Combinatorial Dimension in Fractional Cartesian Products

Combinatorial Dimension in Fractional Cartesian Products Ron Blei, 1 Fuchang Gao 1 Department of Mathematics, University of Connecticut, Storrs, Connecticut 0668; e-mail: blei@math.uconn.edu Department of Mathematics, University of Idaho, Moscow, Idaho 83844 1103; e-mail: fuchang@uidaho.edu Received 3 September 00; revised 18 December 003; accepted 5 July 004 DOI 10.100/rsa.0058 Published online 9 November 004 in Wiley InterScience (www.interscience.wiley.com). ABSTRACT: The combinatorial dimension relative to an arbitrary fractional Cartesian product is defined. Relations between dimensions in certain archetypal instances are derived. Random sets with arbitrarily prescribed dimensions are produced; in particular, scales of combinatorial dimension are shown to be continuously and independently calibrated. A combinatorial concept of cylindricity is key. 004 Wiley Periodicals, Inc. Random Struct. Alg., 6, 146 159, 005 1. DEFINITIONS, STATEMENT OF PROBLEM The idea of a fractional Cartesian product and a subsequent measurement of combinatorial dimension appeared first in a harmonic-analytic context in the course of filling analytic gaps between successive (ordinary) Cartesian products of spectral sets [, 3]. (Detailed accounts of this, and much more, appear in [5].) Succinctly put, combinatorial dimension is an index of interdependence. Attached to a subset of an ordinary Cartesian product, it gauges precisely the interdependence of restrictions to the set, of the canonical projections from the Cartesian product onto its independent coordinates. We can analogously gauge the interdependence of restrictions to the same set, of projections from the Cartesian product onto interdependent coordinates of a prescribed fractional Cartesian product. We thus obtain distinct indices of interpendences associated, respectively, with distinct fractional Cartesian products. A question naturally arises: What are the relationships between these various indices? Correspondence to: R. Blei 004 Wiley Periodicals, Inc. 146

COMBINATORIAL DIMENSION IN FRACTIONAL CARTESIAN PRODUCTS 147 To make matters precise, we first recall, and then extend basic notions found in Chapters XII and XIII of [5]. Let E 1,..., E n be sets, and let F E 1 E n.(we refer to E 1 E n as the ambient product of F.) For integers s > 0 define F (s) = max{ F (A 1 A n : A i E i, A i s, i [n]}, (1) where [n] ={1,..., n}.fora > 0, define The combinatorial dimension of F is d F (a) = sup{ F (s)/s a : s = 1,,...}. () dim F = sup{a : d F (a) = }=inf{a : d F (a) < }. (3) Next we define the fractional Cartesian products. For S [n], let π S denote the canonical projection from E 1 E n onto the product whose coordinates are indexed by S, π S ( y) = ( y i : i S), y = ( y 1,..., y n ) E 1 E n. Let U = (S 1,..., S m ) beacoverof[n] (i.e., S 1 [n],..., S m [n], and m j=1 S j =[n]), and define a fractional Cartesian products based on Utobe (E 1 E n ) U ={(π S1 (y),..., π S m(y)) :y E 1 E n }. We view (E 1 E n ) U as a subset of E S 1 E Sm, and measure its combinatorial dimension by solving a linear programming problem ([4, 8]): If E 1,..., E n, are infinite sets, and n α U = max x i : x i 0, x i 1 for j [m], i=1 i S j then dim(e 1 E n ) U = α U. (4) Examples 1. (Maximal and minimal fractional Cartesian products) For integers 1 k n, let U be a cover that is an enumeration of all k-subsets of [n], and let V be a cover of [n], all of whose elements are k-subsets of [n] such that, for every i [n], {S V:i S} = k. Then, α U = α V = n/k, and (taking E 1 = =E n = N) we obtain from (4) dim(n n ) U = dim(n n ) V = n k. The ambient product of (N n ) U is ( n k) -dimensional, and the ambient product of (N n ) V is n-dimensional. Generally, if k and n are relatively prime, then the dimension of the ambient product of any n -dimensional fractional Cartesian product is at least n, and k no greater than ( n k).. (Random constructions) Fractional Cartesian product are subsets of ambient products typically of high dimension. To wit, there are no nontrivial fractional Cartesian product in N, whereas for arbitrary α (1, ) there exists an abundance of random sets F N with dimf = α [6,7]. How to produce deterministically F N with dimf = α, where α (1, ) is arbitrary, is an open problem.

148 BLEI AND GAO 3. (An application) From α U = 3/ in the archetypal case n = 3, U = ({1, }, {, 3}, {1, 3}), we obtain that that the number of triangles formed from s given edges in a graph is less than s 3. More generally, following a canonical identification of a complex on n vertices with a cover U of [n], we obtain from (4) that the number of complexes formed from s given simplices is bounded by s α U. These results are closely related to the Kruskal-Katona Theorem (drawn to our attention by Joel Spencer); see [10, 9]. Definition 1. For F E 1 E n, and a cover U = (S 1,..., S m ) of [n], let F U ={(π S1 ( y),..., π S m( y)) : y F}, and write dim U F = dim F U. (5) We consider dim U as a gauge of interdependence of π S1 F,..., π S m F (projections π Si restricted to F). By (4), for infinite F, 1 dim U F α U. (6) We continue to write dim F in the extremal case U = ({1},..., {n}). Problem. Let E 1,..., E n be infinite sets, and let U be a cover of [n]. For F E E n, what are the relations between dim F and dim U F? In this article we analyze the first nontrivial case: n = 3, and U = (S 1, S, S 3 ), where S 1 ={1, }, S ={, 3}, S 3 ={1, 3}. Throughout, for convenience (and with no loss of generality), we take E 1 = E = E 3 = N.. GENERAL BOUNDS Theorem 3. If F N 3 and dim F, then dim F dim U F dim F dim F + 1. (7) We will prove the right-side inequality in (7) (the nontrivial part of the theorem) by the use of Littlewood-type inequalities in fractional dimensions. Let E 1,..., E d be sets. Consider scalar d-tensors on E 1 E d with finite support b = (b x :x= (x 1,..., x d ) E 1 E d ), and define (the d-fold injective tensor-norm of b) b d = b x r x1 r xd, (8) x E 1 E d

COMBINATORIAL DIMENSION IN FRACTIONAL CARTESIAN PRODUCTS 149 where r x1,..., r xd are Rademacher functions indexed by E 1,..., E d, whose respective domains are { 1, 1} E 1,...,{ 1, 1} E d (e.g., p. 19 in [5]). For F E 1 E d, and t 1, define ζ F (t) = sup 1/l b x l : b d 1. (9) x=(x 1,...,x d ) F The relation between the harmonic-analytic measurement ζ F and the combinatorial measurement d F is ( ) t ζ F (t) < d F <, t [1, ). (10) t (See Th. XIII.0 in [5].) To prepare for an application of (10), we define b U = b ijk r ij r jk r ik, (11) for b = (b ijk : (i, j, k) N 3 ) with finite support. We observe i,j,k 1 (N 3 ) U (i 1, i, i 3, i 4, i 5, i 6 ) = Er i1 (ω 1 )r i (ω )r i3 (ω )r i4 (ω 3 )r i5 (ω 1 )r i6 (ω 3 ), where 1 denotes indicator function, and E denotes expectation over ω 1, ω, ω 3. From this we deduce (an instance of Corollary XIII.7 in [5]) Lemma 4. and If b = ( b i :i N N N ) is a real-valued 3-tensor with finite support, b j = b πs1 ( j)π S ( j)π S3 ( j), j N 3, then b U b 3. (1) We require also a decomposition property, which follows from Lemma XIII.1 in [5]: if ϕ l (N n ), and 1 c n (ϕ) = sup ϕ(i) : s N, A v N, A v =s, v [n] s, i A 1 An then there exists a partition {Q 1,..., Q n } of N n such that, for v [n], ϕ(i) 1 Q v(i) c n (ϕ), sup k N i πv 1 {k} where π v is the vth canonical projection from N n onto N. Lemma 5. If ϕ l (N n ) is supported in F N n, then there exists a partition {Q 1,..., Q n } of N n such that for v [n], and p 1, ϕ(i) 1 Q v(i) d F (p) 1/p ϕ q, (13) where 1 p + 1 q = 1. sup k N i πv 1 {k}

150 BLEI AND GAO Proof. For S N, and s-sets A v N (i.e. A v =s), v [n], we estimate (by Hölder s inequality) 1 s i A 1 An ϕ(i) 1 s F (A 1 A n ) 1/p ϕ q. Then c n (ϕ) d F (p) 1/p ϕ p, and the lemma follows from the decomposition property above. Lemma 6. If F N 3,a, and d F (a) <, then ( ) 4a ζ FU <. (14) 3a + 1 Proof. Following Lemma 4 and the definition of ζ FU (a), we need to verify that if b = (b ijk : (i, j, k) N 3 ) is a scalar 3-tensor with finite support, and b U 1, then b ijk 4a/(3a+1) K, (i,j,k) F where K depends only on F and a. To this end we use duality, and the fractional mixed norm inequalities ( ) 1/ b ijk λ, i,j k i,k ( ) 1/ b ijk λ, j j,k ( ) 1/ b ijk λ, (15) i where λ>1is an absolute constant; see Lemma XII.1 in [5]. Suppose θ is an element in the unit ball of l a 1 4a (N 3 ), and that θ is supported in F. Apply Lemma 5 with p = a and ϕ = θ 4, thus obtaining a partition {Q 1, Q, Q 3 } of N 3, such that, for v = 1,, 3, θ(i) 4 1 Q v(i) d F (a) 1/a. sup l N i πv 1 {l} For l N, and v = 1,, 3, let F lv = F Q v π 1 v {l}. For convenience, designate T 1 = {, 3}, T ={1, 3}, T 3 ={1, }. For each l and v, apply Lemma 5 with p =, F = F lv (viewed as a subset of N ), and ϕ = θ 1 Flv (viewed as a function on N ), thus obtaining partitions {P l1v, P lv } of F lv, such that sup l N,i N j N sup l N,i N i N (( θ 1 Pl1v ) π T v)(i, j) d F (a) 1/a, (( θ 1 Plv ) π T v)(i, j) d F (a) 1/a. (16) (In applying here Lemma 5, we used d Flv () 1.) For u = 1,, and v = 1,, 3, let P uv = l N P luv,

COMBINATORIAL DIMENSION IN FRACTIONAL CARTESIAN PRODUCTS 151 and then write b i θ(i) = i N 3 b i θ(i) 1 P uv(i). (17) j N3 u {1,},v {1,,3} By applying to each of the six sums on the right side of (17) the mixed-norm inequalities (15), together with (16) (via Cauchy-Schwarz and Hölder), we obtain b i θ(i) i N 3 6λd F(a) 1/4a. Therefore (by duality and definition of ζ FU ), ( ) 4a ζ FU 3a + 1 6λd F (a) 1/4a. Proof of Theorem 3. Let A, B, C be arbitrary s-subsets of N. Then FU (s ) F U (A B) (B C) (A C) = F (A B C). Maximizing over A, B, C we obtain FU (s ) F (s), which implies ( a ) d FU d F (a), a > 0, and hence the left inequality in (7). ( If a > dim F, then d F (a) <. By Lemma 6, ζ 4a ) ( FU 3a+1 <. By (10), a dfu a+1) <, which implies dim F U a, and hence the right-side inequality in (7). a+1 3. CYLINDRICAL SETS For F N 3, v = 1,, 3, and k N, let Also, for E N and k N, we denote F v (k) = π 1 v {k} F. E 1,k ={(j : (j, k) E}, E,k ={(j : (k, j) E}. Definition 7. F N 3 is cylindrical in direction v (v = 1,, 3) if for all a d F (a) < sup{d F v(k)(a 1) : k N} <.

15 BLEI AND GAO F is cylindrical if it is cylindrical in at least one direction, and doubly-cylindrical if it is cylindrical in at least two directions. We say F N 3 is a cylinder if F = E H for E N (base) and H N (height). For such F, dim F = dim E + dim H. Note also that cylinders are obviously cylindrical, but cylindrical sets need not be cylinders. Theorem 8. (i) If F N 3 is cylindrical with dim F, then dim F dim U F 1 dim F 1. (18) (ii) If F N 3 is a cylinder with infinite base and infinite height, then 1 dim U F = dim F 1. (19) (iii) If F N 3 is doubly-cylindrical with dim F, then dim U F = dim F. (0) Proof. (i) The left inequality always holds (Theorem 3). We proceed to verify the inequality on the right, under the assumption that F is cylindrical in direction 3. Suppose a > dim F, and sup d F3 (k)(a 1) K <. (1) k N Let A, B, C be arbitrary s-subsets of N, and define H ={k : max{ B 1,k, C 1,k } s 1/(a 1) }. Then, H s (a )/(a 1), and 1 F (i, j, k)1 A (i, j)1 B (j, k)1 C (i, k) A s (a 3)/(a 1). () k H i,j k H From (1) we obtain 1 F (i, j, k)1 A (i, j)1 B (j, k)1 C (i, k) k / H i,j k / H k / H 1 F (i, j, k)1 B (j, k)1 C (i, k) i,j Ks a a 1 d F3 (k)(a 1) max{ B 1,k a 1, C 1,k a 1 } ( B 1,k + C 1,k ) k / H Ks a 3 a 1.

COMBINATORIAL DIMENSION IN FRACTIONAL CARTESIAN PRODUCTS 153 Combining this with (), we obtain F U (A B C) = i,j,k 1 F (i, j, k)1 A (i, j)1 B (j, k)1 C (i, k) s (a 3)/(a 1) + Ks (a 3)/(a 1). Therefore, FU (s) ( + K)s (a 3)/(a 1), which implies dim U F 1 a 1, and hence the right-side inequality in (18). (ii) We can assume F = E N with dim E > 1, and proceed to verify dim U F 1/(dim F 1). Fix 1 < a < dim E. Then, for arbitrarily large positive integers s, there exist s-sets A 1 N, A N, such that E (A 1 A ) =Ms a, and M > 0 is a large as we please. Let A = E (A 1 A ), B = A 1 [m], and C = A [m], where m is the integer satisfying m 1 < s a 1 m. Then, F U (A B C) = A m Ms a 1. Therefore, ψ FU (Ms a ) Ms a 1, which implies d FU ( 1 ) =, and hence the desired a conclusion. (iii) For v = 1,, we assume (F cylindrical in directions 1 and ), and proceed to show sup d F v(k)(a 1) K < (3) k N dim U F dim F. (4) Fix a > dim F, and let A, B, C be arbitrary s-subsets of N. We decompose A into two disjoint sets G and H, such that max i G 1,i s and max i H,i s (e.g., G = i {A,i : A,i > s}, H = i {A,i : A,i s}). We write F U (A B C) = F U (H B C) + F U (G B C), (5) and rewrite the first term on the right side of (5) as F U (H B C) = i,j,k 1 H (i, j)1 B (j, k)1 C (i, k)1 F (i, j, k) = i 1 H,i (j)1 B,i (k)1 C (j, k)1 F (i, j, k). (6) j,k

154 BLEI AND GAO Let D ={i : B,i H,i }. By applying (3) in the case v = 1, we estimate 1 H,i (j)1 B,i (k)1 C (j, k)1 F (i, j, k) 1 H,i (j)1 B,i (k)1 F (i, j, k) i D i D j,k j,k K H,i a 1. (7) i D For i / D, let m i be the largest integer such that m i B,i + 1, (8) H,i and decompose B,i into pairwise disjoint sets E 1,..., E mi such that E u H,i for u [m i ]. By applying (3) and (8), we estimate 1 H,i (j)1 B,i (k)1 C (j, k)1 F (i, j, k) m i 1 H,i (j)1 E u(k)1 F (i, j, k) i/ D i/ D j,k K i/ D u=1 j,k B,i H,i a + H,i a 1. Combining with (7), and then using H,i s, i B,i =s, i H,i s, we conclude that ( F U (H B C) Ks (a )/ H,i + ) B,i + H,i i D i D c Ks a/. By a similar argument (based on (3) in the case v = ), we obtain an identical estimate for the second term on the right side of (5). Combining the two estimates, we deduce F U (A B C) 4Ks a/. Therefore, d FU ( a ) 4K, which implies (4). 4. RANDOM CONSTRUCTIONS Next we produce random sets (cf. [6,7]), demonstrating that the dim-scale and the dim U - scale implied by Theorem 7 are continuous, and are independently calibrated: Theorem 9. (i) For all x [, 3] and y [ x, x 3 x 1 ], there exist cylindrical sets F N3 with dim F = x and dim U F = y. (ii) For all x [, 3], there exist doubly-cylindrical sets F N 3 with dim F = x and (hence) dim U F = x. The proof uses random constructions based on the following instance of the Prokhorov- Bennett probabilistic inequalities (e.g., [1]):

COMBINATORIAL DIMENSION IN FRACTIONAL CARTESIAN PRODUCTS 155 Lemma 10. If (X i : i N) is a sequence of (statistically) independent {0, 1}-valued random variables with mean δ, then, for all n N, ( ) n ( ) t P X i nδ > t exp, t (0, nδ), (9) 8nδ i=1 ( ) n ( P X i nδ > t exp t ( )) t 4 log, t nδ. (30) nδ i=1 In what follows below we use the notation A K m to mean m/k A Km. Lemma 11. For every α (1, ) there is n 0 (α) = n 0 > 0, so that for every N n 0 there exists E [N] such that, for u = 1,, and for all s-sets A [N], B [N], E u,i N α 1 for all 1 i N, (31) E (A B) 5s α. (3) Proof. Fix N 1, and let {X ij : (i, j) [N] } be a system of independent {0, 1}-valued random variables with mean N α. Consider the random set E ={(i, j) : X ij = 1}. The probability that E fails to satisfy (31) is no larger than ( ) ( ) N N P X ij N α 1 > N α 1 N N + P X ij N α 1 > N α 1, i=1 i=1 which, following (9), is no larger than 1/3 for all N n 0 for sufficiently large n 0. Also, the probability that E fails to satisfy (3) is no larger than N P X ij s N α > 4s α, s=1 A =s, B =s which, following (30), is no larger than s=1 (i,j) A B j=1 j=1 N ( ) ( N N exp( ( α)s α log s s )) < 1 3 for all N n 0 for sufficiently large n 0. We thus conclude that E satisfies the requirements of the lemma with probability at least 1/3. Lemma 1. For every α (1, ) and every β [0, α, 1) there is n 0 (α, β) = n 0 > 0, so that for every N n 0 there exists F [N] 3 satisfying the following: (i) There exists E [N] with E N α, such that, for all k [N], {(i, j) : (i, j, k) F, (i, j) E} 4 N α β ; (33)

156 BLEI AND GAO (ii) for all s-sets A [N] and B [N], and for all k [N], F (A B {k}) 5s α β, (34) (iii) For all s-sets S [N] [N], and for all C [N] with C s 1 1/α, F (S C) 5s 1/α β/α. (35) Proof. Choose E as in Lemma 11. If β = 0, then we simply let F = E [N]. For β (0, α 1), let {X ijk : (i, j) E,1 k N} be a system of independent {0, 1}-valued random variables with mean N β. Then, by argument similar to the one used to prove Lemma 11, we conclude that F ={(i, j, k) : X ijk = 1} satisfies the requirements of the lemma with positive probability. Proof of Theorem 9. (i) For x =, take F = N {1}, and for x = 3, take F = N 3.For x (, 3), let α = x and β = α + 1 x. Then, 1 <α, and 0 β<α 1. Let y 1 n 0 = n 0 (α, β) be as in Lemma 11, and for each integer n such that 3 n n 0, let F n be the set F obtained from Lemma 11 for N = 3 n. Define F = Claim 1: dim F = x. For all s-sets A N, B N, C N, where F (A B C) = n=n 0 {(3 n,3 n,3 n ) + F n }. ((3 n,3 n,3 n ) + F n ) (A B C) n=n 0 = F n ((A 3 n ) (B 3 n ) (C 3 n )) n=n 0 n=n 0 5s x n, s n = max{ (A 3 n ) [3 n ], (B 3 n ) [3 n ], (C 3 n ) [3 n ] }, and the inequality follows from (34). Note that x s x n s n n=n 0 n=n 0 ( A (3 n +[3 n ]) + B (3 n +[3 n ]) + C (3 n +[3 n ]) ) n=n 0 ( A + B + C ) x = 3 x s x. x

COMBINATORIAL DIMENSION IN FRACTIONAL CARTESIAN PRODUCTS 157 We thus conclude that F (s) 5 3 x s x. (36) On the other hand, choosing A = B = C = 3 n +[3 n ], we obtain from (33) F (A B C) = (3 n,3 n,3 n ) + F n = F n 1 4 (3n ) α β+1 = 1 4 (3n ) x. Therefore, F (3 n ) 1 4 (3n ) x, which together with (36), implies dim F = x. Claim : dim U F = y. We let A N, B N, C N be s-sets, and estimate F U (A B C). To this end, let H ={k : B 1,k > s 1/α or C, k > s 1/α }. Then, H s 1 1/α (because ( B 1,k + C,k ) = B + C =s). We note that F U (A B C) is less than or equal to F (C,k B 1,k {k}) + F (A {k}). (37) k / H k H To estimate the first sum in (37), we observe that for every k N, F (C,k B 1,k {k}) = ((3 l,3 l,3 l ) + F l ) (C,k B 1,k {k}), where 3 l k < 3 l+1. By (34), ((3 l,3 l,3 l ) + F l ) (C,k B 1,k {k}) = F l ((C,k 3 l ) (B 1,k 3 l ) (3{k} 3 l )) 5(max{ B 1,k, C,k }) α β. Therefore, F (C,k B 1,k {k}) 5(max{ B 1,k, C,k }) α β k / H k / H To estimate the second sum in (37), denote 5 k / H (max{ B 1,k, C,k })(s 1/α ) α β 1 5 s s 1 (1+β)/α = 50s (1+β)/α. (38) A (l) = A ((3 l,3 l ) +[3 l ] [3 l ]), H (l) = H (3 l +[3 t ]). Then F (A {k}) = k H = (3 l,3 l,3 l ) + F l (A (l) H (l) ) l=1 F l ((A (l) (3 l,3 l )) (H (l) 3 l )). (39) l=1

158 BLEI AND GAO By (35), each summand in (39) is bounded by { 5 A (l) (1/α) (β/α) if H (l) A (l) 1 1/α, 5 H (l) ( (1/α) (β/α))/(1 1/α) if H (l) > A (l) 1 1/α. Therefore, F (A {k}) k H 5 A (l) (1/α) (β/α) + 5 H (l) ( (1/α) (β/α))/(1 1/α) l=1 l=1 ( (1/α) (β/α) ( 5 A ) (l) + 5 A (l) l=1 l=1 5 A (1/α) (β/α) + 5 H ( (1/α) (β/α))/(1 1/α) ) ( (1/α) (β/α))/(1 1/α) 5(1 + α/(α 1) )s (1/α) (β/α), (40) where the last inequality holds because H s 1 1α. By combining (37), (38), and (40), we conclude that FU (s) (50 + 5(1 + α/(α 1) ))s 1/α β/α = 5(11 + α/(α 1) )s y, (41) which implies dim U F y. To obtain the opposite inequality, for n > n 0, let m be the integer such that m 1 < (3 n ) α 1 m. Let A = (3 n,3 n ) + E n, where E n = E is obtained from Lemma 1(i) for N = 3 n, and let B = (3 n,3 n ) +[3 n ] m and C = (3 n,3 n ) +[m] [3 n ]. Then A 3 αn, and B = C < 3 αn + 3 n. Applying (33), we obtain F U (A B C) = {(i, j, k) (3 n,3 n,3 n ) + F n : (i, j) A, (j, k) B, (k, i) C} = {(i, j, k) F n : (i, j) E n, k [m]} 4 (3 n ) α β m 1 4 (3αn ) y. Therefore, FU ( 3 αn ) 1 4 (3αn ) y, which implies dim U F y. Claim 3: F is cylindrical. We have shown that dim U F = α β + 1. For every k N, π 1 3 {k} F = only if 3 l k 3 l, in which case π 1 3 {k} F = π 1 3 {k} ((3 l,3 l,3 l ) + F l ) = (3 l,3 l,3 l ) + π 1 3 {k 3 l } F l. By (34), d π 1 (α β) 5 for all 1 k 3 {k} F 3 l 3l. Therefore, d π 1 3 {k} F(α β) 5 for all k N, and hence F is cylindrical in direction 3. (ii) The construction of (random) doubly-cylindrical sets with the desired dimension follows a blueprint similar to the one followed in (i). Indeed, if x (, 3), then for all N n 0 for sufficiently large n 0, by using independent {0, 1}-valued random variables with

COMBINATORIAL DIMENSION IN FRACTIONAL CARTESIAN PRODUCTS 159 a prescribed mean, we can produce (random sets) F N [N] 3 with the following properties: (a) for every positive integer s N, s-subsets A and B of [N], and every l [N], F N (A B {l}) s x 1, F N (A {l} B) s x 1, F N ({l} A B) s x 1 ; (b) for every l [N], F N ([N] {l} N x 1, F N ([N] {l} [N] N x 1, F N ({l} [N] N x 1. Then, F = n=n 0 (F 3 n + (3 n,3 n,3 n )) is cylindrical in each of the three directions, and dim F = x/. The verification is similar to the proof given in (i), and is omitted. REFERENCES [1] G. Bennett, Probability inequalities for sums of independent random variables, J Amer Statist Assoc 57 (196), 33 45. [] R. Blei, Fractional Cartesian products of sets, Ann Inst Fourier, (Grenoble) 9 (1973), 79 105. [3] R. Blei, Combinatorial dimension and certain norms in harmonic analysis, Amer J Math 106 (1984), 847 887. [4] R. Blei, Fractional dimension and bounded fractional forms, Mem Amer Math Soc 57 (1985), 1 70. [5] R. Blei, Analysis in integer and fractional dimensions, Cambridge University Press, Cambridge, 001. [6] R. Blei and T. W. Körner, Combiantorial dimension and random sets, Israel J Math, 47 (1984), 65 74. [7] R. Blei, Y. Peres, and J. Schmerl, Fractional products of sets, Random Structures Algorithms 6 (1995), 113 119. [8] R. Blei, and J. Schmerl, Combinatorial dimension of fractional Cartesian products, Proc Amer Math Soc 10 (1994), 73 77. [9] G. Katona, External problems for hypergraphs, Combinatorics, J. H. Lint and M. Hall, Volume, 13 4. Math Centrum, Amsterdam, 1974. [10] J. B. Kruskal, The number of simplices in a complex, Mathematical Optimization Techniques, 51 78. University of California Press, Berkeley, 1963.