math3z0. set theory - middle east technical...
TRANSCRIPT
Math3z0. Set Theory
Eduard Emel'yanov
CONTEI\T
Lecture 1 Language, axioms, and elementary constructions
of set theory
Lecture 2 Elementary consequences of axioms
Lecture 3 Cartesian products, relations
Lecture 4 Order relations
Lecture 5 Functions
Lecture 6 Natural numbers
Lecture 7 Equipollence and cardinal numbers
Lecture 8 Hierarchy of cardinal numbers
Lecture 9 Arithmetic of cardinal numbers
Lecture 10 Well-ordered sets
Lecture 11 Principle of transfinite induction, ordinals
Lecture 12 Arithmetic of ordinal numbers
Lecture 13 Some assertions that are equivalent
to the axiom of choice
Lecture L4 Tychonoff theorem
Lecture 15 Filters, measures? and bases
0.7. LECTURE 1
0.1 Lecture 1
Langueg€: axioms, and elementary constructionsof set theory
The main aim of this course is to give a short introduction toset theory for undergraduate students who do not have possi-
bilities to study this branch of mathematics properly and do
need to use basic techniques of set theory in other courses likealgebra, analysis, topology, etc. There are dozens of text-booksin set theory of different level from very introductionary liketo more advanced. The level of our course Math320 is inter-mediate. Basically, our presentation of set theory follows tothe classical textbook Set theory. With an ,introduction todescript'iue set th,eory of Kazimierz Kuratowski and An drzejMostowski, althought it has some minor differences with thisbook in denotations and terminology. We refer to this textbookas to IKM]. We try to find a way of presentation of materialthat is neither too popular nor very formal. The main accent
is made on solving problems directed on developing basic skillsof applications of methods of set theory in neighbor branches
of mathematics.
Nowadays the concept of a set is subject to study even inhigh school programs. In each area of mathematics we have
to play with sets using certain rules. The modern set theory(it is created by works of two generations of mathematicians
a century ago) unifies these rules and simplifies use of them
4
in common mathematics. During our course) we omit manyhistorical comments on earlier stages of set theory and send thecurious reader to standard textbooks in set theory. In manyplaces our presentation is not very rigorous for the benefit ofthose students who need to save strength and time to be able tosolve more exercises. Exercises given with solutions (: proofs),we call usually by theorems or lemmas.
In this lecture we fix the language of set theory and presentthe system of axioms which is known as ZFC-system. Afterthis we discuss some elementary consequences of axioms thatallows develop some elementary constructions like relations,functions, etc. Next lectures, among other things, devotedto natural numbers, cardinal numbers, principle of transfiniteinduction, ordinal numbers? equivalent forms of the axiom ofchoice and several important applications of set theory to anal-ysis, algebra, and topology.
First of all we need to specify the language. It consists ofsymbols fr,U, z) ..., X,Y, 2,..., which will be called uari-ables,, of symbols: A for conjunction, V for d,isjunct,ion,for negat'ion,, -> folimplicat,ion,: for equ,iualence) [ (or V)for general quanti,fi,er, V (ot 3) for eristential quanti,fier. Inour course we use only formulas of th,e class K (following theterminology of [KM]), which are defined as follows.
1. Atom'ic forvnulas n: U, n e U, Z("), where Z(") means
that "tr is a set" are of class K.
0.1. LECTURE I s
2. If O and V belongto the class K then OVt[, On U/, * -* V,Q - V, and .-O do.
3. If O belongs to the class K and u is any variable then theexpressions ! O and A O belong to K.
ua
If X is a set, we shall use the abbreviations V O(") andr€X
A o(") for ! l(" e X) - o(")l and I l(" e X) ^
o(r)] re-rex rspectively.
Formulas which have no free variables are called sentences.
Formulas which are true 'independently of the meaning ofatomic formulas out which they are constructed and indepen-
dent of the values of the free variables are calle d theorems ofthe pred,'icate calculus.
Now we are in the position to give the list of axioms.
Axiom L (Axiom of Extensionality). A A and B arethe sets hau'ing the same elements, th,en tltey are ,identical.
In symbols:
Alr € A- r € Bl (A- B).r
Axiom 2 (Axiom of the Empty Set) . There erists a
set (denoted bU a ) such that no rb an element of a .
Symbol'ically:
VA@ / P)Pr
Here we denote by (" / P) the formula '.(r e P).
6
Axiom 3 (Axiom of Unions) . Let A be a family of set;s
(: set of sets). There erists a set U A such that yis anelement of U A i,ff (- i,f and only i,f ) r ,is an element ofsom,e set X belong'ing to A. Symbol,ically:
"€uy'-Vtt"€x) ^(x€A)1.X
Axiom (Axiom of Power Sets) . For euery set A,there erists a set (denoted by P(A)) whi,ch cons,ists of allsubsets of A:
(xeP(A))=(xeA).Axiom 5 (Axiom of Infinity) . There erists a sets Asati,sfgi,ng the condi,ti,ons: a e A, and if X e A then thereerists Y e A such that Y cons,ists eractly of all elementsof X and the Y i,tself :
Y@eA)^AXA{ @eY):-t("e x) v (*-x)r))
Axiom 6 (Axiom of Choice) . For euery fami,ly A of dis-joint non-empty sets, there erists a set B whi,ch has eractlyone element'in common w'ith each set belong,ing to A:
{ A lx#ol^16+Y)-(xn Y-rlt }-I x,rea )
V A VAlw € Bn E)- (y- *)lBEeAr A
0.7. LECTT"]RE 1
Later, in Theorem 14, we show that the di,sjo'intness cond,itionin the axiom of choice can be omitted.
Axiom 7 (Axiom of Replacement for a Formula A).Let Q(r,y) be a formula of class K wi,th two free uariables.If for euery r there erists eractly one A such that Q(*,A),then for euery set A there erists a set {Q}" A whi,ch con-ta'ins those and only those elements E for whi,ch the cond,i-t'ion Q(*,A) holds for some r e A:
(
{+y+ro @,y)- (y:,,,
} ..+
AVAltu € B)- Vo(,,u)lAB a r€A
Axiom 8 (Axiom of Regularity). If A is a nonemptysetth,enthere erists a set X suchthatX e A and X)A-a.
The list of the axioms of the set theory is completed now.Remark that Axioms III, IV, VI, VII, and VIII are axioms ofcond'it'ional eristence. Axiom I is the principal tool in statingthe uniqueness of set-theoretical objects. The collection ofaxioms: I + II + III + IV + V + VII + VIII is called ZF-system (the Zermelo - Fraenkel system) and the collectionof all axioms I - VIII is denoted by zFC and is called theZFC-system.In the next lecture, we consider several elementary consequences
of axioms of the ZFC-system (: axioms of the set theory).
8
A.2 Lecture 2
Elementary consequences of axioms
Theorem 1 (The existence of a pair.) For arb,itrary aand b, there erists a un'ique set whose only elements are aand b.
Proof: Let A be a family of sets such that g e A and thereis at least one non-empty set belonging to A. Such a familyexists by Axiom V. Consider the formula Q(r,y):
l@ - a) ^
(a -o)l v l@ I a) ^
(y - b)1.
Then the set {o,b} is the set {O}"A, whose existence is pro-vided by Axiom VII. The uniqueness follows from Axiorn I. n
The set {a, b} is called an unordered pa,ir of elements a andb. If a - b, we call {a, a} a s,ingleton containing one elemento and write {o}
Theorern 2 (The existence of a union.) For arb,itrarysets A o,nd B, there erists 0, set C srrch that
(r e C)- l@ € A) v (r € B)1.
Proof: Take the set C - U{r', B}, whose existence is pro-vided by Axiom III. fl
This set is denoted by AU B. It is uniquely determined by Aand B accordingly to Axiom 1.
0.2. LECTURE 2
Theorem 3 (The existence of an unordered triples,etc.) For arb'itrary a, b, and c, there ,is a set C whoseelements are eractly a, b, and c.
Proof: C-{o,b}U{c}. IDefinition 1 The set (a,b) : {{"},{o,b}} ,is called anordered pair whose absciss a 'is a and whose ordinateis b.
Theorern4 (o,b) -(c,d) iffo-c and,b:d,.
The proof is left to the reader as an exercise.
Theorem 5 For any set A and for any formula Q of theclass K, there erists a set whi,ch conta,ins the elements ofA sat'isfying the formula Q and whi,ch conta'ins no otherelements. Thi,s set'is denoted by {t e A: O(r)}.
Proof: If A[(" e A) * -O(r)] then the empty set satisfiesfr
the statement of the theorem. So we may suppose O(o) to be
true for som e a € A. Consider the following formula
v(r,a)= lo(") n(y -")J vl-o(r) n("-o)1.For every fr, there exists eractly oney such that V(r, g). Theset {V}//A satisfies the statement of the theorem. n
Theorem 6 (The existence of the intersection.)eaerA non-empty fami,lA A of sets, there erists a un'iqueconta'ini,ng just those elements wh'ich e,re con'Lrnon to allsets of the fami,ly A.
Forsetthe
10
Proof: Consider the formula
o(") =n[(xe A)-(*ex)].X
Then the required set {r e [J A: Q(r)] exists by Theorem 5.
This set is denoted by nA. fI
Theorern 7 There 'is no set X such that X e X.
Proof: Suppose that such a set X exists. Consider a singleton
{X}. Accordingly to Axiom VIII, there exists an elementY € A-txlsuchthatYtrA- A. Butthe onlyoneelement of A is X. Hence X n {X} - a and X # X,,whichcontradicts to our assumption. n
Theorem 8 There is no set of all sets.
Proof: Suppose X to be the set of al1 sets. Then X € Xwhich contradicts to Theorem 7. fThe following theorem obviously restates Theorem 8.
Theorem 9 There 'is no set Z such that
At(x€ z)-(x /x)lX
We give a proof of the theorem that does not depend on reg-
ularity axiom.
Proof: If such Z exrsls, then taking X - Z we obtain a
contradiction lQ € Z)- (Z / Z)l.n
0.3. LECTI.]RE 3
0.3 Lecture 3
11
Cartesian products, relations
The Cartes'ian product of two sets X and Y is defined to be
the set of aII ordered pa'irs (*, A) such that r e X and y e Y .
To show its existence remark that
(r,y) - {{ r},{ tr,y}} € P(P(X u )'))
and that the Cartesian product
x x Y - {t € P(P(X u }.)) : O(f)},
where O(r) _ V V (t - (r,y)). The existence of X x YreX yeY
is guarantied by Axiom IV, Theorem 2, and Theorem 5. TheCartesian product is uniquely determined by its componentsby Axiom I. The Cartes'ian product operationis neither com-mutative nor associative operation. However it is distributiveover the union, intersection, and difference of sets. Consultany standard textbook in set theory for the details.
Definition 2 ArA subset R g X xY ,is called a (bi,nary)relation. The set
dom(ft) -Dt-{reX,V@,il e E}a
i,s called the dornain of R (or the left domain). The set
rang(R) - D,: {a e y' V @,y) € E}r
i,s called, the range of R (or the right domair).
12
Definition 3 Giuen a relation R q X x Y, the relation
{@, *) e Y x X , (*, a) e -R} ,is called the inverse of Rand 'is denoted by Ri (or by A-t ).
Definition4 LetS q XxZ andftg ZxY. Thenthe relat'i,on {(*,r), U,l@., r} e S A (r,,y) e E]} q X x Y
'is called the composition of R and S and ,is denoted by
RoS.
obviously dom(R o ^9) e dom(,s) and rang(.R o
^9) e rang(E).The operation "o" is associative but it is not commutative ingeneral. The composition operation satisfies also the formula
(RoS)': RioSiand certain distributivity properties, e.g.
(n u S) o T - (R o r) u (^9 o ").
But in general (ftn,S)of + (Eo?)n(5'o"). In thefollowingwe shall write rRA instead of (r,A> € R.
A relation .R g X x X is called an equ,iualence relati,on Iffor all n,U,z € ^F'(R) :- dom(fi) U rang(ft) the followingconditions are satisfied:
L) rRr (refl,eri,ui,ty)
(in other words the di,agonal Iy of X x X is a subset of R);2) rRy yHr (ty*metry)(in other words Pt'
3) rRA n y&z rRz (trans'itiui,ty)(inotherWords Ro Rg,R) .
A3, LECTURE 3 13
Definition 5 Let C be a set. A fami,ly A C P(C),iscalled apartition of C i,f a / A,UA:C, andfor anyX,Y € Aei,therX:Y orXnY: g (i.".thesetsbelong'ing to A are pairwise disjoint/.
There is a close relation between partitions and equivalencerelations given in the following two theorems, whose proofs are
Iefb to the reader.
Theorem 10 If A 'is a part'iti,on of C then the relationRa defi,ned by the formula
rLau-vl@€Y)^(a€)')lYeA
'is an equ'iualence relat'ion, whose field 'is C, i,.e.
C - F(Re):- dom( A,i U rang( Ae). z
Theorem 11 For any equ'iualence relat,ion R u,ith a fi,eldC + o thereerists apart'iti,onA of C suchthatR- Re. l
In both theorems the object, whose existence is claimed, are
unique by the Axiom I.
Thus, there is one-to-one corT espondence between all equiv-alent relations over a nonempty field C and the family of allpartitions of e .
If ft - Rt then elements of the partition A are called equi,u-
alence classes of ,R, the famlly A itself is called the quot'ientof C uith respect to R and denoted by C I R.
L4
A.4 Lecture 4
Order relations
Definition 6 A relati,on R g X x X,is called an orderrelation if i,t'is reflexive, transitive and antisymmetric. Thelast condi,t'ion nxeans that
rRyAyRr-(r:y)We call a relat'ion R a pre-order relation if R ,is onlyrefleriue and trans'it'iue.
Remark that our terminology is a bit different with that one
accepted in the book IKM]. The main reason for this is thefollowing. In our course we never use the term "quasi-orderrelation".
Instead rRA we usually write r {p A or just ralso say that the field of R is ordered (pre-ordered) withoutexplicitly mentioning R. But of course, sets may be orderedby many different relations.
relati,on
orasuch elements that either rcalled comparable . Otherwise
Definition 7 Let X be a set ordered (pre-ordered) by o
r and A are called incomparable.
Definition 8 An order relation(resp., a pre-order relati,on)
,f anA tuo elements 'in the fi,eld of
X o,Te
y o,re
o, linear pre-ordering)are cornp0,ra,ble.
0.4. LECTURE 4
Definition I Letorder relati,on) on o,
early ordered (resp.,
chain 'in (X, S).
Example 0.4.L Let A be a set.
of all coners of A. The relati,on R
15
set X. Then eny subset Y g X lin-ti,nearly pre- o rderi,n g) b y
Consider the fo*ily Ctg Ca x Ct is defi,ned by
v elements of X tllerez. SA*bol'ically,
Ar,UeX
Exercise 3 Show that the relat'ion R from Erample 0.4.Iis directed.
PIRPz= A V 6 gY).XePzYePt
Exercise 1 Show that the relat'ion R on Cd ,is a pre-orderrelat'ion wh'ich i,s not order relat'ion ,in general. Namely,construct a set A and a subset
^9 g Ct such that S conta,ins
di,fferent elements X , Y such that X RY and Y RX .
Exercise 2 Construct A and ,S q C 4 such that there ,is acha'in i,n S conta'ini,ng two d,ifferent elements and there ,is
no cha'in 'in S of more than two elements.
Definition 10 A set X ordered (pre-ordereQ bU S ,is sai,d
to be directed i,f for euery pair tr,eristsz € Asuchthatr
Vtt*zeX
16
Exercise 4 Show that any I'inearly ordered set ,is d,irected.
Exercise 5 Show that any equ'iualence relation n q X x X'is a,n order relat'ion iff R
Definition 11 Gi,uen ordered (pre-ordered) set A. Itssubset B g A i,s called cofinal (resp., coinitial) wi,thD g A i,f for eaerA r € B there eri,sts y € D such thatr3y(resp.,A3r).
Exercise 6 Let A be an ordered set. Cons'ider the relation
that P(A) 'is pre-ordered but not necessari,Iy ordered setw'ith respect to <,.
Exercise 7 Show that the set of atl real numbers ,is cofi,nalbut not co'in'it'ial wi,th the set of all natural numbers.
If on an ordered set (A,, S) there exists an element a satisfying
Af"€ A-rrrcalled a greatest element (resp., a lowest element) of A.
Definition 12 An element y of an ordered setA - (A, <) 'is called a maximal (resp., a minimal)element of A Lf Al(" e A) A (y 3 r) * (r - y)l
(respect'iuely, Al(" € A) A (r < A) -, (r - y)l).r
Remark that a maximal (minimal) element may not exist, andif it exists, it may be not unique.
A.4. LECTURE 4 L7
Definition 13 Let A be an ordered set, let T be a sub-set of A. An element u € A 'is sa'id to be o least up-per bound or a supremum of T ,f A (t
teTA l( A(t
u€A t€Test lower bound (an infirnum) 'is s,im,ilar.
Remark that a supremum (the infimum) does not always exist.However, when it exists, it is unique.
18
0.5 Lecture 5
Functtons
A relation R C X x X is called a function if
A l*Ran rRz (z- il!.fr,U,2
If dom(/) of a function f is X and if its range is a subset ofY, then / is called a mapping of X into Y. The set of allmappings of X into Y is denoted by Yx
Exercise 8 Show that there'is a one-to-one cor"respondence
between {a , {a}}x and P (X) for euery set X .
Remark that for any function / and for a{rJr r € dom(/),an element y such that ,fy (we write A : f @)) is uniquelydetermined by r. It is called the value of f at r.
Exercise 9 Show that for all f , g e Yx the follow,ing for-mula
(f _s)= avr")_s@)l
Ir,olds.
A function / is called one-to-one (or injective) if
AnL,nzedom(/)
The following two theorems are obvious.
0.5. LECTURE 5 19
Theorern L2 If f e Yx then fn gY x X i,s a funct,ionitr f is one-to-one. fl
Theorem 13 If f e Yx and g e 7v then the relationg o f g X x Z'is a function and g o f e ZY. IExercise 10 Sh,owthat arelat'ion RC X xX suchthatIx e R'is a functi,on i,ff R - Ix.
Definition 14 A funct'ion g i,s sa,id to be an extensionof a functi,on f ,f f e g. In th'is co,se we also caII f therestriction of S.
Exercise 1L Show that for gi,uen functi,ons f and g thecond'iti,on f g g is equiualent to
Theorern L4 If A 'is a non-en'Lpty fami,lA of non-emptysets, then there erists a functi,on f € (U A)A safi,sfgi,ng
f(X) € X foreueryX € A.
Proof: Denote A* : U ({X} x X) . A* is a union of non_XeA
empty disjoint sets. By the axiom of choice, there exists a set
B g A* which has exactly one element in common with everyset {X} x X. Remark that {X} x X g Ax X g Ax (UA)for each X e A. Thus B g Ax (UA) is a relation satisfyingdom(B) - A and ((X.,al e B) A((X,z| e B) -> (a - 4.
[dom(/) q,com(g)] n A U@)- s(*)).r€dom (f)
Hence B is a required function. fl
Definition15 LetT I a be aset andl,etfll a forall t € T . The generalized Cartesian product WF, u
the set of att functions T L U e such that f (t) e Ft foreuerA t € T. That i,s,
l[r, ,:{, .(Ur}) ",n tr(t). *r }t€r t \ter / rcr )
Exercise 12 Show that i,f F o - a for some ts e T then
I7n-a.teT
Exercise 13 Show that ,.f n # o fo, all t € T then
ilnla.teT
20
The set Y" is called Cartesian power of the set Y. Obvi-ously, If a e Y then the function equal lo a identically on 7belongs to Yr .
0.5. LECTURE 5 2I
Additional exercises to Lectures 1-5
In exercises L - 7 we use the following denotation. Givensubsets A and B of a set l), then A: B - AU (D \ B)
1 Compute A, (B U C).
2ComputeA (B.C).SCompttteAn(B,C).
4ComputeAn(B\C)
5 Compute A: A.
6Compfi,eA:4.
TComprteA:0.
8 Show that nA g x g u Ait x e A.
e Show that U(A u B) - (U A) u (U B).
10 Show that n(A n B) 2 (n A) n (n B) If An B I a.
11 Solve the system
IAnX: B
\Aux: c)where A, B, and C are sets and B g A g C.
L2 Solve the system
I A\X _ Blx\a- c )
where A, B, and C are sets and B g A, A)C : O.
22
13 Solve the system
IA\XtAux- C)
where A, B, and C are sets and B g A g C .
14 Show that A - B e (A \ B) u (B \ A) : a,, where Aand B are sets.
15 Show that the system
[A.x - aIB.(D\x) _ a 1
possesses a solution if and only 1f B g X g D \ A, where Dis a set such that AU X e D, B g D.
16 Given sets A, B, C such that AU B g C. Show that thesystem
[Anx - a[8.(C \x) - s )
possesses a solution if and only If B g C \ A.
17 Show that (A\ B), C -(A r C) \ (B x C).
18 Give an example showing that the projection of an intersec-tion may be different from the intersection of the projections.
19 Compute (A, B) o (C x D).
20 Show that f M is a non-empty family of equivalence rela-tions with common field C. then ) M is also an equivalencerelation with the field C.
21 Prove that f (A) \ /(B) e f (A\ B)
0.5. LECTURE 5
22 Prove that f (A n f-t(B))
23
23 Prove that there are sets A, B ,and C such that A x (B *C)#(Ax B) x C24 Show that (A rB) u (C, D) e(A u C), (B u D).
25 Show that (Au B) u (C u D) : (A* C) u (B x C)u(Ax D)u(B x D)
26showthat(AuB) * C-(A* C) u(BxC).2Tshowthat Ax (B\C):(A*B) \(AxC).28 Let A,B I a and (A x B) u (B x A) - C x D. Show
that A- B
zsProve that (gR,) o A
Bo Prove that Ao (g R,) s,[} @o Rr)
31 Show that the inclusion
(nR,) o ag n
(Rno a)
may be proper.
32 Let X + a. Is the family of all binary relations on X a
group with respect to operations o and -1? Why?
33 Prove that [(B + s) =+ (Bo # s)).
34 Prove that gA g P@ x B).
35 Show that there is a bijection between ABuc and AB x AcifBnC:a.1
24
36 Prove that
UnArj-nUAtr(i)iel jeJ jell iet
37 Show that there is a bijection between "'r?r'')
und n BA,
if sets A1 arepairwise disjoint. teT
38 Show that the relation Rr U Rz is symmetric if R1 and .R2
are.
39 Let Rr and Rz be both symmetric. Prove that ftr o Rz issymmetric iff Rr o Rz : R2 o R1.
0.6. LECTI]RE 6
0.6 Lecture 6
25
Ir{atural numbers
For any set X the set X' : X U {X} is called the successorof X. In the following we denote g by 0.
Theorem 15 There erists eractly one fami,ly IV of sets
such thet(i) 0 € l/;(rr) X € ff -) Xt € ,,nrl;
(iii) tf K set'isfies (r) end (i,i,) then,^f q K .
Proof: The infinity axiom guaranties existence of at least one
family satisfying (i) and (ii). Let R be such a family and
o- {ttft:oe snfllxe s-, x'.-sr ]t x --')Obviously, I[ ': f-l O is required family. n
In the following, we denote ,n/ by N. The elements of N are 0,
{0}, {0, {0}}, etc. This sets will be denoted often by 0, I,2,etc. If n € N, the operationn'will be denoted also by n+ 1.
Any set K satisfying (i) and (ii) is called inductive.
Lemma 0.6.L For all m,n e N the follow,ing formula
me n-+m'cn (1)
holds true.
26
Proof: Let m € N be arbitrary. Consider the set
K*:{n:m€n-+m'gn}.To prove (1), it suffices to show that N q K* or to showthat K* is inductive. Condition (i) clearly holds since theantecedent of the implication m € @ -> m' c o ts false. Toprove (ii), let n e K* and m e nt . Hence n'L - n or m e n.In the first case m' : n' C n' . In the second case m, C n bythe definition of K* and hen ce n't' e n C nt . Consequentlyn' e K*, which proves the lemma. n
The proof of lemma 0.6.1 is an example of a proof byinduction.
Exercise 14 Proue by i,nduct,ion that n / n for alln € N. Remark that th'is proof does not requ'ire the regu-Iari,ty ariom.
Lemma 0.6.2 For aII m,n € N the follow,ing formula
mt: nt -> m- n
holds tnr,e.
Proof: m': n'-) mby(1) -+m'gnVmSimilarly we can prove
It follows from Theorem 15 and Lemma 0.G.2 that the set Nsatisfies Peano's axioms, namely:
€nl
NC
(2)
--+m€n--+m€nVm--+ m g n V m- n ---+ m C rL.
m. fJ
0.6. LECTURE 6 27
(A) zero is a natural number;(B) every natural number has a successor;
(C) natural numbers having the same successors are equal;
(D) zero is not a successor of any natural number;(E) any set of natural numbers which contains zero and whichcontains the successor of any its element coincides with the set
of all natural numbers.
Exercise 15 Proae that for arb'itrary m)n € N eractlE oneof the follow'ing three formulas
rrL € n) m
holds. ( Hint: see page 91 in IKM\.
A function whose domain is N is called a sequence. Se-
quences are denoted by usually (but not always) by (on)70,(br,)rr.x, (b")r?:1r, etc. Sometimes, functions with domain
{0, 1 ,2, ...,m} are called finite sequences.
There are several typical schemes of inductive definition.We only mention four of them and send the reader to IKM] formore details.
(A) Let Z be a set and z € Z. Let e E Tzxt\. Then there isa un'ique sequence of elements of Z / satisfying the followingconditions:
d(0)
28
(B) Let f , Tzxt\*/ and g e ZA. Then there exists a uniquefunction 0 € TxxA satisfying the conditions:
OQ, o) - g("), O(n', a) - f@(r, a),n, a)
foreveryaeA.
(C) Let z € Z and, h € Zc'N, where C is the set of allfinite sequences whose terms belong to Z . Then there exists aunique sequence / such that:
d(0) : z, d@) - h(01",,fr),
where Qlr, is the restriction of 0 on the set nt: {0, 1,... ,n}.(D) Let g e ZA and H € Trxt\xA, where ? is the set ofall functions whose domains are included in N x A and whose
values belong to Z . Then there exists a unique function / eZNXA such that:
OQ, a) : g(a), 0(n' , a) - H (61@,xA), n, a).
For examples of use of schemes (A) - (D), we refer the reader
to pages 94-97 of [KM] .
Finite and infinite sets
Definition 16 A set X 'is called finite if there erists aone-to-one sequence wi,th doma'in n € N and range X . Inth'is case we write lXl - n. If X 'is not fini,te we say thatX is inftnite.
0.6. LECTT]RE 6 29
Theorem 16 If f is a one-to-one nxappi'ng of X onto Ythen lxl - n i,ff lYl - n.
Proof: If g is a one-to-one sequence of. n terms with range Xthen f " g is a one-to-one sequence of n terms whose range is
Y.nLemma 0.6.3 If f i,s one-to-one functi,on with dom(/) -XU{o}, rang(/) -Y U{b}, anda/X,b/Y,thenthereerists a one-to-one functi'on g such that dom(g) - X and
rang(g) - Y .
Proof: Let f (0) - b1
and it suffices to take
Then the function g is
- o,1. If b1 - b then a1 - a,
+ since f is one-to-one.
r+a1 nr
and f'(b)o,1 - bn (b)
given by
If(r)ir\b1 ifs(r)
Theorern L7 A Vl: * and lyl - n then the set X i,s
a one-to-one 'image if Y iff m : rr.
Proof: The sufficiency is obvious. We shall prove the necessity
by induction by n.
For n, _ 0 the theorem is obvious. Suppose that it is true
for some n and let / be a bijection (one-to-one and onto
mapping) from Y into X, where lyl - n' and lxl - m.
Since X + o, we may assume that n'L - p' : p U {p}. Since
n' : nU {n}, accordingly to the lemma, p is one-to-one image
of. n. By the inductive hypothesis, p - n. Hence TTL : p' : n' ,
which proves the theorem. I
30
Corollary 0.6.1 (Drawer principle of Dirichlet.) IflXl - *, lyl - n, ffi ) TL, and f is a function such thatf (X) - Y, tflen f is not one-to-one. fl
Exercise 16 Show that ,f lxl : * and lyl - n then m In i,ff there erists a setYt e Y such thatlYrl - *.
Exercise L7 Let lxl - n'1, lyl - rL, and X aY - O.Sholu that lX U Yl: m + rL.
Theorem 18 The set N 'is i,nfi,n'ite.
Proof: Suppose that lNl - n. Take a subset X of N such thatX - {0, 1)... jn} : n/, Then lXl - n'. By the assumption,there is a bijection { from n onto N. Take the restrictiong : {lr, of its inverse on X. Then g is a one-to-one mappingfrom nt tnto n. Let lg(n')l - k. By the Exercise 16, k I n.Take a bijection 0 between g(n') and k. Then the composition
f - 6o g is aone-to-one function such that f ("') - k < n In/, which contradicts to Dirichlet's drawer principle. n
Corollary 0.6.2 The range of a one-to-one ,infinite se-quence'is i,nfinite. n
0.7. LECTURE 7
O.7 Lecture 7
31
Equipollence and cardinal numbers
Definition 17 A set A zs equipollent to a set B ,fthere erists a bi,ject'ion from A onto B. We write A ,--, Bor lAl - lBl.
Exercise 18 Show that the equipollen ce ,is refl,eriue, syn'L-
metric, and trans'it'iue.
Exercise 19 Sh,ow that two natural numbers are equi,pol-
lent iff theg are equal.
Definition 18 We denote bg A the relational type of therelational system (A, A x A).
The following theorem is obvious.
::Theorem 19 Let A, B be sets. Then A ,,, B i,ff A:B J
Definition 19 We call A by the power of A, or by thecardinal number of A.
Theorern 2O The collect'ion Card of all cardinals(- cardinal numbers) 'is not o, set. f,
In the next section, we will get this theorem as an immediatecorollary of Theorem 28.
Accordingly to Exercise 19, we may and do denote the finitecardinals (- power of finite sets) by natural numbers.
32
Definition 20 A set A ,is sa,id to be countable if it'is e'ither fi,ni,te or equ'ipollent to the sef N of all naturalnumbers. We denote the cardi,nal number o/ N by s.
For the proofs of the next four elementary theorems we referto any standard textbook in set theory (for example, to ourtextbook IKM]). The readers should try to create those proofsthemselves before consulting textbooks.
Theorern 2t Euery non-enxpty countable set ,is the rangeof an i,nfi,ni,te sequence. tr
Theore rn 22 Th,e range of any i,nfini,te sequence,is count-able. n
Theorern 23(1) lA i,s countable and B g Al + lB i,s countablel.(2) lA'is countable and B i,s countablel + lAuB ,is countablel.(3) t/ ,is countabtle and, B i,s countabl{+ iar B ,is countable].n
Theorem 24oo
If An 'is countable fo, euery n then U ,is
n:0countable. n
It is interesting and quite surprising fact that Theorem 24 can-not be proved without the axiom of choice.
0.8. LECTT]RE 8
0.8 Lecture 8
33
I{ierarchy of cardinal numbers
Up to no\M we know only finite cardinals 0, 1,2,,..., and the
cardinal o - N. Now we shall create an important method for
constructing other cardinals.
Theorem 25 (Diagonalization theorem) If the doma'in
of a funct'ion F 'is contai,ned 'in a set A and i'f the ualues
of F are subsets of A then the set
Z-{redom(F) :r/r(")}'is not a ualue of F.
Proof: We shall show that F(t) + Z for every r e dom(F').
Suppose and F(t) : Z for some t e dom(F). Then
It e z)=lt / 21,
that is a contradiction. nThis simple theorem is one of the most impressive and powerful
results in the set theory.
Corollary 0.8.1 Let A be a set. Then P(A) + B foreny B g A.
Proof: Otherwise, there is a function (a bijection indeed)
whose domain is B and whose range is P (A). This contradicts
to the Diagon altzafr,ion theorem. fI
The proofs of the following two theorems are left to the reader
as exercises. (Hint: consult [KM] on page 175.)
Theorern 26 Let A be a set. I{o sets A, P(A), P(P(A)),., Pn(A), . . . are equipallent. n
Theorem 27 Let
A V A lz#flXEAYCA ZCXto any X € A nor to eny subset of X. n
Tllere erirts
Theorems 26 and 27 give us a tool for constructing of manynew distinct cardinals. Thus, by Thgorem 26, aII cardinals in
thesequence s:N, c -P(N),..., P"(N),...aredistinct (andinfinite). Now, by Theorem 27, the set ,S :- U 2"(N) is not
equipollent to any of sets in the sequence uUo#.')fhen we mayrepeat our application of Theorem 26 lo the set ^9 instead ofN again etc. This idea gives the hierarchy of cardinals. Butthe structure of this hierarchy if very complicated. It basedindeed on possibility to compare sets. This comparability inmost cases depend on the axiom of choice.ability in most cases
depend on the axiom of choice.
Theorem 28 set U which, fo, euery set
the fo*ily A of sets h,aue the property:Then the un'ionu A 'is not equ,ipollent
By Corollary 0.8.1,
any subset of U (J.
(J. Hence A is a set
no
X , conta'ins Y such that X:Y for euery set X .
Proof: S,rppose that such a set tl exists.
the set A:- P(U U) is not equipollent toHence, A rg no! equipollent to any of Y €such that A+Y for every Y e U. The obtained contradictioncompletes the proof. n
0.8. LECTURE I 35
Exercise 20 Sh,ow that Theorem 20 i,s a corollarE of The-orem 28. (Hint: use the replacement ar,iom.)
Let us give a proof of the following theorem.
Theore rn 29 There erists no set containi,ng all sets as
subsets.
Proof: Assume V to be a set such that X g V (and hence
X € P(V)) for every set X. This contradicts to Corollary0.8.1.
Another way to prove is in getting contradiction of the factthat X e P (V) for all sets X with Theorem 9 !
Exercise 2 L Proue that the set NN es uncountable.
Exercise 22 Proue that for any ,infi,nite set A the setPn (A)of all fini,te subsets of A i,s equ'ipollent to A.
Exercise 23 Show that for ang set X there ,is a subsetY g P(X) which 'is equ'ipollent to X.
f t
-
36
0.9 Lecture I
Arithmetic of cardinal numbers
Now we describe shortly operations of addition, multiplication,
and exponentiation for cardinals. Then we mention main prop-
erties of these operations and define the inequality-relation on
cardinals.
Definition 21 Let m and Lb, card'inals. Take sets Mand I{ such that M- m and -l{- n.
1) The power of (M x {M})U (lf x {If}) is called, the surr-
of m and n and 'is denoted bY m * n.
2) The power of M x l[ i,s called the product of m and n
and 'is denoted bY m x n.
A The power of MN i,s called the n-th power of m and 'is
denoted by m".
It is easy to see that the sets A x {A} and B x {B} are
equipollent to A and B and disjoint.
Theorem 30 Let n1, n2, and ns be card'inals. Then
(1) nt * nz - rt2 * nr.
(2) tt * (n, a nr) : (nr 1nz) * ns'
(3) nt ' rr2 : tlz ' l1r'
(a) tt ' (t, 'nr) : (ny nr) 'rr3.
(5) 1 'rr1 : ttl.
0.9. LECTURE 9
(6) nl'*ng(7) (nt 'rt2)nt - nlt 'n;'.(s) (nl')n'(9) n| - tl1 .
(10) 1nt
37
The proof is left to the reader as an exercise.
(Hint. see pages 179-180 in IKM] ).
Exercise 24 Proue that, for any set X,2x - P(A).
Exercise25 Prouethats.s_ q andn.s- &, where:o: N and n € N.
Definition 22 Let m and n be card'inals. Take sets Mand IV such that M- * and,ltr- r. We say thatm ( ni,ffthere'is an'injection f , M,-+ IV. If m < n andm*nthen we write m < n.
Theorem 31 Let m, n, and p be cardi,nals. Then(o) l(* < n)
^ (n < p)l =+ lm < pl.
(b) (m < n) =+ (*+p < n+F).(r) (m < n) =+ (*.p ( n.p).@(m<n) +(*P<nP).(") (m S n) =+ (p* S p").
38
The proof is left to the reader as an exercise.
Exercise 26 Show that we nx&A replace t(<" with *<" onlyi,n the i,tem (a) of the theorem aboue.
The Cantor - Bernstein theorem
Theorern 32 (The Cantor - Bernstein theorem)If m ( n and n S m tlten fil: tt.
There are many proofs of this fundamental theorem. Our proof
is based on the following important lemma.
Lemma 0.9.1 Let Y be a subset of a set X. If there
erists an'inject'ion f : X -> Y then sets X andY are
equ'ipollent.
Proof of thelemma: X )Y I f(X). DenoteZ:-y \ /(X), S'- () f"@). Denote the mappins s : X +Y
n:0as follows:
s(r): { }r,t ll :: *\,s
0,9. LECTURE 9 39
v
#txt
UI'[
The proof will be completed if we show:
t)g(x)-Yand2) g is one-to-one.
1)' g(") *oo
: g(su (x\ s)) - sr u f(x\s) - U f"@)u f(x\s)n:0
oo
-zu[J f"!t)uf(x\s) -zul(s) uf(x\^s) -n:l
tJId
TJ.H
40
- z u f(x) - f(x)u [y \ /(x)] - Ysince f (X) gY.2), The function g is one-to-one on each of the sets S andX \ ,S so it is enough to show that 9(S) a g(X \ S) _ s.Remark that Z. f (X) - s. Hence s(X \ S) - f (X \,S) -f(x)\/(s) - f(x)\t/(s) uzlande(,s)ng(x \s) -sn(/(x) \t/(s) uzl) -sr n(/(x) \s) :@ rNow we are in a position to complete the proof of the theorem.
Take sets M and l[ such that M - rn and ,Af- n. Since
m < n and n < m, there are two injections 6 : M .+ l[ and
t, N + M. Denote X:- M,Y ,: d(M),, f ,: Qotlt andapply the lemma. Then
Corollary 0.9.1
A.IIf Ag B q C and,]: C then U
Exercise 27 Let f : X -> Y be a function. Show that::f (x) < x.
Later we prove that any two cardinals are comparable. Thisfact is even much more nontrivial than the Cantor - Bernsteintheorem. Combination of these two results means that "(" is
a linear order on cardinals.
Cardinals o and c. Continuum hypothesis
0.9. LECTURE 9 4L
The cardinal c : 2o, where o - N, is called the power ofcontinuum. Although most of properties of cardinals o and
c follow from general theorems on cardinals stated in Section
3.2, we reprove some of them below.
Proposition 0.9.1
a) c* c - 2. c - 2 - 2o : 2r+a : 2s - c.
b) Showthatnl o<c forany n€N. Indeednlsando < c by the d'iagonal'izat'ion theorem.
c) c : 2q - 2o+q : 20 .2o : c . c, and by i,nduction c : cn
for euery n € N.
At-2q<noI oo< co:(2o)o-)o'o:2s -c. Hence,bythe Cantor - Bernste'in theoreffi, c: oo. (Here n > I.)
Theorem 33 Th,e set of all sequences of natural numbershas power c.
Proof: Denote this set by A. Then A - ff 4 U B, wheren:0
An is the set of all sequences of length n and B is the set ofall infinite sequences. Then, by using Proposition 0.9.1 d),
:--C: OO -B <R*O. O: B U U AN:A<Ntr - C.
n:0
: : :Theorern 34 If A: c aTLd, B - s then A\ B - c. n
tr
42
For the proof consult IKM] on page 189.
Corollary O.9.2 The set of all transcendental numbershas power c. I
Theorem 35 The sef IRN of alt (i,nfinite) sequences ofreals has power c.
:Proof: RN - co -- c by Proposition 0.9.1 d). n
Theorem 36 The set of all cont,inuous function from Rfo IR has power c. !
For the proof consult [KM] on page 190.
Theorern 37 The sel IRR of alt funct'ion from R to Rhas power 2t.
proof: RR _ cr : (2"), _ 2s.c _ 2r. J
The conjecture that every cardinal b such that s ( b ( c
is equal either to s or to c cannot be proved (or disproved) inthe Z FO-system of axioms of the set theory. This conjecturethat there is no other cardinals between o and c is called theContinuum Hypothesis. For details see, for example IKM]on page 290.
0.7A. LECTURE 10
0.10 Lecture 10
The relationalof ordered set
ordered sets A
Wbll-ordered sets
43
Let us remind some definitions.
Definition 23 A set A wi,th a relation R C A x A,iscalled an order"4 set i,f(1) aRa for att a e A;(2) aRb A bRa a
(3) aRb n bRc -* aRc.An ordered set (A, R) ,is called an Iinearly ordered set(: the order relat'ion R i,s called a linear ordering) ,f(/,) Ya,b € A [(o Rb) v (bRa)] .
type (A, Rl of (A, R) is called the order typeA. In the case if (A, n)and B to be similar. For examples of ordered
sets, we refer the reader to page 20L of [KM] .
Exercise 28 Study eramples 1 7 there and ,inaestigate
which pa'irs of ordered sets i,n the eramples 1 7 possess
common order types.
Definition 24 An(A, R) is so,id to be &y € A. The defi,nition
element r of a linearly ordered setfirst element of A ,f rRA fo, allof o, last element is analogo,u,s.
44
Theorem 38 In any fini,te non-eTrlpty subset of a l,inearlyordered set there 'is a first and a last element. J
Proof is lefb to the reader as an exercise.
Theorem 39 Two I'inearlg ordered sets (A, R) and (8, ^9)
possesses the same order type i,ff there erists o, b'ijecti,on
f ,- B such that
rRa - f @)S f (a) (Y*,a e A). n
The proof is obvious. Theorem 39 may be reformulated as
follows.
Theorern 4O (A, R> - (8, ,S) i,ff there erists a bi,jecti,on
f of A onto B such that rRy -- f (r)Sf (y) for alltr)y € A. n
Exercise 29 Show that two fini,te I'inearly ordered set ares'im'ilar i,ff they are equi,pollent.
For the notions of a successor, predecessor, an initialsegment, d final segment, and an interval, etc. (in lin-early ordered sets) are obvious (see, for example, [KM] on page2A4).
Exercise 30 Solue ererc'ises 1 and 2 from [KM] on po,ges
244-205,
A,7A. LECTURE 10 45
Definition 25 A l,inearly ordered set (A, R) ,is called awell-ordered set i,f euery non-empty subset of A conta,insa fi,rst element (with respect to R).
Example 0.10.1
a) any fi,ni,te l'inearly ordered set 'is well-ordered;
b) N is well-ordered set;
c) Z is not well-ordered;
d,) the set {1 *lT:, 'is well-ordered,;
e) the Cartes'ian product of fini,tely many well-ordered sets
'is well-ordered by th,e lericographi,cal ordering;
j) euery subset of a well-ordered set ,is well-ordered.
We remind that for an ordered set (A,, R) the ordered set
(A,R-t) is called an inverse to (A, R) .
Exercise 31 Show that a relat'ion R ,is a l,i,near orderingi,ff P-t ,is.
Exercise 32 Let R be a I'inear ordering. Show that both
R and, I{-r are well-ord,erings i,ff E * fini,te.
Denote by a- (N, <) and by L,)*
Theorem 4L If o, Iinearly ordered set A is not ?,ilell-
ordered then 'it' contains a subset of type (r)*.
The reader should consult for instance page 225 of IKM] forthe proof.
Exercise 33 Show th,at any ,ini,ti,al segnxent (i,.e. a subsetX such that r € X and yordered set (A, <) is of the form O(o) ,- {* € A : r < a}for some a € A. Is thi,s for (A, <) ?
0.11. LECTURE 11
0.11 Lecture 11
47
?'S
Principle of transfinite induction, ordinals
Let (A, <) b. a well-ordered set. We denote for every r e Aby O(r) theset {a e A,a l rAa # r}: {A e A,a < r}.
Exercise 34 Show that each i,ni,ti,al segnxent of a well-orderednon-enxpty set A 'is of the for* O("). Is thi,s true for thel'inearly ordered set (A
Definition 26 A subset B of acalled hereditary i,f for euery r
lo(") g Bl (r € B).
Theorern 42 (on Transfinite Induction) The onlyhered'itary subset of a well-ordered set A i,s the set A it-
A O(") is called the method of transfiniter€A
setf .
Proof: Suppose B gA,, B # Aisahereditarysubset. ThenA \ B is non-empty. Hence there is a first element r e A\ B.Then O(*)gB butr /B whichisacontradiction. n
For many formulas O it can be proved that the set {r e A :
O(")) is hereditary; consequently {r € A : O(r)} - A or
A O(") is the theorem. The method of proving theoremsreAof the form
induction.
well-ordered set (A,<)€A
Theorern 43 If o functi,on f : A -- A is ,increas,ing on &
well-ordered set (A,<), whi,ch means that r 1y - f (") <f (A), then r I f (") for eueru r e A.
Proof: Our aim is to prove that the set B :- {r : r S f (*)}is hereditary. Lef O(r) g A Then for every y e O(r) it holdsthata<r. Hencef(a)</(r) and A ly<f(a)<f(*)l
YeO(r)since O(") g B.In other words, ly e O(") + y < /(r)], or
f(r) € A\O("). Hence r 1f (")1or tr e B. Wehaveshownthat the set {r : r < f (")} is hereditary. By the principle oftransfinite induction, it coincides with A. A
Corollary 0.11.1 If the well-ordered sets (A, R) and(8, ,S) are s'im'ilar (: haae the same order type) then thereerists only one funct'ion whi,ch establ,ishes thei,r s,im,ilari,ty.
Proof: Let f : A -- B and g, A -+ B be two such functions.Both the function are increasing bijections. Hence g-L o f :
A --+ A is increasing in y'. By Theorem 43, A lrRg-t " /(")].reAHence A lg(") S f (")1. The switching of roles of f and g gives
reAus A [/(") Sg(")]. Hence f = s. z
r€A
Corollary 0.11.2 No well-ordered set i,s si,mi,Iar to anA
of i,ts i,n'it'ial segments.
Proof: If the sets A and O(r) were similar then there is an
increasing function f , A -+ O(r) (which establish the simi-Iarity between A and O(*)). This contradicts to Theorem 43
since f (r)
0.11. LECTURE 11 49
Corollary 0.11.3 No two d,'istinct ,ini,tial segn'Lents of awell-ordered set are similar. n
Theorern 44 (the Cantor theorem) Let A and B be
two well-ordered sets. Then eitller(i) A(rr) A- @ fo, some b € B, or(tit) B
Fbr the proof we send the reader to pages 227-228 in IKM].
Corollary 0.11.4 If sets A and B are well-ordered then::
card,'inakV and,E orc comparable, ,i.e. e,itherA < g orB<A.n
Definition 27 An order type of a well-ordered set 'is
celled o,n ordinal number (- &n ordinal) .
Exercise 35 Show that the power of the set of all ordertypes of countable well-ordered sets es strictly greater thato: N.
Definition 28 We sag that an ord,inal a is less thanan ord'inal p i,f ang set of type a i,s s,imi,Iar to a segmentof a set of type 0. We denote tlti,s relation by o < P orp>a.
The following theorem is a direct consequence of Theorem M.
50
Theorern 45 Let a, P be ord,,inals (symbol,ically a, g €ord) . Then (a< p)v (*- p)v(g<o). n
Exercise 36 Show that the onlA one of three poss,ibil,ities
rnaA occurin Theorem 45.
Let a € Ord, we denote by W (") the set of all ordinals less
than a. Consult for the proofs of the following three theoremsour textbook [KM] on page 229.
Theorem 46 W(t) - e,. f
Theoren 47 Euerg set of ord,inals i,s well-ordered set. J
Theorem 48 For euery set Z of ord,inals there erists anord'inal whi,ch 'is greater th,an all ord,inals belong,ing to Z . a
Corollary 0.11.5 There erists no set of all ord,inals. J
Definition 29 By ordinal numbers (-- ordinals) weunderstand the order types (- relat,ional types) of well-ordered sets.
We already know that for well-ordered sets (X, R) and ()r, S)either X is isomorphic to an initial segment of Y, or Y is
isomorphic to an initial segment of X or X ,-., f .
Definition 30 We say that an ord,inal a es less than anord'inal P ,f a - X,,R, P -Y,,R and X ,is ,isomorph,ic (-s'imi,lar) to an 'in'it'ial segment of Y. We write a < P orp > a. We write a < P'instead of l@ < P) v (o - P)1.
0.17. LECTT]RE 11 51
As we mentioned before (Def. 30), the relation "<" on ordinalsedoys the law of trichotomy.
A[(",0e ord) =+l(o< p)v (o- p)v (a> p))].ar0
Remark that, without using of Axiom of choice it is impossibleto prove the law of trichotomy for cardinals.
Let us collect the elementary properties of ordinals.
Theorem 49 Giuen e,) p,1€ Ord, then
Proof is left to the reader as an exercise.
By Theorem 49, the relation "<" is a linear order on Ord.
Theorem 50 Let a, B € Ord and e, : A, 0 - B. AAnl 81 g B then a
Consult for the proof page 50 of [KM] .
Theorem 51well-ordered by
The set W(*) :- {l € Ord : 1
Proof: Let o - A. Then, for any a e A,, the ordinal O(A) isless than a. Hence M e W (a). It follows from Theorem 49
that the mapping a --+ O(A) is an isomorphism of A ontoW ("). Hence W (") is well-ordered set and W (") - a. n
52
Theorern 52 In any non-enxpty set of ord,i,nals there er-'ists a first element.
Proof: Let a e Z e Ord. If a is not the smailest in Z thenZ ) W (") I a. Since W (") is well-ordered, there exists asmallest 0 e Z nw(a). Obviously 0 is a smallest element ofZ.tTheorem 53 Let Z C Ord be a set. Then
Proof: By the replacement axiom, there exists the set K -{W(") : a e Z}. Take SI - UK. By Theorem 52, the set
,S is well-ordered. Let 7 - 51. By Theorem 50, a - W (*) <Ei - 7 forevery a € Z. Hence a 1 0 t- "y+Lfor all a e Z. a
Corollary 0.11.6 Ord ,is not a set. n
Corollary 0.11.7 There ,is a smallest ord,inal whi,ch i,sgreater than all ord'inals belongi,ng to a g,iuen set. n
V A@leOrd aez
Corollary 0.11.8 If o set Z e Ord has the property
0then Z € W(r) fo, some a€ Ord,
Proof: Take a to be the smallest ordinal in Ord \ Z. ThenZ gW("). If Z # W(") then there exists g e W(*)\2.Hence P < o, which is a contradiction to W(")\Z S Ord\Z. I I
0.11. LECTURE 11
Definition 31
53
A function 0 uhose domain r,s W(*) ,is
called a transfinite sequence of type o (* o-sequence/.
Definition 32 An ordinal a 'is sa'id to be a limit ordinalif it has no direct predecessor (- ",if * # P + 1').
Theorern 54 Each ord'inal a can be uni,quely represented'inthe form )+n where ) zs al'imi,t ordinal and n € N.
Consult for the proof page 231 of [KM]
We say that a is cofinal with a limit ordinal B lf a is a limitof an increasing sequence /:
a- lt%d({) :- min(Ord \ rans(d) ).€<,
Theorem 55 An ord'inal a i,s cofinal wi,th a li,mi,t ord,i,nal
P i,tr W (") conta'ins a subset A such that A i,s cofinal wi,thW(*) andA: g. n
Consult for the proof pages 23L 232 of [KM] .
Theorem 56 (Or Definitions by Tlansfinite Induction)G'iuen a set Z and an ord'inal a, let Q denote the set of aII
{-sequences fo, € I a wi,th ualues i,n Z .
Then for each functi,on h e Zo there erists one and, onlyone transfin'ite (a + I)-sequence f such that
/(€) - h(f I'rer)
for euery € S a. tr
54
Consult for the proof IKM] on pages 233 - 234.
In applications of the theorem (we shall study several of themlater in our course), the function h is often defined by threeformulas:
(1) h(a) : A,
(2) h(il - F(.6lwG) if / is an ({+1)-sequence, in other wordsif the type of O is not a limit ordinal, and
(3) h(d) - c( U dl wut) or h(g) - c( n dlw4t)) if the typeq<^
,\ of O is a limit ordinal.
Here A is a given set and F, G are given functions.
For applications of Theorem 56 consult for example IKM] onpages 235 37.
rt<^
0.72. LECTT]RE 12
0.L2 Lecture L2
55
Arithmetic of ordinal numbers
Definition 33 Let a- A, p
1) The sum a*0 'is defined as AtU F.1, where A1)Br - Aand At: A and Bt - B.
2) The product a - B i,s defined as TWB, where A AB is the set A x B equ'ipped wi,th the lexicographicalorderingr (ot,bt) < (or,br) if q l az orl(q - or)A(bt <br)1.
Exercise 37 Show that both sets AtU Br and AA B arewell-ordered.
Definition 34 We'identi,fy the natural numbers wi,th cor-respondent ord'inals (i.r.N g Ord) bAn-n forn e N.The order type N tue denote bA ,. The ord,er tApe of the
first uncountable setW(a) i,s called the first uncountableordinal o,nd 'is denoted by u)1.
Exercise 38 Show that both, the ord'inals a, uL are lirnitordinals (2.e. they cannot be represented as € + 1) and w'is not cofinal w'ith u1.
Exercise 39 Proue that both operations a, + P and, a . Bare assoc'iat'iue and enjoy
"-(P +7) - a. 0 + a-7. Proaethat,'in general, a + 0 + P * a and a - B # 0 . " by meanof 2. a :? f? = w -2 and2 + w :? #? - a +2.
56
Definition 35 The natural exponentiation is definedby 'induct'ion as follows:
e,o :- L, on'fr : en . e,.
Let T be a well-ordered set. Let fl be a well-ordered set forevery t e T. Consider the set F- - il e with the lexico-
teTgraphical ordering: f {rc* g if for the first element ts ofthe set {r e r : f (t) # s(t)} it holds that f (td < g(tl).
Exercise 40 Proue that F i,s well-ordered by the lerico-graph'ical relat'ion.
The set S - U n x {t} with the ordering ftx {t} <s gqx {q}teT
Lf t < q or t: e and f1 I gq is called the ordered sum ofordered sets 4 and is denoted by I n.
teT
Exercise 41 G'iuen well-ordered sets T and Ft for aII t eT. Show that D n i,s a well-ord,ered, set w,ith respect to
(g. teT
Exercise 42 Let a - h and P - 4 and, T - 2. Show
that a * p -D[ and, a . g : hA,, .,",) .
tez \r€t /Definition 36 Let a, € Ord for any 11 < ^y. The or-dered sum I art 'is defined by
q<^y
I art- (I w@r),.r). rt<^y \aew(r) /
A.12. LECTURE 12
Exercise 43 Show that 0o * 0r * e2: D or.q<3
57
Definition 37 Let T - ^y and a, € Ord for any r7 €W (l). The ordered product il, t, 'is defi,ned by
q<7
il art- (il w(*r),.,,") .
q<1 \ry€w(r) /
Exercise 44 Show tltat ao. ar. a2- dJ: fr or.11<4
Definition 38 Let a, P be ord'inals. Then the ord,inalaP is defi,ned as aP
q<p
Exercise 45 Let a,,n € Ord and n e N. Show that an'in the sense of the last defin'it'ion i,s the san'Le as the n-thpower of a that we defined 'inducti,uely.
58
0.13 Lecture 13
Some assertions that €rre equivalent to the axiomof choice
Here we give the proof of the equivalence of several classical
assertions to the axiom of choice.
Lemma 0.13.1 (ZL) Zorrt's Lemma. An ordered set,
i,n whi,ch any cha'in'is bounded from aboae, possesses a mar-'imal element.
(ZL) + (AC): Let Abe a set, and let & be a non-emptyset for every a € A. Our aim is to show that there exists achoice function for the family {X"}.,ee.Consider the set S of all "partially defined choice functions" ,
i.e. ,S - {OlA f dom (O) I U Xo and 0(a) € Xo for all
a € dom(/)). The set ,S, ortff.a by inclusion, enjoys thecondition of ZL. Indeed, if C g S is a chain then U O is a
QeCfunction belonging to ,S that is an upper bound for C .
By the Zorn lemma, there exists a maximal element f e ,S.
By maximality of f ,, dom(/) - A. Consequently, / is therequired choice function. n
Axiom I (AC) Axiom of Choice. Let {X,}oe4 be a
fami,ly of non-empty sets indered bA o set A. Then there
U X* such tllat f (a) € Xo fo,erists a funct'ion f : A ---+
euery a € A.aeA
0.13. LECTURE 13
Our next aim is to show that the axiom of choice impliesfollowing assertion.
Theorern 57 (WP) Zewnelo's Theorem (We1l-Order-ing Principle). For euery set there ,is a relat,ion thatwell-orders the set.
(AC) + (WP): Let M be a set. We may assume thatM + a. Take a choice function f for the faml\y P(M) \ {a},and let o, / M. By the theorem on definitions by transfiniteinduction (: Tiansfinite Recursion Theor.-), there exists atransfinite sequence (Xe)e.Ord of elements of M U {o} such
that
Y.
Let a be the first ordinal satisfying Xo - a. Then M -{Xr}^,.1 and the transfinite sequenc" (xr)r46 consists of dis-tinct elements of M.
Hence M is well-ordered by the relation u I u Iff u - Xr,, : Xr, and jt < ^lz K o. n
Lemma 0.L3.2 (TTL) The Teichmiiller - Tukey Lem-ma. Euery property of fini,te character has a marimal setpossess'ing that property.
Here we call by u property any family contained in P (X) ,
59
the
where X is a set. LelU gP(X) b. a property (of subsets of
60
X). Then U is said to be of finite character when A e tlif and only if all finite subsets of A belong to U.
(\VP) =+ (TTt): Let X be a set, andletl/ g P(X) be
a property of finite character. Take any set A e U. By theZermelo theorem, the set X\A admits a well-ordering. Hence,
we may represent X \ A as a transfinite sequence (16)6.0 con-
sisting of distinct elements.
By Transfinite Recursion Theorem (- theorem on definitionstransfinite (a + 1)-
U A.y rJ {re} € U7<€
if U A^y u {re} /U,7<€
Ao: U a*.{<o
ThesetsA6poSSeSSthepropertyUbyrecursionif{<Also Ao e U, since for any finite subset B g Ao lhere exists
€o<asuchthat BgA€0.
The set Ao is a maximal set which has the property U sinceall elements of X \A : {16}6." are already jointed to,4, were
it possible. !Now let us complete the proof of the equivalence by showing
re exists a
o- Aandby transfinite induction) , thesequence (Ae)e<* such that A
( U A.u {'e}A€ lUA.
t ''r<€
whenever €
if
that the Teichmtller - Ttrkey lemma implies theZorn lemma.
0.13. LECTURE 13
(TTL) + (ZL): The property "to be a chain" in an orderedset X - (X, S) is, obviously, a property of finite character.
Starting with A : a ) apply (TTL) to that property. Thisgives us a maximal chain, say B.
By the assumption of (ZL), &try chain in X is bounded fromabove. Thus, there is rs e X satisfyingr < 16 for all r e B.The element 16 is a maximal element of X. If not, then thereis 11 e X such that ro I 11 and ro I rt. This leads to acontradiction with maximality of the chai n B , since the chainB U {"t} would be strictly greater than B. n
61
62
0. L4 Lecture L4
Tychonoff theorem
First of all, Iet us recall some elementary background on topo-logical spaces.
Given a set X, d collection r q P(X) of subsets of X is calleda topology on X if0) a er,(2)X€r,(3)A,BeriAaBer,(4) A (A, € r) + U A^t € r.
zel telIn this case, X - (X,r) is called a topological space. Theelements of r are called open sets and their complements arecalled closed sets.
A topological space (X, r) is called compact if every familyU g r) such that UU : X, contains a finite subfamily fsuch that U f - X . It is obvious (by using de Morgan's laws)that X is compact iff every family of closed subsets which has
the finite intersection property has a non-empty inter-section. AIso, we shall call by a neighborhood of r e X aset A such that r € U g A for some (J e r.Let {X"}oer be a family of topological spaces Xo: (Xo,ro).
Definition 39 The collectioTl rB g P ("U, X") d,efi,ned,
bat '1
0.74. LECTT]RE 14 63
f € Ts i,ff there erists a fini,te set Ts g T and a familg{Uo),,rro such that f (*) €U* e ro for euery a €To,'is called the base of the Tychonoff topologybase of the product topology) on n X".
a€T
Let Rs be a fo*ily of subsets of X wdth
(- the
n
It is an easy exercise to show that the family r generated
by rs by using of operationstrary unions possesses all ofogy. This topology is called
a€Txoand (g xo,') is c
of topological spaces Xo _ (Xo.,ro). Among examples ofCartesian products of topological spaces let us mention: then-dimensional Eucledian space IRt; the Cantor setC - t0, 1)N, where {0, t} is considered as a discrete space,
i.e., (o,r\ -P({0, 1}); the Baire space NN, where rry - P(N).Our aim now is to prove the following very important theorem.
Theorem 58 (Tychonoff's compactness theorem)If for eaery a e T the space Xo ,is compact then thespace P - II X" 'is also conxpo,ct (relati,ue to the product
aeTtopology).
In the proof of this theorem, w€ make use of the followinglemma.
Lemma 0.L4.Lthe fi,ni,te 'intersect'ion property. Then tlr,ere erists a nlar-'imal fami,ly n g P6) conta'ini,ng Rs wh'ich also has the
fini,te'inter s ect'i o n prop erty .
of finite intersections and arbi-properties (1) (1) of a topol-the Tychonoff topology on
alled the Cartesian product
Proof of the lemma: Remark that the property F of sub-sets of P(X) given by: A e F iff )Af I a for every finiteAt e ,4., is a property of finite character. Thus, the existenceof a maximal family ,R is a consequence of the Teichmiiller -Tukey lemma. IWe shall make use of the following two properties of maximalfamilies with the finite intersection property.
(r) If A,B e R then A) B e R.
Suppose the contrary. Then R U {A n B} does not have thefinite intersection property. Thus Ru{AnB} contains a finitesubfamily with empty intersection. Clearly A) B belongs tothis subfamily. Hence there exists a finite subset R' g,R such
that (An B) n n Y - @ which contradicts the assumptionYeRl
that .R has the finite intersection property since Rt U {An B}is a finite subfamily of R.
(ii) # A g X and A.Y I a for euerEY e R then A e R.
In fact, if A / R then R U {A} does not have the finite in-tersection property. Thus A) n Y - O for some finite
YeRlR' e R. The intersection n Y belongs to n by (i) Hence
yeRl
n Y - Ys e R. Thus, we have AnYo: Z which contra-YeRldicts the assumption that A)Y + @ for every Y e R.
Proof of the Tychonoff theorem: Let .Rs be a familyof closed subsets of P - n Xo with the finite intersection
aeTthere exists a maximal family R,property." By the lemma,
every a € T. We shall prove that f € n Y.YeR
A.74. LECTT]RE 14 65
Ro e R with the finite intersection property. For the proof ofthe theorem, it now suffices to show that n Y + a, where
YeRby Y we denote the closure of Y (i.e. Y - n{C : C is closed
and Y q C))
For arbitrary Z g P,Iel Zodenote the projection of Z inlo Xoand let Ro : {Yq : Y e R}. It is easy to see that the familyRo has the finite intersection property. From the assumptionthat every X" is compact we infer that l-l R" : n Y" + g.
YeRIt follows that there ir / € P such that f (*) € n Y" for
YeR
Take Y € R and a neighborhood U of f . We have to showthatLfnY#s.Let U be the open neighborhood determined by a finite set
^9 g 7 and open sets Go e Xo (o e ,S), wherc f (a) €. Gofor
a e ,S. Letting (Jo: {S , g@) e G"}, we have LI - )rroIf Z is an arbitrary set belonging to .R then f (") e Z"; thusGo)Z" + o. This means that there exists zo e Go such thatfor some function g € Z we have g(a) : zal hence g € (Jo.
Thus for any Z e R we have [Jo) Z + o.
By (ii), it follows that (Jo €.R (since Uo) Z # a for everyZ € ft). Bv (i) , (I : n Uo e R. This implies U )Y +g. Thus every ,r.ighb;.ffood containing / has elements incommon with Y, consequently / € Y for all Y € R, and
bb
hencef e nYlo. Then n Y- n YTo,whatisYeR Yefu YeRo
required. n
0.15 Lecture 15
Filters, measures, and bases
Definition 40 Let X be a non-ernpty set. A fami,Iy f eP (X) 'is sa'id to be a filter on X i,f(L) o / F and X e F;(2)A,BeF+A.BeF;(3) A€F, AeBeX+BeF.
Remark that by the definition any filter F g P6) possesses
the finite intersection property. By the lemma, which wasproved in the previous section, there is a maximal family U gP(X) possessing FIP such that F g [/. The family [/ satisfies
obviously (t) , (2), and (3). Indeed, the properties (1) and (2)
hold true by FIP, and (3) holds true by the maximality of t/.Thus, we obtained the following
Theorem 59 AnU filter possesses an ertens,ion to a rno,tr-
'imal fiIter. n
We denote by F(X) the family of all filters on X. Thus,f (X) g Pe(X)). The family F(X) is ordered by inclusion.
0.75. LECTURE 15 67
Theorem 60 Let Af be a d'irected subset of F(X) (- "fo,euery Ft, fz e Al' there erists Tt € AI such that Fr e Feand Fz e Fz"). Then there erists Fo : sup"A/ e F(X).Moreouer, sup AI - U If .
Proof: It is enough to show that UIf e F$). Obviouslya / U,A/ and X e UI/. Let A, B € UIf then (since ,A/is directed) A, B e F for some F e Af . Hence An B e F.Then AnB e UI/. Let A € U"A/and Ag B g x.Takeany F e Af suchthat Ae F. Then B € f g Uyt/ nTheorem 60 gives us one more proof of Theorem 59. Indeed,
Theorem 60 states that any chain in F(X) is bounded fromabove. Having this, we infer Theorem 59 directly from theZorn lemma.
Theorem 61 Let F e F(X) then the follow,ing cond,i-
t'ions are equ'iualent:
a) F 'is marimal;b)foranyAgX e'itherAe F orx\Ae F.
Proof: a) +b). Assumethat A/f andA":X\A/ffor some A e F. It is clear that A+ o and A" + a since Fis a filter. Consider the following family:
Tt:{C eP(X):AuC e f}.Obviously a / h since A / f and X e -Fr since X e F.If CyC2€ftthen AUCteF andAUCze F. Hence
Au(CrnCz) -'(AuC)n(AuCz) € F. Thus CtfiCz e Fr.
68
If C e FtandCg B qXthen AUC € f and,hence
A U B € F. Thus B € Ft So Fr is a filter that extendsF. Moreover Fr is a proper extension of F since A" / .F butA" e Ft. This contradicts to the maximality of f .
a) =+ b). Let fr e F(X) and F 9fr. Take A e Fr\f .
Then Ac € f by the condition b). Hence Ac € Fr andg
Definition 41 Any filter F satisfyi,ng b), 'i.e., f e F(X)and
(A g X) =+ (A € F) v (A'€ F),'is called o,n ultrafilter.
It can be also proved that a filter F e f 6) is an ultrafilteriff F satisfies:
Au B € f =+ (A € f) v (B € F) (*)
for ev€ry A, B € P(X).
Definition 42 A filter fa e F(X) g,iaen bA Fa - {y gX : A q y) 'is called the principle filter generated by A.AnA fi,Iter whi,ch 'is not pri,nc'iple 'is called a free filter.
It is clear that every principle filter F{o} e F(X), where o €X is an ultrafilter. If the set X is finite then all ultrafilters onX are of form F{ot for some o, € X.Let X be infinite. Consider the following filter fo : {Y g X :
X\y is finite]. Extend fo to an ultrafiIter F. Then .F cannot
A.fi. LECTURE 15 69
be of the form F{o} since {o} e F{o} and X \ {o} € Fo g f .
is not countablyand define
Thus F is a free ultrafilter.
There are many interesting and important applications of free
ultrafilters in mathematics. Let us mention only one of them:the construction of a finitely additive measure on P(N) which
additive. Take a free ultrafilter U € -F(N)
tu''): {l ll 1;1It is left to the reader to show that pu(A) is finitely-additiveand is not countably additive on P(N) .
We finish this section with the following theorem.
Theorern 62 (Otr Hamel bases) Euery uector space
X , that conta'ins at least two elements, possesses a l,inearbas'is (: Hamel basi,s).
Proof: Consider the following property P e P@):
A € P <+ A is linearly independent.
Obviously, P is a property of the finite character. By theTeichmiiller - Ttrkey lemma, there exists a maximal familyin P, say B € P. The family B of vectors of X is linearlyindependent and maximal. Hence there are no proper linearlyindependent extensions of B. This means that B is a basis inX.n
7A
Additional exercises to Lectures 6-15
1. Let (X,, <) b. an ordered set. Prove that any non-emptysubset of X possesses a minimal element iff any chain in X is
well-ordered.
2. LeI X be a set. Prove that X is finite Itr X is not equipollentwith any proper subset of X.
3. Prove that the set Pn"6) of all finite subsets of an infiniteset X is equipollent with X.
4. Prove that the cardinality (- power) of any pairwise disjointfamily of letters 7 on the Eucledian plane is countable.
5. Prove that an infinite linearly ordered set (X, <) has theorder type e) : N if any of its initial segme rft, O(r) - {* eX : r < o) if finite.
6. Prove that any countable linearly ordered set is similar toa subset of Q.
7. Let a € Ord be alimit ordinal and let njm € N such thata * n - a * m. Prove that TL - TrL.
8. Prove that (l.o < ^y.P) + (a < B), where e,) g,^y € Ord.
9. Provethat (o.l < P.7) =+ (a < B), whereo, 0,^f € Ord.
10. Prove that l@, - a + P) A (P + 0)l + (P - r1), wheree,) p,^f € Ord.
11. Prove that l(ta, 0,1 € Ord.
L2. prove that a7+t : Q,0 .dl , where a,9,7 € ord.
0.75. LECTURE 15 7T
13. Prove that @0)t - *0r,, where a,0,1 € Ord.
L4. Let (A,fi) be an ordered set. Prove that there exists alinear ordering L on A such that R g L.
15. Prove that n. o - n if o - N and n > n for al| n€ N.
16. Prove that n2 - n if n is an infinite cardinal number.
L7. Give an example of a cardinal number n satisfying (nn)" -nn.
18. Prove that n* m - n iff n* o. rTt : n where o : N.
19. Prove that -W
<D,=Eiel i€I
20. Provethatm*n-niffk.m*n-nfor aIIk e N.
2L. Assume that cardinal numbers tr1, rtt1 satis* nt ( mt forall t € T. Prove that I fi1
t€T t€T
22. By using of the result of Problem 21, show that for anycardinal n the cardinal nq cannot be represented as a sumoo
I n* of a strictly increasing sequence rr ( flz I1-_1,!_ -f
...ofcardinals.
23. Show that the equationu(t+r) - u(t)+z(s) for all t, s eIR. possesses a solution in the class of all functions u : IR + IR.
24. Show that any order admits an extension to a linear order.
25. Show that there is no countably additive non-zero finitemeasure defined on all subsets of the unit circle of the Eucle-
dian plane.
72
26. Let X be a finite set, F - rr. Show that there are nldifferent well-orderings on X.27. Prove (without using of the axiom of choice) that theZorn lemma implies the existence of a maximal (with respect
to inclusion) disjoint family in any family of non-empty sets.
28. Prove (without using of the axiom of choice) that the ex-
istence of a maximal (with respect to inclusion) disjoint familyin any family of non-empty sets implies the Zorn lemma.
29. Show (without using of the axiom of choice) that lhe Zornlemma implies the trichotomy principle: for every sets
A, B there exists a one-to-one mapping either from A tnto Bor from B into A.
30. Show that the trichotomy principle implies the mappingprinciple: for every non-empty sets A, B there is a functionthat maps either A onto B or B onto A.
31. Show that the mapping principle implies the axiom ofchoice.
32. Show that every vector space with at Ieast two elements
has a basis.
33. An ordered set (L,<) is called a lattice if every subset
{*,AI g L possesses a least upper bound. Prove that a lat-tice in which every chain is bounded from above has a uniquemaximal element.
34. Describe all linearly ordered sets possessing the followingproperty: for every a,b e L - (L,<) the set lo,bl - {r e L:alr<blisfinite.
0.15. LECTURE 15 73
35. Let F gP6) br a filter and,4 U B € F. Show thatthereexistsafilter Ft) .F suchthat Aefror B e Fr.
36. Show that any maximal filter is a prime fiIter by make ofuse the previous problem.
37 . Show that the cardinality of the set of all principle filtersonNisc.
38. Show that any ultrafilter is a maximal filter.:
39. Let X be a finite set, X - n € N. Show that there isexactly n ultrafilters on X.
40. Show that ZL is equivalent to HMD: in any ordered set
there exists a maximal chain.
4L. Show that any filter is contained in an ultrafilter.
42. Prove that the family P(N) contains 2c different ultrafil-ters.
43. Prove that the family P(N) contains o - N differentprinciple ultrafilters.
44. Prove that any family of sets possessing the finite intersec-
tion property (- FIP) is contained in a maximal family withFIP.
45. Let f e P6) b. a maximal family with FIP. Show thatif Z gX suchthat Z)Y la foreveryY €F then Z e F.