Bourbaki

Exercise 14, § 5

Tag 03MQ, permanent. Cite it as [Bourbaki, Tag 03MQ].

Theory of Sets, III, 239

Bourbaki marks this one as harder than the rest.

Let E be a finite set with nn elements, let (aj)1⩽j⩽n(a_j)_{1 \leqslant j \leqslant n} be the sequence of elements of E arranged in some order, and let (Ai)1⩽i⩽m(\mathrm{A}_i)_{1 \leqslant i \leqslant m} be a sequence of subsets of E.

(a) For each index jj, let kjk_j be the number of indices ii such that aj∈Aia_j \in \mathrm{A}_i, and let Si=Card (Ai)\mathrm{S}_i = \mathrm{Card}\,(\mathrm{A}_i). Show that

∑j=1nkj=∑i=1msi.\sum_{j=1}^{n} k_j = \sum_{i=1}^{m} s_i.

(b) Suppose that, for each subset {x, y}\{x,\, y\} of two elements of E, there exists exactly one index ii such that xx and yy are contained in Ai\mathrm{A}_i. Show that, if aj∉Aia_j \notin \mathrm{A}_i, then Si⩽kj\mathrm{S}_i \leqslant k_j.

(c) With the hypotheses of (b), show that m⩾nm \geqslant n. (Let knk_n be the least of the numbers kjk_j. Show that we may suppose that, whenever i⩽kn,j⩽kni \leqslant k_n, j \leqslant k_n, and i≠ji \neq j, we have aj∉Aia_j \notin \mathrm{A}_i and an∉Aja_n \notin \mathrm{A}_j for all j⩾knj \geqslant k_n.)

(d) With the hypotheses of (b), show that m=nm = n if and only if one of the following two alternatives is true : (i) A1={a1, a2, …, an−1}\mathrm{A}_1 = \{a_1,\, a_2,\, \ldots,\, a_{n-1}\}, Ai={ai−1, an}\mathrm{A}_i = \{a_{i-1},\, a_n\} for i=2,…,ni = 2, \ldots, n; (ii) n=k(k−1)+1n = k(k-1) + 1, each Ai\mathrm{A}_i has kk elements, and each element of E belongs to exactly kk sets Ai\mathrm{A}_i.

Solution

No solution has been written yet.

Cited by

Nothing.

Cites

Nothing.

Transcribed from the printed volume. Not translated.