the ordinal-least-upper-bound of any set of cardinals is a cardinal; for any set I, and cardinals λ is a cardinal.

Transcription

1 11 Regular cardnals In what follows, κ, λ, µ, ν, ρ always denote cardnals. A cardnal κ s sad to be regular f κ s nfnte, and the unon of fewer than κ sets, each of whose cardnalty s less than κ, s of cardnalty less than κ. In symbols: κ s regular f κ s nfnte, and for any set I wth I < κ and any famly A of sets such that A I < κ for all I, we have A < κ. I (Among fnte cardnals, only, 1 and 2 satsfy the dsplayed condton; t s not worth ncludng these among the cardnals that we want to call "regular".) To see two examples, we know that ℵ s regular: ths s just to say that the unon of fntely many fnte sets s fnte. We have also seen that ℵ 1 s regular: the meanng of ths s that the unon of countably many countable sets s countable. For future use, let us note the smple fact that the ordnal-least-upper-bound of any set of cardnals s a cardnal; for any set I, and cardnals λ for I, lub λ s a cardnal. I Indeed, f =lub λ, and <, then for some I, <λ. If we had, then by I <λ, and Cantor-Bernsten, we would have λ, contradctng the facts that λ s a cardnal and <λ. Thus, f <, then, whch means that s a cardnal. The condton of regularty can be stated n the followng equvalent manner: the nfnte cardnal κ s regular ff (1) for each set I wth cardnalty less than κ, and for each famly I of ordnals, each less than κ, we have that lub < κ. I 12

2 Indeed, assumng that κ s regular, and assumng that I < κ and < κ for all I, we have that < κ, thus, lub = < κ ; but, for any cardnal κ, I I and any ordnal, < κ mples that < κ (why?); thus, lub < κ follows. I Conversely, assume that (1) holds, and show that κ s regular. In partcular, κ s a lmt ordnal. Let I <κ and A <κ for each. Let λ = A ; λ s an ordnal less than κ. Hence, by (1), λ def = lub λ < κ. As we noted above, λ s a cardnal. Let I µ=max(λ, I ) ; snce λ, I are both cardnals less than κ, µ s a cardnal <κ. But then I µ and A =λ λ µ. Hence, f µ s nfnte, we have A A µ µ=µ < κ ; I I 1 and f µ s fnte, then A I s fnte, and so agan A < κ. I Ths shows that κ s regular. In fact, we can formulate regularty n terms of ordnals, wthout (overtly) referrng to cardnals. Let us say of a lmt ordnal δ that t s regular f t satsfes (1') for any ordnal less than δ, and for each famly ι ι < of ordnals ι, each less than δ, we have that lub ι < δ. ι < Note that f κ s an nfnte cardnal, then t s a lmt ordnal, and f t satsfes (1), then t satsfes (1'), snce <κ mples that <κ. Conversely, I clam that f the lmt ordnal δ satsfes (1'), then t s a regular cardnal. δ s nfnte. Assume (1') for such a δ. Let <δ. If we had δ, then we would have an ndexng ι ι < of all ordnals less than δ, and so, also usng that δ s lmt, lub ι =lsub ι = δ, contradctng (1'). Ths shows ι < ι < that δ s a cardnal. But now lookng at formulaton (1), f we have an arbtrary set I wth 13

3 I <δ, we can ndex I by an ordnal <δ, and transform any nstance of (1) nto one of (1'). We have shown that an equvalent defnton of "regular cardnal" s gven by the condton (1'), together wth the condton that δ s a lmt ordnal. + We have a large class of regular cardnals. Let us wrte κ for the least cardnal greater than κ. If κ=n ω, then κ + =n+1, and f κ=ℵ, then κ + =ℵ +1. A successor cardnal s one of the form κ +. The fact s that all nfnte successor cardnals are regular. The reason s that f I + + < κ and A < κ for I, then A κ, and I κ, thus A I I A I κ κ κ = κ < κ. 1 On the other hand, e.g. ℵ ω = lub ℵ n s not regular (we say sngular for not regular). n<ω Namely, now ω = ω = ℵ < ℵ, and for each n ω, ℵ ω n = ℵ n < ℵ ω, but ℵ = ℵ n ω = ℵ ω ; thus, the famly ℵ n n ω s a counter-example to the regularty n ω condton for ℵ ω. We have now seen that the cardnals ℵ and ℵ +1 are regular. We just saw that for =ω, ℵ ω s sngular. Are there lmt cardnals, that s, cardnals of the form ℵ δ for a lmt ordnal δ that are regular? One sees that f so, then we must have ℵ δ =δ : the reason s that the set {ℵ :<δ} s cofnal n ℵ δ : lubℵ = ℵ δ (why?); therefore, f δ< were the case, we <δ would have a contradcton to formulaton (1') of the regularty of ℵ δ. However, f we take the smallest δ for whch ℵ δ =δ, κ=ℵ δ turns out to be sngular; thus, the above does not have a converse! Consder the followng sequence: κ = ℵ κ = ℵ ( n<ω ) n+1 (κ ) n defned recursvely. We have κ <κ 1 snce <ℵ. By an easy nducton, we show that κ n <κ n+1 for all n<ω. Let us defne λ = lub κ n. I clam that ℵ λ = λ. We have n<ω 14

4 ℵ λ = lub ℵ = lub ℵ κ snce λ=lub κ n (exercse: show that f for <λ n<ω n n<ω <γ, and γ=lub, then lub = lub ); but ℵ κ =κ n+1, and so I I <γ n ℵ λ = lub κ n+1 = λ. However, λ=ℵ λ s not regular: we have λ = lsub κ n and ω<λ. n<ω n<ω It turns out that, usng the usual axoms of set-theory, one cannot prove that there are regular lmt cardnals. A regular lmt cardnal s called a weakly naccessble cardnal. A strongly naccessble, or smply naccessble, cardnal κ has, by defnton, the addtonal property that λ for all λ<κ, we also have that 2 <κ. It s easy to see that lmt cardnals wth ths latter property are exactly the ones of the form beth δ for a lmt ordnal δ ; these latter are called strong lmt cardnals. Thus, an naccessble cardnal s the same as a regular strong lmt cardnal. In secton 13, we wll take a look at naccessble cardnals -- despte the fact that we cannot prove ther exstence wth the axoms so far lsted. In fact, the exstence of naccessble cardnals wll be seen as a reasonable new axom of set theory. There s an mportant calculaton of the cardnalty of a partcular knd of set nvolvng regularty. We consder the followng stuaton. We have (B, ) where () B s a set and () s a famly = f I of partal operatons on B, that s, each f s a J J functon wth doman contaned n B, dom(f ) B for a partcular "arty" J (an arbtrary set, gven for each I ; usually, J s a natural number) and range(f ) B. As a remnder, J B = {f: f s a functon, dom(f)=j, range(f) B }. ) We are gong to wrte f: A B to mean that f:dom(f) B and dom(f) A. J Thus, the partal J -ary operaton f may be dsplayed as f : B B. 15

5 Examples. (A) The noton of group s a very mportant one n mathematcs. A group s a set 2-1 B together wth a bnary operaton :B B, a unary operaton ( ) :B B, and a -ary operaton e B, where these data are requred to satsfy the denttes -1-1 (x y) z = x (y z), e x = x e = x, x x = x x = e. Ths fts our setup -1 above, wth I = 3, J = 2, f =, J = 1, f = ( ), J =, f = e (note that a -ary operaton f:a A s bascally the same as a dstngushed element of A : A ={ }=1, and to gve a functon f:{ } A s the same as to gve the element e=f( ) A ). In ths case, all operatons are total, that s, dom(f ) = B 2, etc. A smlar example s provded by each of a number of other knds of algebrac structure: lattce, Boolean algebra (these we wll see n some detal later), rng, feld, etc. (In fact, n the case of a feld, dvson s a genunely partal operaton.) (B) The noton of lmt n analyss can be consdered as a partal operaton. If x ω s an nfnte ( ω-type) sequence of reals, x = x n n<ω, then lm x = lm x n may not exst, n but f t does, t s unquely determned. Let lm: ω be the partal operaton whose doman s the set of those sequences of whch the lmt exsts, and whose value, at such a sequence, s the lmt of that sequence. In ths case, the arty of lm s the nfnte ordnal ω. Lm can also be defned on ( ) ω. That s, f f ( ) ω,.e., f f s a sequence f = f of functons f : n n<ω n from the reals to the reals, then lm f s the functon f: for whch f(x) = lm f n (x) ; lm f s defned ff the lmt n lm f (x) exsts for all x n ("pontwse lmt"). n Below, we wll consder the systems (,lm), (,lm) as examples of (B, ) of the general stuaton; n both cases, I = 1 = {} and f = lm. 16

6 Let us return to a general "system" (B, ) as above. Relatve to (B, ), a closed set s any subset X of B for whch J x X dom(f ) f(x) X J whenever I, x B. For nstance, f (B. ) s a group, then a closed set s what s usually called a subgroup of the gven group: a subset of the (underlyng set of the) group -1 whch s closed under, ( ) and contans e. In the "lmt" examples, closed sets are what are usually called closed: closed under lmts (n the functon-space example, ths s only one possble knd of closedness; there are also other "topologes" on that space). It s clear that the ntersecton of any famly of closed sets s closed (exercse; the empty ntersecton s taken to be B tself). Therefore, for any subset A of B, we may take the ntersecton of all closed sets contanng A, and call t the closure, or -closure of A ; - -( ) notaton: A, or A. - In the example of type (A) (groups, Boolean algebras,...), A s called the subalgebra generated by A ; n the examples under (B), t s called the closure of the set A. - What we want to do s calculate the cardnalty of A, n terms of some cardnaltes gven wth the stuaton. Let us denote: λ = max(ℵ, I ), def µ = max(ℵ, A ) def (that s, λ = I unless I s fnte, n whch case λ = ℵ ; smlarly for µ ), and κ = the least nfnte regular cardnal for whch def κ > J for every I. 17

7 We use the notaton <κ ρ ν = lub{ν : ρ < κ, ρ any cardnal} def We have, wth these notatons, the estmate formulated n the next - Theorem. Wth (B, ) any system of partal operatons, we have, for the cardnalty A of the closure A of a subset A of B, that - <κ A max(λ,µ) (2) wth κ, λ and µ as determned above. - - Of course, we always have that µ A (why?); the two estmates determne A qute sharply (see also more on ths below). <κ Before we turn to the proof of (2), we note three propertes of the operaton ν. One s <κ ν ν (for κ 2 ), (3) whch s obvous; the other s <κ κ ν (for ν 2 ); (4) the thrd s <κ ρ <κ (ν ) = ν ( κ nfnte regular, ρ<κ, ν 2 ). (5) 18

8 ρ ρ <κ For (4), note that ρ < 2 ν ν for every ρ < κ ; but κ s the least cardnal > all Cantor's <κ ρ < κ ; t follows that κ ν. (5) s slghtly more nvolved. We clam that ρ <κ ρ µ (ν ) = { (ν ) : µ < κ} ; here, we have made a dstncton between exponentaton "of sets" and that of "cardnals"; µ means the set of all functons ρ σ ; ν s a cardnal (n partcular, an ordnal), the result of cardnal exponentaton appled to the cardnal-arguments ν and µ. In ths dsplay, on the <κ left sde, we have the set of all ρ-type sequences of ordnals < ν ; on the rght, we have a unon of sets of ρ-type sequences. ρ σ The clam s clearly the same as to say that for any ρ (ν <κ ξ ξ<ρ ), there s µ < κ ρ µ µ such that (ν ). But, for each ξ<ρ, ν ξ for some µ < κ, snce ν <κ ξ ξ<ρ ξ ξ s the unon of the sets ν µ, µ < κ. Snce κ s regular, ρ < κ and each µ ξ < κ, we have that lub µ ξ < κ (see the second defnton of regularty). Wth µ = lub µ ξ, µ s a ξ<ρ ξ<ρ µ cardnal, ν ξ ν µ for all ξ < ρ, and thus ν µ ξ for all ξ < ρ. We have that ρ µ (ν ) as we wanted. ξ ξ<ρ On the bass of the last clam, we have <κ ρ ρ <κ ρ µ (ν ) = (ν ) = { (ν ) : µ < κ} as requred for (5). µ ρ µ ρ <κ <κ (ν ) = ν κ ν = ν µ<κ µ<κ µ ρ=max(µ, ρ)<κ (4) - To prove (2), we gve a formula for the set A tself frst. Defne, by recurson on the ordnal 19

9 < κ, the sets A A as follows: = A A = A f < κ s a lmt ordnal; and < J A = A {f (a) : I, a dom(f ) A }. +1 Clearly, A A B whenever < < κ. We clam that - A = A. (6) <κ For one thng, the rght-hand-sde s contaned n the left-hand-sde: by nducton on < κ, we show that A - A. The nducton-steps for = and lmt are obvous. For = +1, note that snce A - A (nducton hypothess), and snce A - s closed, f (a) - A J whenever I and a dom(f ) A ; thus, by the defnton of A, A A as desred. Secondly, we need to show that the unon A s closed; snce A - s the least closed set <κ contanng A, and the unon contans A, t wll follow that A - A, whch s the other <κ contanment we need. To show that the unon s closed, let I, and J a = a ( A ) dom(f ) ; we want that f (a) j j J A. Snce <κ <κ a - j A, there s j < κ such that a j A. We now have the famly j j J of j ordnals less than κ. By the defnton of κ, we have that J < κ ; also, κ s regular. Therefore, by the second verson of the defnton of "regular", = lub{ j :j J } < κ def 11

10 (second use of regularty of κ!). Then, of course, a A for all j J, snce A A. Thus, j j J a dom(f ) A, and by the defnton of A, f (a) A, and as a consequence, f (a) A. <κ Havng the formula (6), t s easy to estmate - A. By nducton on < κ, we show that <κ A max(µ,λ). By (3), ths s true for =. If the asserton holds for all < < κ, and s a lmt ordnal, then A <κ <κ A max(µ,λ) = max(µ,λ) < < <κ <κ <κ κ max(µ,λ) = max(κ,max(µ,λ) ) = max(µ,λ). 1 (4) J Fnally, for = +1, the set U = {f (a): I, a dom(f ) A } has cardnalty def <κ J <κ ρ I (max(λ,µ) ) λ (max(λ,µ) ) nd. hyp ρ = def J <κ <κ <κ = λ max(λ,µ) = max(λ,max(λ,µ) ) = max(λ,µ). (5) (3) <κ <κ <κ Snce A = A U, A max(λ,µ) + max(λ,µ) = max(λ,µ) as desred. Our nducton s complete. Fnally, 111

11 - A A <κ = A κ max(λ,µ) = <κ <κ max(κ,max(λ,µ) <κ = max(λ,µ) <κ (4) as wanted for (2). Ths completes the proof of the Theorem. <κ If, n partcular, λ µ and µ s of the form ν, then <κ <κ <κ <κ <κ ρ <κ <κ max(λ,µ) = µ = (ν ) = sup (ν ) = sup ν = ν = µ. Thus, we ρ<κ ρ<κ (5) obtan: <κ - f A = ν for some ν 2, and I A, then A = A. <ℵ Note that for all nfnte ν, ν = sup ν n = sup ν = ν. Thus, f κ = ℵ, then for all n<ω n<ω nfnte A, we have that A <κ = A, and we conclude that - f κ = ℵ, A s nfnte, and I A, then A = A. Let us turn to the examples above. In the example of type (A), we have that κ = ℵ, because all operatons are fntary, that s, the artes J are fnte cardnals. Also, λ=ℵ. Thus, f (B, ) s an nfnte group (say), and A B s any nfnte subset of B, then A -, the subgroup generated by A, s of the same cardnalty as A. If B = ν, then for any µ ν, we may choose a subset A B such that A = µ (why?); we thus obtan that an nfnte group has subgroups n each nfnte cardnalty less than or equal the cardnalty of the group. Of course, the same concluson holds for Boolean algebras, etc. ℵ <κ In the examples under (B), we have κ = ℵ. Now, ν = ν. Also, now 1 112

12 λ = max(ℵ,1) = ℵ. Wth (,lm), we get a trval concluson, namely that - ℵ ℵ ℵ ℵ - ℵ A max(ℵ, A ) (2 ) = 2 (trval snce A, and = 2 ). But, ℵ - ℵ wth (,lm) we obtan, smlarly, that A, A 2 mply that A 2 as well, whch s "non-trval" now snce see later...) ℵ 2 = 2 (for the cardnaltes related to, Here s another example. Let us fx a postve nteger n. The class (n fact, set) of Borel subsets of n s the least subset X of n ( ) whch contans all open sets, and whch s closed under countable unon and complementaton: whenever A X for all ω, then A X n -A X A X ; I ℵ (Borel sets are mportant n measure theory). We have that =2 : the number of Borel ℵ sets s 2. (It mmedately follows that there are non-borel sets; why?) Ths result comes out of the above general theorem n the followng way. Let A be the set of all open subsets of n. Let B= n, I={, 1} ; J =ω ; J =1 ; f :B ω 1 B s the operaton of countable unon: f ( A )= A ; f :B ω 1 B s complementaton: f 1 (A)=B-A. Then t I s clear that s nothng but A, =A, the closure of A n the gven system of operatons. ℵ ℵ Clearly, λ=ℵ. We know that A =2 ; thus, µ=2. κ = ℵ 1, the least regular cardnal above J =ℵ. Therefore, ℵ <ℵ ℵ ℵ ℵ ℵ ℵ 1 max(ℵ,2 ) = (2 ) = 2 = 2. The most obvous feature of the theorem s that n the estmate for the cardnalty of the 113

13 closure A, the cardnalty of the set B does not play any role. And ndeed, B can be of any sze, n fact, even a proper class. The verson of the theorem when B s a (possbly proper) J class, rather than just a set, s almost the same as the orgnal; the defnton of the class B remans the same; now, each f s possbly a proper class; the only change s that now we cannot talk about the collecton of the operatons f ( I) n the same sense as before; but we take a class whose elements are of the form (, a) for I, wth I a set, and J stpulate that f = {a: (, a) } s a Functon whose doman s a subclass of B. def The proof of the theorem remans esssentally the same. Fnally, we refne the dstncton between regular and sngular cardnals by ntroducng the cofnalty of a cardnal κ ; as t wll turn out, a cardnal s regular f and only f ts cofnalty s tself. We formulate the basc noton n terms of ordnals. For every ordnal, we have that = lsub ; s the least strct upper bound of all ordnals less than. It may happen < that can be wrtten n the form = lsub (7) ξ ξ<γ for a system of ordnals ξ ndexed by an ordnal γ smaller than ; the least ordnal γ for whch there s a representaton of the form (1) of s called the cofnalty of, and t s denoted cf() ; we have just seen that cf(). When =+1 s a successor ordnal, then we can take γ=1 and = ; ths shows that cf()=1 ; the cofnalty of all successor ordnals s 1. Let us note that f (7) holds, then we can change the ξ to some ξ such that the sequence ξ ξ<γ s (not necessarly strctly) ncreasng: ξ<ξ <γ mples that ξ ξ ; and (7) stll holds, wth some γ γ : = lsub. (8) ξ ξ<γ To ths end, we defne =lub ξ ζ for all ξ<γ. It s clear that the ξ are ncreasng, and ζ ξ 114

14 also that ξ for all ξ<γ. If n fact < ξ for all ξ<γ, then = lsub ξ lsub ξ ξ<γ ξ<γ and so (8) holds wth γ =γ. If, on the other hand, there s ξ<γ such that =, then we ξ let γ be the least such ξ ; then < for each ξ<γ, and thus ξ = lub ξ lsub ξ, ξ<γ ξ<γ and (8) agan holds. Note that f, n addton to the above, γ=cf(), then γ s necessarly equal to γ, snce γ <γ and (8) would contradct the "least-ness" of γ. Thus, f γ=cf(), the we have a representaton (7) of n whch the ξ are ncreasng. The characterzaton (1') of regular cardnals shows that (9) for a lmt ordnal, cf()= mples that s a cardnal, n fact a regular cardnal. In fact, more generally, (1) for a lmt ordnal, cf() s always a regular cardnal. Ths follows from the dentty cf(cf())=cf() (11) that we prove as follows. Let γ=cf() and δ=cf(γ). We have equaltes = lsub( ξ ) and γ = lsub(γ ζ ) ; ξ<γ ζ<δ and as we sad, we may assume that ξ ξ whenever ξ<ξ <γ. But then = lsub( γ ) : ζ<δ ζ 115

15 ndeed, clearly, s a strct upper bound of all the γ ( ζ<δ ); and f <, then there ζ s ξ<γ such that ξ (why?); and then there s ζ<δ such that ξ γ ζ, thus ξ γ, ζ and so γ -- whch shows that s a the least strct upper bound of all the ζ γ ζ ( ζ<δ ). But then, by the defnton of γ=cf(), t follows that γ δ ; that s, (9) holds. Now, (1) follows from (11) and (9). Exercse: Let γ=cf(). Then there exsts a representaton of n the form of (7) n whch the are strctly ncreasng: < whenever ξ<ξ <γ. ξ ξ ξ We have defned the cofnalty of any ordnal, and n partcular, of any cardnal; and we saw that t s always a regular cardnal. For nstance, f s a countable lmt ordnal, then cf()=ω ; ths s because cf(), and cf() s an nfnte cardnal, and there s only one such, namely ω=ℵ. What we have thus seen s that every countable lmt ordnal can be wrtten n the form = ls ub n, and we may also assume that the n here are n<ω strctly ncreasng. For an ordnal wth =ℵ, the possbltes for cf() are three: t can be 1 (when 1 s a successor), t can be ℵ, and t can be ℵ. Obvously, cf(ω +1)=1, 1 1 cf(ω +ω)=ℵ and cf(ω +ω )=ℵ (we may wrte ω for ℵ when "we use t as an ordnal"; smlarly for ω ). Exercse: show that f s a lmt ordnal, then cf(ℵ )=cf(). Thus, we have cf(ℵ ω )=cf(ω)=ℵ, cf(ℵ ω )=cf(ω 1 )=ℵ 1, and 1 cf(ℵ ℵ )=cf(ℵ ω )=ℵ 1. ω

16 12 Models of the axoms of set theory Let us lst the axoms of set theory ( X, Y,... are varables for classes, x, y,... are varables for sets): Every set s a class. Every element of a class s a set. Axom of Extensonalty: x(x X x Y) X=Y. Axom Schema of Class Comprehenson: for any meanngful property P(x) of sets x, there s a (unque) class X such that x(x X P(x)). X s denoted by {x:p(x)}. Axom of Regularty: all sets are pure: x X([ y(y X y X)] x X ). Axom of Empty Set (AES): the class def = {x:x x} s a set. Axom of Par Set (APS): for any sets x, y, the class {x, y} = {z:z=x or z=y} s a set. def Axom of Subset (AS): any subclass of a set s a set: X x mples that X s a set. Axom of Unon Set (AUS): for any set x, x = {y: z.y z & z x} s a set. def Axom of Power Set (APoS): for any set x, (x) = {y: y x} s a set. def Range(F). Axom of Replacement (AR): for any Functon F, f Dom(F) s a set, so s Axom of Infnty (AI): the class = {x: X([ X & y(y X S(y) X)] x X} s a set [here, def S(x) = {x, {x}} ]. 117

17 (Global) Axom of Choce ((G)AC): there s a Functon C wth Dom(C)=V-{ } such that for all x Dom(C), C(x) x. The axom system we descrbed above s called the Morse-Kelley (MK) class-set theory. To be sure, our formulaton s, n one respect, less precse than the offcal verson of MK: the term "meanngful" n the formulaton of class-comprehenson should be made more specfc to read "formulated n frst-order logc"; we wll say more about ths below. On the other hand, there are two other, related, formal axomatzatons of set theory: Zermelo-Fraenkel (ZF) set theory, and Godel-Bernays (GB) class-set theory. As these terms ndcate, n ZF, the only knd of entty s "set"; n GB, as n MK, we have both sets and classes. The formulatons of both ZF and GB depend more senstvely on the concepts of frst order logc than that of MK. Although MK and GB may seem related, on account of both beng a theory of sets and classes, and ZF may seem separate snce t only talks (drectly) about sets, as a matter of fact, ZF and GB are very closely related, beng essentally of the same deductve power, whereas MK s stronger than the prevous two. Henceforth, we assume the axoms of (class-)set theory, n the formulaton gven above (that s, MK). We want to nvestgate what subclasses, and preferably, subsets, of V, the unverse of (all) pure sets, may serve as models of all or part of the axoms of set theory. We take a subclass A of V ; the ntenton s that A should play the role of V. We re-nterpret the noton of "set" as to mean: beng an element of A. To be a class should mean: beng a subclass of A. Now, the very frst axom: "every set s a class", requres that x(x A condton that we called the transtvty of A. x A), whch s the Henceforth, we assume that A s a transtve class. For the "new" classes, the classes n the sense of the model gven by A, one beng an element of the other should retan ts orgnal meanng, the orgnal -relaton. In summary, what we nvestgate s ths. Gven a transtve class A, and gven certan axoms of set theory, we ask whether t s the case that, under the nterpretaton n whch " a s a set " means " a A ", " a s a class" means " a A ", and " a belongs to b " means " a b ", the axoms are true. The model we are lookng at conssts of the subclasses of A ; the 118

18 elements of A, the "sets" n the model, are partcular subclasses of A. Furthermore, we want to nvestgate the sad questons wthn our (MK) class-set theory, that s, usng those axoms freely n the total unverse of sets and classes. Let us wrte (A) for the totalty of all subclasses of A ; the "model" we are talkng has (A) as ts doman of ndvduals. To be sure, when A s a proper class, (A) s not a class ; ts "elements" are sometmes proper classes. Thus, t s not possble to talk drectly about (A) wthn our class-set theory. Stll, there s no problem wth the meanng, wthn MK class-set theory, of sayng, about any partcular statement referrng to classes and sets, that t s true n the model (A). When the statement contans a quantfer X on classes, the understandng s that ths should be nterpreted as X(X A...) ; and the quantfer X should be read, when meant n the sense of the model provded by A, as X(X A &...). Note that X(X (A)...) s ntutvely the same as X(X A...), snce X (A) s ntutvely the same as X A, even though X (A) s not somethng meanngful wthn MK. On the other hand, when A s a set (and ths s n fact the case we are mostly nterested n), then there s no problem of the knd descrbed above at all. Now, (A) s the same as the set (A) ; and we can wrte X (A) just as well as X A. If a partcular axom s true n the model, we say the class so, we wrte (A) Φ when Φ stands for the axom. (A) s a model of the axom; f The frst few observatons tell us that, under the stated (meager) condtons, (A) s automatcally a model of the axoms "Every element of a class s a set", Extensonalty, Class Comprehenson, and Regularty. The frst, "Every element of a class s a set", when nterpreted n the model that "every element of a subclass of X s an element of X ", whch s certanly true. (A), says The second, Extensonalty, says, n (A), that X=Y ". (1) "f X, Y are subclasses of A, and f for all x A, x X ff x Y, then Ths holds snce each of x X A or x Y A mples that x A, and thus 119

19 "f for all x A, x X ff x Y " s the same as "f for all x, x X ff x Y "; and so, (1) becomes an nstance of Extensonalty understood n the orgnal unverse of sets and classes. The thrd, Class Comprehenson, says that for any meanngful property P(x) of sets x n A, there s a (unque) subclass X of A such that x A.(x X P(x)). But, gven P, we can consder the property Q of sets n general defned by Q(x) P(x) & x A. Consder X = {x: Q(x)} = {x: x A & P(x)} = {x A: P(x)} def gven by Class Comprehenson n the orgnal unverse. By ts defnton, X A, therefore, X s a "class n the sense of the model A ". But also, x A ( x X P(x) ), whch shows what we want. Ths smple proof s, alas, msleadng. The root of the trouble s that we do not really have a clear dea what a "meanngful property" s. Later on, we wll learn about frst order logc n explct detal, and nterpret "meanngful property" by "a property expressble by means of frst order logc n terms of the concepts of set, class and membershp". In fact, although we do not yet have theoretcal defntons concernng frst order logc, we do have a workng knowledge about t. When nterpretng Class Comprehenson n the model (A), we wll encounter 12

20 the stuaton llustrated by the followng example. Consder the property P(x) " x s a natural number". Ths s gven by P(x) X([ X & y(y X S(y) X)] x X). P s not qute expressed n terms of the prmtves "set", "class", and "membershp"; but ths can be easly remeded, by usng u = v.v u and z = S(y) w(w z (w y or w=y)). The phrase X can therefore be replaced by u(( v.v u) & u X) : X u(( v.v u) & u X) ; smlarly, S(y) X) z( w(w z (w y or w=y)) & z X). 4 4 Usng these defntons, we have P(x) X([ u(( v.v u) & u X) & 12 y(y X z( w(w z (w y or w=y)) & z X))] x X). (1') P now s expressed n terms of the sad prmtves. Now, consder the case that we want to take the class of natural numbers n the model A. Ths may be sad to be gven by the property Q(x) x A & P(x), the one we used above. 121

21 But there s another nterpretaton, one that s the really ntended one. Namely, what we really want s, more explctly, the class of sets n A that are natural numbers n sense of the model A. How can we descrbe the property R(x) x s a natural number n the sense of the model A? (2) Ths has a natural answer: nterpret each of the quantfers X, u, v, y, z, w as quantfcaton over ""class" n the sense of the model (A) " (n the case of X ), and ""set" n the sense of the model (A) " (n the case of the rest). The result s that we should replace the above quantfers by the respectve expressons X A, u A, v A, y A, z A, w A. Of course, wrtng X A means X.X A u A means u.u A &...; v A means v.v A...;... ; etc. The result of the sad replacement s the rather large expresson R(x) X A([ u A(( v A.v u) & u X) 12 & y A(y X z A( w A(w z (w y or w=y)) & z X))] x X) (3) (A) ( (A)) Let us wrte P (x) (although we should really wrte P (x)...) for ths property R(x), and call t the relatvzaton of P to the model (A). 122

22 Unfortunately, the notaton P (A) (x) s, stll, slghtly msleadng; P (A) (x) s gven only when the expresson for P n frst order logc s gven; what f the same property P can be gven by two dfferent expressons whch, when undergong relatvzaton, gve non-equvalent expressons for the relatvzatons? When dealng wth logc n earnest, we wll have to face ths problem; for the tme beng, we gnore t. We have concluded that the property R n (2) s P (A) (x). Thus, when we want to see that, n the sense of the model take s the class (A), the class of natural numbers exsts, what we have to {x A: P (A) (x)} rather than {x A: P(x)}. Of course, there s no problem wth the exstence of the class n queston n the model (A) ; the sad expresson denotes, agan, a subclass of A. Fnally, the valdty of the fourth of the four axoms mentoned above, Regularty, s left as an exercse to verfy. Next, we dscuss the meanng of each of the next seven axoms: those of Empty Set, Par Set, Subset, Unon Set, Power Set, Replacement and Infnty n the full model based on the (transtve) class A. Let us call these, to have a smple term, the set-exstence axoms. Note that each set-exstence axom asserts that a certan class, formed as a comprehenson-term on the bass of some data satsfyng some condtons, s a set. In other words, n each case, we have a property P(w), a property P for an undetermned set w, P formulated n terms of some gven data; the asserton s that {w:p(w)} s a set. To be sure, P s not always formulated usng only the prmtves of set theory ("class", "set" and "membershp"), but we can remedy ths as was llustrated above. Now, accordng to what we sad above, the correspondng set-exstence axom, " {w:p(w)} s a set", when nterpreted n 123

23 the sense of the model have that (A), s to mean that, provded the data are n the model A, we (A) {w A: P (w)} s an element of A, (4) where we referred to the relatvzaton P (A) (w) of P descrbed above. We are to see what ths last statement (4) means, for each of the seven set-exstence axoms. In many cases, we wll establsh that (5) for all w A, P(w) P (A) (w). When (5) holds, we say that the property P s absolute (wth respect to relatvzaton to A ). Under the condton that P s absolute, (4) becomes (6) {w A: P(w)} A. A qualty of P that s stronger than (5), but whch, nevertheless, s often present, s that (7) for all w V, P(w) P (A) (w). In ths case, (A) (8) {w A: P (w)} = {w: P(w)}, and (4) becomes (9) {w: P(w)} A. Now, to the ndvdual set-exstence axoms. AES : now, P(x) s x x. Ths contans no quantfers; thus P (A) (x) s just the same as P(x). Therefore, (7) trvally holds. Hence, to say that (A) AES s to say that {w: P(w)}, that s,, s an element of A. Snce A s transtve, t s easy to see that ths 124

24 always holds provded A s non-empty. We summarze: (A) AES A. APS : we are gven x, y A ; the property P(z) s z=x or z=y. The same argument apples as n the prevous case to show that (7) holds. Thus, (A) APS means that, under the condtons that x, y A, (9) holds. The concluson s that (A) APS A s closed under takng par-sets: x, y A {x, y} A. AS : we are gven a set x A, and a class X A such that X x ; the property P(w) s w X ; note that " X s a set" s the same as " {w: w X} s a set". The same argument apples as n the prevous two cases to show that (7) holds. Concluson: (A) AS A s closed under takng subsets of ts elements: x A, y x y A. (A) AUS : we are gven x A ; P(y) s z.y z & z x. Therefore, P (y) s z A.y z & z x. I clam that (7) holds. The fact that P (A) (y) mples P(y) s tautologous. Gven that P(y), that s, y z & z x for a sutable z, snce x A and A s transtve, we nfer that z A ; ths shows P (A) (y). In concluson: (A) AUS A s closed under takng unon-sets: x A x A. APoS : We are gven x A ; P(y) s y x z.(z y z A.(z y z x) ; P (A) (y) s z x). For any y A, P(y) f and only f P (A) (y) ; ths s mmedate from the transtvty of A. We have shown (5), the fact that the subset-relaton s absolute. (Note, however, that we have not shown (7).) The term n (6), 125

25 {y A: P(y)} = {y A: y x)} = (x) A. We have (A) APS x A (x) A A. Note, on the other hand, that under the condton that A AS, that s, x A, y x we have that (x) A = (x) ; thus, y A, f (A) AS, then (A) APS A s closed under takng the power-set of ts elements; x A (x) A. Also note that f (x) A, then (x) A (transtvty), hence, (x) A = (x) ; f for all x A, (x) A, then (A) APS. AR : As a prelmnary remark, let us note that for F A, that s, for F a class n the model, to say that F s a Functon n the sense of the model s the same as to say that F s a Functon: the concept of Functon s absolute. Ths wll be seen by nspectng the equvalence F s a Functon x F. y x. z x.[ w x(w=y or w=z)) 1 & u y[( t y.t=u) & u z & v z. s z.(s=u or s=v)]] 2 21 & x F. y x. z x[ w x(w=y or w=z)) 1 u y[ t y.t=u [ v z[ s z.(s=u or s=v) x F. y x. z x.[ w x (w=y or w=z )) [ t y.t =u 5 [ v z [ s z (s =u or s =v ) v=v ]]]]]]] (the frst part says that every x n F s an ordered par (u, v)={y={u}, z={u, v}} ; the second part says that f x=(u, v) F and x =(u, v ) F, then v=v ). 126

26 Indeed, snce each quantfer n the expresson s bounded,.e., of the form a b or a b, and F A, the expresson, when relatvzed to the transtve class A, wll not change ts meanng: x F.Q(x) relatvzed to (A) s x A[x F Q(x)] ; and x A[x F Q(x)] x[x F Q(x)] x F.Q(x) snce F A ; smlarly, y x.q(x, y) relatvzed to y A[y x Q(x, y)], and (A) s y A[y x Q(x, y)] y[y x Q(x, y)] y x.q(x, y) snce x A and so x A. Next, consder the fact that v Range(F) x F. y x. z x.[ w x(w=y or w=z)) 1 & u y[( t y.t=u) & u z& s z.(s=u or s=v)]] 2 21 Wrtng P(v) for the rght-hand-sde, we see that (7), the stronger verson of absoluteness holds. We may wrte that Range (A) (F), the Range of F n the sense of the model (A) (A), s Range(F) ; Range (F) = Range(F). A smlar nspecton wll show that Dom (A) (F) = Dom(F). Now, AR n the model means that F A & F s a Functon (A) & Dom (A) (F) A (A) Range (F) A. Under the hypothess that F A, we can drop the superscrpt A from all three postons, and we obtan that (A) AR f and only f for all classes F A, F s a Functon & Dom(F) A Range(F) A. If, n addton, (A) APS, that s, A s closed under takng par-sets, then we have that 127

27 x, y A mply that (x, y) A, and therefore, X A, Y A mply that X Y A. Thus, f F s a Functon, and Dom(F) A, Range(F) A, then F Dom(F) Range(F) A. Ths shows that n the last equvalence, we can drop the condton " F A ", and we obtan: f (A) APS, then (A) AR f and only f for all classes F, F s a Functon & Dom(F) A Range(F) A. AI : We wll ndcate the proof of the fact that f (A) AES, APS and AUS, then (A) AI ff A. Assume that (A) AES, APS and AUS. Accordng to what we showed above, ths means that A, and x, y A mples that {x, y}, x A. We frst clam that we have A. In fact, we have the stronger statement that V A. Ths ω s left as an exercse to prove. We show that (A) = ; that s, for the property P(x) that defnes the class as {x:p(x)}, we have the relaton (7), and as a consequence, (8). Recall that P(x) was (A) descrbed n (1'), and P (x) n (3). The proof of the ncluson (A), that s, x A & P(x) P (A) (x), s left as an exercse. For the converse ncluson (A), the man pont s that X n the quantfer X A n (3) can be nstantated by X=, snce A ; therefore, f x A s such that x (A), that s, P (A) (x), then the statement (...) after the quantfer X A n (3) 1 1 s true for X= ; the facts A and y A S(y) A can be appled to show that [...] 2 2 n (3) true (ths expresses that X= satsfes X & y(y X S(y) X) n the sense of (A) (A) ); now, x follows; ths shows. 128

28 Fnally, we turn to the axom of choce. GAC : We show that f (A) APS, then (A) GAC (of course, here t s assumed that A s a transtve class, and that GAC holds n the unverse). In fact, f C s a global choce functon, one whose exstence s asserted by GAC as read n the unverse, then C A s a global choce functon n the model (A). Indeed, by the concept of Functon beng absolute (see the dscusson of AR above), we have that C A s a Functon n the sense of (A). By two uses of the fact that (A) APS, t follows that a A-{ } and b a mples that (a, b) A. Hence, t s mmedate that Dom (A) (C A) = Dom(C A) = A-{ }. Of course, when a Dom(C A), then (C A)(a)=C(a) a. Ths proves what we clamed. Let us draw some conclusons from our analyses. We see that the condton for each of the seven set-exstence axoms s related to a closure condton. Let us elaborate. Let us call a predcate P(x, X) of a set-varable x and a class-varable X a closure-predcate f t s monotone n ts second varable: for all x, X and Y, P(x, X ) & X Y P(x, Y). A class X s sad to satsfy the closure condton assocated wth the closure predcate P = P(, ), or, X s closed under P, or agan, X s P-closed, f for all x, P(x, X) x X. The man fact s that, for a closure-predcate P, the ntersecton of any number of classes closed under P s agan closed under P ; f Q(X) s a predcate on classes, and 129

29 X(Q(X) X s P-closed) {x: X(Q(X) x X)} s P-closed}. The proof of ths s easy, but very mportant; t s left as an exercse. (At ths pont, I menton that the present formulaton of the noton of "closure predcate", wth the last-stated property, was gven by Peter Green.) In partcular, the ntersecton of all P-closed classes s P-closed. Let us note the easly seen fact that f P and P are closure-predcates, then P(x, X) 1 2 defned as P(x, X) P (x, X) & P (x, X) s agan a closure-predcate. 1 2 Now, let us defne P (x, X) x= AE P (x, X) y X. z X. x={y, z} APS P (x, X) y X. x= y AUS P (x, X) y X. x y AS P (x, X) y X. x= (y) APoS P (x, X) F.(F s a Functon & Dom(F) X & x=range(f)) AR P (x, X) x=. AI Inspecton shows that each of these predcates s a closure-predcate. We have X s P -closed AES X ; X s P -closed y. z(y, z X APS {y, z} X) ; X s P -closed y(y X AUS y X) ; X s P -closed y. x((y X & x y) AS x X) ; X s P -closed y(y X APoS (y) X) ; X s P -closed AR F(F s a Functon & Dom(F) X Range(F) X ) ; X s P -closed AI X. Above, we saw that, for any transtve class A, beng P AES -closed s equvalent to (A) AES ; smlarly for the axoms APS, AUS and AS, n relaton wth the 13

30 closure-predcates P, P and P, respectvely. The stuaton wth APoS s APS AUS AS dfferent; the condton on X s y(y X (y) X X) s not a closure condton (exercse). However, we saw that, under the condton that A s P -closed, (A) AS APS ff A s P APS -closed. The rest of the axoms are smlarly clarfed n terms of closure condtons. Always assumng that A s transtve, we have that If A s P -closed, then (A) APS AR ff A s P AR -closed; f A s P - and P -closed, then (A) AI ff A s P -closed. AES APS AI Note beng transtve s also a closure condton: X s transtve X s P TR -closed where P (x, X) y(y X & x y). TR Let us denote the conjuncton of the eght closure-predcates P, P, P, P, P, P, P and P TR AES APS AUS AS APoS AR AI by P SE, for "Set Exstence" ( P TR s taken as a conjunct n P SE too); P SE s the set-exstence closure predcate. We conclude the followng 131

31 Proposton. For any class A, SC(A) s a transtve model of Morse-Kelly class-set theory f and only f A s closed under (the) set-exstence (closure-predcate P ). SE It s worth rememberng that, on the bass of our analyses, we have smlar statements for a varety of combnatons of the axoms of set theory. 132

32 13 Inaccessble cardnals In the last secton, we characrerzed the classes A that gve rse to a model (A) of Morse-Kelley class-set theory by closure under the combned set-exstence predcate P. SE Our frst man goal n ths secton s to relate the same condton to the noton of naccessble cardnal. (1) Theorem For any class A, (A) s a model of all axoms of MK class-set theory f and only f ether A=V or A=V for an naccessble cardnal θ. θ By the last secton, an equvalent statement s ths: (1') Theorem For any class A, A s closed under P f and only f ether A=V or SE A=V for an naccessble cardnal θ. θ To establsh the Theorem n ts second form, we prove some prelmnary facts. (2) Lemma. Suppose that s an ordnal, and for each <, B s a set such that B B and B < B for all < <. Then B = lub B. (3) < < Proof. We prove the asserton by transfnte nducton on the ordnal. Suppose that Ord, and for all <, the asserton wth replacng holds. If =, then both sdes of the equaton are equal to. Assume that = +1. Ths case s also trval; on both sdes, we have "maxma". In more detal: lub B = B ; on the other hand, B = B ; and so < < 133

33 B = B =lub B. < < For the rest of the proof, let be a lmt ordnal; let B < be a system of sets as stated n the hypotheses of the Lemma. Let κ = B (<). We have κ <κ for < <. Let us defne the sets C = B for each <. Note that C = and C =B for def γ +1 γ< all < ( < mples that +1< ). Clearly, we have C C for < <. (4') Another fact s that for a lmt ordnal < (f there s any such), we have that C = C = B = B = C ; (4) γ γ+1 γ γ γ< γ+1< γ+1< γ< For (4') and (4), we say that the system C < s contnuous. Now, let any ordnal <, and let λ = C def. When =γ+1, then λ = B γ =κ γ. Now, let < be lmt. Snce the restrcton of the orgnal system B γ γ< to ordnals γ<, the system B γ γ<, satsfes the same assumptons as the orgnal, by the nducton hypothess apples and we can conclude that, for a lmt ordnal <, λ = C = C = lub κ = κ = κ = λ. (5) γ γ γ γ+1 γ γ< γ< γ< γ< γ< On the cardnals λ now we have the nequaltes λ <λ for < <, and also the contnuty relaton (5). Fnally, note that and lub λ = lub λ = lub κ, +1 < < < C = C = B +1 < < < 134

34 Therefore, f we can show that?: C = lub λ, (6) < < the desred equalty (3) wll follow. What we have done so far was to reduce the general case of the lemma for a lmt ordnal to the case when the system of sets nvolved s contnuous. Let us extend the defnton of the sets C and the cardnals λ to = by C = C, and λ = lub λ = λ. Thus, (4) and (5) hold for = too. < < < By recurson on, we defne a bjecton f :λ C, such that, n addton, we also have that for < <, f f (that s, f (ξ) = f (ξ) for ξ dom(f ) ). The fact that λ = C gves a bjecton f :λ C. (In fact, f =.) Suppose that we have defned f, and +1< ; we defne f. I clam that +1 C -C =λ. Indeed, f C -C =µ, then λ +µ= C + C -C = C =λ But ths s possble only f µ=λ : f we had µ λ, then we would have +1 λ +µ=λ λ ; and f µ λ, then λ =λ +µ=µ ; ths shows that the clam s vald By an dentcal argument, λ -λ =λ Therefore, we have that the sets C +1 -C and λ +1 -λ are equnumerous, and thus we can fnd a bjecton g:(λ -λ ) +1 (C +1 -C ). Fnally, put f +1 =f g ; that s, for ξ λ, f +1 (ξ)=f (ξ), and for x λ -λ +1, f +1 (ξ)=g(ξ). Clearly, f +1 :λ +1 C +1, and f f +1, and as a consequence, f γ f +1 for all γ<+1. When s a lmt ordnal, we defne f = f γ. Snce we have that f γ f γ for all γ< γ<γ <, and each of the f γ s a 1-1 functon, we have that f s a 1-1 functon. The 135

35 doman of f s dom(f )= γ λ γ =λ (see (5)). The range of f s γ< γ< ran(f )= γ C γ =C (see (4)). Therefore, f s a bjecton, γ< γ< f :λ C. The constructon of f also ensures that f f for all γ<. γ Ths completes the recursve constructon of the functons f for all wth the stated propertes. Snce we have f :λ C, we have proved (6) as desred. (7) Corollary. V ω+ =beth for all Ord. Ths follows from the lemma, by an easy transfnte nducton (exercse). (8) Proposton () For any Ord, V s closed under P TR, P AUS and P AS. () For any >, V s closed under P AES. () For any >ω, V s closed under P AI. (v) For any lmt ordnal, V s closed under P APS and P APoS. (v) Let θ be a strong lmt regular cardnal (an naccessble cardnal). Then V θ s closed under P AR. Proof. (): We have that V s transtve, x V x V, and y x V y V ; the frst of these facts was ponted out before; the other two are left as easy exercses (hnt: use transfnte nducton). () and (): obvous. (v): exercse. 136

36 (v): Frst, let us note that beth θ =θ. Snce θ s a strong lmt cardnal, θ=beth for some lmt ordnal ; of course, θ. If we had <θ, then θ would be sngular: θ = beth = lub beth, wth each beth <θ. Therefore, we must have =θ. Thus, < beth =θ. θ Next, I clam that for all x V, we have x <θ. From x V, t follows that x V θ θ for some <θ ; therefore, x V, and x V beth by (7), and so x <beth =θ. θ Now, we can show that V θ s closed under P AR. Let F be a Functon such that D def = Dom(F) V θ and Range(F) V θ. For every x D, we have F(x) V θ, and therefore, there s x <θ such that F(x) V. We have the system of ordnals x, wth each less than θ, ndexed by the set D whose cardnalty s D x x D x <θ (snce D V θ ). By the regularty of θ, we conclude that def = lub x < θ. Snce x x D mples that V V, we get that F(x) V for every x D. Ths says that x Range(F) V, and Range(F) (V )=V +1 V θ as desred. The last proposton mples the "f" part of the Theorem: f A s V or V θ for an naccessble cardnal θ, then (A) s closed under the sad eght closure condtons. But we also get, for nstance, that (9) Proposton Let A=V for a lmt ordnal δ>ω. Then (A) satsfes all δ axoms of set theory except possbly the Axom of Replacement. (1) Lemma () V, but V ; V Ord = ; V () V = (V ). < () Defne the so-called rank functon r:v recurson on the well-founded relaton as follows: <. Ord by r(x) = lsub r(y). y x 137

37 We have that r(x) Ord, and r(x) s the least such that x V. (v) x V (v) x V r(x) <. r(x) V. Proof: (), (): exercses. (): We prove that for any Ord, x V r(x) ; ths wll suffce. We proceed by -nducton on x. Thus, we take x, and assume that for any y x and Ord, y V r(y). We have: x V y x.y V y x.y (V ) < ( ) y x. <.y (V ) y x. <.y V y x. <.r(y) y x.r(y)< nd.hyp. def. of r(x) lub r(x). y x (v): x V x (V ) <.x (V ) <.x V < <.r(x) r(x)< () (v): x V r(x)< r(x) V. (v) () Proof of the "only f" part of (1') Theorem. Assume that condtons n queston. I assert that (A) satsfes the closure 138

38 (11) x A mples r(x)+1 A. We prove ths by -nducton on x. Suppose x A, and for all y x, we have r(y)+1 A. We have r(x)=lsub r(y) = lub (r(y)+1). Defne the Functon F to y x y x have doman equal to x ( A ), and to satsfy F(y)=r(y)+1 for each y x ; F(y) A for all y x ; that s, Range(F) A. Therefore, by A beng P AR -closed, Range(F) A. Usng that A s P -closed, we have AUS Range(F) A. But Range(F) = (r(y)+1) = lub (r(y)+1) = r(x) ; y x y x thus, r(x) A. Snce A s closed under par-sets (under P APS ), and unon-sets (under P AUS ), u A mples S(u) A. Therefore, r(x)+1 A as asserted. Next, I clam that (12) A Ord V A. Ths s done by nducton on Ord. For =, the asserton s true snce A s P AES =closed ( A ). For a successor ordnal, we use that u A mples (u) A (closure under P APoS ), and for a lmt ordnal, we use that A s P AR - and P AUS -closed; the detals are easy. I can now demonstrate that A = V. (13) A Ord Suppose frst that x A. Then, by (11), def = r(x)+1 A Ord. Snce x V (see (1) Lemma (v)), x belongs to the rght-hand sde. Suppose x belongs to the rght-hand sde: there s A Ord such that x V. By (1) Lemma (v), r(x)+1. By A beng transtve, r(x)+1 A. By (12), V A, hence V A ( A s transtve). Snce x V, x A follows. r(x)+1 r(x)+1 r(x)+1 Consder the class θ=a Ord. Snce A s transtve, θ s a transtve class of ordnals. 139

39 Ether θ=ord, or there s some Ord-θ. But n the latter case, for all >, Ord-θ, by the transtvty of θ ; therefore, Ord-θ Ord-, and θ ; thus, θ s a set, a transtve set of ordnals; hence, θ s an ordnal. We see that ether θ=ord, or θ Ord. Let us also note that, n case θ Ord, θ s a lmt ordnal. Ths s because <θ A +1 A +1<θ. By (13), A = V. When θ=ord, A = V. Otherwse, snce θ s a lmt ordnal, θ A=V θ. We show that, n ths case, θ s a strong lmt regular cardnal. Recall that κ s a strong lmt cardnal means that κ=beth δ for a lmt ordnal δ. Ths s the same as κ>ℵ and for all λ, λ<κ, λ a cardnal mples 2 λ <κ. Frst, for the regularty of θ, n the "purely ordnal" formulaton (') n 11. θ s a lmt ordnal. Suppose <θ, and ι <θ for each ι <. By the defnton of θ as A Ord, we have A, and ι A for ι <. Consder the functon F whose doman s A, and for whch F( ι )= ι A. By A beng P AR -closed, range(f)={ ι : ι } A ; thus, lub = ι range(f) A by A beng P AUS -closed. lub ι A mples that ι ι lub ι < θ as desred. ι Fnally, for θ beng a strong lmt cardnal. θ>ℵ follows from A ( A s P AI -closed). Suppose κ<θ, κ a cardnal. Then κ A, and (κ) A ( A s P APS -closed). Let λ= (κ) ; we have some f: (κ) λ. We want to see that λ A, or equvalently, that λ<θ. Suppose otherwse; θ λ. But then, we can let -1-1 X = f [θ]={f ():<θ} (κ) ; by A beng P -closed, X A. We have f X:X θ, dom(f AS X)=X A, and range(f X)=θ A ; from whch by A beng P AR -closed, t follows that θ A Ord=θ, contradcton. Therefore, we must have λ<θ. The proof s complete. Theorem (1') says that for a class beng closed under set-exstence s equvalent to beng the same as V θ, the θth stage n the cumulatve herarchy, for θ a strongly naccessble 14

40 cardnal, or θ=ord, where, by conventon, V = V. Ord Gven any closure predcate, t s natural consder the least class closed under t. Let U denote the least class closed under P SE. What do we know about U? Ether, U equals V (n partcular, U s a proper class), and there s no naccessble cardnal (f there s such θ, V θ V, V θ s P SE -closed, contradctng U beng least such); or else, U s a set, n fact, U s V θ for the least naccessble cardnal θ (there cannot be any naccessble θ <θ, snce then V θ V θ, V θ s P SE -closed (by (1')), and V θ s not least among such classes). Let us wrte θ for the least naccessble cardnal f any. We have (14) Ether there s no naccessble cardnal, and U s V (frst alternatve) or there s at least one naccessble cardnal, and U s V, for θ the least θ naccessble (second alternatve). Whch one of the two alternatves s the case? Whatever the answer, we can prove that (15) Proposton (U) "there s no naccessble cardnal". To establsh ths, we need (16) Lemma Let A be a class closed under set-exstence. For any a A, a s an naccessble cardnal n the sense of the model cardnal (n the sense of the unverse). (A) f an only f a s an naccessble Proof of (15) from (16). Suppose (U) "there s at least one naccessble cardnal". Ths means that there s a U whch s an naccessble cardnal n the sense of the model (U). By (16), a s an naccessble cardnal. Ths excludes the frst alternatve n (14). But under the second alternatve, we have U = V θ, and then a V θ mples that a<θ (see (1)()), whch contradcts θ beng the least naccessble. We have reached a contradcton, whch completes the proof. 141

41 Outlne of the proof of (16). We have that a s an naccessble cardnal Ord(a) & a s lmt ordnal & a>ω (17) F((Functon(F) & Dom(F) a & Range(F) a) Range(F) a) (18) y z G((y a & Ord(z) & Functon(G) & Dom(g)= (y) 1 & Range(G)=z & G s a bjecton) z a) (19) 1 Lne (18) expresses condton (1') n 11. Lne (19) expresses that for all cardnals κ<a, we κ have 2 <a. To be sure, what t says drectly s that for all ordnals y, z, f (y) z, then z<a -- but ths s an equvalent statement of the above, gven that a s a cardnal by the prevous condtons. Remember that Ord(a) a s transtve and trchotome b a. c b.c a & b a. c a(b c or b=c or c b). a s lmt b a. c a(b a) As we saw n the dscusson of AR n 12, the fact that these expressons contan only bounded quantfers, ensure, on the bass that A s transtve, that the predcates n queston are absolute: for a A, (A) Ord (a) Ord(a) ; (A) ( a s lmt) a s lmt Further, we saw n loc. ct. that, for reasons dentcal to the ones just cted, the predcate (A) Functon(F) s absolute, and Dom (F) = Dom(F), (A) (A) (A) Range (F) = Dom (F), F = F for F A. Beng a bjecton s also 142