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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

The terms system, (a,b)(a,b)-system and Δ(>a)\Delta(>a)-system are those of Theorem I (p. 85): a system is an indexed family whose sets need not be distinct.

Theorem II (p. 86, quoted). "For every a,ba,b such that a,b⩾1a,b\geqslant1 there exists a (ab+1,b)(a^{b+1},b)-system which does not contain any Δ(>a)\Delta(>a)-system."

Remark 2 (p. 86) says the system is constructed explicitly. Remark 1 (p. 86) draws from it that Theorem I(ii) is best possible, and the paper says (p. 86) that by Theorem II, Theorem III is best possible except for a factor between 11 and b!b!.

Proof pointer

P. 89. Take sets AA, BB with ∣A∣=a|A|=a, ∣B∣=b|B|=b, and let FF be the set of all maps from BB to AA. The system has index set A×FA\times F, and the member indexed by (t,f)(t,f) is the graph {(x,f(x)):x∈B}\{(x,f(x)):x\in B\} of ff, which does not depend on tt. So each of the aba^b distinct graphs occurs aa times. In a Δ(>a)\Delta(>a)-subsystem, for each x∈Bx\in B two members agree at xx by pigeonhole, so the kernel contains a point over every xx and all members are the same graph; since that graph carries at most aa indices, two indices of the subsystem coincide, a contradiction.

Read depth

Claims checked: Theorem II, Remarks 1 and 2 and the construction on p. 89 were read clause by clause on the page images of the print. Nothing here is independently reviewed.

Dependencies

None.

Source. P. Erdős and R. Rado, Intersection theorems for systems of sets, J. London Math. Soc. 35 (1960), 85--90, doi:10.1112/jlms/s1-35.1.85; the edition read is named on the source card.

Bears on

  • Problem 20: with b=nb=n and a=k−1≥1a=k-1\ge1, the construction's distinct sets are the (k−1)n(k-1)^n graphs of maps from an nn-set to a (k−1)(k-1)-set, nn-element sets of which no kk form a sunflower, so f(n,k)>(k−1)nf(n,k)>(k-1)^n. The theorem's count ab+1a^{b+1} counts each set aa times, which a family of distinct sets does not allow.