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

R e V ( - d.v+p) = - (/~ + 1) -<_ - 1;

(2.5) thus

(2.6) (2.7)

I 1 + d, I =>,s>0 In(n) _

,1 2 i-I

YI 1+ v=mp+l

z- 1

and by (2.6) III n l _<_(1 + (I z - 1

Iz

v = 1,2, ... z-1

II v=mp+l

,-1 .

Thus [-[<2n) is uniformly bounded for every n > 1 and for every z belonging to a fixed bounded point-set. (2.8) ]l-I~')i = f l k=l

tP(z)+kl2

p-~¥E

f i ( 1 + 2kReP(z)+lP(z)12-lP(1)[2.)

--

k2+lP(1)l 2

k=l

By a well known theorem on infinite products (2.9)

~0

if ReP(z) < 0

[ oo

ifReP(z) > 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 ] __
for R e f ( z ) < O for R e f ( z ) > O .

By (1.7) this proves the theorem. REtaARtC. A generalization of Theorem 2 can be made to the situation where R is the union of an increasing sequence of bounded closed sets R~ the complement of each of which is an unbounded domain. This result will prove the existence of an [F*,P,]-transformation that is effective for ~ in the entire MittagLeffler star of (1 - z)- 1. It has to be mentioned that the IF*, P,]-transformations are row-finite. Because of the length of proof we only state the following result too: THEOREM 3. Let D be an union of a finite number of simply-connected bounded domains having Jordan boundaries. Let z = 1 lie on the boundary, and let E be a closed subset of the complement of D. Then there exists an [F*,Pn]-transformation, which sums the geometric series to the sum (1 - z)-1 for every z ~ D and does not sum if for every z ~ E. REFERENCES 1. Betel, B., 1928, Leans sur les series divergentes, 2nd ed. Paris. 2. Cooke, R. G., and Dienes, P., 1937, The effective range of generalized limit processes, Proc. London Math. Soc., 12, 299-304.

228

AMRAM MEIR

3. Jakimovski, A., 1959, A generalization of the Lototski-method of summability, Mich. Math. d. 6, 277-290. 4. Okada, S. Y., 1925, Llber die Ann~herung analytischer Funktionen, Math. Z., 23, 62-71. 5. Russel, D. C., 1959, Summability of power series on continuous arcs outside the circle of convergence, Bull .dead. Roy. Belg, 45, 1006-1030. 6. Teghem, T., 1958, Sur des extensions d'une m6thode de prolongement analytique de Borel, Co[loque sur [a thForie des suites, Bruxelles, (1957), 87-95. 7. Vermes, P., 1952, Convolution of summability methods, J. d'Analyse Math., 160-177. 8. Vermes, P., 1958, Summability of power series at unbounded sets of isolated points, Bull. Aead. Roy. Belg, 44, 830-838. 9. Walsh, T. L., 1935, Interpolation and approximation by rational functions in the complex domain, Amer. Math. See. Coll. Publ. vol. XX. THE HEBREW UNIVERSITYOF JERUSALEM

EXISTENCE OF A UTILITY IN INFINITE DIMENSIONAL PARTIALLY ORDERED SPACES* BY

YAKAR KANNAI ABSTRACT

An example is given of a preference order on a space of denumerable algebraic dimension that has no utility, and meeessary and sufficient conditions for the existence of utilities in various linear spaces are given. 1. Introduction. " U t i l i t y " is a numerical function representing a preference order, which is useful in solving optimization problems. Until recently, utility theory was restricted to complete preference orders;the extension to orders that are not necessarily complete was made in [1], but the results of that paper are restricted to finite-dimensional spaces of alternatives. The infinite-dimensional case is important for certain types of economic models, e.g. those involving an infinite time horizon or continuous time, commodities that may be available in a continuum of quantities, continua of prices or of geographic locations, etc. It is the object of this paper to investigate to what extent the results of [1] can be extended to infinite-dimensional spaces of alternatives. In particular, a problem raised in [1] will be answered. Let X be a real linear space. We assume that on X there is defined a transitive and reflexive relation called preference-or-indifference and denoted by ~ . If x ~ y and y ~ x, we shall say that x is indifferent to y and write x ,,~ y. If x >- y but not x ~ y, we shall say that x is preferred to y and write x >- y. The relation ~ will be called a partial order. (We shall not assume that ~ is complete). We (1.1) (1.2) (1.3)

assume that the following conditions hold: x~y impliesx+z~y+zforallz~X; x~y and~>0implies~x~y; x >- kz for all positive integers k implies not z >- 0.

A real linear functional u defined on all X will be called a utility if x ~ y implies u(x) > u(y), and x >- y implies u(x) > u(y). A vector valued linear function v defined on all X will be called a multi-dimensional utility if x ~ y implies v(x) ~ v(y) and x >-y implies v(x)>-v(y). Here the order on the vector space Received November 27, 1963. * Research partially supported by the U.S. Office of Naval Research under contract No. N 62558-3586. 229

230

YAKAR KANNAI

[December

(the space of values of v) is the lexicographic order, i.e., the vector v = (v~ .... , v,) is preferred or indifferent to w=(wl, ...,wn)if o = w or if v~>wifor the first coordinate i such that v~v~wi. (The lexicographic order satisfies (1.1), (1.2)but not(1.3)). Aumann [1] proved that if X is a finite-dimensional Euclidean space, the assumptions (1.1)-(1.3) imply the existence of a utility. He gave an example of a partial order satisfying these assumptions without having any utility, if X is the set of all infinite sequences of real numbers, and raised the problem of the existence of a utility if X has countable dimensionality (algebraic dimensionality). An example will be given here of an order on the space with countable dimensionality which does not have even a multi-dimensional (finite-dimensional) utility and, of course, does not have a (numerical) utility. A necessary and sufficient condition for the existence of a utility in the countable dimensional case will be given, and a sufficient condition for the existence of a utility if X is a seperable normed space will also be given. The author wishes to thank Mr. M. Perles for a number of very helpful conversations, and in particular for the present form of the example in section 3. 2. Some preliminaries. The following are simple consequences of the assumptions (1.1)-(1.3). (We denote the zero element of X by 0): (2.1)

x ~ y if and only if x - y ~ 0; x ~,-y if and only i f x - y ~ 0 ;

(2.2)

x ~ y and z ~ y and 0 < a < I imply x + (1 - a)z ~ y; x ~- y and z :>-y and 0 < ~ < I imply x + (1 - ~)z ~-y;

(2.3)

x ~ 0 if and only if - x ~ O, x >- 0 if and only if - x ~: O.

Set T = { x : x ~ - O } , S = { x : x

~0).

(2.4) If

A, B c X , s e t A + B = ( a + b ; a ~ A , b ~ B } , - A = ( x : - x ~ A }

T and S are convex cones, 0~ T, 0 E S and T c S.

(2.5)

{x : x ~ 0) -- S ~ ( - S) and is a linear subspace of X.

We remark that a real linear functional u defined on X so that u(x) ~_0 ff x E S, u(x) > 0 if x ~ T, is a utility. (It is obvious that a utility satisfies this). Similarly, a vector valued linear function v defined on X is a multi-dimensional utility if and only if x ~ S implies v(x)~ O, x ~ T implies v(x)),-0 (preference in the lexicographic order). TI-mOREMA. A necessary condition for the existence of a utility for the partial order ~ is that in every linear topology on X in which every linear functional defined on X is continuous, (2.6) * A

( - T) C3 g = ~ (which implies of course that ( - T) t 3 / ' = 12i).* superior bar denotes closure.

1963]

INFINITE DIMENSIONAL PARTIALLY ORDERED SPACES

231

Proof. Assume, on the contrary, that there is a point x e ( - T ) n ,-¢. From (2.3) it follows that u(x) < 0. In every neighbourhood U(x) of x there is a point y with y ~ S and so u(y) > 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

-<_£ + ,=, 2'[lup, II, + 1 ,=,~

(

llxll,---ll

,l], + 1

II ll 2'

'"

ll= |z

\j=:l

"=

x,=./ for /

) i_>m

.

234

YAKAR KANNAI

The first term on the right hand of this inequality is a constant for a fixed x, and the second term converges. For every x ~ T, u(x) > 0, since up.(x) >=0 for all n and there is at least one k with x ~ Vpk,so that upk(x) > 0. Thus u is a utility. From theorems A and B and by the above remarks it follows that (2.6) is both necessary and sufficient in the countable dimensional case. TrmOgEM C. I f X is a separable normed linear space, ~ is a partial order on X satisfying (1.1), (1.2), (2.6) then there is a (bounded) utility. (This includes the case of lp spaces, 1 <=p < oo.) Proof. By a theorem of Klee ([3 theorem (2.7)]) there is a functional u such that x e $ implies u(x) ~_ O, and u(x) > 0 for x ~ $ and x ~ - $ (hence u(x) = 0 for

xESn(-S)). It is easy to see, by the method of proof of theorem A, that (2.6) is also necessary for the existence of a bounded utility. P~'ERENC'ES 1. Aumann, R. L, 1962, Utility theory without the completeness axiom, Econometrica, 30, 445-462. 2. Hausner•M.••954•Mu•tidimensi•na•uti•ities•inDecisi•nPr•cesses•edit.byThra•••C••mbs and Davis, John Wiley, New York, pp. 167-180. 3. Klee, V. L., 1955, Separation properties of convex cones, Proc. Amer. Math. Soc., 6, 313-318. 4. K6the, G., 1960, Topologische Lineare Riiume, Springer Verlag.

UNAMBIGUOUS POLYHEDRAL GRAPHS* BY

BRANKO GRUNBAUM ABSTRACT

The existence of unambiguous d-polyhedral graphs is established for every d.

1. A graph G is called d-polyhedral provided G can be realized by the vertices and edges of a d-dimensional convex polytope I-3]. In general, a d-polyhedral graph may be dimensionally ambiguous, i,e., it may be also d'-polyhedral for d' ~ d (though this can not occur for d < 3 [3]). A polyhedral graph is unambiguous provided it is not dimensionally ambiguous, and provided for every two convex polytopes realizing the graph a biunique correspondence exists between their vertices in such a way that a set of vertices of one of the polytopes determines a face of the polytope if and only if the corresponding vertices of the other determine one of its faces. Recently, Klee [-5] disproved one of the conjectures of I-3] and established the existence, for every d, of d-polyhedral graphs which are not dimensionally ambiguous. Klee's proof is based on a new condition for d-polyhedrality. The aim of the present paper is to give a simpler proof and a slight sharpening of Klee's result, by proving the following THEOREM. For every d there exist unambiguous d-polyhedral graphs. The author is indebted to Victor Klee for many long and interesting conversations on polyhedral graphs. 2. Before proving the theorem, we collect some well-known definitions and facts, and state a few easily established assertions. If P is a d-dimensional convex polytope in Euclidean d-space E d, F a ( d - 1)face of P, and A a point, we shall say that A is beyond F provided A belongs to the open halfspace which has F in its boundary and which does not meet P. The following statements are easily established: (i) If P is a d-dimensional convex polytope and if A is a point of E d not belonging to P, there is a (d - 1)-face F of P such that A is beyond F (Weyl [6]). (ii) If A and B are vertices of a d-dimensional convex polytope P, joined by an edge of P, and if P0 is the convex hull of the vertices of P different from A,

Received January 3, 1964. * R~oarch ~tlpported in part by the National Science FoUndation, U. S. A. (NSF--(3P-378). 235

B. GRONBAUM

236

[Deeomber

then either A is beyond some (d - 1)-face of Po incident to B, or Po is (d - 1)dimensional. (iii) If an open halfspace H contains at least two vertices of a convex polytope then H contains an edge of the polytope. (iv) If a vertex V of a convex polytope P is beyond exactly one face F of the convex hull of the other vertices of P, then V is joined by an edge of P to each vertex of F. For the following notions and facts see Gale [1, 2], Klee [4], and the references given in those papers.

A cyclic polytope C(d, n) is the convex hull of n distinct points on the "moment curve" in E a, n >- d + 1, d > 2, given parametrically by (t, t2,t 3, ...,td). It is well known that C(d, n) is a d-dimensional polytope with n vertices, which is neighborly in the sense that every s < [d/2] of its vertices determine an ( s - 1)dimensional face of C(d,n). In particular, for d ~ 4, every pair of vertices of C(d, n) determines an edge. All the (d - 1)-faces of C(d, n) are (d - 1)-simplices. Their number re(d, n) is given by d+l

m(d,n) =

>

+ n-d

n-d

Let #(d, n) denote the maximal possible number of (d - 1)-dimensional faces for d-dimensional polytopes with n vertices. Obviously p(d,n)~_ m(d,n); it has been conjectured that equality holds in this relation for all d and n >=d + 1. Itis known that #(d,n) = m(d,n) if either n < d + 3 or n =>[(d + 1)/2] 2 - 1. Let C(d,n) be a cyclic polytope with vertices {Vi :i = 1,2, ...,n}; beyond each of its re(d, n) (d - 1)-faces Fj we take a point W~ sufficiently near to the centroid of F j, in such a way that in the convex hull K(d, n) of (V/:i = 1, ...,n}U(Wj :j = 1, ...,m(d,n)} each Wj is joined by an edge only to the vertices V~incident to Fj (then all the edges of C(d, n) are also edges of K(d, n)). We call K(d, n) a Kleetope derived from C(d, n). The graph of vertices and edges of K(d,n) shall be denoted by K*(d, n), its nodes by Vi*, 1 < i < n, and WT, 1 <=j < re(d, n) . . . . 3. We shall prove the theorem by establishing the following assertion:

1963]

U N A M B I G U O UPOLYHEDRAL S GRAPHS

237

For all n and d, such that d >=4 and n + 1 >=max {2d,[d/2]2}, the d-polyhedral graph K*(d, n) of the Kleetope K(d, n) is unambiguous. Proof. (i) K*(d, n) is not dimensionally ambiguous. Indeed, since each of the nodes W* is d-valent, K*(d, n) is not d'-polyhedral for d' > d. Assuming K*(d, n) to be realizable by a (d - 1)-dimensional polytope P, let Po be the convex hull of the vertices ~, 1 _< i < n, of P corresponding to the nodes Vi* of K*(d,n). By the above, Po has at most/l(d - 1, n) = m(d - 1, n) faces of dimension d - 2. By (i) above, each vertex Wj of P (corresponding to the node Wj of K*(d, n) is beyond at least one of the (d - 2)-faces of P0. Since no two vertices Wj determine an edge of P, (iii) implies that no two of those vertices may be beyond the same ( d - 2)-face of Po. Therefore m(d - 1, n) ~ re(d, n), in contradiction to the value of m(d,n) and the assumption n > 2 d - 1 . Since n > 2 d - 1 implies also re(d, n) > m((d - s), n) for every s > 1, the same reasoning shows that K*(d, n) is not (d - s)-polyhedral. Thus K*(d, n) is not dimensionally ambiguous. (ii) Let P' and P" be two d-dimensional polytopes realizing K*(d,n), with vertices V'i, Wj and V~, Wj. Let P0 be the convex hull of the vertices Vi of P', and P~ the convex hull of the vertices V[' of P". In each of P', P~, the vertex corresponding to the node W* of K*(d,n) is beyond at least one of the (d - 1)faces of P~ resp. Po, and vertices corresponding to different nodes W* are not beyond the same (d - 1)-face. Since P' and P" have each at most m(d, n) faces of dimension d - 1, each vertex Wj or Wj is beyond exactly one ( d - 1)-face of P~ resp. P6. By ((iv) above, that face has as vertices exactly those V'i's resp. V~' 's which correspond to nodes V* connected to the given W* by edges of K*(d,n). Therefore each (d - 1)-face of P~, and of P~, is a (d - 1)-simplex, and the correspondence of V: and W) to V:' and W~ shows that K*(d, n) is unambiguous. v

I/

tv

v

1

4. REMARKS. (1) The assertion of §3 can be established for some additional values of d and n. Thus, K*(5,6) is unambiguous. The argument is similar to the above, with the addition that in the present case Po = C(4, 6) and therefore all its 3-faces are simplices. It follows that each Wj is beyond at least two 3-faces of Po, which is impossible since C(4,6) has only 9 such faces. Even the 11-node 5-polyhedral graph, obtained from K*(5,6) by deleting one of the nodes W*, is not dimensionally ambiguous. (2) It is some of interest to note that although K*(5,6) is unambiguous, the graph of the polytope polar to K(5, 6) (in E s) is 4-polyhedral. (3) For a d-dimensional convex polytope C let K(C) denote the Kleetope derive from C, i.e. the polytope obtained from C by adjoining above each of its (d - 1)-faces a sufficiently flat pyramid. Let K*(C) denote the graph of K(C). CONJECTtrRE. For every C, the graph K*(C) is unambiguous.

238

n. GRUNBAUM

REFERENCES 1. Gale, D., 1964, Neighborly and cyclic polytopes, Proc. Symp. Pure Math., 7, 225-232. 2. Gale, D., 1964, On a number of faces of a convex polytope. Canad. J. Math., 16, 12-17. 3. Griinbaum, B. and Motzkin, T. S., 1963, On polyhedral graphs, Proc. Syrup.Pure Math., 7, 285-290. 4. Klce, V., 1963, On the number of vertices of a convex polytope, Math. Note No. 304. Boeing Sci. Res. Labs., June, 1963 (42 pages); to appear in the Canadian J. Math. 5. Klee, V., A property of d-polyhedral graphs, Math. Note No. 319. Boeing ScL Res. Labs., August, 1963 (7 pages); to appear in the J. Math. and Mech. 6. Weyl, H., 1934/35, Elementare Theorie der konvexen Polyeder, Comm. Math. Helv. 7, 290-306. UNIVERSITYOF WASHINGTON~SEATTLE AND ThE I-~Bm~w U N ~ S I T Y OF JERUSA~M

SCISSOR CONGRUENCE* BY

LESTER DUBINS, MORRIS W. HIRSCH AND JACK KARUSH ABSTRACT It is shown that certain simple figures can not be cut by scissors into pieces

that can be reassembled to form certain other simple figures. Bolyai has shown that every convex polygon of unit area can be cut by a finite number of line segements into a finite number of pieces which can then be rearranged to form the unit square (see, for example, [1]). We show that the only convex bodies that can thus be rearranged are polygons even if the scissors are permitted to cut along arbitrary Jordan curves. Similarly, a circle of radius two cannot be cut by Jordan scissors into pieces that can be reassembled to form four circles of radius 1. Variants of the problems studied here already occur in Euclid and have been studied up to recent times. (For reference, for example to work of Banach and Tarski, see [4] and [5]). This paper, though self-contained, is, in a sense, a sequel to that of Rodrigues [2]. An easily stated result is PROPOSITION 1. Suppose that E and E' are strictly convex planar bodies. Then E and E' are scissor-congruent if and only if they have the same area and their respective boundaries B and B" are scissor-congruent. Definition of scissor congruence. A topological disc D is the image of the unit disc under a homeomorphism of the plane onto itself, or equivalently, is the interior and boundary of a simple, closed Jordan curve. Let intD = interior of D, bd D = boundary of D, extD = exterior of D = complement of D, and let ~ be the empty set. Throughout the first 15 lemmas these notations are used. (a) Dt,--', Dn, D_ i , ' " , D_ n are 2n topological discs (included in a fixed 2-dimensional Euclidean plane n). (b) T1,..., T~ are n rigid motions (of n onto itself). (c) E+ is the set-theoretic union of D~ for i > 0; E_ is the union of Dl for i < 0. (d) K+ is the boundary of E+; K_ is the boundary of E_.

Received January 24, 1964. * The authors' workwas partiallysupportedby NationalScien¢~FoundationGrant Cr-14648 for Dubins, and Grant (3;-11594 for Hirsch. 239

240

L. DUBINS, M. W. H1RSCH AND J. KARUSH

[December

(e) E = E + U E _ ; K=K+kJK_. (f) J, is the boundary of Di for all i. Throughout the first 15 lemmas these four assumptions are implicit. (i) The image of Di under Ti is D_i, for i = 1, ..., n. (ii) int D i n int Dj = • for i > j > 0; int Di ~ int Dj -- • for i < j < 0. The joint assumptions (i) and (ii) are abbreviated to: E+ is scissor-congruent to E_. The next assumption is automatic in the most interesting case in which E+ and E_ are themselves topological discs (as in Proposition 1) or even finite unions of disjoint topological discs (as in Theorem 1). (iii) J,v'IJjV'IK+ is a finite set for i > j > 0; J, n Jj n K_ is a finite set for i < j < 0. The following assumption is of no mathematical importance and is made mainly to simplify some ensuing notation. (iv) E + n E_ = O. An arc A is the image of a connected subset A' of the circumference C of a circle under a homeomorphism (of the plane onto itself) except that the empty set, C, and one-point sets are not considered as arcs. A is open irA' is open in C. If K + is the disjoint union of a finite number of arcs A 1, "", A, and a finite number of points and K_ is the disjoint union of a finite number of arcs A_ 1, "", A_, and a finite number of points and for each i, 1 < i _< r, A_ ~ is the image of A i under a rigid motion R~, then K+ is scissor-congruent to K_.

Circles and squares are not rectifiably scissor-congruent. A proof that a circular disc S cannot be partitioned into pieces with rectifiable boundaries which can be rearranged so as to form a square S' will be given here. Though it seems impossible to modify this proof so as to apply to the case of nonrectifiable boundaries, we present it now because it is so simple and the underlying idea pervades the more complicated proof of Theorem 1. It may however be skipped without logical loss for it is superseded by the general argument below. If A is an arc on the boundary of a disc D, say that A is convex relative t o / ) if the line segment joining every pair of points of A is a subset of D; say that A is concave relative to D if the line segment joining every pair of points of A is disjoint from the interior of D. For each disc D, and each point x on the boundary of D, letfo(x ) = + 1 [respectively - 1] if there is an arc A containing x in its interior that is congruent to a subarc of the circular boundary of S and which is convex [concave] relative to D; letfD(x) = 0 otherwise. If the boundary of D is rectifiable, thenfo may be integrated with respect to the arc length measure determined by the boundary, obtaining thereby a number #(D). More generally, these definitions are applicable if D is any set such as E+ above. The measure # is easily seen to be invariant under rigid motions - - # ( D ) = tI(M(D)) for all isometries M " and to be

1963]

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additive --/~(D t W D2) =/.t(Dl) + g(D2) whenever the intersection of D 1 and D 2 consists of at most a finite number of arcs. These two properties of/~ imply that p(Dz) = #(D2) whenever Dz and D 2 are scissor-congruent. Since p ( S ) = 2zcR where R is the radius of S, and/1(S') = 0, S and S' are not scissor-congruent if only rectifiable cuts are admissible. This argument obviously applies to cubes and balls in any finite dimensional Euclidean space. The main proof. For any x, let N(x) mean neighborhood of x. The following topological properties are obvious for the unit disc and consequently hold for any topological disc D. For any x in the boundary of D and any N(x): (a) N(x) n i n t D # ¢ ; (b) N(x) n ext D # ~ ; (c) (N(x) - { x ) ) ~ bdD # ~; (d) There is an N'(x) c N(x) such that N'(x) n intD and N'(x) n extD are connected. Note the following. If Ax,A2,'" is a sequence of disjoint arcs c J, where J ( c plane) is homeomorphic to the unit circle C, then diameter A, ~ 0 as n ~ ~ . Since the homeomorphism between C and J is uniformly continuous it suffices to show this for J = C. But if J = C, the sum of the diameters of the A, is less than or equal to the sum of the arc lengths of the A, which, in turn, is at most 2n. This implies that no Jordan arc contains infinitely many, pairwise-disjoint, congruent subarcs. Notice also that K is the boundary of E. LEMMA 1. For all i, Ji -

U J~ c

K.

Proof. Let x e Ji - U Js. Since U JJ is dosed, for every e > 0 there is a j,i

j~i

discD(x) of radius less than e centered at x, such that D(x) oJs=t~, j # i, so

D(x) = (D(x) n int D~) W ( D(x) n e x t Ds). Since D(x) is connected, either D(x) c i n t D s or D(x) c e x t D s. Of course, D(x) n intD i # • and D(x) nextD~ # ~. Recall that intD i n intD s = ~ for j # i; consequently D(x) ~ extDj for j # i, and hence D(x) ~ extE # tI); so

xeK. DEFINITION. v =

U

i~j~k i~k

(s, n s j n

u

U (s, n i~j

n K).

As is not difficult to verify, (ii) together with (iii) imply: (v) V is a finite set. (At the slight cost of a not very important redundancy, (v) can be assumed instead of verified.)

242

L. DUBINS, M. W. HIRSCH AND J. KARUSH

LEMMA 2. Let A be an arc, A c ( J i - V ) . j#i.

Then A c K

[December

or A c J j f o r

some

Proof. By Lemma 1, A = ([..J A n Jj)to (A c~ K), so A is a union of sets closed in A whose pairwise intersections are in A n V = t~. Since A is connected, Lemma 2 is obvious. DEFINITIONS. For i > 0, T_~ = T~-~. It is convenient to introduce two transformations T and R defined for certain ordered couples (x, i). For x ~ Ji, let T(x,i) = ( T i x , - i ) . For x e J i n J j - V, with i ~ j , let R ( x , i ) = (x,j); since x ~ V, j is unique, so R is well-defined. Denote the identity map of a set X onto itself by Ix. LEMMA 3. (a) ~(R), that is, the domain of R, is the set of all (x,i) such that x ~Ji and xq~K t3 V. (b) ~(R), that is, the range of R, is the same as ~(R), and R 2 = 1~(~). (c) :~(T) = ~ ( T ) , and T 2 = l~tz). (d) ~((TR)~T) = ~((TR)kT), and ((TR)kT) 2 = l~to-x)kr), for each k ~_ O. Proof. (a) follows from Lemma 1; (b) and (c) are immediate; and (d) follows from (b) and (c), using induction on k. LEMMA 4. Assume: (a) x and y are in K, (b) ( x , i ) ~ ( ( T R ) k T), and (c) (y,j) ~ ~(( T R) h T). Then (TR)*r(x,i) = (TR)hT(y,j) only if k = h and (x,O = (y,j).

Proof. Suppose k < h. Then (TR)hT = (TR)*T(RT) h-k. By Lemma 3(d), (x, i) = (RT) h- k(y,j). If k = h, then (x, i) = (y,j). If k < h, then (x, i) e ~ ( R ) = ~ ( R ) by Lemma 3(c), so x ~ K by Lemma 3(a), which is a contradiction. DEFINITmN. W = {x: x ~ K and there exist i, k, v, i' such that v E V, (x,i) ~ ~((TR)* T) and (TR)* r(x, i) -- (v, i')}. LEMMA 5. W is finite. Proof. For every x ~ W, the set of aU (v,i') such that for some (i,k), (i,k,v,i') satisfies the conditions of the definition of W is nonempty. By Lemma 4, distinct x's correspond to disjoint sets. Hence, the cardinality of W does not exceed that of {(v,j) : v ~ V}, which is finite, since V is finite. LEMMA 6. K - ( V tO BO is a finite union of disjoint open arcs, each included in some Jl, and this decomposition is unique.

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Proof. Clearly, K = U J,, so K - (V L) W) = U [K t3 (J, - ( v u IV))]. Since i

t

V U W is finite, Ji - (V u W) is uniquely expressible as a finite union of disjoint open arcs (its connected components). By Lemma 2, every open arc in the decomposition of Ji - (V U W) is either included in K or is disjoint from K. Therefore, K C3( J ~ - (V U W)) has such a unique decomposition, and since the K rh (Ji - (V U W)) = (K - (V U W)) r3 dt are disjoint sets closed in their union K - (V L/W), the result follows. DErIrnXIOS. (1)

p(x) = X' if x,X' e K - ( V

U W) and there exist i, i', k such that

(x, i) e ~(( T R)kT) and ( T R )t T(x, i) = (x', i').

Here is a proof that p is well defined and p2 = l~tp). If ( x , i , k , x ' , i ' ) and (x,j,h,y',j') satisfy (1), then so do (x',i',k,x,i) and (y',j', h,x,j) by Lemma 3(d). Since both (x, i) and (x,j) are in ~ ( T ) , x e Jt ¢3 Jj. But x e K - V, so i = j . By Lemma 4, x' = y', so p is well-defined, and p2 = l~t~). D~INmON. Let ~¢ be the finite set of open arcs described in Lemma 6.

LV2at,IA 7. I f A e ~/, then A c ~(p), and, therefore, ~(p) = K - (V u W). Moreover, there exist io, il, ...,i k such that for 0 < r < k: (a) -- i, ~ i,+ l ; (b) A c J~o ; and, letting A, denote T~ ... TtoA, (c) A , ~ J - l , o J t , + , ; (d) p(A) = a k c J_,,, N (K - (V L) W)). Proof. Fix A e ~¢. Then A is a subset of some unique Ji, say i = io. Let (A,i) be the set of all (x,i) for which x c A . Then (A, i o ) c ~ ( T ) , and T(A, io) = (Ao, - io), where A o = TtoA is congruent to A. The immediate goal is to show that if (r, Ar, - i,) satisfies (2) or

(A,io) c 2((TR)'T) and (TR)" T(A, io) = ( A , - i~), then either A, c J _ ~ N ( K - ( V U W ) , (r + 1,A,+I, - i,+1) satisfies (2),

where A, c J _ t n J~,÷~, - i, ~ i,+~, and A,+ 1 = T~,+~A,. Suppose (r, A , , - i,) satisfies (2). Since A O W is empty, A, N V certainly is empty; since Ar is congruent to A, A, is an arc; and A, c J _ . . By Lemma 2, A, = K or A, c Jt,+, for some i,+1 # - i,. In the latter case, ( A , , - i,) c ~ ( R ) = 2 ( T R ) , so (A, io) ~ 2 ( ( T R f + I T ) and (TR)~+IT(A, io) = TR(A,, - i , ) = T(A,,i~+I) = (Ti,+ , A , - i , + t ) = ( A , + I , - i~+1). In the former case, it suffices to show that A, o W is empty. Suppose that it is not empty, and contains, say, w.

244

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[December

Then there exist j, k, o, j ' such that v ~ V, (w,j) ~ ~ ( ( T R ) k T ) , and (TR)kT(w,j) = (v,j'), which implies that w e J~ and that (w,j) ~ ~ ( ( T R ) k + l T ) . On the other hand, by Lemma 3d, ( w , - i , ) c ~ ( ( T R ) ' T ) and ( T R ) r T ( w , - ir)e(A,io) , which implies that ( w , - i,) q~((TR~+IT)). Since w e J _ ~ n K V, j = - i r. Consequently (since ~ ( ( T R ) ~ T ) is decreasing in k), k = r, which implies that (v,j') ~ (A, io). Since this contradicts A n V = O, the immediate goal has been achieved. The lemma will be proved once it is shown that (2) cannot hold for all r, for then k + 1 can be taken as the first r for which (2) does not hold. If (2) holds for all r, then A, c J - i , , a n d A , is congruent to A for all r > 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)

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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 <=k/2, D~om

and D-t~,~

are on the same side of A2,,

D~om

and Di~.. ÷, are on opposite sides of A2,, :~

O2m+l io a n a D_ t2., + ~are on opposite sides o f A2rn + I :e-

D2m+l to a n a- Dt2,.+2 are on the same side of

A 2 m + l ::~

2m+2 Dto and D _ ~ . ÷, are on the s a m e side of A2m + 2"

Since k is even if A ~ K+ _ and is odd if A ~ K + + or if A c K _ _, the L e m m a follows.

246

L. DUBINS, M. W. HIRSCH AND J. KARUSH

[December

It is now easy to obtain various results of intuitive interest by specializing E+ and E_. For example, Lemma 14 is immediate after this preliminary. LEMMA 13. I f E+ is a finite union of disjoint strictly convex bodies, (each of which is necessarily a union of some of the Di) then K+ + is empty. Proof. Suppose K+ + ~ • and let A = K+ +. In the notation of the above proof, D~o and D-tk are on opposite sides of p(A). Since Dto and D-tk are connected, there exist D and D' in the assumed decomposition of E+ such that Dto = D and D-tk C D'. Plainly A c bdD because A = A N bdE+ = A ~ (bdD U bd(E+ - D)) = A n bd D; similarly, p(A)c bdD'. Let Dk = Tt~ ..- T~oD. By Lemma 10, Dk and D' are on opposite sides of iv(A). Let x E p(A). For any N(x), there is a y ~ (N(x) - {x}) n p(A). Since Dk and D' are strictly convex, ½x + ½y ~int Dk I"1int D' which is in contradiction with D ~ and D' being on opposite sides of p(A). L~MMA 14. I f E+ and E_ are finite unions of disjoint strictly convex topological discs, then K+ is scissor-congruent to K_. Proof. Immediate from Lemma 13. Other conclusions are similarly easy to derive. For example, the only convex body that is scissor-congruent to a polygon is itself a polygon. In particular, the circle is not scissor-congruent to the square. Though Jordan arcs can have positive 2-dimensional Lebesgue measure, it is easy to verify: LEMMA 15. I f K has two-dimensional Lebesgue measure zero, then E+ and E_ have the same area. LEMMA 16. Let E+ and E_ be finite unions of disjoint compact convex planar bodies. I f E+ has the same area as E_ and K+ is scissor-congruent to K_, then E+ is scissor-congruent to E _ . Proof. The scissor congruence of K+ and K _ , implies the existence of convex arcs A1, "-',An, At, -'-, A" and rigid motions M1, ...,M, such that: K+ = u At ; K_ = u A~ ; Mr(At) = A~ ; At n A~ and A " n A/consist of at most one point each for i ~ j . Let Pt, ...,P, and P~, ...,P" be the end points of the arcs As, ...,A, and A'~,...,A" respectively. Suppose first that E+ and E_ are themselves convex bodies. The Pt and P~ are the vertices (not necessarily in order) of convex polygons P and P' inscribed respectively in the convex bodies E+ and E_ respectively. Each arc At together with the chord joining its end points determines a sector St of E+. Plainly Mt maps St onto S~. Consequently, St and S't have the same area as do their unions U St and U S [ Therefore, E+ - U St has the same area as E _ - u S~', that is, the interiors o f P a n d P ' have the same area. Since P aild P' have the same area, Bolyai's theorem applies to show that P and P' are scissor-

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congruent. Since a scissor congruence of P and P' and isometries of S~ onto S~ for all i determines a unique scissor congruence of E+ with E _ , the proof is complete if E+ and E_ are convex. The argument is easily modified to handle the general case. T~mOR~M 1. Suppose that E+ and E_ are finite unions of disjoint compact strictly convex planar bodies. Then E+ and E_ are scissor-congruent if and only if they have the same area and their boundaries are scissor-congruent. Proof. Apply Lemmas 14, 15, and 16. Since the only convex body whose boundary is scissor-congruent to a circle, •s a congruent circle, one gets

A circular disc is scissor-eongruentto no other strictly convexibody. For a slight generalization of Theorem 1 and of the italicised statement appearing after Lemma 14, introduce two definitions: an arc is elementary if it is either strictly convex or a straight line segment; a convex body is elementary if its boundary consists of a finite number of elementary arcs. COROLLARY.

PROPOSITION 2. An elementary convex body E+ is scissor-congruent to a convex body E_ if and only if E_ is elementary, E_ has the same area as E+, and the strictly convex portion of the boundary of E+ is scissor-congruent to that of E_. We are grateful to Glen Bredon for showing us how to remove "strictly" from the Corollary, and to Branko Griinbaum for subsequently pointing out to us that Proposition 2, together with an interesting result of Blaschke 1-3, Chapter II, §6], implies a considerable further improvement of the Corollary, namely: THEOREM 2. I f an ellipse E+ is scissor-congruent to a convex body E_ then there is a rigid motion carrying E+ onto E _ . We do not know whether any convex body other than an ellipse is scissorcongruent to no convex body other than itself. Nor do we know how to formulate and prove a theorem in the spirit of Proposition 1 to the effect that a cube in three dimensions is not scissor-congruent to a ball. REFERENCES I. Boltyanskii,V.G., 1963, Equivalent and EquideeomposableFigures, D.C. Heath and Co., Boston. 2. Enriques, Federigo, 1912, QuestioniRiguardantile Mathematiche Elementari, Volume 1, Articoli di U. Amaldi, "Sulla teoria della equivalenza", Bologna, Nicola Zanichelli,pp. 145-198. 3. Fejes T6th, L., 1953, Lagerungen in der Ebene, auf der Kugel und im Raum, Springer. 4. Hadwiger, H., 1957,Vorlestmgeu fiber Inhalt, Oberfl~eheund Isoperimetrie, Springer. 5. Sierpinski, Waelaw, 1954, On the Congruence of Sets and Their Equivalence by Finite Decomposition, Lucknow University. UNIVEI~ITY OF CALIFORNIA BERKELEY, CALIFORNIA

A PROPERTY OF PLANAR CONVEX BODIES BY JACK G. C E D E R ABSTRACT It is proved that from each interior point of a planar convex body emanate

three distinct vectors terminating on the boundary, the sum of any two of which also terminates in the boundary. Some other related results are obtained. In this article we prove that from each interior point of a planar convex body emanate three distinct vectors terminating on the boundary, the sum of any two of which also terminates o n the boundary. (See Figure 1). We also show the related result that through each interior point of a planar convex body pass the boundaries of three distinct translates which cover the body in the sense of property fl below. (See Figure 2). In addition, we investigate the nature of the set of points on the boundary which can be end points of such triples of vectors and their pair sums. Finally, we pose some related unsolved problems. We begin by saying that a planar body C has:* property ct at p ~ C ° if there exist distinct points x,y, z eBdC such that (x - p) + (y - p), ( x - p) + (z - p) and ( z - p ) + ( y - p ) also belong to BdC. (See Figure 1).

Figure 1

property fl at p ~ C ° if there exist distinct translates, C1, C2, and C3, such tha n~=iBdCi={p} and the set B d C n B d C ~ n B d C j for each i:/:j consists of exactly one point. (See Figure 2). There exists a natural relationship between the three points of property ~t and the three translates of property fl as given by the following lemma: LEMMA 1. (1) I f property ~ is satisfied at p e C ° with x, y, and z, then property fl is satisfied with the translates C + ( p - x), C + ( p - y ) and C + ( p - z ) . Received January 9, 1964. * The symbols C ° and BdC denote the interior and boundary of C respectively. 248

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(2) I f C is rotund (ie, there are no line segments in BdC) and property fl is satisfied with the translates C + ai(i = 1,2,3), then property ct is satisfied with the points p - a~ (i = 1, 2, 3).

Figure 2 Proof. The p r o o f of (1) is obvious. For the p r o o f o f (2), let us assume that p = 0 and that C + a 1, C + a 2 and C + a 3 are in consecutive counterclockwise order around BdC. Define xl by {xi} = BdC n Bd(C + ai) n B d ( C + a~+l). Then to obtain property c~ it clearly suffices to show that x~ = - a i + 2 (rood. 3). Without loss of generality we can assume that we have x2 < - a l < xl < - a 3 or - a l < x2 < - a a _~ xl, where the order is taken to be counterclockwise around BdC. Consider the first case (the second case will be similar) and the resulting two pairs of two parallel and equal (in length) line segments ['0, - a3"] and [ x 2 , x 2 - a 3 ] ; [ 0 , - a l l and [ x l , x l - a l ] . From the convexity of C and the absence of line segments in BdC it follows that these four segments must form a parallelogram, in which case x I = - a 3 and x2 = - a ~ . This in turn forces x3 = - a2, proving the Lemma. It is easily seen that the relationship of part (2) in the Lemma need not hold for non-rotund bodies (e.g., a parallelogram). Now we prove our main result:

TrmogE~ 1. Properties ~ and fl are satisfied at each interior point of a planar convex body. Pr~mf. By Lemma 1 it suffices to show only property ~ at each interior point. We will first prove this for a rotund body and then pass to the limit for the general case. Suppose C is rotund and that p = 0 E C °. Let x~ be an arbitrary point in BdC. Then there will be exactly two chords of C equal and parallel to the segment

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[0,xt]. Taking the chord on the counterclockwise side o f xi we will obtain a

Yl ~BdC such that x~ + yx ~BdC. Repeating this process (counterclockwise around BdC) we also obtain a z I ~ BdC such that Yl + zl e BdC, and a x2 ~ BdC such that zl + XzZ BdC. I f x~ = x2, then we have our desired triple o f points. Otheroo oo wise we continue this process to obtain sequences {X .}.=1,{Y.}.~=1 a n d { Z .}.=1 such that for each n, x. + y., y. + z. and z. + x. ÷ ~ belong to BdC. Now, since x: # x2 and C is not a parallelogram, we obtain two cases: either xl < x2 < zl or Yl < x2 < xl. Case I. In case xl < x2 < zl, let us prove by induction that for each k

x1< ~_ x~ < _~

Xk+ l

< zt

Yl < Yk ~- Yk+l < xt and

zl
.~ Zk+ 1


For k = 1 we have x~ < x2 < Zl by assumption and the inequalities y~ < Y2 < Xl and z~ < z2 < y t follow f r o m this. N o w assume that the above proposition holds for all i < k. Then f r o m the fact that zt < zk+: < yt it clearly follows that x~ <xk+~ < x k + 2 < z l . F r o m this in turn follow the other two inequalities Y ~< Y t + x < = Y=~ + 2 < x l a n d z ~ = zk+~ =< zk+2 < Y l , which proves the proposition f o r / c + 1. Hence, each o f the three sequences {x.}.~= 1, {Yn}~=1 and {z.}.~ 1 is a m o n o t o n i c sequence contained in a closed proper subarc o f BdC. Therefore, the sequences will converge to distinct points x, y and z respectively. Taking limits we also obtain x + y, x + z and y + z ~ BdC. Thus, property • is satisfied in Case I Case II. In case y: < x 2 < xt we first need to prove the following lemma: LEM~A 2. If there are three consecutive translates of the rotund body C passing through p e C ° (i.e., there exist translates C 1, C2 and C3 such that n~=l BdC ~= {p} and BdC~n BdCi+ I n BdC =/=A for i = 1,2) which fail to cover BdC (i.e., B d C - u~= tCt ~ 0), then property fl is satisfied at p. Proof. Suppose p = 0 and Ct is the first translate in the counterclockwise direction f r o m the arc BdC - U~31 C~. F o r x ~ BdC let C(x) be the first translate o f C in the counterclockwise direction from x which has [0, x] as a chord. Clearly each translate can be so expressed. Also let x' denote the other point of intersection o f BdC with BdC(x). Let al be such that C 1 = C(al). Put bl = a~ and cl = bl. Then C: C(bl) and Ca C(ci). Then put a 2 = cl, b 2 = a 2 and c2 = b2. ® Continuing in this way we obtain sequences {a .}~_-~, {b.}.~ ~ and {c~}.~ 1 such that a~ + 1 = c', b. + 1 = a" + 1 and c. + ~ = b'. + x. The assumption that Cx, C2 and Ca fail to cover BdC means that a~ < a2 < c~. Now (using the fact that each two distinct translates of C whose boundaries contain 0 intersect in two distinct arcs in BdC, no one of which is contained in the other) employing the same argument as we did in Case I we see that the sequences {a.}.= t, {b.}.= ~ and {c.}.~= t converge to distinct a, b, c respectively. Consequently, the three sequences of translates {C(a.)}~= 1, {C(b.)}.~=~ and {C(c.)}.~=~ converge to the translates

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C(a), C(b) and C(c) respectively which obviously satisfy property fl, finishing the p r o o f of Lemma 2. Now consider the three translates C - x 2, C - y x , and C - z 1. Then {Zl} =Bd(C-x2)('l Bd(C-yl) and the two points o f B d C n Bd(C-Zl) lie outside the arc BdC n ((C-x2) U ( C - y l ) ) . Then clearly C - x2, C - Yl and C(xl) will constitute three consecutive translates passing through 0 which fail to cover BdC. Hence, by Lemma 2, property fl, and therefore property ~ by Lemma 1, is satisfied at 0. Thus, property ~ holds in Case II. Now we consider a general convex body C and let p = 0 e C °. Then we can easily find a sequence of rotund bodies {C n}n= 1 such that 0 e C O ~ C for each n and C is the limit of {Cn}~= 1 in the Hausdorff metric. For each n there exist xn, yn and z~ in BdC for which property ~ holds with respect to Cn. Since C is sequentially compact, we can then find a subsequence k such that {Xkn}n°°=1, {Yk,}n°~=1 and {Zkn}nc°=1 converge to points x, y, z respectively in ndC. Moreover, x, y and z are all distinct. For suppose x = y, then since 0 e arc (xkn, Yk) c_BdCk, we have, by taking limits, 0e BdC, a contradiction. Also by taking limits we have x + y, y + z and x + z in BdC. Therefore, property ~ is satisfied with x, y and z, which finishes the p r o o f of the Theorem. If property ~ is satisfied at 0 by a triple x, y and z, then x, x + y, y, y + z, z and x + z determine an inscribed centrally symmetric hexagon. (See Figure 1)*. Let us denote by V the set o f all points on BdC which are vertices of some such inscribed centrally symmetric hexagon. The set V may not be all of BdC; for example, the vertices of a triangle are not in 11. In the case C is rotund and differentiable, it seems reasonable to conjecture that V = BdC. However, the best we have done in trying to settle this is the following theorem: THEOREM 2. Suppose C is rotund and differentiable. Then ( 1 ) B d C - V consists of isolated points (and hence is countable) and (2) if x e B d C - V , then x' ~ 17, where x' is the unique point z so that [x,z] is a diameter of C. Proaf. Let x be any paint in BdC amnd let Ux be an open arc o f BdC which contains x, but small enough so that no diameter of C intersects it twice. Let y be to the clockwise side o f x in Ux. Denote by [a,b] the other chord o f C equal and parallel to [x,y]. Then we have nine possible cases:

1. 2. 3. 4.

a
5. 6. 7. 8. 9.

y' < a < b = x ' a
* It is well known that in each convex body one can inscribe an affme-regular (hence, centrally symmetric) hexagon. For references to this and other results on such inscribed hexagons s ~ Grtinbaum [1; p. 242].

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Correspondingly, we get nine cases if y is to the counterclockwise side o f x in U=. Now we need three lemmas: LEMMA 3. y • V °.

Suppose one of the first f o u r cases prevails. Then x • V ° and

Proof. Suppose Case I holds and that y is to the clockwise side o f x in Ux. The other cases will be similar. Consider the two arcs arc (x,y) and arc ( b , a ) i n BdC. Next "reflect" one upon the other so that a coincides with y. Considering the direction of the support lines at x, y, x', y', a and b, we see that arc (x,y) must intersect the reflection of arc (b, a) in a point other than x and y. Then we obtain in the obvious way an inscribed hexagon with opposite sides equal and parallel. By the rotundity of C, its vertices, including x and y, must belong to V. Now if we consider x fixed, there will obviously be a neighborhood W of y such that the Case I configuration holds relative to x and any pointin W. Hence, y • V°. Now fixing the y, we can find a neighborhood W of x such that the Case I configuration holds with respect to any point in W and y. Hence, x • V°, which finishes the p r o o f of the Lemma. LEMMA- 4. Suppose there are three distinct y in Ux f o r which either one of cases, 5, 6 or 7 prevails (i.e., b = x'). Then x • V. Proof. We can assume without loss of generality that two o f these points, say Yl and Y2, are to the clockwise side of x. Then it is easily seen that x , y l , y 2 and x' comprise four vertices of an inscribed centrally symmetric hexagon. LEMMA 5. Suppose there exists an open arc N containing x such that for any y • N - { x } , cases 8 or 9 prevail. Then there exists an open arc M containing x such that M - {x} _~ V. Proof. Let A = {y • N - {x} : y ' < b < x'} and B = {y • N - {x}: y' < x' < b}. Let y~ A. Then reflecting arc (y, x') upon the appropriate opposite arc we get y •V. Moreover, the configuration of case 8 will prevail in a small neighborhood of y. Hence, A is open and contained in V°. Now if y • B, the configuration of case 9 prevails in a small neighborhood of y, so that B is open, too. Therefore, if arc (x,z) N, then arc (x,z) is contained entirely in A or B. If arc(x,z) G B, then for x < y < z let [e(y),f(y)] be the other chord which is equal and parallel to [y',x']. Clearly, e is a continuous function of y. And if we reflect arc (y', x') upon arc (e(y), f ( y ) ) we get e(y) • V with x < e(y) < y. Hence, the range of e, restricted to arc(x,z), must be an interval of the form arc(x,w) which is contained wholly in V. Similarly, we take care of the case when arc (z, x) _ N. Hence, there exists an open arc M such that M - {x} ~ V. Now for the proof of part (1) o f the Theorem we let x be any fixed point not in V. Then by applying Lemmas 3 and 4 we can find a neighborhood W of x such that

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for any y e W case 8 or 9 prevails. Then applying Lemma 5 we can find a deleted neighborhood o f x which is contained in V. Hence, the points o f B d C - V are isolated and consequently, B d C - V is countable. For part (2), let x ~ B d C - V and choose y to the clockwise side of x in Ux. Then case 8 or 9 prevails. In case 8, reflecting the arc(y, x') upon the appropriate arc(c,d) we get x ' e F. And in case 9 we reflect a r c ( a , x ' ) upon the appropriate arc(c,d) to get x ' e V. Hence, x ~ B d C V implies x ' ~ V, finishing the p r o o f of the Theorem. There remain some interesting and seemingly difficult unsolved problems relating to the nature of V in BdC. Namely, (1) is V = B d C when C is rotund and differentiable; and (2) in the general case, is V dense in B d C or is B d C - V even countable? Also property ~ can be generalized to any closed curve C in the plane to : C has property ct at p if there exist distinct x, y, z in C such that (x - p) + ( y - p), ( x - p) + (z - p) and ( y - p) + (z - p) are also in C. Some unsolved problems are (3) is there such a point for each closed curve C; (4) if so, what is the nature of such points; and (5) can convex closed curves be characterized as those closed curves having property ~ at each interior point? REFERENCES 1. Gr'tinbaum, B., 1963, Measures of symmetry for convex sets, in: Convexity, Vol. 7 of Proe. of Symposia in Pure Math. A.M.S., pp. 233-270. UNIVERSITY OF CALIFORNIA, SANTA BARBARA