an application of spanning trees to k-point separating

AN APPLICATION OF SPANNING TREES TO K-POINT SEPARATING FAMILIES OF FUNCTIONS W. Edwin Clark Gregory L. McColm Boris Shekhtman Department of Mathematics University of South Florida Tampa, FL 33620-5700, USA

January 23,1992 (Revised May 15, 1996) Abstract. A family F of functions from R n to R is k-point separating if for every k-subset S of R n there is a function f 2 F such that f is one-to-one on S . We show that if the functions are required to be linear (smooth) then a minimum k-point separating family F has cardinality n(k ? 1). In the linear case we extend this result to a larger class of elds including all in nite elds as well as some nite elds (depending on k and n). We also obtain some partial results for continuous functions on R n including the case when k is in nite. The proof of the main result is based on graph theoretic results that have some interest in their own right: Say that a graph is an n-tree if it is a union of n edge-disjoint spanning trees. We show that every graph with k  2 vertices and n(k ? 1) edges has a non-trivial subgraph which is an n-tree. We also establish a determinantal criterion for a graph with k vertices and n(k ? 1) edges to be an n-tree. 1. Introduction.

A family F of functions from R n to R is k-point interpolating if for every ksubset S of R n and every function f : S ! R there is a function f 0 2 F such that f 0jS = f . An unsolved problem of approximation theory is the determination of the smallest dimension, CIk (R n ; R), of a k-point interpolating subspace F of continuous functions from Rn to R [3]. This problem is unsolved even for n = 2. We investigate a related problem that was inspired by this one, namely, suppose we have the less ambitious goal of obtaining a family F of functions such that for each k-subset S of R n there is a function f 2 F such that f is injective on S . How small can F be? Key words and phrases. graph, tree, interpolation, separation, determinant, linear functional, smooth function, continuous function.

Typeset by AMS-TEX

We say that a function f : X ! Y separates a subset S of X if f is injective on S . A family F of functions from X to Y is k-point separating if every k-subset S of X is separated by some f 2 F. Our principal result is the determination of the cardinality LSk (R n ; R) (respectively, DSk (R n ; R)) of a smallest k-point separating family of linear (respectively, smooth) functions from R n to R . We prove that if n; k  2 then LSk (R n ; R) = DSk (R n ; R) = n(k ? 1): Our proof that n(k ? 1) is an upper bound for LSk (R n ; R) is based on graph theoretic results presented in Section 2 which have some interest in their own right: We say that a graph is an n-tree if it is a union of n edge-disjoint spanning trees. Such a graph with k vertices clearly has n(k ? 1) edges. Using a result of NashWilliams [10,11], we show that every graph with k vertices and n(k ? 1) edges has a non-trivial subgraph which is an n-tree. We also establish a determinantal criterion for a graph with k vertices and n(k ? 1) edges to be an n-tree. In Section 3 we prove that LSk (F n ; F ) = n(k ? 1) holds for a large class of elds F , including all in nite elds and even some nite elds (depending on n and k). This result implies in particular that DSk (Rn ; R)  n(k ? 1). So to prove equality here we only need to establish n(k ? 1)  DSk (Rn ; R) which we do in Section 4. Finally, in Section 5, we prove some results on CSk (R n ; R), the cardinality of a smallest k-point separating family of continuous functions. We nd that CSk (R 2 ; R)  k for k  2 and CSk (R n ; R) > b k2 c(n ? 1) for n; k  2. We also consider the case where k is in nite. We prove that for any ordinal number  such that @ < jRj, if n is nite, then CS@ (Rn ; R ) = LS@ (R n ; R) = @+1. 2. Graph Theoretic Results

2.1 n-Trees. In this section G = (V; E ) will denote a graph with vertex set V and edge set E . We allow multiple edges but no loops in our graphs. Recall from graph theory that a tree is a connected acyclic graph. We say that G is an n-tree if there is a partition fE1; : : :; Eng of E such that for every i the graph (V; Ei) is a spanning tree of G, i.e., G is a union of n edge-disjoint spanning trees. (The concept of an n-tree, but not the name, was introduced by Nash-Williams [10,11]. Our notion of an n-tree should not be confused with that of Harary and Palmer [7, 3.5, p.73] which is an entirely dierent concept.) We will need a theorem of Nash-Williams. Recall that a forest is an acyclic graph, viz., a disjoint union of trees { a forest is connected if and only if it is a tree. For every subset X  V , let EX denote the set of all edges of G both ends of which lie in the set X . Nash-Williams' Theorem. ( [11]) A graph G = (V; E ) is decomposable into n forests if and only if for every non-empty subset X of V we have

(1)

jEX j  n(jX j ? 1):

Theorem 1. If jE j = n(jV j ? 1) and jV j  2, then the graph G = (V; E ) contains

a subgraph with at least two vertices which is an n-tree. 2

Proof. Our theorem follows by induction on jV j. If jV j = 2, then E contains n edges. Each edge joins the two vertices in V and therefore determines a spanning tree. So G is a union of n spanning trees and is therefore itself an n-tree. Assume jV j  3 and the theorem holds for graphs with less than jV j vertices. Suppose there is a proper subset X of V with jX j  2 and jEX j  n(jX j? 1). Then by removing some edges from EX if necessary we obtain a subgraph (X; E 0) of (V; E ) satisfying jE 0j = n(jX j? 1) and jX j < jV j and by induction we are done. Otherwise, for all proper subsets X of V with jX j  2 we have jEX j < n(jX j?1). By hypothesis (1) holds when X = V and since we have no loops, (1) holds trivially when jX j = 1. Hence (1) holds for all subsets X of V . We conclude from Nash-Williams' Theorem that G decomposes into n forests, say, Fi = (Vi ; Ei), i = 1; : : :; n. This means that V = V1 [


[ Vn , E = E1 [


[ En , and Ei \ Ej = ? for i 6= j . A forest Fi with !i components and pi vertices has pi ? !i edges. Since the edges of the forest partition the edges of G we have

n(jV j ? 1) = (p1 ? !1 ) +


+ (pn ? !n ): Since pi  jV j and !i  1 it is clear that pi = jV j and !i = 1 for all i. Hence each Fi is a spanning tree and consequently G is itself an n-tree. By induction, we are done.  2.2 A Determinantal Criterion for n-trees. We let G = (V; E ) be a graph with p  2 vertices and q edges. We assume q = n(p ? 1), n  1. This is obviously necessary for G to be an n-tree. We x an ordering v1 ; v2 ; : : :; vp for the vertices of G and an ordering

e1 ; e2 ; : : :; eq for the edges of G. For each such graph we will de ne a q  q matrix M (G) with entries which are polynomials in several variables over the rationals so that det M (G) 6= 0 if and only if G is an n-tree. In case G is an n-tree the polynomial det M (G) will be a kind of generating function for the ordered partitions of G into n spanning subtrees. First we de ne the q  p matrix A = [aij ] where 8 >
?1; if ei has endpoints vj , vk where k < j : 0; otherwise Fix ` 2 [p] = f1; : : :; pg. Let K be the q  (p ? 1) matrix obtained by deleting the `-th column of A. We shall see in the next lemma that the choice of ` is not important for our purposes. This lemma is well-known (see, for example, [1], Propositions 5.3 and 5.4) and so we omit its proof. 3

Lemma 1. Let T be any (p ? 1)-subset of [q ]. Let KT denote the (p ? 1)  (p ? 1) submatrix of K whose row indices are in T . Let GT be the spanning subgraph of G with edge set fei ji 2 T g. Then det KT 2 f0; 1; ?1g and det KT = 1 if and only if GT is a spanning tree in G. In particular jdet KT j is independent of `. Let fmij ji = 1; : : :; n; j = 1; : : :; qg, be a set of nq real numbers which are

algebraically independent over the eld Q of rational numbers. Set

Xi = diag[mi1 ; mi2 ; : : :; miq ]

(3)

for i = 1; 2; : : :; n and de ne M = M (G) to be the matrix in block form:

M = [X1K jX2K j : : : jXnK ]:

(4)

Here G is a graph with p vertices and q = n(p ? 1) edges and K is the q  (p ? 1) matrix de ned above. Hence XiK is a q  (p ? 1) matrix and M is a q  q matrix. Note that each row of A and therefore of K corresponds to an edge of G and so to simplify notation we identify the edge ei with row index i of K . We say that the (p ? 1)-set T  [q] = E is a tree if the edge-induced subgraph is a tree. Let (T1 ; T2; : : :; Tn ) be an ordered partition of E such that each Ti is a spanning tree of G. Let Ti = fji1; ji2; : : :; jip?1g: Form the monomial

m(T1 ; T2 : : : ; Tn) =

n Y i=1

miji1 miji2 : : : mijip?1 :

For distinct partitions we obtain distinct monomials; hence these monomials are linearly independent over Q . Theorem 2. If G is a graph with p vertices and q = n(p?1) edges then det M (G) 6= 0 if and only if G is an n-tree. If G is an n-tree, then (5)

X

det M (G) =

T ;T2 ;:::;Tn )

(T1 ; T2; : : :; Tn )m(T1 ; T2; : : :; Tn )

( 1

where (T1 ; T2; : : :; Tn ) runs over all ordered partitions of E into disjoint spanning trees and (T1 ; T2; : : :; Tn ) 2 f1; ?1g for all (T1 ; T2; : : :; Tn ). Proof. This theorem is a consequence of Lemma 1 and the following generalization of Laplace's expansion for determinants. Let M be any q  q matrix and let R; C  [q], then MR;C denotes the submatrix of M whose rows have indices in R and whose columns have indices in C . Laplace's expansion of det M for our purposes may be stated as follows (see, e.g., [9], Th. 12, p. 564): For a xed ordered partition (C1 ; C2) of [q]

(6)

det M =

X

R ;R2 )

(R1 ; R2; C1; C2)det MR1;C1 det MR2;C2

( 1

4

where (R1; R2) runs over all ordered partitions of [q] into two sets such that jRi j = jCi j, i = 1; 2, and (R1 ; R2; C1; C2) 2 f1; ?1g. This generalizes by induction to the following. If (C1 ; : : :; Cn) is any xed partition of [q], then (7) det M =

X

R ;:::;Rn )

(R1; : : : ; Rn; C1; : : :; Cn )det MR1 ;C1 : : : det MRn ;Cn

( 1

where (R1; R2; : : :; Rn) runs over all ordered partitions of [q] into n sets such that jRij = jCi j for all i and (R1 ; : : :; Rn; C1; : : :; Cn ) 2 f1; ?1g. To apply this to the q  q matrix M = [X1K jX2K j : : : jXnK ]; we take C1 = f1; 2; : : :; p ? 1g C2 = fp; p + 1; : : :; 2(p ? 1)g ::: Cn = f(n ? 1)(p ? 1) + 1; : : :; n(p ? 1)g: Note that Ci has p ? 1 elements corresponding to the column indices of Xi K . Now if Ri  [q] = E is a (p ? 1)-set with Ri = fj1 ; : : :; jp?1g we obtain det MRi ;Ci = det (diag[mij1 ; : : :; mijp?1 ]KRi ): Hence by Lemma 1 (

det MRi ;Ci =

(8)

 mij1 : : : mijp?1 ; 0;

if Ri is a spanning tree otherwise.

From (8) and (7) we immediately obtain (5). Then since the monomials m(T1 ; : : :; Tn ) are linearly independent, det M (G) 6= 0 if and only if G is an n-tree.  In this paper we are primarily interested in families of functions from R n to R . However some of our results hold when R is replaced by other elds. For background material on eld theory see, e.g., [8]. We now introduce a de nition that allows us to extend our results to a wider class of elds. The elements mij may be replaced by eld elements which are not algebraically independent over the ground eld: Definition Let X be a nite subset of a eld F . For each subset S of X de ne 8Y < s; if S 6= ? (S ) = : s2S 1 ; if S = ? ?


Let Xk denote the set ?of
all k-subsets of X . We say that X is k-boolean independent if for all functions  : Xk ! f0; 1; ?1g   X (S )(S ) = 0 =) (S ) = 0 for all S 2 Xk . S 2(Xk ) 5

Note that if X is k-boolean independent and k0  k, then X may not be k0 boolean independent. Example 1. Let F = K (x1 ; : : :; xn ) where x1 ; : : :; xn are algebraically independent over K . In this case X = fx1; : : :; xn g is k-boolean independent for 1  k  n. Example 2. Let F be a eld and let F0  F1  F2 


 Ft = F be a chain of sub elds where Fi = Fi?1 (i) for i = 1; : : :; t and [Fi : Fi?1 ] = ni  2. Let X = f1 ; : : :; tg. Since for each i, f1; i; i2; : : :; ini?1 g is a basis for Fi over Fi?1 , it follows that f(S )jS  X g is linearly independent over F0 . In particular, X is k-boolean independent for all 1  k  t. For example: (a) Let F = Q (ppp1 ; ppp2 ; : : :; pppn ) where p1 ; p2; : : :; pn are distinct primes. p1 ; p2 ; :p: :; p pn g is k-boolean The set X = f p independent for 1  k  n p since for each i, pi 2= Q ( p1 ; p2 ; : : :; pi?1 ) (see [12]). (b) Let F = GF (pn ) where p is any prime and n = n1 n2 : : :nt , ni  2 for all i. Then there is a tower of sub elds F0  F1  F2 


 Ft = F where F0 = GF (p), F1 = GF (pn1 ), F2 = GF (pn1 n2 ), : : : , Ft = GF (pn ). Then for i = 1; : : :; t there exists i 2 F such that Fi = Fi?1 (i ). Hence X = f1; : : :; tg is k-boolean independent for 1  k  t. Example 3. For integers n  k  1, let X = fm1 ; : : :; mn g be a set of positive integers chosen so that   n mk < m i+1 bnc i 2

for i = 1; 2; : : :; n ? 1. It is easy to prove by induction on t that X is a t-boolean independent subset of Q whenever 1  t  k. Example 4. Let F contain an element x which is transcendental over the prime eld of F . Let m1 ; m2; : : :; mn be a sequence of positive integers satisfying m1 + m2 +


+ mi < mi+1 ; for 2  i  n ? 1: It is easy to see that any integer m = mi1 +


+ mik where i1 <


< ik uniquely determines the summands mi1 ; : : :; mik and since the powers of x are linearly independent over the prime eld, it follows that the set X = fxm1 ; xm2 ; : : :; xmn g is k-boolean independent for 1  k  n. Proposition 1. If F is an in nite eld and n is a positive integer, then F contains an n-subset which is k-boolean independent for 1  k  n. Proof. If F has characteristic 0, then Q is a sub eld of F and the result follows from Example 3. Suppose F has characteristic p. If F is algebraic over Zp then for any nite sub eld K of F take an element  2 F ? K . Now  is algebraic over K since Zp  K . Hence K () is a nite sub eld of F that contains K properly. Thus we may produce an arbitrarily long chain of nite sub elds of F and the proposition follows from Example 2. If F is not algebraic over Zp then F contains an element which is transcendental over Zp and the result follows from Example 4.  6

Corollary 1. Theorem 2 holds if in the de nition of M (G) we replace the set of real numbers mij by any set of n(p ? 1)-boolean independent elements of cardinality

n2(p ? 1) in any eld. 3. k-Point Separation with Linear Functionals. Let U be a vector space over a eld F and let U  be the dual space of U . We de ne (9)

LSk (U; F ) = minfjFj : F  U  and F is k-point separatingg:

We say that LSk (U; F ) is unde ned if there is no k-point separating family of linear functionals in the dual space of U . For any eld F , n-dimensional ?k
vector space U over F and positive integer k, in [2] it is shown that if (n ? 1) 2  jF j then LSk (U; F ) is well-de ned and  

LSk (U; F )  (n ? 1) k2 + 1: From this we see that LSk (U; F ) may be unde ned only when F is nite. This is the case, for example, when k > jF j. For more details on the case when F is nite see [2]. Theorem 3. Let F be a eld which contains a set of cardinality n2 (k ? 1) which is t-boolean independent for 2  t  n(k ? 1) and let U be an n-dimensional vector space over F . Then for n; k  2, we have LSk (U; F ) = n(k ? 1). We divide the proof of this Theorem into two lemmas. Lemma 2. If F and U are as in Theorem 3, then for n; k  2, LSk (U; F ) is de ned and LSk (U; F )  n(k ? 1). Proof. Let fmij jj = 1; : : :; n(k ? 1); i = 1; : : :; ng be a set of cardinality n2(k ? 1) in F which is t-boolean independent for 2  t  n(k ? 1). Take U = F n and for each j = 1; : : :; n(k ? 1), de ne fj 2 U  by

fj (x1; x2; : : :; xn) = m1j x1 + m2j x2 +


+ mnj xn : We claim that these n(k ?1) linear functionals are k-point separating on U . Suppose not. Then there is a set V = fp1; : : : ; pk g of k distinct points in U such that for each j , fj restricted to V is not injective. We construct a graph G with vertex set V and edge set E = ff1; : : :; fn(k?1) g. The incidence relation is de ned as follows: For each j 2 f1; 2; : : :; n(k ? 1)g, since fj is not injective on V , we can select integers aj ; bj such that 1  aj < bj  k and (10)

fj (paj ) = fj (pbj ):

We say that paj and pbj are endpoints of the edge fj . Note that dierent edges may join the same vertices, but there are no loops. 7

This gives us a graph G with k vertices and n(k ? 1) edges. By Theorem 1, G contains a subgraph H = (VH ; EH ) with at least two vertices which is an n-tree. By renumbering we can take

VH = fp1 ; : : :; ptg; and Write

EH = ff1; : : : ; fq g;

pa = (xa1 ; xa2 ; : : :; xan); Now in coordinates (10) becomes for j 2 [q]

2tk

q = n(t ? 1): a = 1; 2; : : :; t:

m1j xa1j + m2j xa2j +


+ mnj xanj ? m1j xb1j ? m2j xb2j ?


? mnj xbnj = 0: We think of (11) as a homogeneous system of n(t ? 1) linear equations in nt variables xij . Since the pi are distinct and the fj are linear we may replace pi by p0i = pi ? pt , for all i. Then p0t = 0, so we have reduced the system to a system of n(t ? 1) equations in n(t ? 1) variables zji = xij ? xtj , i = 1; : : :; t ? 1, j = 1; : : :; n. Now if we order the variables : z11 ; z12 ; : : :; z1t?1 ; z21 ; z22 ; : : :; z2t?1 ; : : :; zn1 ; zn2 ; : : : ; znt?1 then the matrix of coecients of the linear system is equal to the matrix M (H ) of Theorem 2. By Corollary 1, since H is an n-tree we have det M (H ) 6= 0. This implies that the homogeneous system has only the trivial solution. It follows that p0i = 0 for all i. Hence pi = pt for all i. This contradicts the fact that the pi are distinct.  We recall some de nitions and results from linear algebra: Let U be an ndimensional vector space over a eld F . An ane subspace X of dimension w in U is a coset X = x + W where W is a (linear) subspace of U of dimension w. X is a hyperplane (respectively, line) if X is of dimension n ? 1 (respectively, 1). The following are well-known: (A) If L is a line and H is a hyperplane in U , then either (1) L is contained is some (unique) hyperplane parallel to H , or (2) L intersects each hyperplane parallel to H in precisely one point. (B) If H1 ; : : :; Ht are hyperplanes in U then the intersection H1 \


\ Ht is either empty or an ane subspace of dimension  n ? t. Let F be a family of functions from X to Y and let S  X . We say that S spoils F if for each f 2 F there exist x; y 2 S , x 6= y, such that f (x) = f (y). Clearly F is k-point separating if and only if F is not spoiled by any subset S of X with jS j  k. Lemma 3. If U is an n-dimensional vector space over a eld F , then LSk (U; F )  n(k ? 1) provided that LSk (U; F ) is de ned. In particular, this is the case when F

(11)

is in nite. Proof. Let F be any family of linear functionals on U of cardinality n(k ? 1) ? 1. It suces to show that there is some subset S of U of cardinality  k which spoils F. 8

The idea of our proof is to rst nd two points p; q which spoil at least n?1 functionals in F. Then we generate successively additional points p1 ; p2; : : :; p` ; : : :; pk?2 so that as each new point p` is generated we make sure that n additional functionals in F are spoiled by the three points p; q; p`. Then since (n ? 1)+(k ? 2)n = n(k ? 1) ? 1, all of F will be spoiled by fp; q; p1; p2; : : :; pk?2 g. Clearly we may assume that each functional f 2 F is not zero, so f ?1(c) is a hyperplane for c 2 F . We begin by selecting any (n ? 1)-subsetT F00 of F. Then since 0 2 f ?1(0) for all f 2 F00 , we have by (B) above that I = f 2F00 f ?1 (0) has dimension at least 1. Let p; q 2 I , p 6= q, and let L0 be the line containing p and q. Let F0 = ff 2 FjL0  f ?1(c) for some c 2 F g: Note that F0 contains at least n ? 1 functionals in F and is spoiled by fp; qg. Now assume that we have found for ` < k ? 2 the following: (1) A subset F` of F such that (i) F0  F` and (ii) jF` j  (n ? 1) + `n. (2) A subset fp; q; p1; : : :; p`g of U which spoils F` . If jF` j  n ? 1 + (` + 1)n we take F`+1 = F` and p`+1 = p` . So we may assume that jF` j < (n ? 1) + (` + 1)n  n(k ? 1) ? 1: We select D0  FnF` such that jD0 jT= n ? 1 if jFnF` j  n ? 1; otherwise take 0 D = FnF` . Then the intersection J = f 2D0 f ?1(f (p)) contains p and by (B) has dimension at least 1. So there is a line L1  J with p 2 L1 . Let D = ff 2 FjL1  f ?1(c) for some c 2 F g: Now there are two cases: I. F` [ D = F and II. F` [ D 6= F. In case I let p`+1 be any point on the line L1 dierent from p. Then D is spoiled by fp; p`+1g and hence all of F is spoiled by fp; q; p1; : : :; p`+1 g. In case II take g 2 Fn(F` [ D) and let d = g(q). Now by de nition of D, L1 is not contained in any hyperplane g?1(c) parallel to g?1(d), so by (A) above g?1(d) intersects L1 in exactly one point, call it p`+1 . We claim p`+1 6= p and p`+1 6= q. If p`+1 = p then the line L0 = pq lies in g?1(d): this implies that g 2 F0 , but F0  F` and g 2= F` . If p`+1 = q, the lines L0 = pq and L1 = pp`+1 are equal. This implies that D = F0 . Then D0  D  F0  F` which is contrary to the de nition of D0 . It follows that F`+1 = F` [ D [ fgg is spoiled by fp; q; p1; : : :; p`+1 g and jF`+1 j  (n ? 1) + `n + (n ? 1) + 1 = n ? 1 + (` + 1)n, as desired.  4. k-Point separation with smooth functions. In this section we will determine the cardinality DSk (Rn ; R) of a smallest family of smooth (C1 ) functions from R n to R that separates every set of k distinct points in R n . Our references for this section are Fleming [6] and Spivak [14, 15]. Given f1; : : :; fm : R n ! R we write F = (f1; : : : ; fm) for the mapping F : R n ! R m de ned by F (x) = (f1(x); : : :; fm(x)) for x 2 R n . We identify the derivative DF (x): R n ! R m with the the Jacobian matrix [ @f@xi(jx) ]. The rank of F at x is the rank of DF (x). We require the following three lemmas. 9

Lemma 4. If F : R n

! R m is smooth on an open set U , there is a non-empty open

set V  U on which F has constant rank. Proof. Let x be a point of U at which the rank of DF (x) attains its maximum, say, k. This implies that DF (x) has a k  k non-singular submatrix. Since the determinant is a continuous function it follows that the same submatrix is nonsingular in a neighborhood V of x and hence DF (X ) has rank k in V .  Lemma 5. (Proposition 12, page 65 of [15]) If F : R n ! R m is smooth and of constant rank k on an open set U , then for each x 2 U , M = U \ F ?1 (F (x)) is a manifold of dimension n ? k. In Lemma 5 the tangent space to M at x is the kernel of DF (x). Lemma 6. (Theorem 5-2, page 111 of [14]). A subset M of R n is a k-dimensional manifold if and only if for each x 2 M there is an open set U of R n containing x, an open set W in R k and a 1-1 smooth function f : W ! R n such that f (W ) = M \ U , and Df (y) has rank k for each y 2 W . The function f in Lemma 6 is called a coordinate system around x. In this case the tangent space to M at x is the image of Df (y) where f (y) = x. Theorem 4. For all n  1 and k  2, DSk (R n ; R ) = n(k ? 1). Proof. Since linear functionals on R n are smooth, by Proposition 1 and Lemma 2,

DSk (R n ; R)  n(k ? 1): It therefore suces to prove that any family F of smooth functions on R n of cardinality nk ? n ? 1 is spoiled by some set of at most k points in R n . For G  F, let g1; : : :; gs be the distinct elements of G listed in any xed order. De ne FG : R n ! R s by FG = (g1; : : :; gs): Note that y 2 FG?1 (FG(x)) ? fxg () fx; yg spoils G: By Lemma 4, since F is nite, we can nd a non-empty open set U of R n such that for all G  F, FG has constant rank throughout U . Without loss of generality we take U = R n . Our plan is to show there exists a partition of F: (12)

F = F0 [ F1 [


[ Fk?2

such that (i) jF0j = n ? 1 and jFj j = n for 1  j  k ? 2; there exists a point a 2 R n ; and there exists a nested sequence Lj , 0  j  k ? 2, of submanifolds of R n of positive dimension such that for 1  j  k ? 2: (ii) a 2 Lj  Lj?1 and for each b 2 Lj ?fag, fa; bg spoils F0 and if j  1 there is a point cj = cj (b) 2 R n such that fa; b; cj g spoils Fj . 10

Once we have established the existence of such a partition we may complete the proof by taking b 2 Lk?2 ?fag. Then b 2 Li ?fag for all i so by (ii) there exists for each i  1 a point ci such that fa; b; cig spoils Fi [ F0. Hence fa; b; c1; c2; : : :; ck?2 g spoils F, as desired. To prove the existence of such a partition we consider rst the easy case in which rank(FG) < n ? 1 for every (n ? 1)-subset G of F. In this case we can take (12) to be any partition satisfying (i) and let a be any point of R n . Write Fi = FFi for all i and set Li = F0?1 F0 (a); 0  i  k ? 2: Since rank(F0) < n, by Lemma 5, Li is a manifold of positive dimension. Hence if b 2 Li ? fag then fa; bg spoils F0 . Under our current assumptions for i  1, rank(Fi ) < n , so Mi = Fi?1 Fi (a) is also a manifold of positive dimension; if ci 2 Mi ? fag then fa; cig spoils Fi , so fa; b; cig also spoils Fi . It remains to consider the case where there is an (n ? 1)-subset G0 of F with rank(FG0 ) = n ? 1. In this case, in (12) take F0 = G0 and let the rest of the partition be chosen arbitrarily subject to (i). Again write Fi = FFi and let a be any point of R n . We de ne the Li 's recursively as follows: First let L0 be the 1-dimensional manifold F0?1 F0 (a). Clearly if b 2 L0 ? fag then fa; bg spoils F0 . Assume we have already selected Lj , 0  j < i  k ? 2, satisfying the conditions in (ii). Notice that each Lj is 1-dimensional. To de ne Li , we consider three cases: Case I. rank(Fi) < n. In this case we set Li = Li?1 . Then M = Fi?1 Fi (a) is a manifold of positive dimension, so there is an element ci 2 M ? fag and hence fa; cig spoils Fi . Case II. There exists f 2 Fi such that the mapping F = FF0[ff g has rank n ? 1. In this case M = F ?1 F (a) is a manifold of dimension 1 containing a and contained in L0 . By assumption, a 2 Li?1  L0 . Since M and Li?1 are both 1dimensional submanifolds contained in the 1-dimensional manifold L0 and a 2 M \ Li?1 using Lemma 6 one may show that Li := M \ Li?1 is a 1-dimensional manifold. Note that Li contains a and is contained in Li?1 . If b 2 Li ? fag, then fa; bg spoils F0 . Since b 2 M ? fag, fa; bg spoils F0 [ ff g. On the other hand, M 0 = FF?i1?ff g (FFi?ff g (a))

is a manifold of positive dimension, so any point ci 2 M 0 ? fag serves to make fci ; ag spoil Fi ? ff g. Then fa; b; cig spoils Fi . Case III. rank (Fi) = n and rank FF0[ff g = n for all f 2 Fi . Let F0 = ff1; : : :; fn?1 g and Fi = fg1; : : :; gn?1; gng: Then F0 = FF0 = (f1; : : :; fn?1) and Fi = FFi = (g1; : : :; gn?1; gn) In addition, we de ne: F = FF0 [fgng = (f1; : : :; fn?1; gn) and G = FFi ?fgng = (g1; : : :; gn?1): 11

In this case Fi and F have rank n and F0 and G have rank n ? 1. Thus the manifold K = G?1 G(a) is 1-dimensional. Now it suces to establish the following : (13) There is a 1-dimensional manifold Li with a 2 Li  Li?1 and a 1-dimensional manifold K0  K such that for any x 2 Li , the level set gn?1gn (x) contains exactly one point from each of the manifolds Li and K0. Once (13) is established we can complete the proof as follows: Let b 2 Li ? fag. Then b 2 L0 ? fag so fa; bg spoils F0 . By (13) Li \ gn?1gn (b) = fbg and K0 \ gn?1gn (b) = fcg for some point c. Now there are two possibilites: b = c or b 6= c. Suppose b = c, then b 2 K0 ? fag so fa; bg spoils g1; : : :; gn?1. Then we take ci to be any element of gn?1gn (b) ? fbg and fb; ci g spoils gn . Hence fa; b; cig spoils Fi . On the other hand, if b 6= c, then fb; cg spoils gn. And c 2 K0  G?1 G(a), so fa; cg spoils g1; : : :; gn?1 unless c = a. But if c = a, then a and b are distinct elements of Li \ gn?1gn (b) which contradicts (13). So if we set ci = c, fa; b; cig spoils Fi and the proof is completed. It remains to prove (13). Since Fi has rank n throughout Rn , by the Inverse Function Theorem there are open sets U and V with a 2 U such that Fi maps U dieomorphically onto V . To simplify the proof we identify U with V via Fi . This allows us to assume that the functions gj are simply the coordinate projections: gj (x1 ; : : :; xn) = xj for (x1; : : :; xn ) 2 U . From above both K and L0 are 1-dimensional manifolds containing a. Since Li?1 is a manifold of positive dimension contained in L0, it also has dimension 1 and by assumption contains a. It follows from Lemma 6 that there are coordinate systems : ' : I1 ! K and : I2 ! Li?1 where I1 and I2 are open intervals in R chosen so that 0 2 I1 \ I2,

'(0) = (0) = a = (a1; : : :; an); and each of D'(t), t 2 I1, and D (t), t 2 I2 have rank 1, and in particular are non-zero. We write

Then

'(t) = ('1(t); : : :; 'n (t));

for i 2 I1 ;

(t) = ( 1(t); : : :; n(t));

for i 2 I2 :

for i 2 I1; D'(t) = ('01(t); : : :; '0n(t)) D (t) = ( 10 (t); : : :; n0 (t)) for i 2 I2: We wish to establish that '0n (t) 6= 0 for t 2 I1 and n0 (t) 6= 0 for t 2 I2 . Note that since gn is the projection on the n-th component, for all y 2 U the tangent space to gn?1gn (y) is the hyperplane of all points (x1; : : :; xn ) 2 R n such that xn = 0. Thus to show that n0 (t) 6= 0 it suces to show that D (t) is not in the tangent space to gn?1gn (y) when y = (t). Suppose v := D (t) is in the tangent space to gn?1gn (y) and y = (t). By the comment following Lemma 5 we have Dgn (y)v = 0. By the same comment since v 12

is in the tangent space to L0 we have DF0 (y)v = 0. It follows that DF (y)v = 0. But DF (y) is non-singular since F has rank n and hence v = 0. This contradicts the fact the D (t) is non-zero for t 2 I2 and thus proves that n0 (t) 6= 0 for t 2 I2 . By the same reasoning, since rank Fi = n we can prove that '0n (t) 6= 0 for all t 2 I1 . This shows that the mappings 'n : I1 ! R and n : I2 ! R are both injections. Since 'n (0) = n (0) = an , the nth component of a, there is an open interval I in R such that an 2 I  'n (I1) \ n (I2). Let J1 = '?n 1 (I ) and J2 = n?1(I ). Finally if we set K0 = '(J1 ) and Li = (J2) then (13) holds. This completes the proof.  5. k-Point separation with continuous functions. This paper was originally motivated by the following question. Suppose that F is a family of continuous functions from R n to R that separates every k-element set of points in R n . How small can F be? Unfortunately, we were unable to determine this number, CSk (R n ; R), for all k, n. Nevertheless, we know from the linear case in Section 3 above that if n, k > 1, then CSk (R n ; R )  n(k ? 1), and we suspect that equality holds. All we can say at present is that the lower bound on CSk (Rn ; R) increases at least linearly with k and n. Proposition 2. If n > 1, then CSk (R n ; R )  k for all k > 0. Proof. We prove the proposition by induction on k. As there is no continuous injection from R n to R , the proposition holds for k = 2. We claim that if the proposition holds for k, then it holds for k + 1. Since CSk (R n ; R)  CSk (R 2 ; R) for n  2 we can assume that n = 2. Suppose that CSk (R n ; R)  k, and that f1; : : : ; fk : R n ! R are continuous. We will nd a (k + 1)-set that f1 ; : : : ; fk fail to separate. By induction, there exists a k-set fp1; : : : ; pk g  R n that f1; : : : ; fk?1 fails to separate. If this set is not separated by fk either, we are done; otherwise, without loss of generality, we can suppose that fk (p1 ) < fk (p2 ) <


< fk (pk ). There are two cases. If k = 2, draw three disjoint arcs, call them A1 , A2 , A3 from p1 to p2 , and by the Intermediate Value Theorem, there exists q1 2 A1, q2 2 A2, and q3 2 A3 such that f2(q1 ) = f2(q2 ) = f2(q3 ). If any two of f1(q1 ), f1(q2 ), f1(q3 ) are equal, say f1 (q1 ) = f1(q2 ), then fq1 ; q2 g spoil f1 and f2 . So without loss of generality, suppose that f1(q1 ) < f1(q2 ) < f1(q3 ). We have two subcases. If f1 (q2 ) = f1 (p1), then fp1 ; q2; q3 g spoil f1 and f2 . On the other hand, if f1(q2 ) 6= f1(p1 ), say f1 (q2 ) > f1(p1 ), then by the Intermediate Value Theorem, there exists q30 2 A3 between p1 and q3 such that f1 (q30 ) = f1 (q2 ), and fq2 ; q3 ; q30 g spoils f1 ; f2; similarly, if f1 (q2 ) < f1 (p1), there exists q10 2 A1 such that fq1 ; q10 ; q2g spoils f1 ; f2. If k > 2, there exists an arc from p1 to p3 that misses p2 . By the Intermediate Value Theorem, there exists pk+1 on this arc such that fk (pk+1) = fk (p2). But then fp1 ; : : : ; pk+1 g spoils f1; : : :fk .  Proposition 3. For all k; n  2 , CSk (R n ; R ) > b k2 c(n ? 1). Proof. First, note that there is no continuous injection from R n to R n?1 . This follows immediately from the Borsuk-Ulam Theorem [13, p.266]. It suces to prove that CS2` (R n ; R) > `(n ? 1) for `  1. 13

Suppose that we have `(n ? 1) continuous functions f1; : : : ; f`(n?1) from Rn to R : we will nd 2` points that spoil them. For each i, let Fi : R n ! R n?1 be de ned by Fi (x) = (f(i?1)(n?1)+1 (x); : : : ; fi(n?1) (x)): As Fi is not injective, there exists xi , yi such that Fi (xi ) = Fi (yi ), and thus, for each j , (i ? 1)(n ? 1) < j  i(n ? 1), fj (xi ) = fj (yi ). Then fxi : i = 1; : : :`g[fyi: i = 1; : : :`g spoils f1; : : : ; f`(n?1) .  We suspect that it should not be too dicult to prove that CSk (R n ; R) > (k ? 1)(n ? 1). Showing that the lower bound is n(k ? 1) may be more dicult. We conclude with a curious observation. Given n, CSk (R n ; R) increases linearly (and hence polynomially, not exponentially) in k. And yet, as we shall see, CS@0 (Rn ; R) is bigger than @0 . This suggests that the in nite analogue of this Theorem should have CS (Rn ; R) increasing slowly above , but not so rapidly that CS@0 (Rn ; R) = c, where c = 2@0 . The result is some empirical evidence against the Continuum Hypothesis, (Note: no evidence can be more than empirical, for Cohen [4 and 5] proved that the Continuum Hypothesis is neither provable nor disprovable from standard set theory, i.e., the Zermelo-Frankel axioms. Whether or not the Continuum Hypothesis is \true" depends on its utility, philosophical justi cation, aesthetic appeal, etc. Some authors have argued that in \reality", the continuum is \much" larger than the integers.) Recall that if @ is a trans nite cardinal number, then @+1 is the next highest cardinal number. Theorem 6. If 1 < n < @0 and @ < c, then CS@ (R n ; R ) = LS@ (R n ; R ) = @+1 . Proof. First, CS@ (Rn ; R ) > @. Given some continuous functions f : R n ! R , < @, we choose a sequence of pairs of points p , q , such that for each , f (p ) = f (q ) and for any , , fp ; q ; p ; q g contains four distinct points. (This can be done by recursion on : if p , q have been chosen for all < , then fewer than c points have been chosen so far. If f is constant, choose any thus far unchosen p , q ; otherwise, choose o , r such that for some xed c, f (o ) < c < f (r ), and then imagine c mutually disjoint arcs from o to r . By the Intermediate Value Theorem, each arc contains at least one point p such that f (p) = c, and as at most j j < c of these have been chosen so far, we can choose two more, and call them p and q .) This set of @ points spoils ff : < @g. On the other hand, LS@ (R n ; R)  @+1 . To prove this we construct a family of @+1 linear functions from R n to R that separate all @ -subsets of R n . Let H be a set of @+1 real numbers. For r 2 H de ne fr : R n ! R by setting

fr (x1; x2; x3; : : :; xn ) = x1 + rx2 + r2x3 +


+ rn?1 xn : Let F = ffr : r 2 H g. By Vandermonde's determinant if r1; : : :; rn are distinct elements of H the homogeneous linear system fri (x) = 0, 1  i  n, has only the trivial solution x = 0. It follows that if p and q are distinct points of R n , the set F(p; q) = ff 2 Fjf (p) = f (q)g 14

has at most n ? 1 elements. Now we claim that F separates every set of @ points. If not, there is a set S of @ points not separated by F. Hence for each f 2 F there is a pair of distinct points p; q 2 S such that f (p) = f (q). That is, (14)

F=

[

fp;qg2(S2 )

?


F(p; q) ?


where S2 is the set of all two element subsets of S . But since S2 has cardinality @ and each set F(p; q) has cardinality n, the right side of (14) has cardinality @ and we have a contradiction.  References 1. N. Biggs, Algebraic Graph Theory, Second Edition, (Cambridge University Press. Cambridge, 1993). 2. W. E. Clark, 'Separating sets with parallel classes of hyperplanes', Bulletin of the ICA 13 (1995) 65{82. 3. F. R. Cohen and D. Handel, k-regular embeddings of the plane, Proc. Amer. Math. Soc. 72 (1) (1978) 201{204. 4. P. J. Cohen, 'The independence of the continuum hypothesis Part I', Proc. Nat. Acad. Sci., USA 50 (1963) 1143{1148. 5. ||||-, 'The independence of the continuum hypothesis Part II', Proc. Nat. Acad. Sci., USA 51 (1964) 105{110. 6. W. H. Fleming, Functions of several variables ( Addison-Wesley Pub. Co. , Inc., 1965). 7. F. Harary and F. M. Palmer, Graphical Enumeration ( Academic Press, 1973). 8. T. W. Hungerford, Algebra ( Holt, Rinehart and Winston, Inc., 1974). 9. S. MacLane and G. Birkho, Algebra (The MacMillian Co., 1974). 10. C. St. J. A. Nash-Williams, 'Edge-disjoint spanning trees of nite graphs', J. London Math. Soc. 36 (1961) 445{450. 11. |||{, 'Decomposition of nite graphs into forests', J. London Math. Soc. 39 (1964), 12. 12. R. L. Roth, 'On extensions of Q by square roots', Amer. Math. Monthly 78 (4) (1971) 392{393. 13. E. H. Spanier, Algebraic Topology (McGraw-Hill Book Co, 1966). 14. M. Spivak, Calculus on Manifolds ( W. A. Benjamin, Inc., 1965). 15. |||{, Dierential Geometry, Volume I (Publish or Perish, Inc., Berkeley, 1979). Department of Mathematics, University of South Florida, Tampa, FL 33620-5700, USA email addresses: [email protected], [email protected], [email protected] 15