Israel Journal of Mathematics, Volume 1, Issue 4 (1963)

BARGAINING SETS OF COOPERATIVE GAMES WITHOUT SIDE PAYMENTS BY

BEZALEL PELEG ABSTRACT

In this paper an analogue of the bargaining set M~0 is defined for cooperative games without side payments. An existence theorem is proved for games of pairs, while it is shown by an example that no general existence theorem holds. One of the important questions in the theory of bargaining sets is the question of existence of stable payoff configurations. In [3] and [5] it is proved that for every coalition structure B, in an n-person game with transferable utilities, there exists a payoff vector x such that the individually rational payoff configuration (x,B) ~ M~ °. In this paper we investigate the validity of the above theorem for cooperative games without side payments. We find that it is valid for games of pairs*, while for games with non-trivial coalitions which contain more than two players, it is not always true. It is still possible that basically different generalizations of M[ ° will lead to existence theorems. As this paper belongs both to the areas of bargaining sets and cooperative games without side payments, the reader is referred to introductory papers in both fields: [2] in the first field, and [1] in the second. §1. DEFINITIONS. Let N be a finite set and let B be a subset of N. A B-vector x n is a real function defined on B whose value at i ~ B is x ~. The superscript N is omitted. E B denotes the euclidean space o f all the vectors x B. We write x A > yB if x ~>- y~ for all i ~ B; x B> y Bis interpreted similarly. We now give the definition of a cooperative game without side payments in characteristic function form: DEFINITION 1.1. An n-person game is a pair (N,v), where N is a set with n members, and v is a function that carries each subset B of N into a subset v(B) of E B so that (i) v(B) is closed and convex; and (ii) i f xB~ v(B) and x ~ > yn then

y B~ v(B). N is the set of players and its members will be denoted by the numbers 1,..., n. v is the characteristic function; we assume that it satisfies v({i}) = {x ~: x t < O} for all i e N, and v(B) DiXV({i}) , for all B = N. Received November 7, 1963

* Our method of proof is similar to those in a detailed version of [3], (to appear in Studies in Mathematical Economics, Essays in Honor of O. Morgenstern, M. Shubik ed.) 197

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[December

Let (N,v) be an n-person game. For B c N we denote ~(B) = {x ~ :xS~v(B), x B > 0, there is no yB~ v(B) such that yB > xB}. A coalition structure (c.s.) is a partitition of N. DEriSiON 1.2. An individually rational payoff configuration (i.r.p.c.) is a pair (x, B), where B is a c.s. and x ~ E s satisfies: x s ~ ~(B) for all B ~ B. An i.r.p.c, represents a possible outcome of (N, v). DEFINITION 1.3. Let (x,B) be an i.r.p.c, and i , j ~ B ~ B , i:fij. An objection of i againstj in (x, B) is a Q-vector y~ that satisfies: i ~ Q, j ¢ Q, yk > x k for all k ~ Q, and y Qe 6(Q). DF.FINtTION 1.4. Let (x,B) be an i.r.p.c, and yQ an objection of player i against player j in (x, B). A counter objection o f j against i is an R-vector z Rthat satisfies: j e R, i ~ R, z ~ > x k for all k e R, z k __>y~ for k e R :3 Q, and z R e ~(R). An i.r.p.c. (x,B) is stable if for each objection in (x,B) there is a counter objection. The set of all stable Lr.p.c." ' SiS"called the bargaining set ~vl 1"~~) . Let (x, B) be an i.r.p.c. An objectionin (x, B) is justi[ied if it cannot be countered. Let i,j ~ B ~ B, i :/=j. We write i ~ j in (x, B), i f j has no justified objection against i in (x, B). We also denote by X(B) the set of all the payoff vectors y such that (y, B) is an i.r.p.c., and by E~ the set {y : y ~ X(B), i ~ k in (y, B) for all k e B - {i}}. If B c N we denote by [B [ the number of members of B. An n-person ~game (N, v) is a game of pairs if v(B) = x v({i}) whenever B c N and 2

IBI

§2. Existence theorem for the bargaining set M~ i) o f games o f pairs.

Let (N, v) be a game of pairs. We remark that if B ~ N then ~(B) is homeomorphic to a closed interval; so if B is a c.s. then X(B) is homeomorphic to a cartesian product of closed intervals. The following lemma is not difficult to prove LEMMA~ 2.1. Let B = {i,j} be a subset of N; the function

x~(x s) = max {y~ : yB ~ 6(B), yJ = x j } is defined and continuous f o r 0 < x j < max {yl : y8 e 6(B)}. Ln~iMA 2.2. Let B be a c.s. and i e B ~ B ;

then E~ is a closed subset of X(B).

Proof. If I BI 2 then* E, = X¢n); so only the case I BI -- 2 is left. Without loss of generality B = {1,2} and i = 1. We shall prove that E~ is dosed by showing that X ( B ) - Et is open relative to X(B). Let x o ~ X ( B ) - E~. 2 has a justified * Since x ~ ----0 implies that x ¢ E~.

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objection yQ against 1 in (x o, B). Without loss o1' generality Q = {2, 3}. Since 1 has no counter objection to yQ we must have: (a) Xo1 > 0; (b) either x~ > max {xx :x (1'j}¢ fi({1,j})) or x~ > xJ(xt), for all j ~ N-{1,2,3} ; (c) either x~ > max {x t: x O,3)~ ~({1, 3})} or y3 > x3(x~). Since all the functions of x0 that appear in (a), (b) and (c) are continuous, we can find a set F, open in X(B), that contains Xo, and such that if z ~ F then (a), (b) and (c) are satisfied with z in place of Xo and also y2 > z 2 and y3 > z 3. So ya is a justified objection of 2 against 1 in (z,B): it follows that F = X(B) - El, which shows that X(B) - Et is open relative to X(B). Let B be a c.s. and B ~ B . We denote U s = t'3f~sE f. Also, if x ~ X ( B ) , we denote VB(xs-B) = {ya :(yB, xN-S ) ~ Us}. LEMMA 2.3. Let (x,B) be an i.r.p.c. If B ~ B then VB(xN-B) is homeomorphic to a closed interval. Proof.* If I BI ~ 2 then VB(x s-B) consists of one point; so only the case [B I = 2 is left. Without loss of generality B = {1,2}. We denote Gi= {yB:(yB,xS-B) EE~}, i = 1,2. G 1 and G2 are non-void closed subsets of fi(B) and G1 r i G 2 = Vn(xS-B). If a point yaeG~ then every point z s ~b(B) that satisfies z ~< yi is also in G~. So to prove that Gt r3 G2 is homeomorphic to a closed interval it is sufficient to show that G 1 n G2 =/= 0. Since ~(B) is connected we shall complete the proof if we shall show that b(B) = GI k3 G2 . Assume that yS ~ fi(B) - (G1 W G2). 1 has a justified objection z~ against 2 in ((yS,xS-B),B) and 2 has a justified objection z2Ragainst 1 in the same i.r.p.c. We have [g[ = [Q[ = 2. If R c3Q = ¢ then z~ is a counter objection to z2R. If R c3 Q :~ ¢ then it contains a single player j. In this case if z ] ~ z ] then zxeisa counter objection to z2R, and if z~ > zl then z2g is a counter objection to z~. So the assumption 5(B) - (Gl U G2) va ~ leads to a contradiction and the proof is completed. THEOREM 2.4. Let B be a c.s. in a game of pairs; then there always exists a payoff vector x such that the i.r.p.c. (x,B) ~ ~1~° . Proof. For x ~ X ( B ) let T(x) =

x VB(xS-~). Since the sets Us, B ~ B , are

BeB

closed, T is upper semi-continuous. Lemma 2.3 implies that for each x ~ X(B) T(x) is homeomorphic to a cartesian product of closed intervals. By the fixed-point theorems of Eilenberg and Montgomery [4] T has a fixed point, i.e. there is a payoff vector Xo e X(B) such that xo e T(xo). From the definiton of T it is clear that (Xo, B) e / ~ o . We now give an example which shows that Theorem 2.4 cannot be generalised to games with non-trivial coalitions which contain more than two players. * See [2] Lemma 7.2. for the proof for games with side payments where z3(B)is an interval

200

BEZALEL PELEG

EXAMPLE 2.5. Let (N, v) be a 4-person game given b y : v({1, 2})•= = {x~1'2':xl+x2 0: (3)

(;K

+ /~/2)/2 _ E

there is equality if and only if 2K is the unit sphere E. The mixed volume V(K1 ..... K,) is monotonic increasing in each convex Received Novoraber 13, 1963; revised version received January 27, 1964. * This work was supported in part by a grant from the National Science Foundation, NSF-G 19838.The author is indebted to the referee for fruitful comments. 201

202

WM. J. FIREY

body K , of. [1]. We write Wp(K)for the mixed volume with KI . . . . . and the remaining Ks set equal to K. From (3) we have for p < q :

Kp = E

Wp([aK + /~/~]/2) >= W.([2K + X/a]/2), with equality if and only if 2K = E, because in the ease at hand the monotonicity is known to be strict, of. [1], p. 43. In the plane this yields (4)

2A([2K + i¢/2]/2) ~ L([AK +/~/2]/2) ~ 27r

since in this case

We(K) = A(K), WI(K) = L(K)/2,

W2(K) =

where A and L are the area and perimeter. From Steiner's formula we have

rain A([2K + ~7/2]/2) = min[A2A(K) + 2A(K, 1~) + A(~)/22]/4 >- (A(K, 2~) + ~/[A(K) A(g)])/2, and min L([2K +/~/2]/2) = min [2L(K) + L ( £ ) / 2 ] / 2 __> ~/[L(K)L(i~)], the minima being taken over 2 > 0. By Minkowski's inequality: (5)

A(K, i~) ~_ ~I[A(K)A( I~)].

We replace the terms in (4) by these minima and use (5) to get (6)

2 A(K, 1~) ~_ ~/[L(K) L(J¢)] _~ 2n.

From the cases of equality in (3), we see that there is equality in (6) if and only if K = rE for some r > 0. In a similar fashion, in Euclidean 3-space we have, for the mixed surface area and total mean curvature

4~S(K, 1¢) > ~I[M(K)M(I~)] > 16n2, with equality if and only if K = rE for some r > O. REFERENCE 1. Bonnosen, T. and Fenchel, W., 1934, Theorie der konvexen K6rper, Berlin. OREGONSTATEUNIVERSITY, CORVALLIS,OREGON

REAL TIME COMPUTATION* BY

MICHAEL O. RABIN ABSTRACT

We introduce a concept of real-time computation by a Turing machine. The rclativo strengths of one-tape versus two-tape machines is established by a now mothod of proofs of impossibilityof actual computations. In the formulation of computations by Turing Machines it is assumed that the problem (say a numerical value of an argument for which a function value is to be computed) is given on the machine-tape and the machine proceeds, to perform its computation. No a-priori bound is imposed on the number of steps (the "time") needed for completion of the computation. The functions computable in this way are precisely all recursive functions. It is of great interest from the point of view of a general theory of computation to gain insight into computation procedures where there is some limitation on the time allowed for computation. One natural limitation is to require that if the problem (the input data) consists of n symbols then the computation will be performed in n basic steps, one step per input symbol. We may assume that the input sequence is entering the machine one symbol at a time and that the machine performs one of its atomic moves per input symbol. We again let our machines be Turing Machines which may, however, have more than one tape. Computations which are performed in this way will be called real-time computations (by a Turing Machine). Note that our systems may serve as mathematical models for computers (with auxiliary tapes) which are used for what is called "real-time" control. If, in particular, the result of a computation on every input sequence is always 0 or 1 then we can view the machine is defining a set, namely, the set of those input sequences which yield 1. Informally we can also say that the machine recognizes for every input sequence whether it is in the set defined by the machine or not. There are several results, notably by Yamada [2], about real-time computation by a Turing Machine. These are mainly along the~lines that certain computations are possible in real time. The concept of real-time computation is generalized Rec~ivodJanuary 24, 1964 * This research was supported in part by National Science Foundation grant GP-228 to Harvard University.This paper was writton while the author was visiting at th¢ Computation Laboratory of Harvard Universityduring the summer of 1963. 203

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[December

in an interesting way by Hartmanis and Stearns in [1]. In contrast with the case of finite automata there is no neat intrinsic characterization of the class of sets which are real-time definable. In fact, rather than attempt a complete characterization we should probably contend ourselves with insight about feasibility and nonfeasiblity of certain problems in real time. Our main result is that there exists a recognition problem which can be done in real time using two tapes but cannot be done in real time using a single tape. This result has obvious implications concerning the relative strengths of computers with one or more tapes when used as real-time control devices. In Section 6 we discuss the difficulty inherent in proofs of impossibility of certain computations. By way of illustration we give an example of a problem which somewhat unexpectedly can be done in real time on a single tape (Theorem 3). The result about relative strength of one-tape versus two-tape real-time computation is but one example in this dit~cult area of assessment of "degree of difficulty of a computation." We hope that some of the ideas in our proof, especially the concept of a bottleneck in a computation, may generalize to apply to other problems in the same area. 1. Real-time Turing Machines. The model for real-time computation that we employ is the one used by J. Hartmanis and R. Stearns [1] and byYamada [2]. A multi-tape Turing Machine over the input alphabet Y. is a finite automaton M having a finite set S of states and a working alphabet W = {cq, "",~n}. One of the states, call it So, is distinguished as the initial state of M. A subset F ~ S is singled out as the set of designated final states. The machine has k two-way infinite linear work-tapes t~, ..., tk which are divided into squares. Furthermore, there is a reading printing head which at any given time scans one square on each of the work-tapes. M is capable of receiving inputs a ~ ~. The working alphabet is always assumed to contain a blank symbol and at least one other symbol so that 2 < n. The operation of the Machine is specified by a function

M(tr, s,oq,,...,%) = (s',X1, ...,X~, %,,...,ejk) where a e Z, s, s' e S, ~i,% ~ W, X~ e {0,1, - 1}. We shall refer to this function as the machine-table of the k-tape Turing Machine M. The interpretation is that if the input is tr and M is in state s and is reading cctr on the tape tr, 1 < r < k, then M will go into state s', print ~j. on the square scanned on tr, and move each tape tr one square left, or one square right, or not at all, according as to whether X, equals 1, - 1, or 0. This action of M is called an atomic move. RE~ARK. When M prints a symbol ~ e W on a square it first of all erases the contents of that square. Printing the blank symbol of W simply means erasing the contents of the square. The set of all finite sequences on the 'alphabet ~ will be denoted by ]~*.

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DEFINITION 1. A sequence x = o'~ ...o-p ~E* is said to be accepted by M if, when started in So and with blank work tapes, M will go under the input sequence x through its atomic moves and end in a state in F (i.e., the state of M at the pth time unit is designated). The set of all sequences accepted by M is called the se ~ defined by M and is denoted by T(M). A set T _ ~* is called real-time definable (recognizable) if there exists a multitape Turing Machine M such that T = T(M). In particular Tis called k-tape real-time definable if for some M with k work tapes, T = T(M). It is quite clear that the adjective "real-time" is appropriate for this mode of operation. If x is a sequence of length p and requires p time units to feed into M then by time p we know, by looking at the state of M, whether x is accepted. Thus, there is no time delay between receipt of data and its processing. We restrict our attention to recognition problems. A simple analysis, however, will show that real-time computation problems can be easily reduced to recognition problems. Thus, our restriction involves no loss of generality. 2. The set T 2 . Let E = {a,b,O,l,oe, fl}. Words on {a,b} will be called ab words and the set of ab words will be denoted by A. Words on {0,1} will be called 01 words and the set of 01 words will be denoted by Z. If x = tYttr2 " ' " tYn_tffn then, by definition, x* = a, tr._ x ... a2al. DEFINITION 2.

T2={uwu*Iu~A,veZ} u

{uvl~v*[ueA,veZ}

.

LEMMA 1. The set T2 is real-time definable by a two-tape machine. Proof. We shall describe the mode of operation of a two-tape machine M for which T2 = T(M). The reader can verify that this mode of operation can indeed be realized by a suitable machine-table. As the ab word u is coming in, M will print it on its first tape. When the 01 word v comes in, M will print it on its second tape. According as to whether the input following uv is a or fl, M will start tracing back its first or second tape. M will end in a designated state if and only if the sequence w of inputs following a (or/3) coincides with the sequence being traced backwards on the first (second) tape. THEOREM 1. The set T z is not real-time definable by a one-tape machine. Consequently two-tape real-time computation can do more than one-tape realtime computation.

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3. Preliminary lemmas.. To prove that T2 is not real-time definable by a one-tape machine assume by way of contradiction that the one-tape machine M does define T2 in real time. Let the number of states of M be m and the number of letters in its working alphabet be n. Throughout the following Sections 3-5 M will always designate this fixed one-tape machine for which T2 = T ( M ) . DEFINITION 3. If M has input w then the work space t(w) of M on w is the sequence of tape squares covered by the motion of M while having the input sequence w. If x is a sequence of squares on the tape or a sequence of symbols then l(x) will denote the length, i.e. the number of elements, of x. Let x be an input sequence, by the coding of x we shall refer to the sequence of symbols in the squares of the work space fix), the state of M, and its position on the tape, at the end of the input x. LEM~tA 2. There exists a numerical constant c > 0 such that f o r every u ~ A and every integer i > 0 there exists a v ~ Z such that l(v) = i and ci < l(t(uv)).

Proof. There are 2 ~ sequences v e Z such that l(v)= i. Since the input uv may be followed by fl, if vl ~ v, then uvl and uv must be coded differently. Otherwise, uvflv* and uvtflv* will both be accepted by M. Let l(t(uv)) < k for all v ~ Z , l(v) = i. Then there are at most n k. k" m different codings of the inputs uv. Hence 2~< n k . k . m . If i is large this forces k to be large so that we may assume that k m 0 (depending only on M ) such that for every u e A and every integer i > flu) there exist a sequence v ~ Z , l(v) = i, such that a) ci < l(t(uv)), b) no more than ~th of the squares of t(uv) are covered by M more than d times. Proof. Let us choose a sequence v ~ Z, l(v) = i, for which a) holds. Let dl be a number such that more than ~th of the squares of t(uv) are covered by M more than dl times. Then the total number of moves of M exceeds d ~ l ( t ( u v ) ) > ~dlci. But since M operates in real time the number of moves of M by the input uv is exactly l(u) + l(v) < 2i. Thus, ~dlci < 2i and dl < lO/c. The number d = [(10/c) + 1] satisfies b). 4. Bottleneck squares. The proof of our main theorem rests on the idea that in working on certain input sequences the machine M develops bottleneck squares

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on its work-tape through which information cannot flow in sutficient quantity. The idea o f a bottleneck is made precise in the following: DEFINITION 3. Let u e A, v e Z . A square B on t(uv) is called a bottleneck square of t(uv) if 1) under input uv the machine passes through B no more than d times (where d is as in previous Lemma 3), 2) B lies outside the work space t(u), 3) the length of the section of t(uv) determined by B which does not contain t(u) exceeds l(u) + 1.

I

t(uv)

I

B

/ tl

I

l(u) + 1 < l(tt) Figure I

For every u e A there exists a v E Z such that the tape t(uv) has a bottleneck square. [,EMMA 4.

Proof. Let i be an integer such that 5 l ( u ) + 5 < ci and also l(u)< i. By Lemma 3 there exists a sequence v ~ Z such that ci < l(t(uv)) and fewer than -~th o f the squares o f t(uv) are covered more than d times. Now l(t(u)) < l(u) + 1 < (ci/5)< l(t(uv))]5. Dividing t(uv)into 5 equal parts (there is a trivial modification of the argument if l(t(uv)) is not divisible by 5) we see that either on the left or on the right end of t(uv) there is an interval o f length } l(t(uv)) which does not contain any squares of t(u). In this interval consider the }th oft(uv) which does not run to the end. Since fewer than }th squares of t(uv) are covered more than d times by M, there is a square B in this }th of t(uv) which is covered at most d times. There are at least l(t(uv))/5 >(ci/5)> l(u)+ 1 squares between B and the end o f t(uv). Thus B is a bottleneck square. 5. Proof of main theorem. Let u e A and v e Z be such that t(uv) has a bottleneck square B. To fix ideas let us assume that B is to the right of t(u). As the input uv is coming in,there is a first time that M enters the right-most square E of t(uv) (see Figure 1). Let w e Z be the initial section o f v such that uw is the sequence leading to the first visit of M at E. Thus t(uv) and t(uw) have the same right-hand end square E and B is also a bottleneck square of t(uw). Denote the square immediately to the right o f B by R. By a passage o f M through B we mean either a move of M from B to R or a move from R to B. The state of M during a passage is the state M has when it reaches R in the first case, and the state M has when it reaches B in the second case.

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Note that passages of M through B do not include atomic moves of M in which it starts on B and stays on B. Under the input uw the machine M will first cover the tape t(u) and then, under the w portion of the input, move to the square E. Let pl,P2,..-,p,, be the consecutive passages through B (r = 1 is not excluded). The passage Pl is a move from B to R, P2 is a move from R to B, etc. Let the state of M during the passage p~ be si, 1 < i < r. The scheme of the bottleneck square B is the r + 1 tuple (e, sl, ...,s,) where e is 1 if B is to the right of flu) and e is - 1 if B is to the left of t(u), and sl, ..., s, are as above.The notions of passage and state during a passage are modified in an obvious way when B is to the left of t(u). Now the number r of passages through B is at most d. Thus, there are at most N

N=2.m+2.mZ+...

+ 2.m a

different schemes of bottleneck squares, where m is the number of states of M. Let g be a number such that N < 2 g. For each u ~A, l(u)= g, let v ~ Z be a 01 sequence such that t(uv) has a bottleneck square Bu and let w denote the section of v leading to the first visit of M to the end Eu of t(uv). There must be two different sequences ul, u2 ~ A, l(Ul) = l(u2) = g, such that the bottleneck squares B~I and B,,2 have the same scheme, say (1,st,...,sr). Note that e = 1 which means that B~, is to the right of t(u~),i = 1,2. Let UlW1 =

U181"'Sn1""~'n2"'Sn,.'"Sn.+l

U2W2 =

U 2 t l "'" t m l " ' " tm2 "'" t m r "'" t i t + 1

where e , t ~ {0,1}, enl is the input when M visits B~I during the first passage, e,, is the input when M visits Bul during the second passage, and so on up to end; similarly for tin:tin2,'", in the second sequence u2w2. After receiving the input en,+~ (tim,+,) M visits for the first time the right-hand end-square E~I(Eu2). We come now to the main point of our argument, In the sequence UlWl replace, for each odd 1 < i < r - 2 , the segment en~+~...e,~+:~ by the sequence tree + 1"'" trot + 1- 1" Furthermore, replace en, + 1"'" e,,, +, by tin, + 1"'" t,~, +r Call the resulting sequence u~w~. Note that all the changes were made in the Wl portion of u~wl. Now, ulwl and u2wz have the same scheme of states in the passages of M through B~, and B~2, respectively, and our changes in ulwl were made only in the inputs between visits to Bu:while M was on the right of B~,, or after the last visit to B~I. One can see by finite induction over 1 < i < r + 1 that UlW'~ again has the same scheme (1, sl, Sz,"., s,) and that at each input e,~, j odd and 2 0 such that

-oo 0 so that 23/2c

(8) Then

rl =

e-U2/2du


1 satisfying (9) will do. A little computation shows this requirement as equivalent to (10)

C~(87~) -1/4 max 0__1, A1 A2... A, ~ P. Now, since A1... An- 1 is an ideal in M, it follows that A 1A2... A n c A I A 2 . . " A,_ 1 and therefore ( A I A 2 . . . A n_ t)r -- ( A t ... An)r" On the other hand, each Ai is nilpotent, being generated by a single element, hencethereexistanintegerm > 2suchthatA1A2... A n" = OandA1A2...A~ '-x ~: 0; consequently, A~- 1 ~ (AtA2... An)r whereas A,m- 1 ~ (Ax ... A,_ 1)r. This leads to the increasing chain (A1) r = (A1A2) , c ... c ( A 1 A 2 ... A,)r... which is a contradiction. The p r o o f of the theorem follows now readily. If the chain condition holds for annihilators then S = L(S) for a nil subring S, and since N(S) is nilpotent it follows that N(S) = L(S) and thus S is nilpotent. I~MARK. If a ring R satisfies the chain condition (B) for two sided ideals A i of R, then the same p r o o f shows a well-known result of the author that for these rings R, L(R) = N ( R ) is nilpotent. This follows immediately from the fact that the Ai chosen in the proof of lemma 2 can be chosen to be ideals in R. BIBLIOGRAPHY 1. Herstein, I. N. and Small, L. W., 1963, Nil rings satisfying certain chain conditions, Notices Amer. Math. Soc., 10, p. 662. 2. Jacobson, N., 1956, Structure of Rings, Amer. Math. Soc. Colloquium Publ. No. 37, Ch. VIII, p. 193. THE HEBREWUNIVERSITYOF JERUSALEM

THE SHAPE OF THE SIDES OF AN ARC BY

JAMES M. SLOSS ABSTRACT

Associated with certain oriented Jordan arcs is a region which is called the side of the arc. Circles centered on the arc and passing through one of the end points of the arc have an envelope which permits one to find analytically the shape of the side of the arc. A condition, stated geometrically, is imposed on the arcs considered, to insure that the circles have an envelope. An analytic condition is then imposed (Theorem 3) to insure an envelope. As an example a parabola is given. In this note we shall be concerned with finding the shape of the left and right neighborhoods of a Jordan arc L with a prescribed direction, as described in Muskhelishvili [-1]. We define the left neighborhood (or left side) S + of an arc L with end points La and Lb as follows: a point p is in S + if there exists a point t o on L a n d an open disc Do centered at to for which Do - L i s the union of two disjoint open sets, one on the left, D +, and one on the right, Do, as we traverse L i n the prescribed direction, and p is in D~-. The right neighborhood S - is defined in an obvious way. It will be our aim to find the shape of S ÷. In this note we shall impose a further restriction on L, namely: (C1) every circle centered on L and passing through La or Lb is divided into at most two disjoint sets by L. Note that the arc y = 2 sin nx, 0 < x < 1 does not satisfy (C1) since the circle centered on L which is tangent to the line x = 1 and passes through (0,0) is cut by L in four points. We now define the set E ÷. A point p is in E + if there exists a point tl and an open disc D~ centered at t~, whose closure passes through L~ or Lb, D~ - L is the union of two disjoint open sets D~" and D~-, for which p is in D~. THEOREM 1.

S + = E+.

Proof. S + ~ E + clearly. To see that E + _~ S + let p be in S + and let the disc containing p, centered at to on L, and divided by L into two disjoint sets, be denoted by D 0. Let the circular boundary of Do be denoted by Bo. We shall show as a preliminary result that Do cannot contain either L~ cr Lb. To see this, assume that Do contained L~ or Lb. Then since Do is divided into two disjoint sets by the non-intersecting continuous arc L, L must intersect Bo in at least two points b~ and b2. Without loss of generality, assume bl lies between Received November 1, 1963 217

218

JAMES M. SLOSS

t'December

to and Lo on L. Since La is an interior point of Do, there is a circle about to, smaller than Bo, which contains La in its interior, and bl and b 2 in its exterior. Such a circle is cut by L in at least three distinct points and thus (Cl) is violated and we see that Do cannot contain La or L~. If B 0 contains La or Lb we are finished, since then p is in E +. On the other hand, if Bo contains neither L a nor Lb, we know from the above paragraph that Do UBo contains neither La nor Lb. Increase the size of Do (maintaining its center at to) until Bo cuts La or Lb or both. Then D o becomes a D1 with D + _ D + and thus p is in E +. Thus E + = S +. We shall now introduce a second condition: (C2) Let L be a Jordan arc with end points Lo and L b. Given any point t on L there exists a circle yt(e) centered at t and of radius s > 0 such that the extended line joining any point of Linterior to ?t(s) and passing through t does not contain Lo or L~. Note that the interval [0,1] does not satisfy (C2). DEFINITION. The class of arcs that satisfy (C1) and (C2) and which have a continuously turning tangent will be denoted by .~. THEOP~M 2. If L~.~, the circles centered on L and passing through La(Lb) have an envelope. Proof. This follows since any two such circles, if sufficiently close together, intersect in one other point than Lo since if not they would be tangent and their centers would lie on a line through La. But this is impossible since Le.~. A similar argument holds for circles passing through Lb. The condition (C2) is defined geometrically. We shall now be concerned with finding a subset of LP that is characterized analytically. LEMMA. Let g(s) be a continuous map from [a*, t~**] to L* and let g(So):/= 0 f o r tr* < So < a**. Then there is a unit vector A and a circle ?so(e*) about g(so) of radius s* > 0 such that g(s) is not perpendicular to A for g(s) in ~so(e*). Proof. By contradiction. Let e. o 0 as n --, ~ . Then if A is an arbitrary, but fixed unit vector, there is an s. with g(s.) in yso(S.)such that g(s.) is perpendicular to A and thus g(s.). A = O. Since the 5. ~ 0 it follows that g(s.) --, g(So) and thus g(s,).A ~ g(So)'A. But for each n, g(s.)'A = 0 and thus g(So)'A = O. But A was arbitrary and l a I= 1. Thus g(So)= 0 which is a contradiction.

THEOREM 3. Let the Jordan arc L: y = y(s), 0 0 be such that y(s o - 6*) is the first point of intersection of L with ~,~o(e) as we travel along L from Yo to La and 6** > 0 bc such that y(s o + 6**) is the first point of intersection of Lwith T~o(e)as we travel along L from Yo to Lb. Without loss of generality, we may assume that yC")(s)is defined and continuous on Io = (So - 6", So + 6**) c [0, ~]. Let y(s*) = y* be any point of L within 7so(e). Then Z(s) =

y* -

Yo (s - So) + Yo, So < s

min Ilxll/max x ~ bdC

Ilxll

x e bdC

where b d C is r e l a t i v e to t h e s u b s p a c e s p a n n e d by C. D v o r e t z k y [11 p r o v e d t h a t : For every positive integer k and every e > 0 there exists a number

N(k,e) (e.g., N(k,e)=exp(215e-2k210gk)), so that for n > N, every convex body (compact convex set with non-empty interior) in E" which is symmetric about the origin there exists a subspace E k with ~(C ~ E k) < e. I n a r e c e n t p a p e r [2] D v o r e t z k y r e m a r k s t h a t the s a m e r e s u l t h o l d s i f we c o n s i d e r t h e p r o j e c t i o n C IE k of C i n t o E k i n s t e a d o f C n E k, since

~(clEb = ~(c* n ~ ) w h e r e C* is the p o l a r b o d y o f C. H o w e v e r he states as an u n s o l v e d q u e s t i o n w h e t h e r t h e r e is a n N'(k,8) so t h a t f o r n = N ' t h e r e exists an E k for which b o t h

c~(C n E k) < 8 a n d ct(ClE k) < 8. T o give a n a f f i r m a t i v e a n s w e r to this q u e s t i o n we p r o v e the f o l l o w i n g . Tt-I~OREM. For each pair of positive integers k, 1 and every ~ > 0 there exists an N(k,l,8) so that for n > N and any l-tuple of convex bodies C1,...,C t in E n symmetric about the origin, there exists a subspace E k so that

~(C i n E k) < 8 Here N ( k , l , 8 ) = N(k,8) and N ( k , l + 1,8) < N(N(k,l,~),8). Received February 6, 1964. 221

i = 1,...,1.

222

E.G. STRAUS

Proof. For I = 1 this is Dvoretzky's theorem. Assume the theorem true for l" Then for n > N ( N ( k , l, e), 8) there exists an E N¢k:'O so that ct(Ct n E mk't'~)) < e and by the induction hypothesis applied to C~ = C i ~ E N~'t:), i = 2 ..... l + 1; there exists an E k c E mk'l'~) so that ct(C" n E ~) = a(C~ ~ E k) < 8 for i = 2 ..... I. On the other hand we have ~(C1 n E k) < ~(C1 n E N¢k:'*~) < 8 so the result holds with N(k, l + 1, e) = N ( N ( k , l, ~), e). Dvoretzky's question is now answered in the affirmative for Ct = C, C2 = C*, I = 2. The bound computed here grows very rapidly since it involves/-fold iteration of an already very rapidly increasing function of k and 1 [8. A second question raised in [2] can be answered in the negative. Dvoretzky proves that it is not possible to give a uniform positive lower bound for the Haar measure of the set of all k-planes (k >=2) which intersect a convex body C in E"in a set of asphericity < 8. His example is an ellipsoid of revolution with a very large axis on its axis of revolution. He asks therefore whether such a uniform lower bound could exist if asphericity is replaced by unellipsoidality, that is the minimum asphericity of all affine transforms of the set. As an example of a body for which this is not the case we consider the union of two spherical caps: x 2 + ... + x . -2 1 + ( x . -

2 1 + 3)2_< 1, x 2 + ... + xn_ t + ( x . + 1 - 3) 2- O.

lira

,-,oo

Also, the convergence to 0 is uniform in every point-set where ReP(z) < - e, with e > 0 fixed. (2.9), (2.7), (2.4) and (1.6) prove the theorem. EXAMPLE. (i) The Lototski-transform definited by [F, d, = n - 1] sums the geometric series for Rez < 1, and does not sum it for Rez > 1, [3]. Here

P(z) = z - 1. (ii) If P(z) = e~r(z - 1) with a suitable real ? we obtain as domain of summability any given half plane, the boundary of which is a straight line passing through z = 1. (iii) If P(z) = e'~(z - 1) (z - ~ - ifl), with real ~, fl, ?, we obtain as domain of summability the "inside" or "outside" of hyperbolas passing through z = 1. Next we prove the following theorem: THEOREM 2. Let R be a set that contains the point z = 1 and whose complement consists either of the point oo or of an unbounded domain. Letf(z) be an analytic regular function on R satisfying

1963]

ANALYTIC CONTINUATION BY SUMMATION-METHODS

(2.10)

Re f(1) = 0 .

227

Then, there exists a sequence of polynomials {P.(x)} (n_~ 1, Pn(1) # 0) such that the [F*,Pn] transformation sums the geometric series to the value (1 - z ) - l for every z ~ R for which Re f ( z ) < 0 and does not sum it f o r z e R f o r which Re f ( z ) > O. Proof. By the well-known theorem of Walsh I9] for every k > 1 there exist

polynomials Qk(z) satisfying (2.11)

I Qk(Z) - f ( z ) I < k - 1

for z ~ R , [z I < k, and (2.12)

Qk(1) =f(1)

k = 1,2,...

Pk(Z) = Qk(Z) + k

k = 1,2,...

Define (2.13)

By (2.11), (2.12) and (2.13) for any fixed z (]z ] __ O. Hence by the continuity of u, u(x) > 0, a contradiction. This theorem shows that condition (1.3) is necessary for the existence of a utility in R ~. Otherwise there are x , y , z with x >-kz, z ~ O. Hence ( x / k ) ~ - z and by (2.1), x / k - z >-0. But in the usual topology on R ~ ( x / k ) - z ~ - z, so that - z e S, contradicting (2.6). (Every linear functional defined on R n is continuous in the usual topology). In R n (1.3) implies (2.6) ([1, page 456]). A partial order is called pure if x ~ y implies x = y. A partial order is pure if and only if x ,,, 0 implies x = 0, or equivalently, if and only if S = T U {0}. 3. An example. Let X be the space of the sequences of real numbers which have only a finite number of members different from zero. X has denumerable dimensionality and up to isomorphism is the only such space. The algebraic dual space of X is the space o f all real sequences. Developing an earlier example of mine, M. Perles gave the following example of a pure partial order on X which satisfies (1.1)-(1.3) and has no utility. In fact, this order has no finite-dimensional utility. Denote by e~ the i-th unit vector. Let en,1 = el, e , , ~ = - ( n 1)ei-1 + e~, i = 2,..., n. (e~,i is the vector o f X whose (i - 1)-th coordinate is - (n - 1) and whose i-th coordinate is 1, all the other coordinates being zero). For example: e4 1 = (1,0,...,0 .... ),e4,2 = ( - 3,1,0,...),e4.3 = (0, - 3 , 1 , 0 . . . ) , e44 = ( 0 , 0 , - 3 , 1 , 0 .... ) Define P , as the set o f all linear combinations o f the form ~t"= i ~e,,t with ~, > 0, ~, > 0. We shall show presently that U ~ - - 1 P , and [..J~=IP, are convex cones in X. It is obvious that each P , is a convex cone. e~ ~ P,, since e i = ( n - 1)~-Xe, 1 + ( n - 1)~-Ze,.2 + ... + ( n - 1)e, ~-1 + e , i. Let x e P , , y ~ P m and assume n > m. Then n

X =

~

ra

~ e n i,

Y = ~. ~lem.l

1=1

-- t=1 ~,

/=2

(~,+

p,)e,,., + (n - m) t =~2 p'

n

+

with ~.,/~, > O, ~t,/~J --> O.

|~1

1=1

~

e~en.t

|~m+l

and so x + y is contained in P..

i=m+l

1( n - - 1 ) t - ~ - l e

,,.k

232

YAKAR KANNAI

[December

DefineT U,=~ ,,andsetx~-yifandonlyifx-y~T. Weobtainapure order satisfying (1.1)-(1.3), (0 ¢ T). It is obvious that (1.1) and (1.2) are satisfied. Set E" = {x ~ X ; xi = 0 for i > n}. It suffices to prove (1.3) for x, z ~ E", n = 1,2,.... By theorem A, it suffices to show that the partial order, reduced to E n has a utility. Define u,, ~ E n* by

u,(x) = ~ nkxk. k=l

F r o m the definitions of T and the Pi's it follows that x ~ T t3 E" if and only if

x~Pi, i = 1,...,n. Now u,,(e~j) > 0 for 1 < i < n, 1 < j < i. Hence u,(x) > 0 for x E T c3 E" and therefore u, is a utility on E" and our order satisfies (1.3). T has no finite-dimensional utility. Suppose that v(x) is such a utility, v(x) = (q~l(x), ..., ~b,,(x)), ~bi(x) are linear functionals on X, and without loss of generality let m be the minimal dimension of a possible multi-dimensional utility for this order (this includes the case m = 1). Then ~b~ is not identically zero. Every unit vector e~ is contained in T. Hence there is an ek with q~l(ek) > 0 so that ~bx( - ek) < 0. Let n be an integer, n > k + 2. For every ~ > 0, the vector

ak.n,8 =

--

ek

1 + -~'~_

lek+l

--

e(n - 1)e,-1 + ee,

n---Z-]- .... , - e(n - 1), 8,0,... k k+l

n-1

)

n

is contained in P, and therefore in T, so that ~bl(ak,,,~) > 0, (if ~bl(x) < 0 then 0 is preferred to v(x) in the lexicographic order). By letting e--* 0 it follows that ~x(-ek + (1/(n-1))ek+l)>= O, and by letting n ~ oo it follows that ~bx(- ek) __>0, a contradiction. (It is clear that every linear functional on X, reduced to E", is continuous on g"). 4. The main existence theorems. Let X be the space of the real sequences which have only a finite number o f members different from zero, ~ a partial order on X satisfying (1.1)--(1.2). We may assume that ~ is pure. Otherwise divide by E = {x :x ~ 0} which is a linear subspace of X. On the quotient space X/E, induces a pure partial order satisfying (1.1)-(1.2). The quotient space is isomorphic to a linear subspace of X, which is either a Euclidean space or has denumerable dimensionality, i.e. is isomorphic to X. The first case is settled in [1], and it follows that (1.3) is a necessary and sufficient condition for the existence o f a utility. (It is obvious that a utility defined on X/E induces in a natural way a utility defined on X.) In order to settle the second case, let us topologize X in the following way: a typical neighbourhood of zero is the set of all x ~ X such that [ x~ [ < e~for a given g oO • sequence of positive numbers ( t)i = x

1963]

INFINITE DIMENSIONAL PARTIALLY ORDERED SPACES

233

Every linear functional defined on X is continuous in this topology. For if u ~ X* (the algebraic dual space o f X), then u is represented by a sequence o f real numbers us where us = u(e~). Let an e > 0 be given. Define e~ = e/2iui if us ~ 0, e~ = 1 if ui = 0. Let x be a vector in X. For every y E X, I(Y - x)~l < ei implies [ u(y) - u(x)] < e, hence u is continuous. (The topology induced by any one of the lp norms does not have this property). We remark that this topology is separable, since in every E " there is a dense sequence, and the union o f these sequences is dense in X. Moreover, the induced topology on E 4 has a countable basis for every n (the induced topology coincides with the usual topology). Hence, if . . / / c X and to each x ~ .///there corresponds an open set Ux which contains x, then there is a sequence {x~} of points o f . / / s u c h that ~1/= 1,.)~Ux,. We may construct this sequence by first covering ..//r3 E 4 and then taking the union of these sequences (union over n). We are now able to state and prove the following theorem: TI-mOR~MB. Let ~ be a pure order on X and let ( - T) t3 ~ = ~ above topology. Then there is a utility on X .

in the

Proof. Let p be any point of T. Following Klee ([3]), we assert that there is a neighbourhood Up o f p such that [Up to T] (the convex hull o f Up and T) does not contain 0. Otherwise there are q e U p , a c T , 0 < ~ < 1 such that ~q + (1 - ~)a = 0, for every Up. Then - ~q = (1 - ~)a, - q = (1 - ~)/~)a ~ T (since T is a cone) so that - p e T, contradicting ( - T) t3 T = ~ . [Up to T] is a convex set with non-empty interior, and 0 ( [ U p to T]. Hence there is a nonzero linear functional up which supports it, i.e., x e [Up to T ] implies up(x) > 0 ([4 page 191]). Since p is an internal point o f [ U v to T], up(p) > 0. There exists a neighborhood Vp of p with up(y) > 0 for y e Vp, because up is continuous. By the remark made above, there is a sequence Pi of points of T such that T = U s Vv,, since T = U p Vv" Up, I E 4 is a linear functional on E 4 and so is bounded there. Let us denote its Euclidean norm on E 4 by ~uv, 1[4. Set oo

u(x) = 2

4=,

u,.(x) 24

llup.ll4 + 1

The series converges pointwise for each x c X, since there is a positive integer m with x ~ E m, so that x ~ E * for all i > m. F o r every n > m,

u,,(x)

•=

2'llu,,l],+1

m-, -,=,

"-'

lup,(,,)l 2'llu,,,H,

+ +

1

lup,(x)l

Ilu,,ll, llx[I, .= ~g= 2,1lu

- 0; and, by Lemma 4, the sequence (Ao, - io), (AI, - il), '.., is disjoint. There then is a strictly increasing sequence rl, r2, "" with - ir~ = i for some i and all j, A~j c J~ f o r j = 1,2, ..., and the A,~'s are disjoint and congruent. But this is impossible. LEMMA 8. For every A ~ ~ , p(A) ~ ~ . Proof. Since p2 = la~p), obviously & ( p ) = ~(p) and p is 1 - 1. By Lemma 7, ~ ( p ) = K - (V u W), and, for every A6 ~/, p(A) is congruent to A and p(A) c J~ for some i, so that p(A) is an open arc. Thus the decomposition K - (V U W) = p(K - ( V u W)) = U p(A) is of the type described in Lemma 6, and the assertion follows. ~t~ Lemmas 7 and 8 show that p is a scissor congruence of K onto itself. They also show that the rigid notions comprising this scissor congruence are in the group generated by T~,..., T~. The scissor congruence p decomposes into components in a natural way: DEFINITION. K+_ is the set-theoretic union of all A 6 ~ such that A o K + and p(A) c K _ ; similarly for K _ +, K + +, and K _ _. Plainly, K+ - ( V u W) = K + _ u K+ + ; and K _ - (V u W) = K_ + u K _ _ . Moreover, since p(A) ~ ~ and p2(A) = A for all A ~ ~¢, (a) p(K+_) = K _ + , (b) p(K++) = K++, and (c) p ( K _ _ ) = K _ _ . DEFINITION. p+_ = p restricted to K + _ . Our next goal is to present a simple intuitive property of the scissor congruence p+ _ in Lemma 12. But rigor seems to demand two definitions as well as preliminary Lemmas 9, 10, and 11. DEFn,nTIOtqS. Let D and D' be topological discs, let their respective boundaries be J and J ' , and let A be an open arc with A = J n J ' . Say that D and D' are on the same side of A if, for every x 6 A,

1963] (3)

SCISSOR CONGRUENCE

245

there exists N ( x ) such that N ( x ) - A c (int D N int D') u (ext D n ext D').

Say that D and D' are on opposite sides of A if, for every x e A, (4)

there exists N(x) such that N(x) - A c (int D • ext D') U (ext D n i n t D').

The proofs of the next three lemmas are not difficult. LEMMA 9. D and D' are either on the same or on opposite sides of A. LEMMA 10. I f int D ~ int D' is empty, then D and D' are on opposite sides of A. I f D c D', then D and D' are on the same side of A . LEMMA 11. I f D and D' are on the same side of A, and D and D" are on opposite sides of A, then D' and D" are on opposite sides of A. l f D and D' are on opposite sides of A, and D and D" are on opposite sides of A, then D' and D" are on the same side of A. DEFINITION. For each A e ,~, let Ta denote the rigid motion T~k... T~o described in L e m m a 7. O f course, Ta(A) = p(A). LEMMA 12. The scissor congruence p+_ preserves sides of arcs. T h a t is, if (a) A ~ , (b) A c J i f o r some i > O, (c) p(A) c J j f o r s o m e j < O, then Ta(D~) is on the same side of p(A) as is D j . The scissor congruence K+ + -~ K+ + reverses sides of arcs, as does the scissor congruence K _ _ -~ K _ _ . Proof. Let A ~ . ~ , and let i0, "",ik be as in L e m m a 7. Since - i, # i,+j, clearly int D_~r nintDir+ 1 = @, so by L e m m a 10, D_~. and Dr. ÷ 1 are on opposite sides of Ar for 0 < r _< k. Let Di'o = T~r"" T~oDto for 0 _ r < k. As will now be shown, Dio and D_ ~. are on the same side of Ar for r even and on opposite sides for r odd. By definition Dt° = T~o(Dio) = D-io ; therefore D°o and D-to are on the same side of A o. N o w use L e m m a 11 repeatedly to see that, for 0 < m