# Combinatorial Mathematics V by C. H. C. Little

By C. H. C. Little

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Extra info for Combinatorial Mathematics V

Example text

1) is easy. If ne [n] lies in a block by itself, the remaining blocks forma set partition of [n-1]. Otherwise n lies in one of the k blocks of a set partition of [n-1]. We shall use another bijection to list a1l set partitions of [n]. A restricted growthfunction on [n] (or RG function) is a vector (v 1, v2 , ... , v0 ) satisfying v 1 =1 and vi~max{v 1 , . ,vH}+l. The15RGfunctionson [4] are 1111 1211 1223 1112 1212 1231 1121 1213 1232 1122 1221 1233 1123 1222 1234. 1 There is a bijection between RG functions on [n] and set partitions of [n].

R·c 1)) u { 1}. ) Because these two subcollections are disjoint, This completes Step (2). 10) + ... i'(1)) u { 1}. 2, Eq. 7) implies e:=D + ... + e: =D. 16) lol'Cl)l ~ 1ar·ci>1. 13). 10) musthold, and the proof is complete. Notes Three good general references for posets are [Ai], [Be] and [Gre-K1] . Spemer theorems are included in §3 ofChapter Vill of [Ai]. They are also a central topic of [Gre-K 1]. D. The entries for BD given in Exercises 8 and 10. For f>D they are Exercises 9 and 22, for and '\IÂ.

1 implies that P bas the Spemer property. 1. We merely identify those chains in the proof. Suppose L0 ~ 0, and choose any a e ao =a, (ao), ... , at= fk(ak_1). a1 = f 1 We now have a chain Lo- Let ao <· a 1 <· ... <· ak. If ak is in the image of gk+l• let ak+l be the unique element in ~+l such that gk+ 1(ak+l) = ak. We continue in this manner until either (a) (b) we encounter an a;_ which is not in the image of gi+l' or we reach am e Lm. Let C 1 = ao <· a 1 <· .. <· a1 be this chain. Now repeat the preceding construction until the elements of L0 are exhausted.