Ð-ß22 Ñœ<9 Ð# Ñœ ÐDÑœ0ÐDß!Ñ . Since ( -ß ÑÁÐDß!Ñ0 , is not one-to-one. ì (Note: we did not even use the full strength of the assumption that f is continuous. We needed only to know that 9< and were continuous a weaker statement, as you should know from advanced calculus. This means there cannot even be a 1 1 map of the plane to the line which is continuous in each variable separately— a condition which is weaker than assuming f is continuous.) A similar argument shows that there cannot be a one-to-one continuous map 0À Ò!ß "Ó# Ä Ò!ß "Ó . Therefore, in particular, there cannot be a continuous bijection from Ò!ß "Ó# to Ò!ß "Ó . However, in the “opposite” direction, it is possible to construct a continuous map 0À Ò!ß "Ó Ä Ò!ß "Ó# which is onto . Such a map is called a space-filling curve , and such a map is often constructed in an advanced calculus course. Can there be a continuous onto map from ‘‘# Ä ? 44 11. More About Subsets We have already seen that there is more than one “size” of infinite set: some are countable, others are uncountable. Moreover, we have seen uncountable sets of two different sizes: ‘‘ and ‘ . In fact, there are infinitely many different sizes. We will see this using the power set operation c . But first we will prove a result about subsets of countable sets that is frequently useful. Theorem 11.1 The set YÐEÑ of all finite subsets of a countable set E is countable. Proof If EE is finite, it has only finitely many subsets. So assume is countable and infinite. For convenience, we assume E is the set of all prime numbers (Be sure you understand why this is no loss of generality! ) Then each finite F©E is a set of primes and we can define 0ÀY ÐEÑÄ by "Fœgif 0ÐFÑ œ the product of the primes in FFÁg if By the Fundamental Theorem of Arithmetic, 0ÐEÑì is one-to-one; so Y is countable. Recall that ]E denotes the set of all functions from E into ] . When ] œ Ö!ß "× , we will also use the notation #EEE for the set Ö!ß "× . Thus, # is the set of all “binary functions” with domain E . The following theorem gives us two important facts. Theorem 11.2 For any set E , 1) #µE c ÐEÑ 2) EÐEÑEµ is not equivalent to c (so Î#E Ñ "B−Fif Proof 1) For each F © E , define ;;FF À E Ä Ö!ß "× by ÐBÑ œ . !BÂFif The function ;F is called the characteristic function of F . E Define 9cÀÐEÑÄ# by 9 ÐFÑœ ;F . The function 9 pairs each subset of E with its characteristic function. The function 9 is clearly one-to-one because different subsets have different characteristic functions. The function 9 is also onto since every function 0 À E Ä Ö!ß "× is the characteristic function E of a subset of E À 0 œ;9F œ ÐFÑ , where F œ ÖB − E À 0ÐBÑ œ "× . So c ÐEÑ µ # . 2) Suppose We use the notation ccc#88"ÐEÑ for Ð ÐEÑÑ and, more generally, ccc ÐEÑ for Ð ÐEÑÑ . The preceding theorem says that no set is equivalent to its power set, so cc8"ÐEÑ is not equivalent to 8 ÐEÑ . We can prove even a bit more. 45 Corollary 11.3 No two of the sets in the sequence E , cc ÐEÑ , #8 ÐEÑ , ... , c ÐEÑ , ... are equivalent. 44" Proof For any 4 , there is a one-to-one function 344 Àcc ÐEÑ Ä ÐEÑ given by 3 Ð+Ñ œ Ö+× . Now suppose there is a bijection 0 Àcc445 ÐEÑ Ä ÐEÑ for some 4ß 5 . Then we would have the following maps: 3334 4" 45# 3 45" cc4ÐEÑ Ä 4" ÐEÑ Ä ...... Ä c 451 ÐEÑ Ä c 45 ÐEÑ lÅ ______0 45" 45 " 45 45" Then 345" Àcc ÐEÑ Ä ÐEÑ and 3 45# ‰ÞÞÞ‰3 4 ‰0 À cc ÐEÑ Ä ÐEÑ are both one-to-one. But then the CSB Theorem would tell us that cc45"ÐEÑ µ 45 ÐEÑ , which is impossible by Theorem 11.2(2). ì Therefore, by repeated applications of the power set operation, we can generate an infinite sequence of sets no two of which are equivalentTÐÑßÐÑÐÑ for example, c , #8 , ... , c , ... . In the “exponential” notation of Theorem 11.2(1), we can also write this sequence as , 2 , ## , ... . Thus we have infinitely many different “sizes” of infinite set. Examples 11.4 1) # œ Ö!ß "× µc Ð Ñ µÎß so we conclude that the set of all binary sequences is uncountable. 2) Because cÐÑ is uncountable and because has only countably many finite subsets (Theorem 11.1), we conclude that has uncountably many infinite subsets. Therefore every infinite set has uncountably many infinite subsets (explain the details!). 3) Using the CSB Theorem, we can state another paradox similar to Russell's Paradox. Let F œ ÖÖ,× À , is a set × , so F is the “set of all one-element sets.” Then the maps ccÐFÑ Ä F and F Ä ÐFÑ given (in both directions!) by @ Ä { @ } are one-to-one. Then the CSB Theorem gives that Fµc ÐFÑ . What's wrong? (The paradox is dealt with in the same way as Russell's Paradox. ) 46 Since #µÎ# , we can ask whether is equivalent to some other familiar set. The next theorem answers this question. Theorem 11.5 #µ ‘ Proof We use the CSB Theorem. If 0 − # , then 0 À9‘ Ä Ö!ß "× . Define À # Ä by 99Ð0Ñ œ∞0Ð8Ñ œ ∞ +8 where + œ 0Ð8ÑÞ In other words, Ð0Ñ is the real number whose 8œ"1088 8œ" 10 8 decimal expansion is !0Á1−#Ð0ÑÐ1Ñ .a"# a ... a 8 ... . If , then 99 and have different decimal expansions, and neither expansion ends in a string of * 's (each +!"8 is either or ). Therefore 99Ð0Ñ Á Ð1Ñ, so 9 is one-to-one. To get a one-to-one function from ‘ into # , we begin by defining a function <‘À Ä c Ð Ñ. List the rational numbers in a sequence œ Ö;"#$ ß ; ß ; ß ÞÞÞß ; 8 ß ÞÞÞ × and, for each B−‘< ß let ÐBÑœÖ8− À ;8 B× . Then < ÐBÑ− c Ð Ñ . Since two different real numbers BC and always have a rational between them, the set of rationals B is different from the set of rationalsC in other words, << ÐBÑÁ ÐCÑ , so < is one-to-one. Then we can compose By the CSB Theorem, #µ ‘ . ì In fact, it is fairly easy to generalize Theorem 11.5 replacing #5 with , where 5 is short for Ö!ß "ß ÞÞÞ ß 5 "× . 12. Cardinal Numbers To each finite set E we can assign, in principle, a number called the cardinal number or cardinality of the set). It answers the question: “How many elements does the set have?” The symbol lEl represents the cardinal number of E . For example, lgl œ !ß lÖ!×l œ " , and lÖ!ß "×l œ #Þ In practice, of course, this might be difficult. For example, if EœÖ:− À : is prime and :"!"!! × , then lElœ ? Of course, there must be a correct value for lElß but actually finding it would be hard. (You could make a rough estimate using the Prime Number Theorem see Example 5.1(8), p. 16: "!! "!"!! *( 1Ð"! Ñ ¸ lnÐ"!"!! Ñ ¸ %Þ$% ‚ "! ÞÑ We have already stated informally that nonequivalent infinite sets have “different sizes”. To make this more precise, we assume that to each set EE± there is associated an “object” denoted | and called the cardinal number of E (or cardinal of E for short), and that this is done in such a way that lEl œ lFl if and only if EµF . Note: How to justify this assumption precisely doesn't matter right now. That question belongs in a more advanced set theory course. It turns out thatlEl can be precisely defined in ZFC. Like everything in mathematics, lEl is itself a certain set and, of course, lEl is defined in terms of the given set E . For these notes, it is enough to know that two sets have the same cardinal number if and only if the sets are equivalent. There are standard symbols for the cardinal numbers associated to certain everyday sets: 47 lgl œ! l Ö!× l œ " (Of course, ± Ö+× l œ " for any + , because Ö!× µ Ö+× ) lÖ!ß "×l œ # (Of course, ± Ö+ß ,× l œ # for any + Á , ) ã ll œi!! (So i is the cardinal number for every countable infinite set; forexample, l ‚ ‚ lœl lœÞÞÞœi! ) ll‘ œ- The symbols i- and were chosen by Cantor in his original work. The symbol i is “aleph,” the first letter of the Hebrew alphabet (but Cantor was Lutheran). We read i! as “aleph-zero” (or the more British “aleph-nought” or “aleph-null.”) Cantor chose “c” for ll‘ since - is the first letter of the word “continuum” Ð an old word for the real line ). Because we know that the following sets are equivalent, we can write, for example, that lÒ!ß"ÓlœlÐ!ß"ÑlœlÐ!ß"Ñlœl##$‘‘ lœl lœl ‘ lœ ÞÞÞœ-. What is ll ? Why? (You can give a definite answer for this, but be careful: you can't assume that an uncountable subset of ‘‘ must be equivalent to . See the discussion of CH on pp. 40-42.) 13. Ordering the Cardinals We have talked informally about some infinite sets being “bigger” than others. We can make this precise by defining an order “Ÿ ” between cardinal numbers. Throughout this section, QßRßT are sets with ±Q± œ7ß±R± œ8ß and ±T ± œ: ( R is not necessarily the set natural numbers, ). We say that the sets QßRß and T represent the cardinal numbers 7ß8ß and : . Definition 13.1 1) 7Ÿ8 means that Q is equivalent to a subset of R . ÐWe also write 7Ÿ8 as 8 7ÞÑ 2) 78 ( or 87Ñ means that 7Ÿ8 but 7Á8 . ÐWe also write 78 as 87ÞÑ Thus 78 means that Q is equivalent to a subset of R but Q is not equivalent to R . According to the CSB Theorem, this is the same as saying that QRR is equivalent to a subset of but is not equivalent to a subset of Q . There is a detail to check. The relation 7Ÿ8 is defined using the sets Q and R . Another person might choose different sets, say 7œlQlww and 8œlRl to represent the cardinal numbers. Would that person necessarily come to the same conclusion that 7Ÿ8? We need to check that our definition is independent of which representing sets are chosen or, in other words, that Ÿ has been well-defined . This is easy to do. Suppose QRww and also represent 78Q and . If is equivalent to a subset of R , then there are bijections 01 and and a one-to-one map 2 as pictured below: 48 f QÄQ w 2Æ Æ5 1 RÄR w Then 5œ1‰2‰0" is a one-to-one map 5ÀQ w ÄR w , so Q w is equivalent to a subset of R7Ÿ8w . Therefore the question “Is ?” does not depend on which representing sets we useŸ that is, the relation is well-defined. Theorem 13.2 For any cardinal numbers 7ß 8ß and : : 1) 7Ÿ7 2) 7Ÿ8 and 8Ÿ: implies 7Ÿ: 3) 7Ÿ8 and 8Ÿ7 implies 7œ8 4) 7Ÿ8 and 8: implies 7: 5) at most one of the relations 78 , 7œ8 , and 78 holds 6*) at least one of the relations 78 , 7œ8 , and 78 holds. Proof The proofs of 1) and 2) are obvious. For 3), notice that we are given that each of QR and is equivalent to a subset of the other. Then QµR (by the CSB Theorem), so 7œ8 . 4) If 8: , then 8Ÿ: so, by part 2), 7Ÿ: . If :œ7 , then :Ÿ8 and therefore 8œ: by part 3). But this contradicts the hypothesis that 8:Þ Therefore :Á7 , so 7:Þ 5) By definition of ß7œ8 excludes 78 and 87Þ And if 78 and 87 were both true, then 7Ÿ8 and 8Ÿ7 , so 7œ8 , which is impossible. 6*) The proof is postponed. ì It seems like the proof of part 6*) of the theorem shouldn't be difficult. Informally, you simply pair an element of QR with an element of , and keep repeating this process until either a bijection is created between QR and or until one of the sets has no remaining elements. If one of the sets is used up before the other, then it has the smaller cardinal. In fact, this works for finite sets, but for infinite sets it is hard to make precise what “keep repeating this process until ...” means. To make the argument precise, the Axiom of Choice has to be used in some form. We could develop the machinery to complete the proof here, but it would digress too much from the main ideas. So for use in our, examples, we'll simply assume part 6*) for now. From parts 5) and 6*) of the theorem, we immediately get the following corollary. Corollary 13.3 For any two cardinals 7 and 8 , exactly one of the relations 78ß7œ8ß or 78 is true. Examples 13.4 49 1) If 5œlOl is a finite cardinal, then O is equivalent to a subset of but O is not equivalent to . And if 70ÀÄQ is an infinite cardinal number, there is a one-to-one function . Therefore 5i!! Ÿ7 , so i is the smallest infinite cardinal number. ‘ 2) i-±! ‘ ±Ð Explain Ñ 14. The Arithmetic of Cardinal Numbers We want to define exponentiation, addition and multiplication for cardinal numbers. Of course, for finite cardinals, these operations will coincide with the ordinary arithmetic operations in . We begin with exponentiation. As in the previous section, 7ß 8ß and : will denote cardinals represented by sets QßRß and T . Definition 14.1 8œ±R±7Q Thus, 8QR7 is the number of functions from the set into the set . As with the definition of the order relation Ÿ , we must check that exponentiation is well-defined. (That is, if one person calculates -i! using l‘ l and another person uses l Ð!ß "Ñ l , will they get the same answer?) If 7œlQlœQ |ww | and 8œlRlœlRl , we must show that R QwQ µÐRÑ w , that is, we must produce a bijection 9 ÀRQwQ ÄÐRÑ w . By hypothesis, we have bijections 0ÀQÄQwwQ and 1ÀRÄR . For 2−R , define a function 9Ð2Ñ œ 1 ‰ 2 ‰ 0" − ÐR w Ñ Q w . 9 is one-to-one: Suppose 2 Á 5 − RQw Þ Then for some 7 − Q , 2Ð7Ñ Á 5Ð7ÑÞ Let 7 œ 0Ð7ÑÞ Then 9Ð2ÑÐ7w"w Ñ œ 1Ð2Ð0 Ð7 ÑÑÑ œ 1Ð2Ð7ÑÑ and 9Ð5ÑÐ7w"w Ñ œ 1Ð5Ð0 Ð7 ÑÑÑ œ 1Ð5Ð7ÑÑÞ But 2Ð 7Ñ Á 5Ð7Ñ and 1 is one-to-one so 99 Ð2ÑÐ7ww Ñ œ 1Ð2Ð7ÑÑ Á 1Ð5Ð7ÑÑ œ Ð5ÑÐ7 ÑÞ Therefore the functions 99Ð2Ñ and Ð5Ñ have different values at 7 , so 99 Ð2Ñ Á Ð5Ñ . Therefore 9 is one-to-one. 9 is onto: Exercise. Examples 14.2 1) l Ò!ß "ÓÐ!ß"Ñ l œ l‘‘ l œ - - 2) #œl#lœ-i! , since #µ‘ . Thus, there are - different binary sequences. As remarked earlier, it is not hard to generalize this to show that 5œ-i! is true for any integer 5"Þ Students sometimes confuse the fact that #œ-i! with the continuum hypothesis. The i! continuum hypothesis states that there is no cardinal 7 such that i! 7# œ- . In other words, CH states that #i! is the immediate successor (in terms of size) of the cardinal number ii! . Here is still another way of putting it: let us define " to be the immediate 50 successor of i! () assuming for now that an immediate successor must exist . Then CH simply states that -œi" . To go a little further, the generalized continuum hypothesis (GCH for short) states that for any infinite cardinal 7# , the “immediate successor” is 7 . (Since ZFC cannot prove CH, certainly ZFC cannot prove the stronger statement GCH. In fact, GCH is undecidable in ZFC.) 3) For any set Qß Q is equivalent to ÖÖ7×À 7 − Q× ©c ÐQÑÞ Therefore 7 Ÿ #7 . But 7Á#7Q7 because Q is not equivalent to # . Therefore 7# for all cardinals 7 . It follows that 7 7#7# #7# # # ... ... is an infinite increasing sequence of infinite cardinals. In i i#! ! particular, for 7œi!! , we have i # œ-# ÞÞÞ We now define multiplication and addition of cardinal numbers. Definition 14.3 Suppose 7œlQl and 8œlRl , 1) 7†8 (or simply 78 ) means lQ‚Rl 2) 78 means lQ∪Rl , provided that Q and R are disjoint. Part 2) of the definition requires that to add 78 and , we choose disjoint representing sets M andRÞ This is always possible because if Q ∩R Á g , we can replace Q and R by the equivalent disjoint sets Q ‚ Ö!× and R ‚ Ö"× . You should check that multiplication and addition are well-defined that is, the operations are independent of the sets chosen to represent the cardinals 78 and . Theorem 14.4 For cardinal numbers7ß 8 , : , and ; À 1) 78œ87 1)w 78œ87 2) Ð7 8Ñ : œ 7 Ð8 :Ñ 2)w 7()() 8: œ 78 : 3) :Ð78Ñœ:7:8 If 7Ÿ8 and :Ÿ;, then 4) 7:Ÿ8; 4)w 7:Ÿ8; Proof Theproofs are all simple. For example, we prove 4w ). Suppose 7ß8ß:ß; are represented by the sets QßRßTßUÞ Since 7Ÿ8 and :Ÿ; , there are one-to-one functions 0ÀQÄR and 1ÀT ÄUÞ Define 2ÀQ‚T ÄR‚U by 2Ð7ß :Ñ œ Ð0Ð7Ñß 1Ð:ÑÑÞ Then 2 is one-to-one, so 7: Ÿ 8; . ñ Addition Examples 14.5 51 1) For any 77!œ7, because Q∪gµQ. 2) For finite 8 , 8i!!!! œi i œi because the union of two countable sets is countable. 3) For finite 8, 8-œi! -œ--œ- . -- œ- is true because Ð∞ß!Ñ∪Ò!ß∞Ñœ‘ . Then write the inequalities !Ÿ8ŸiŸ-! and -Ÿ-Ÿ- Ÿ- Add the inequalities and apply Theorem 14.4(4) to get -Ÿ8-Ÿi! -Ÿ-- . But --œ- , so we conclude that -œ8-œi! -œ-- . 4) If 77iœ7 is infinite, then ! . To see this, pick Q so that 7œlQl and Q∩ œg . Q is infinite so there is a one-to-one map 0À ÄQ and we can write Qœ0ÒÓ∪ÐQ0ÒÓÑ . Because 0Ò Ó is countable, so is 0Ò Ó∪ . Therefore we have a bijection 1À0Ò Ó∪ Ä0Ò Ó . We can then define a bijection 2ÀQ∪ ÄQ by BB−Q0ÒÓif 2ÐBÑ œ 1ÐBÑif B − ∪ 0Ò Ó The following two facts about addition are also true, but their proofs require more complicated arguments that involve AC. We omit the proofs and, for now, simply assume 5*)). and 6* 5*) If 78 or is infinite, then 78œÖ7ß8× max : that is, a sum involving an infinite cardinal number equals the larger of the two cardinals. 6*) If 78 and :;, then 7:8; . This result deserves a word of caution. For infinite cardinals, it is not true in general that if 78 and :Ÿ q, then 7:8; ! For example, i! - and -Ÿ- but i-œ--Ðœ-ÑÞ! 7) The World's Longest Song: “i! Bottles of Beer on the Wall" ( It fits the music better with the British pronunciation “Aleph-nought.” ) Multiplication Examples 14.6 1) For any 7 , 7†!œ! and 7†"œ7 . These equations are true because Q‚gµg 52 and Q‚Ö+×µQ . # 2) iœi! ! . # If we view ii†i! as shorthand for !! , the equation is true because ‚µ . But an alert reader might point out that we could also # Ö!ß"× interpret ill! as an exponentiation, that is, as . But this gives to the same result because Ö!ß"× µ‚ . (We can establish a bijection by pairing each 0 −Ö!ß"× with the pair Ð0Ð!Ñß 0Ð"ÑÑ − ‚ . ) 3) If 88!8†iœi is finite and , then !! . To see this, begin with the inequalities "Ÿ8Ÿi ! iŸiŸi!!! Multiplying and applying Theorem 14.4Ð%w Ñ gives # iŸ8†iŸi!!! ## Since i!!! œi!! , we get that i œ8†i œiÞ 4) -## œ - because Ð!ß "Ñ µ Ð!ß "Ñß # 5) If 8 is finite and 8! , then 8-œi! †-œ- œ- . Begin with the inequalities "Ÿ 8 Ÿi! Ÿ- -Ÿ-Ÿ-Ÿ- and multiply to get # -Ÿ8-Ÿi! †-Ÿ- ## Since -œ- , we conclude that -œ8-œi! †-œ- One additional fact about multiplication of cardinals will be assumed, for now, without proof: 6*) If 7 is infinite and 8 Á ! , then 78 œ max Ö7ß 8× . In particular, for an infinite cardinal 77œ7 , we have # . In algebraic systems (such as ™ , or ‘ ) where subtraction is defined, the definition of subtraction is always given in terms of addition: +,œ- is defined to mean +œ,-Þ This shows why subtraction cannot be sensibly defined for infinite cardinal numbers: should we say iiœ!!! because iœi!!! ? or iiœ" !! because iœi" !! ? or iiœi !!! because + iœii!!!? Similarly, division is usually defined in terms of multiplication: , œ- means +œ,-Þ Think about why there is also no sensible definition for division involving infinite cardinal numbers. The following theorem is an excellent check on whether you understand “function notation.” Theorem 14.7 For cardinals 7ß 8ß and : À Ð78: Ñ œ 7 8: 53 Proof We want to define a bijection 9 ÀÐQRT Ñ ÄQ ÐR‚TÑ . If 0 − ÐQRT Ñ , then 0 is a function T Ä Q R . If : − T ß then 0Ð:Ñ − Q R , so 0Ð:Ñ is a function R Ä Q . Therefore, for 8 − R , 0Ð:ÑÐ8Ñ makes sense: it is an element of Q . So we can define a function 99Ð0Ñ − QR‚T by the rule Ð0ÑÐ8ß :Ñ œ 0Ð:ÑÐ8Ñ . 9 is onto : if 1 − QR‚T , let 0 À T Ä Q R be the function defined by 0 ( :ÑÐ8Ñ œ 1Ð8ß :Ñ . Then 0 − ÐQRT Ñ and 99 Ð0Ñ œ 1 because for any pair Ð8ß :Ñ , we have Ð0ÑÐ8ß :Ñ œ 0Ð:ÑÐ8Ñ œ1Ð8ß:Ñ. 9 is one-to-one : if 0ß 1 − ÐQRT Ñ and 0 Á 1 , then for some : − T , 0Ð:Ñ Á 1Ð:Ñ . Because 0Ð:Ñ and 1Ð:Ñ are different functions R Ä Q , there must be some 8 − R for which 0Ð:ÑÐ8Ñ Á 1Ð:ÑÐ8Ñ. But this says that 99 Ð0ÑÐ8ß :Ñ Á Ð1ÑÐ8ß :Ñ , so 99 Ð0Ñ Á Ð1Ñ . ì Examples 14.8 1) -œÐ#Ñœ#iiii†ii! !!!!! œ#œ- ii!!ii!! 2) -œ# Ÿi!! Ÿ- œ- , so i œ- 3) l‘‘ lœ--i-i†-- œÐ#!! Ñ œ# œ# i! i! Caution: We now know that iiœ-! ! , but that - œ- . So we might be tempted to conjecture i! that if 7i! , then 7 œ7 . This is false . In fact, for every cardinal number 57577 , it is possible to find an for which i! and also to find some other cardinal 75 for which 7i! œ7 . The proof of this is a little too complicated to look at now. 54 Exercises E46. Prove or disprove: Let ‘ be the set of algebraic numbers. Then every open interval in contains a point of ‘ . E47. a) Suppose W- is a countable subset of ‘ . Prove that there exists a fixed real number, such that=- is transcendental for every =− S. b) For any set F©‘α‘ and − , we write F α for the set Ö, α À,−F×Þ Find a set E©‘α"α" for which lElœ- and Ð Ñ∩Ð Ñœg for all Á −EÞ E48. You and I play the following infinite game. We take turns (you go first) picking “!" ” or “ ” and use our choices as the consecutive digits of a binary decimal which, when completed, represents a real number in the interval Ò!ß "ÓÞ You win if this number is transcendental; I win if it is algebraic. Explain how to make your choices so that you are guaranteed to win, no matter what choices I make. (Hint: Consider the binary expansions of the algebraic numbers in Ò!ß "Ó . Look at the proof that Ð!ß "Ñ is uncountable.) E49. Find the cardinal number of each of the following sets: a) the set of all convergent sequences of real numbers b) the set of all straight lines jlj∩ЂÑl # in the plane for which c) the set of all sequences 0À Ä that are eventually constant (Note: 0 is eventually constant if there are natural numbers 67− and such that 0Ð8Ñœ6 for all n m. ) d) the set of all differentiable functions 0À‘‘ Ä e) the set of all geometric progressions in ‘ (A sequence ÐB8 Ñ in ‘ is a geometric progression if there exists a real number rÁ! such that +8" œ<+ 8 f 9< every 8 .) f) the set of all strictly increasing sequences 0ÀÄ (Note: 0 is strictly increasing if 6ß 7 − and 6 7 Ê 0Ð6Ñ 0Ð7ÑÞ) g) the set of all countable subsets of ‘ÞÐHint: part f) could be used. Ñ Note: Sometimes one proves lEl œ 7 by giving two separate arguments, one to show lEl Ÿ 7 and the other to show |E 7 | . One of these two inequalities often is easy and all the harder job is to show that the other inequality holds. E50. Find a subset of equivalent to the set of all binary sequences ÐÑa8 , or explain why no such subset exists . E51. Explain why the following statement is true: The set of all real numbers B which have a decimal expansion of the form B œ !ÞB"#$ B B ÞÞÞB 8 !"!"!"ÞÞÞ (8B may depend on ) is countable. 55 E52. Prove that for any collection of sets ÖE- À-A − × , there must exist a set ] such that |]llEl- for every -A − . ÐHint: use the fact that \µÎÐ\ÑÑ c . E53. Prove or give a counterexample for the following statement: If V‘ is an uncountable collection of uncountable subsets of , then at least two sets inV must have an uncountable intersection. E54. Prove that if 7ß 8ß and : are cardinals, then a) 7œ7†78: 8 : b) Ð7 † 8Ñ::: œ 7 † 8 E55. Prove or disprove: -- a) iœ#! i! b) 2Ð# ÑœÐ#Ñ # i! c) if #Ÿ7 2ii!! , then 7 œ- d) if 7 is infinite and #Ÿ8Ÿ#777 , then 8 œ# E56. a) Prove that ‘ has - countable subsets. b) Prove or disprove: If EF and have the same number of countable subsets, then EµF . E57. Prove or disprove: there are exactly - sequences of the form ÐE"# ß E ß ÞÞÞ ß E 8 ß ÞÞÞÑ where each E8 is a subset of . E58. Find all unjustified steps in the following “proof” of the continuum hypothesis: i! ii!! If CH is false, then i7-! for some cardinal 7 . Since -œiŸ7Ÿ! c œ- , we have 7ii7i7!! œc œ# # , so 7 ! # . Therefore Ð7ÑÐ#Ñ i-7--- ! , so 7# , which is impossible because 7#Þ . Therefore no such 7 can exist, so CH is true. E59. Find all unjustified steps in the following “disproof” of the continuum hypothesis: # i!!!!!! i! ii iii! i We know -œ#œ#œÐ#Ñ . However, Ð#Ñi! Ð because #iÑ! . Since ii!!" i!!! ", we have i!! i! œ i . Therefore c i i , so CH is false. 56 E60. Since ‘ is countable and llœ- , we know that is uncountable. But that does not automatically mean that llœ-ÞÐ Unless you assume CH, it could happen that ill-! . Ñ a) Without using the properties 5*), 6*) of cardinal addition or multiplication and without using CH, prove that: if lFlœ- and E is a countable subset of F , then | FElœ-Þ (Hints: Obviouslyi! lF El Ÿ -Þ One way to show lF El œ - : without loss of generality, you can assume BœFÞ‘# . Then consider vertical lines in Of course, there are other approaches.) b) Using a), deduce that llœ-Þ E61. Prove that for any infinite set I , there is an infinite sequence of disjoint subsets III"#$ , , ,... such that ∞ and | | for all n . ( Hint: The multiplication rule in Example 14.6(6*) Iœ8œ" Inn lIlœ I that implies that 7†i! œ 7 for any infinite cardinal 7 . You can assume that rule.) E62. Call a function 0À\Ä] “double-rooted” if | 0" ÐC± ) œ# for every C−] . Find the number of double-rooted functions 0À Ä . E63. Assume the Generalized Continuum Hypothesis (see p. 51). Then True or false (explain ): ccÐEÑ µ ÐFÑ implies E µ F . E64. Say that a pair of sets ÐEß FÑ has property (*) if all three of the following conditions are true: i) E∪Fœ ‚ ii) every horizontal line intersects E in only finitely many points iii) every vertical line intersects F in only finitely many points. We saw in Exercise E44 that such pairs ÐEß FÑ exist. Prove or disprove: there are exactly c different pairs ÐEß FÑ with property (*). E65. Let H œ Ö+"# ß + ß ÞÞÞß + 8ß ÞÞÞ× be a countable subset of ‘% and choose positive numbers 8 for ∞ which Define by Clearly if . %‘‘%88∞Þ 0À Ä 0ÐBÑœ Þ 0ÐBÑŸ0ÐCÑ BŸC 8œ" +ŸB8 Prove that 0H is discontinuous at each point in and continuous at each point in ‘ H . In Theorem 8.5, we saw that a monotone function 0À‘‘ Ä has a countable set of discontinuities; this result is a “sort of” converse. Note that this 0 is continuous from the right at every point.) 57 15. A Final Digression Let W# denote the sphere ÖÐBßCßDÑ−‘ $### À B C D œ"× . We will prove the following surprising (or is it not so surprising ?) result: stated informally If a countable set HW is removed from ## , it is possible to write the remainder WH as the union of two subsets EF and which, when rotated, give the whole sphere W# back again. For a point @œÐBßCßDÑ−W# , we use vector notation and write @œÐBßCß DÑ . Since W@−W@ÂH∪ÐHÑ## is uncountable (why? ), we can choose a point for which œÖ„.À.−H×Þ Then neither @ nor @ is in HÞ Change the coordinate axes so that the D -axis goes through @@ and (so the “north and south poles” of W# are not in HÞ ) For any G©W# , write GÐÑ" to represent the set obtained by rotating G on the surface W# around the DGÐÑG-axis through angle "" . In other words, a point in comes from taking a point in and adding " to its “longitude” on W# . With this notation, the precise statement of what we want to prove is: Suppose HWÞEFW is a countable subset of ## Then there are subsets and of , and there are real numbers α# and such that WHœE∪F## and WœEÐÑ∪FÐÑ α # . First, we claim that we can choose a "1− Ò!ß # Ñ that makes the sets Hß HÐ "" Ñß HÐ# Ñß ÞÞÞ ß HÐ8 " Ñß ÞÞÞ all pairwise disjoint. For any point @Ð@Ñ−Ò!ß#Ñ (other than the north and south poles), let arg1 be the longitude of @ measured from the great circle through Ð"ß !ß !Ñ ( œ the “Greenwich meridian”). HÐ5" Ñ and D(8+ß,−H" ) can intersect only if there are points such that argÐ+Ñ 5""1™ œ arg Ð,Ñ 8 #4 ( 5ß 8 − , 4 − ) (**) argÐ+Ñ arg Ð,Ñ#41 This means HÐ8"" Ñ and HÐ5 Ñ can intersect only if " satisfies the equation " œ 85 . But there are only countably to pick a “5-tuple” of values Ð+ß,ß4ß8ß5Ñ to plug into the right side of the & equationiœi because ! ! . So there are only countably many values " for which a pair of the sets Hß HÐ"" Ñß HÐ# Ñß ÞÞÞ ß HÐ8 " Ñß ÞÞÞ could intersect. Choose any " − Ò!ß # 1 Ñ different from these countably many values; then the sets will be pairwise disjoint. Now, define ∞ X œ H ∪ HÐ"" Ñ ∪ HÐ# Ñ ∪ ÞÞÞ ∪ HÐ8 " Ñ ∪ ÞÞÞ œ8œ! HÐ8 " Ñ and ∞ F œ X H œ HÐ"" Ñ ∪ HÐ# Ñ ∪ ÞÞÞ ∪ HÐ8 " Ñ ∪ ÞÞÞ œ8œ" HÐ8 " Ñ A rotation through Ð """" Ñ moves each set HÐÐ5 "Ñ Ñ onto the set HÐ5 Ñ , so FÐ Ñ œ X . Let EœWX### . Then E∪FœÐWXÑ∪ÐXHÑœWHÞ Since EÐ!ÑœE and FÐ"" ÑœX , we have EÐ!Ñ∪FÐ ÑœÐW## XÑ∪X œW Þ ñ 58 Chapter I Review Explain why each of the following statements is true or provide a counterexample. 1. ‘c‘‘ µÐÑ 2. The continuum hypothesis (CH), which states that #œ-i! , is independent of the other usual axioms of the set theory. 3. If 0À‘‘ Ä is strictly decreasing, then there are at least ( points at which 0 is continuous. 4. There exists a straight line jj in the plane such that contains exactly three points ÐBßCÑ where both BC and are rational. 5. In ‘$ , it is possible to find uncountably many “solid balls” of the form ÖÐBßCßDÑÀÐB+Ñ### ÐC,Ñ ÐD-Ñ % × such that any two of them are disjoint. 6. The continuum hypothesis is true iff the set of all sequences of !" 's and 's has cardinality - . 7. There are - different infinite sets of prime numbers. 8. Let E be the set of all sequences (aa88 ) where − Ö!ß "ß #ß $× and such that Ö8 À + 8 œ 5× is infinite for each 5 œ !ß "ß #ß $Þ Then E is countable. 9. If V‘V is an uncountable collection of uncountable subsets of , then at least two sets in must have uncountable intersection. (Hint: recall that c# œ c ) 10. The set of real numbers which are not transcendental is uncountable. 11. c‘‘ µÐÑ 59 12. Let W# denote the unit sphere ÖÐBß Cß DÑ −‘ $### À B C D œ "× and T œ Ð!ß !ß "ÑÞ There exists a continuous bijection 0ÀW# ÖT×Ä‘ Þ (The values of 0 are in ‘‘ , not # ! ) 13. Assume that for an infinite cardinal 77# , there is never a cardinal strictly between and 7 (the Generalized Continuum Hypothesis ). Then ccÐEÑ µ ÐFÑ implies E µ F . 14. There are exactly - sequences of the form ÐE"# ß E ß ÞÞÞ ß E 8 ß ÞÞÞÑ where each E 8 is a subset of . 15. There is an algebraic number between any two real numbers. 16. Let f‘ œÖ0©# À every horizontal line and every vertical line intersects 0 in exactly one point × . Then lWl œ #- Þ 17. There are #- subsets of ‘ none of which contains an interval of positive length. +, 18. Let >‘>ÀGÒ+ß,ÓÄ by Ð0Ñœ0Ð# Ñ . Then > is one-to-one. 19. Suppose a nonempty set \\œ]‚^]^ can be “factored” as . Then and are unique. 20. Suppose EßF and G are infinite sets and that EµF∪GÞ Then EµF or EµGÞ 60