11!Hí MATHEMATICS : ERDŐS AND ULAM PROC. N. A. S. of decomposiion, properly speaking) conradics he possibiliy of defining a counably addiive real-valu

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1 ON EQUATIONS WITH SETS AS UNKNOWNS BY PAUL ERDŐS AND S. ULAM DEPARTMENT OF MATHEMATICS, UNIVERSITY OF COLORADO, BOULDER Communicaed May 27, 1968 We shall presen here a number of resuls in se heory concerning he decomposiion of a se E in various ways as sum (union) of is subses. These resuls have connecion wih problems on counably addiive measure funcions in absrac ses, bu hey may also bear on he problems of he axiomaics of se heory and generally on foundaions of se heory iself. Some of hese resuls employ he coninuum hypohesis or he generalized coninuum hypohesis. The several _problems which will be presened also pu hese hypoheses in a cerain limeligh. The impossibiliy of defining a counably addiive measure for all subses of a se of power of coninuum (a measure which would vanish for subses consising of any single poin) was firs esablished wih he use of he coninuum hypohesis by Banach and Kuraowski.l Very shorly aferwards, one of us showed he impossibiliy of such a measure for subses of a se of power NI wihou he use of any hypohesis.' The same resul was shown here o hold for ses of higher powers, in fac, for all he accessible alephs. More recenly, hese resuls have been exended o a large class of inaccessibles as well. These resuls show ha his "problem of measure" is closely relaed o fundamenal problems concerning he role of axioms of se heory. Recen developmens have furher clarified hese relaions. Imporan resuls have been obained by Sco, Solovay, Marin, and ohers. The proofs of hese relaions make use of he mehods inroduced by Paul Cohen in proving he independence of he coninuum hypohesis. Boh he resuls of Banach and Kuraowski and he sronger resul of Ulam are obained by exhibiing purely combinaorial schemaa of decomposiion of absrac ses wih cerain properies : B and K show a counable sequence of decomposiions of a se of power of he coninuum, each ino counably many disjoin subses so ha, no maer how one akes a finie number of ses from each of hese decomposiions, he inersecion of all hese finie unions conains, a mos, counably many poins. Sierpinski3 generalized he B and K schema in he following way. There exiss a sequence of decomposiions ino aleph disjoin ses, each so ha if one is seleced from any counably many of hese (no necessarily all), he union of he seleced ses gives he whole of he space, excep perhaps for counably many poins. Decomposiion given by U show, wihou he use of he coninuum hypohesis, he following phenomenon. A se E of power NI can be decomposed counably many imes ino NI disjoin ses in he following way : A "marix" of ses cam be consruced such ha we have counably many rows and noncounably many columns. Ses in each row are disjoin. The anion of ses in any column gives he whole se E excep for possibly counably many poins. As is easy o see, he exisence of such a decomposiion (a sequence 1189

2 11!Hí MATHEMATICS : ERDŐS AND ULAM PROC. N. A. S. of decomposiion, properly speaking) conradics he possibiliy of defining a counably addiive real-valued measure funcion. In his paper we shall show various modificaions of such consrucions. In paricular, we srenghen he resul of Sierpinski. One can decompose E in he nh row ino 2"+ 1 disjoin ses wih he above propery. I is clear ha he impossibiliy of a measure funcion follows because if he measure for hese ses exised, we could selec a se of power less han 1 /s" and heir union would have o measure less han 1. This is a conradicion, since he complemen is counable. This consrucion uses he coninuum hypohesis. Wheher one can do i by using a weaker hypohesis remains an open problem. Should he number of ses in each row be finie and fixed, we show ha one does no need he coninuum hypohesis for his propery o hold (bu of course one does no ge he impossibiliy of a measure funcion from such a "marix"). We shall inroduce a special symbol for decomposiion of ses in such "marix" paerns. Finally, we would like o say ha all he resuls and problems in his paper form only a special aspec of a more general problem which we formulae raher vaguely here : Given a class of Boolean relaionships o be saisfied by unknown ses, all subses of a given se, one wans o find or "consruc" ses saisfying such relaions, which may be counable or noncounable in number. We hope o aack his more general quesion in a paper o be published in he fuure. THEOREM 1. The real line (and in fac every se of power e) can be decomposed in infiniely many ways as he union of k disjoin ses so ha k S = Al (), n = 1,2, m IS- U A 1,(")J < k (1) "~1 for every choice of he ses A,,( "), 1 < l" < k. We prove Theorem 1 wihou he axion of choice. Consider all ses of k - 1 disjoin raional inervals and wrie hem in a sequence {I"0k-1)}, n = 1,2.... The firs k - 1 ses A,0 0, 1 < l < k - 1, of he nh row of our decomposiion marix are he k - 1 inervals of I"<k-1>. A k (" ) is he complemen of he union of he inervals in Now le m U A I.W = F "-1 be a ypical family of ses, one from each row. To prove (1) i suffices o show ha if xl, xz,... xk are any k real numbers, we mus have x, E F for a leas one i, 1 < i < k. To see his, observe ha here is a se of k - 1 raional inervals, say I"1k_1,, which separaes x ir xs,... xk. Bu hen every A, 00, 1 < l < k conains exacly one of he x, or x i E F for a leas one i, as saed. This complees he proof of Theorem 1. I is easy o see ha (1) fails o hold wih k - 1 insead of k, bu we leave his o he reader.

3 VOL. 60, 1968 MATHEMATICS : ERDŐS AND ULAM 1191 THEOREM 2. Le I SI >_ No,, 2 < k, < k.2 :5,..., k. and le k. S= U A <*>, 1-1 n= 1,2,... be a decomposiion of S ino k, disjoin ses. Then here is always an 1,,, 1 < In < k n = 1,2,... so ha IS - U A >_ No. (2) n-1 To prove Theorem 2 we will define elemens x *, 1 < u < - of S (he x*'s are no necessarily all differen, bu here are infiniely many differen ones among hem) and ses A,.(*) so ha x*iza,.(*), 1<u<-,1<n<-,1<1.<k (3) (3) will clearly imply Theorem 2. We consruc he x * and he ses A1,(*) by inducion wih respec o n. Assume firs ha 2 = kl =,... = k, < k,+,. Consider he 2' ses f1á 1<*> U-1 i = 1 or 2. The union of hese ses is S, hus a leas one of hem is infinie, say 2A.,(*)1 n1 = No, (en = 1 or 2). Le x be an arbirary elemen of f1a. (*), pu x = x, _... = x, and A,<*> _ i-1.. m A -,,(*) for n = 1,2,.... Clearly he complemen of UAI.W = UA '-,,(*) a-1 *-1 (which equals f1a..w) is infinie. *-1 Now assume ha we have already succeeded in choosing elemens x,, x=, and ses A,.("), 1 < n < u having he following properies : * * 1S - UA,, ( *)1 > No, S* = S - UA,* ( *) n-1 n-1 (4) x,«a,* ( * ), 1<a<u;1<n<u and finally here are a mos k* - 2 disinc elemens among he x, 1 < i < u. Now we consruc x*+, and A,,+,(*+ 1) so ha (4) and (5) remain saisfied and so ha here are a mos k * - 2 disinc elemens among he x,... X., x*},. Assume firs ha kn+, = k *. Then here are a leas wo ses A,,(*+' ) and Ay*+1> which do no conain any of he elemens x,,...x* (his follows from he fac ha here are a mos k* - 2 disinc x's and he A,(*+' ) 's are disjoin). A leas one of hese, say A,,(*+ 1), has an infinie complemen in S* (i.e., infiniely many elemens of S* do no belong o Pu A,,c*+u A 1*+1<*+u and clearly (4) and (5) are saisfied. Assume nex ha k*+, > k Then here are a leas hree ses A,,(*+' ), (5)

4 1 192 MATHEMATICS: ERDÖS AND ULAM PROC. N. A. S. A,,("+' ), A,, ( "+') which do no conain any of he x i, 1 < i < u. A leas one of he ses, say A,,("+'), has an infinie complemen in S" ; we choose an arbirary elemen x"+1 from his complemen and pu Al,("'" = A.+,c"+u. x"+1 is clearly differen from x l, xg,... x", and (4) and (5) are again saisfied. Thus we have consruced he infinie sequence xl, xs,... and he ses A,,("). (3) is clearly saisfied and i is clear from our consrucion ha here are infiniely many disinc elemens among he x's, hus (2) is saisfied and Theorem 2 is proved. Now we would like o sae he following quesion which we canno solve. PROBLEM I. Le No < ISI < e. Le 2 < kl < ks <..., k" be any sequence of inegers. Does here exis for every n a decomposiion k S = U A,( "~ 1-1 ino disjoin ses so ha for every 1 < l" < k" 1S -- U A,"(") < No? n-1 If c = N1, we will see ha he answer o our problem is affirmaive ; in fac very much more is rue. Bu if we do no assume ha c = Nl, we canno solve his quesion even if we assume ha f S1 = Nl and le our sequence k end o infiniy very slowly. If 181 = c and k" > 2" as saed in he inroducion, we canno expec a posiive soluion wihou some assumpion on he power of he coninuum since his would imply ha here is no real-valued compleely addiive measure on he subses of he eals where poins measure Q. I seems very likely ha if we do no make some assumpion abou he power of he coninuum, hen we canno obain a soluion o Problem I. I may even be possible o show ha if Problem I has a soluion for a fixed sequence k hen a soluion exiss for every such sequence. I will be convenien o inroduce he symbol m --a. (p, q, r, s), which means ha a se S of power m can be decomposed ino he union of p disjoin ses in q ways : S = U A (s), 1 < f < w, 1 <a <~ so ha if we choose any one of he ses A.("') for r differen o, hen S - U A Q, (s' ) 1 < s, (6) m -/ > (p, q, r, s) means ha such a decomposiion is impassible. Sierpinski proved3 ha c (N1, No, No, N1) (7) is equivalen o he coninuum hypohesis. Sierpinski's resul implies ha if we assume c = Ni, he answer o Problem I is affirmaive. We generalized his and proved oher resuls on our symbol, bu do no give hese proofs since Hajnal observed ha all our resuls follow from previous resuls of Erdős,

5 VOL. 60, 1968 MATHEMATICS : ERDŐS AND ULAM 1193 Hajnal, and Milner,4 and Erdös, Hajnal, and Rado.5 We sudied he following symbol exensively : \ q) -' (r r) (8) The meaning of (8) is as follows : Le 181 I = m, I S21 = q ; we spli he pairs (x,y), x E S,, y E Ss ino wo classes. Then here is always a U C S,, V C SR, i u I = s,! V I = r so ha all he pairs (x,y), x E U, y E V are in he same class. (q) -/> (r r) means ha here is a division of he pairs ino wo classes for which such ses U and V do no exis. I is easy o see ha 8 8 (q)-(r r) (9) is equivalen o m -/ > (2, q, r, s). (10) Firs we show ha (9) implies (10). Le IS, i = m, I S21 = q. Le x E S,, 0 E $,, 1 < 0 < w a. The pair (x,,6) is in class I if x E A, (s) and in class II if x E A2(0). (9)implies ha (6) is no always saisfied, hence (10) is proved. Nex we show ha (10) implies (9). In fac we show ha if (9) does no hold, hen m-+ (2,q,r,s). Consider hen a spliing of he pairs (x,p) ino wo classes so ha if U C S,, V C Ss, I UI = s, (V) = s, hen here is always a pair (x,,y,) in class I and a pair (xs,yz) in class II where x, E U, x, E U; y, E V, ys E V. Now we pu x in A, (") if (x,,b) is in class I, and in A= (s) if (x,,6) is in class II. Now we show ha (6) is saisfied. Since p = 2 in (6), a, = 1 or = 2. Wihou loss of generaliy we can assume ha for r values of i, a, = 1. Bu hen if (6) is no saisfied, we would have IS - UA, (I" ) I >- s, (11) i where S runs hrough a se of ordinals V of power r. (11) means ha here is a se U C S,, I UI >- s so ha all he pairs (x,#,), 0, E V are in class II, which conradics our assumpion ha (9) is false ; hence (6) and hus m - (2,q,r,s) is proved. Hence (10) implies (9) and hus he proof of he equivalence of (9) and (10) is complee. Theorem 48 of Erdős-Rado saes ha hus clearly Wi W1 W1 (No) --. (No No), Wi (No) --~ W0 No ( No No)'

6 1194 MATHEMATICS : ERDÖS AND -ULAM PROC. N. A. S. which by he equivalence of (9) and (1U) implies Perhaps N1 -- (2, No, No, No). (12) Na -> (2, N., N,, Ni), bu his is undoubedly very difficul since i is equivalen o one of he mos difficul unsolved problems of Erdös-Hajnal-Rado (problem 12), Na --i N1 Nl (N1) (Wi N1)~ Perhaps N, -o- (2, N1, No, N1), bu his is also very difficul, since (see problem 12) N1 Na (13) --~ N1 N1 (14) No No is also unsolved ; (13) would imply (14), bu (14) also seems very hard. On he oher hand, i follows from Theorem 33 of Erdös-Hajnal-Rado ha Na (2, Ni, No, N~ We will no discuss m -+ (2&r,s) furher, bu refer o reference 5. We sae anoher resul which generalizes Sierpinski's resul : Assume ha 2N1 = N.+,, hen N.+1 -i (N.+,1 N.+,1 N., N.+1). (15) (15) follows from Lemma 14.1 of Erdös-Hajnal-Milner.' A slighly weakened form of his lemma is saed as follows : Assume ha 2"' = N.+,, I Si j _ I Sa 1 = N.+,. The pairs (x,y), x E S, y E Sa can be spli ino N.+1 classes so ha whenever U C S1, V C Sa, I U1 _ N.+,, I V ( = N., here is an x E V so ha every class is represened by he pairs (x,y), y E U. Now we deduce (15) from his lemma. The elemens of S, are denoed by { x,,1, 1 < y < w, +11 hose of Sa by,61 1 < < w,+11 and he classes ino which he pairs are spli are denoed by i al, 1 < a < w.+1 We pu x,. E A,.«1 if he pair (x,,#) is in class a. A simple argumen using Lemma 14.1 shows ha (6) is saisfied wih r = N s = N.+11 which proves (15). Perhaps N.+, i (N,+1, N.+2, N., N.+,) also holds, bu as we already saed we could no even prove Ni --*- (2, Na, No, Ni) Finally we sae a rivial resul. Le m < N, and assume 2m < N,+, Then N, +1 i ( 2, m, m, N, +,). In fac, he following sronger resul holds. Le 181 _ N,+1. Pu S = A 1(0) U Aa (a), 1 < B Then for some ~1 = 1 or 2, U A,%(O) has a complemen of power N,+1. We a leave he simple proof o he reader.

7 VOL. 60, 1968 MATHEMATICS : ERDŐS AND ULAM Banach and Kuraowski, Fund. Mah., 15 (1935). s Ulam, S., "Uber die Massheorie in der Allgemeinen Mengenlehre," Fund. Mah., 16 (1936). $ Sierpinski, W., Hypohése du Coninu (New York : Chelsea Publishing Company, 1956), 2d ed., pp Erdős, P., A. Hajnal, and E. Milner, "On he complee subgraphs of graphs defined by sysems of ses," Aca Mah. Acad. Sci. Hung., 17, (1965). 1 Erdős, P., A. Hajnal, and R. Rado, "Pariion relaions for cardinal numbers," Aca Mah. Acad. Sci. Hung., 16, (1965) ; see also he earlier paper by Erdős, P., and R. Rado, "A pariion calculus in se heory," Bull. Am. Mah. Soc., 62, (1956).