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126 CARDINAL NUMBERS EXERCISES 1. Prove Thcorems 87 and 88, 2. Prove Theorems 89 and 90. 8. Prove Theorems 92 and 93, In the next cight exercises, ossumo that m, 1, p and Q are finito cardinals. 4. Provethatifm+n=pthenms< Pp 5. Prove thatn S mA 6. Prove thatifm
0 thenmp< np, 10. Prove thatilmp), or just as well, the order type of the negalive in- tegers under less than; : o 2 is the order type of (Rat, <), where Rat is the set of rstional numbers; . Mis the order type of (Re, <), wwhero Re is tho set of reu numbers. Each of these aovder types may be characterized abstractly (sce, for instance, Sierpinski [1928]. Seo. 4.1 FINITE ORDINALS AND DENUMERADLE SEF'S 129 Jt should ho mentioned that order types may be defined simply in terms of relations. 1 R is a simple ordering, then 2 E = [8:(88, 5) is similar to (58, £)), but none of the difficulties of (1) aro avoided by this approach. mn view of the difficulties with definition by abstraction of cardinal numbeus or order types, it is fortunate indeed that for the special case of ordinal numbers, shich are elassically order types of well-ordered seis, à device may be adopted which requires no special axiorms to support it. The idea is to choose just one representativo of ench order type which is «m ordinal and cal this representative the ordinal. That is, in the case of well-orderings we arc actually able to construct a definite examplo of each possible wellordering, which we cannot do in the case of arbitrary orderings. The definite representatives we choose are built up from the empty set: 1= (0) 2= to, t0)) = f0,15 3 = 40,10), (0, (03)) = 10,1,2) Each ordinal has all smaller ordinals as members, and the membership relation provides the appropriate well-ordering. This construetion is due to von Neumann [1922]. Tts intuitivo adequacy to provide a representative of cach type of welkordering may be explained by the following informal argument, for which 1 um indebted to Dana Scott. Let R bo a wellordering of 4. We want to give a method of associating a new welkordering relation with Z that does not depend on the particular nature of the clements ot 4. Put another way, we want to define & function, say f, on A which will yield an approprinte definite yepresentative matching the ordering due to R. “The definition procecds us follows. Let «o be the Rirst element of 4. We sel Tao) = 0. Let a, be the next element of 4 under R. We set Ja) = ao) = to). Similarly, for a; Ha) = (Ha), Hay) = to, f0)3, and for any integer à Han) = UHlao), Had «o fla). 132 FINTEE ORDINALS AND DENUMERADLE SETS Cmap, 5 But sinco 4 is complete, every member of z is a member of 4 and (1) reduces simply to: B=iz veel, that is, B=-a, and z is a member of 4. QE.D. THeonuy 5. If 4 and B ore ordinals, then d C Bifandonlyifà EB. vroor. IF 4 C B then since À is complete, by the preceding theorem, AEB IE AEB then sinceBiscomplte AC B. QED. TeEoreM 6. very member of an ordinal is an ordinal. ynoor. Let 4 be an ordinal and BC 4. Since A is complete, BC A and thus 85, which is a subset of &4, connects B, Morcover, since 84 well-orders 4, it well-orders B, and hence is transitive on B. Thus if we have: : z84y&y 84 B, we infer x &4 B, thatisa C B, whence y C E, and Bis complete. Hence, B is an ordinal. Q.E.D. Proofs of the next two theorems are left as exercises. The first one is slightly dificult. TazoRmm 7. If A and B are ordinals thencilher ACBoBCA. 'Trrorem 8. If A and Bare ordinals then exactly one of the following holds: AEBBEAA=B, Tuzonsu 9. If Bis a set of ordinals then UB is an ordinal. »roor. Since B is a set of ordinals and members of ordinals are ordinais (preceding theorem), in view of Theorem 8, &UB comnects UB, To show that UB is complete, let € € UB. Then there is an ordinal D such that, Ce Dand DE B. Sinco Dis complete, C E D, and by virtue of Thcorem 62 of 82.6 from DE B wo infer DC UB, and thus by transitivity of inclusion we conclude: CC UB. Q.E.D. This proof illustrates the typical, straightforward technique for proving. that some set is an ordinal: prove that the set is connected by the member- ship relation and that it is complete. We define strict and weak less than for sets in terms of membership, «and also striet and weak greater than. Sec. 5.1 FINTIE ORDINALS AND DENUMERADLE SETS 133 DEFINITION 5. OD ABifandonlyifBBUÍ(B) EA). The attempt to prove the existence of an infinite sot of objects has a rather bizarre and sometimes tortured history. Proposition No. 66 oL Dedekind's famous Was sind und was sollen die Zahlen?, fest published in 1888, asserts that there is an infinite system. (Dedekind's systems correspond to our sets) His proof is such a beautiful combination of mathematical reasoning and vague cpistemology that I give a free trans- lation of it here: rroor. My world of ideas, Le. the totality S of all things which can he objects of my thought, is infinito. For if s is am clement of 8, then the idea sº, that s can be an object of my thought, is itsclf an element of S, TF one consider the lattor the image g(s) of the element 5, then the hereby defined mapping g o! S has the property that the image 8º is a part of 8; and indeed S' is a proper part of 8, becnuse there are in É clements (e.g, my individual ego) which are dilferent from every such idea 5! and hence are not contained im 8”, Finally, itis evident thatif o and d are dilfcrent elements 05, then their images o” and bº ate also different; consequently the mapping is 1-1. There- fore S is infinite.* Dedekind's ptoof depends, of course, on using his definition of infinito systems (or sets), but what is interesting is his excursion into the cpis- temology of ideas.f In no sense does this proof satisty modern canons. Subtler arguments are to be found in Russell [1903, $339], but Russel later conceded that his arguments too were fallacious.jt *A similar asgument is to be [ound in Bolzano [1851, 413]. +Epistemological criticiams of Dedekind's view are to be found in Russell (1920, pp-:.; 1839-140]. +fRussell boldiy begins $339: “That there arc infinite classes is so evident that it will searecly be denied. Since, however, it is capable of formal prof, it may be as well to prove it.” Sec. 5.2 FINITE ORDINAIS AND DENUMERABLE SETS 139 As far as 1 know the first unequivocally clear recognition that such an axiom is needed is to be found in Zermelo's epoch-makihig paper of 1908.* Zermelo"s formulation is essentially the following: Goes &(vBBcA>[B) EA). He construeted the natural numbers as 0, (0), [fON, (ffO!)I,..., which approach does not as easily generalize to the construction of infinite ordinals as does the one adopted in this book. A few years after Zermelo, Whitchead and Russell postulated an axiom of infinity im Principia Mathematica, and there is an interesting discussion in Ramsey [1926] of whether or not their axiom of infinity may be regarded as a logical truth. Turning back to systematic considerations, we first define Lhe set w of all natural numbers and then leave as an exercise proof of the theorem that « is non-empty. The axiom of infinity is essential for the proof. Deriwrrion 8. w= fá: Aíisanatural number). We then have: Tanonem 23. Ago ifandontyif A isa natural number Tt will be convenient for the soquel to use « to give a “set” formulation of the principle of induction. Tenormm 24, Tf (1) OCA, (ii) forwerynifncAliennicaA, then oC We turn now to our main task of justifying definition by recursion. Tor funetions of onc argument, wc want a theorem something like: For every object « and every set Q' there is à unigue F such that () Fisafunctonono, GD 70) = «, Gi) for every a, Fl) = PÇ). Tt is important to notice that x need not be a natural number; it may be any individual or set (we usc “object as a neutral term), and G need not be a function, nor if it is a function nced it include F(n) ip its domain. Of course, when G is defective in one of these respects, then HF (mn) = 0, by virtue of Definition 39 of $3.4, *Moreover, Zermelo [1909] was the first to recognize that elementary number theory could be developed without the axiom of infinity. 140 FINETE ORDINALS AND DENUMERADLE SETS Cear. 5 Eor functions F of two arguments we want: For any sets G and H there is a unique P such thai for every m and n () Fisafunctionon o X q, Gi) Fm, 0) = Him), (im) (Gm, ni) = Gm, Pim, DD). Note again that G and H need not be functions, although it is natural in defining the standard operations by recursion to have & and HH be simple funetions mapping natural numbers or paits of natural numbers into natural numbers. For example, if É is meant to be the operation of addition, we sct Hm) =m since m+0=m, and (1) OHCorês Fon, n)D) = Pleno. However, (1) is not quite correct; because the successor symbol does not designste a function in our set-thcoretical universo, we do not know divectly lhat G is a proper set-thcorctical function. But this difficulty is easily remedied by the technique of defining a fragment, of the intuitive successor function, corresponding to what we did earlier in the case of identity and membership. Drrinrrion 9. GS = ((B,B): BE 4). - The expected theorem holds: Tuzorkm 25. (BB)ESAOBCA. For simplicity of notation in this section, we further define: Derinrron 10, S = So. Thus G(1) = 0 and G(1) = 2, where we define, as already indicated: Derinirion li. 1= (0), 2= (0,10). And we replace (1) above by Gm, FlGrn, 0) = BUG, 0). We also define at this point n—1 and n—2 for any ordinal. Sec. 5.2 FINITE ORDINALS AND DENUMERABLE SETS 141 DerinrrioN 12. O) IA =08A is an ordinal then A-l = Bifand only if BL= A () AXO8A = 1 &A isan ordinal then A-2 = Bifand only if (BOI = A. Rather than state separate thcorems for functions of one argument, functions of two arguments, ete., we may assert a gencral theorem on recursion for functions of» arguments, In the thcorem we use the notation (2) tmo, ma, Ma 1) for r-tuples, which we now consider. First we define: Derinsrion 13. avisar rtupleof A if and only if xe Ar. In other words, an s-tuple of A is a function from the set of natural numbers Jess than » to the set 4, To justify the notation “tz, y) for couples (i.e, 2-tuples) when this notation has already been used for ordered pairs, we have the following theorem, whose proof we leave as an exercise: Tueorem 26. A?=ÀAXA under the function g such that for any fe 4%, 6) = (0), HD) Because A! is equipollent to 4 X 4, we shall use the same notation for couples and ordered pairs, and this ambiguity will be the source of no diMiculty. In fact, in intuitive developments of set theory it is common to “identify” the two. Without using Lhe notation (2), our fundamental theorem on recursion can be given the following somewhat unintuitive formulation: For any seis G and H ond any r > O ihere is à unique F such that (D Fisafuncion on ar, Gi) Jforeeryf,iffco andfr— 1) =0Othen - FO) = HO ]r— 1), (ii) Jor ceryf and everyn, fSCo and fr— 1) =nthen Pr DU fr 1 nD) = GGlr- DU lt — 1, POD. To reformulate the theorem in a schematic fashion after the manner of (2), we need; Derinrrion Sceema 14. If to... CA then (ro. Ze) =f ifandonyiffisanriuplega & HO = m&...& fr — 1) = ty Also, we modify in a standard manner the notation f(x). 14d FINITE ORDINALS AND DENUMERABLE SETS Cuar, 5 To show that w? is the domain of F, suppose that p is the smallest natura] number such that for some 3, (m, 7) is not in the domain of 7. Then in view of (5), p = 0, and thus (m, p — 1)isin DF, and H € A, but then also (0) PU u Era, 2), Elm, Tlm, p — De A. We may conclude from (10) that tm, p) E DF, contrary to our supposition, We conclude that «? is the domain of F. Finully, we leave the simple proof that F is unique as an exercise, Q.E.D. With this fundamental general theorem proved concerning the existence of a unique lunetion P satisfying the definiens of a recursive definition, we may define the standard arithmetical operations without individual justifying theorems. We simply pick the appropriate functions G and TF. We first define addition of natural numbers by the recursion already indicated. We use the same symbol “+ for cardinul and finite ordinal addition (and later for arbitrary ordinal addition), bnt in any given context it will always be clear which is meant. It may be noted that the symbol for finite ordinal addition: designates a set-theoretical entity, in particular, a certain set of ordered pairs, the first member of each pair being a couple. In contrast, the symbol for cardinal or general ordinal addition does not designate any set. Derinition 16. + =fif and onhy if: () fisafunction on «?, Gi) for every m SGm, 0) = m, (ui) for every m and n Fim, ni) =S((m, n)). To obtain the usual notation, we define: Derisrrion 17. mtn=pifandonyifQmn,p e. As immediate consequences of Definition 16 we have: THEOREM 28. OD m+0=m, O) mtal= (mA The familiar commutative and associative laws of addition are asserted by the next pair of theorems. Their proofs illustrate typical uses of mathe- matical induction (via Theorem 24). Note Lhat the corresponding theorems for cardinal arithmetic did not need to be proved inductively. + Tenorgu 29. mpn=ntm Sec, 5.2 FINITE ORDINAIS AND DENUMERABLE SET'S 145 proor. We need two preliminary results: (D 0+n=n [045] mtn=(mAdl, each of which wc prove inductively, using Theorem 24. To prove (1), we define: (65) Z=tu0+n=al, and we want to show that Z = «, that is, cvery natural number belongs to Z. First, by virtue of Thcorem 28 0cZ. Next assume that n € Z, that is, [64] 0+tn=n. Then, we have: Orat=(0+n)! by Thcorem 28 =! by (2). Henceifr € Z then n! € Z and thus by virtue of Theorem 24 Z=w (since it follows from (1) that Z E « and from Theorem 24 that w € 2). Now to prove (II) we define: Alm) = (n: min = (mn). By two applications of Thcorem 28 m+O=m= (m+O! and thus Dc Am). Now, assume that n € A(m), that is, (3) mtas (mn Then, mta (mta by Theorem 28 =(mAmblo by) = (mn) by Theorem 28, 146 FINITE ORDINAIS AND DENUM ERABLE SETS CHar, 5 and thus a! € A(m) whenever n € A(m). Whence by Theorem 24 A(m)-= q, which establishes (II), and completes our proparation for the main businoss at hand. Wo define: Bm =fmm+rn=n+m). First, wo have: 0+n=n by (1) =n+0 by Theorem 28, and therefore Dc B(m). Now assume m € B(n), Lhat is, (4 “ompacatm Then mitn=(m+ml by (ID) Am by(g ="r+ml by Theorem 28, whence ml € B(n) whenever m € B(n), and we conclude that Btn) = q, the desired result. Q.E.D. The proof of the associative law for addition is similar in structure. Tirorem 30. (mty)tp=m+t(n+p). rroor. We define: Am =tolmen)+p=m+(r+p). Now (rt +o m+mn by Theorem 28 m+td(nr+oO) by Thcorem 28 again, and therefore 0 € Am, n). Let p bein A(m,n). Then (45) mm) +p=m+(n+op), Sec. 5.2 FINTIE ORDINAES AND DENUMERABLE SKYS 147 and we have: (mn) +pl=((mAn-+p by Theorem2s Guto to by (1) =m+(g+p! by Theorem 28 m+ (n+ pl) by Theorem 28, ] n whence pl€ A(m, n), and we may conclude: Atm,n) = q QED, We state without proof a familiar thcorem. Tazorey 31, Ifm < n then there is a unique natural number p such thatm+p=n. We now turn to the definition of multiplication. If we were procceding without sct thcory, we would add to P6 and P7 two more axioms: Ps. Ifxisa natural number then x -O = 0. P9. Jfxandy are natural numbers lhenv-gl= (2) + 2. These two axioms accurately forecast the defmition we use. Derinirios 18. = fifandonhy if: 6) fisa function on w?, (1) for every m Tm, 0) =0, (ii) for every m and n Fm, nl) = fm, n)) + ma Liko Definition 17 we havo: Derinrrion 19, mn = pifand onty if (em, n)p) E + When no confusion will result we designate multiplication by juxtaposition rather than the dot. We leave proofs of the following three thcorems as exercises, Trrorzm 32. mn = nm, TmroREM 33. m(n+p) = mn + mp. Treorux 34. (mp = m(np). A number of further theorems arc stated as exercises. 150 FVINITE ORDINALS AND DENUMERABLE SETS Cuar, 5 18. Give a recursive scheme for and then explicitly define; (a) the Inctorial operation n! ; (b) the Fibonacci sequence mentioned at the end of the section (by slight, reformulation of th recursion given for the sequence it may be brought within the scope of Theorem 27). 19. Prove the familiar elementary facts about the exponentintion operation for natural numbers. 20. Frove lixercises 4-12 of 84.4 for natural numbers. 21. Prove 'Theorom 35. 22. Prove Theorem 36. 28. Prove Theorem 37. $ 5.3 Denumerable Sets. A denumerable set is one which is equipol- lent to the set of natural numbers, Such a set affords the simplest example of an infinito set. We begin with some general theorems about infinite seis. Darinsmox 23. A set is infimite if and only df at is not finito. The first two thcorems may be easily proved by using results in $4.2, TrrorEM 38. If À is infinite and A = B then B is infinite. Tazorem 39. I/AC Band A is infinite ihen Bis infinite. Of somewhat more interest is the foliowing thcorem which provides a necessary and sufficient condition for a set to be infinito. THroREM 40. A sef À às infinite 1f and onty af for every natural number n there is a subset of A equipollent to a. rRoor, [Necessity]. By virtuc of the axiom gchema of separation there isa set C such that for every n nECeonecu&(IBBCAÂÃ&B = n). We show by induction that C = «o, Since the empty set is a subset of every set, ec. Now assume that 2 € €. Then by hypothesis there is a subset B of A such that B Because À is infinite, B x A, forif À = B then 4 =n and À would be finite (Theotem 35). Let x therefore be an element of 4 = B. Then BuÍs| CA and elearly in vicw of Theorem 4 of $4.1 Bufr| = avhence nl g O. sec. 5,3 FINITE ORDINALS AND DENUMERABLE SETS 151 [Suficiency!. Suppose if possible that A is finite. Then (Theorem 35), for some 2 A xa, But by hypothesis there must be a non-empty proper subset B of 4 such thai B=nl Letxc B, then B- (3) =, whenee A is equipollent to ane of its proper subscts, namely, B — fx), and this contradicts Theorem 4608842 QED. Using this thcorem and some carlicr results it is casy to prove: Twrorem 41. The set wo of natural numbers is infinite, À more difficult theorem is the following, which expresses a necessary and sufficient condition for a set to be finite. The proof of sufficiency is the difficult part becauso it involves definition by recursion of à function (Thcorem 27). fTmmnonem 42. 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