Stirling numbers of the second kind and Bonferroni s inequalities

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1 Eem. Math. 60 (2005) /05/ c Swiss Mathematica Society, 2005 Eemente der Mathematik Stiring numbers of the second kind and Bonferroni s inequaities Horst Wegner Horst Wegner studierte Mathematik an der Universität Hamburg. Nach dem Dipom 1966 und kurzer Tätigkeit in der Industrie promovierte er 1970 über ein Probem zu Stiringschen Zahen zweiter Art an der Universität Kön. Seit 1973 ist er as Akademischer Oberrat an der Universität Duisburg tätig, zunächst in der Lehrerausbidung und seit 1982 im Fachgebiet Stochastik. 1 Introduction The number of ways of partitioning a set of n eements into k nonempty subsets is usuay denoted S(n, k). The numbers S(n, k) are caed Stiring numbers of the second kind. It is we-known that S(n, k) = 1 k 1 ( 1) i (k i) n k! i i=0 for n, k N. By an easy appication of Bonferroni s inequaities, it can be shown that the partia sums 1 k! ( 1) i (k i) n, = 0, 1,...,k 1, i i=0 successivey overcount and undercount the number S(n, k) (see Theorem 3).. Auf wie viee Weisen ässt sich eine Menge von n Eementen in k nicht-eere Teimengen zeregen? Die Antwort hierauf iefern die Stiringschen Zahen S(n, k) zweiter Art. Der Zusatz zweiter Art hat historische Gründe. Die Stiringschen Zahen erster Art können z.b. as Koeffizienten von x k in dem Poynom x(x 1)... (x n + 1) und die Stiringschen Zahen zweiter Art umgekehrt as Koeffizienten in der Darsteung von x n as Linearkombination der Ausdrücke (x) k := x(x 1)... (x k + 1) eingeführt werden. In der nachfogenden Arbeit wird mit eementaren kombinatorischen Überegungen, insbesondere den Ungeichungen von Bonferroni, gezeigt, dass die Partiasummen der bekannten geschossenen Darsteung für die S(n, k) abwechsend obere und untere Schranken für die S(n, k) sind.

2 Stiring numbers of the second kind and Bonferroni s inequaities 125 There are ony a few textbooks on combinatorics, which present Bonferroni s inequaities (e.g. [1], [2]), and the representation is sometimes not very satisfactory to the reader (cf. [1]). Therefore, we start with a deduction of Bonferroni s inequaities in a concise and eementary way (see Theorem 1). 2 Bonferroni s inequaities Bonferroni s inequaities are cosey reated to the principe of incusion and excusion. Theorem 1 Let be a nonempty set, and et A 1,...,A n be finite nonempty subsets of. Putting { m := max T : T {1, 2,...,n}, } =, then m 1 and for = 1, 2,...,m = > < n if = m, n if < m, odd, n if < m, even. The proof of Theorem 1 becomes short and cear, if we appy the foowing emma. For this, a usefu device is the so-caed indicator function of a set A { 1 if ω A, 1 A (ω) = 0 if ω A. Lemma Let be a nonempty set, and et A 1,...,A n with ω n put m(ω) := {i : ω }. Then, m(ω) 1 and for = 1, 2,...,n. Proof. Let ω n n 1 (ω) = 1 ( 1) =. For each. For abbreviation we put I (ω) := {i : ω }. Hence I (ω) = and m(ω) = I (ω) 1. Then, for = 1, 2,...,n we have 1 (ω) = 1 (ω) = T I (ω) ( ) m(ω).

3 126 H. Wegner Hence, for 1 n 1 (ω) = ( ) m(ω) ( 1) 1 (( ) ( )) m(ω) 1 m(ω) 1 = ( 1) = 1 + ( 1) 1. Proof of Theorem 1. Obviousy, m 1. For abbreviation, we put A := n. Then, using the notation and the resut of the preceding emma, we obtain ( 1) 1 = ( 1) 1 1 (ω) ω A = ( ( )) m(ω) 1 1 ( 1) ω A = A ( 1). ω A Obviousy, m = max m(ω). Therefore: ω A = m = 0 fora ω A, < m > 0 for at east one ω A. Thus, the resut of Theorem 1 foows immediatey. It is obvious that Theorem 1 coud be obtained as a coroary to the foowing Theorem 2, which shoud aso be mentioned here. Nevertheess, we have presented an individua proof of Theorem 1 to show its simpicity. Theorem 2 Let (, A,) be a measure space, and et A 1,...,A n A with 0 < ( n ) { ( ) } <. Putting m := max T : T {1, 2,...,n}, > 0, then m 1 and for = 1, 2,...,m ( = ( n ) ) > ( n ) < ( n ) if = m, if < m, if < m, odd, even.

4 Stiring numbers of the second kind and Bonferroni s inequaities 127 Proof. Since there is at east one with ( )>0, it is cear that m 1. n Moreover, =. Thus we obtain by the emma for a = 1, 2,...,n and a ω ( ) with 1 (ω) = 1 n (ω) ( 1) R (ω) if ω n, R (ω) = 0 if ω n. Since ( n ) <, the functions 1 n, 1 A, R are -integrabe. i First we consider ( ) for = n. SinceR n = 0, it foows from ( ) by integration with respect to that ( n ) = n According to the definition of m, this means ( ). ( n ) = m ( ), which is the asserted equaity for = m. Now et < m. From the definition of m there is a subset T 0 {1, 2,...,n} with T 0 = m such that ( ) > 0. Hence = and m(ω) = m for a ω. This impies ( ) m 1 R (ω) = 1 fora ω. 0 Hence R d ( ) 1 d = > Integrating ( ) with respect to, the ast resut competes the proof of the asserted inequaities for < m.

5 128 H. Wegner 3 Inequaities for the numbers S(n, k) Theorem 3 Let n, k N with k n. Then 1 = S(n, k) if = k 1, ( 1) (k ) n > S(n, k) if 0 < k 1, even, k! =0 < S(n, k) if 1 < k 1, odd. Proof. Because of S(n, 1) = 1, the statement is vaid for k = 1. Now suppose 2 k n. Then et X, Y be sets with X = n, Y = k. According to the definition of S(n, k), it is evident that the number of surective functions from X to Y is k!s(n, k). For abbreviation we define the nonempty sets A y := { f Y X : y f (X)}, y Y. Then we obtain for the number of non-surective functions (i) k n k!s(n, k) =. A y y Y This suggests to appy Theorem 1. Suppose y 0 Y and define g(x) := y 0 for a x X. Then g is obvious that A y =. Thus, we have (ii) Furthermore, y Y y T y Y \{y 0 } { m := max T : T Y, } A y = = k 1. y T A y ={f Y X : f (X) Y \ T }, and hence (Y A y = \ T ) X = (k T ) n. y T A y. Otherwise, it Thus, for = 1, 2,...,k 1 y T (iii) T Y A y = (k T ) n = T Y (k ) n. By (i), (ii), (iii) and Theorem 1 we obtain thereupon for = 1, 2,...,k 1 = k n k!s(n, k) if = k 1, ( 1) 1 (k ) n > k n k!s(n, k) if < k 1, odd, < k n k!s(n, k) if < k 1, even. Then a simpe rearrangement gives us the desired resut for 1. Since (i) immediatey impies k n k!s(n, k) >0, the resut of Theorem 3 is aso vaid for = 0.

6 Stiring numbers of the second kind and Bonferroni s inequaities 129 References [1] Comtet, L.: Advanced Combinatorics. Reide, Dordrecht [2] Staney, R.P.: Enumerative Combinatorics. Vo. 1, Cambridge University Press Horst Wegner Universität Duisburg-Essen Standort Duisburg Institut für Mathematik Lotharstr. 65 D Duisburg, Deutschand e-mai: wegner@math.uni-duisburg.de