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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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Source. Theorem 3, p. 6, of David Ellis, Irredundant families of subcubes, arXiv:1003.2960v1 (2010), published in Mathematical Proceedings of the Cambridge Philosophical Society 150(2) (2011), 257–272, as identified on the source card. Labels and pages are those of arXiv:1003.2960v1.

Statement

Theorem 3 (p. 6; the paper's heading reads "Bollobás, 1965"). Let a1,…,aNa_1,\ldots,a_N and b1,…,bNb_1,\ldots,b_N be subsets of {1,2,…,n}\{1,2,\ldots,n\} such that ai∩bj=∅a_i\cap b_j=\varnothing if and only if i=ji=j. Then

∑i=1N(∣ai∣+∣bi∣∣bi∣)−1≤1.\sum_{i=1}^N\binom{|a_i|+|b_i|}{|b_i|}^{-1}\le1.

The theorem as printed adds an equality condition: equality can hold only when there are a set Y⊆[n]Y\subseteq[n] and an integer aa with {a1,…,aN}=Y(a)\{a_1,\ldots,a_N\}=Y^{(a)}, the aa-element subsets of YY, and bi=Y∖aib_i=Y\setminus a_i for every ii.

The pairs may have different sizes; no uniformity of ∣ai∣|a_i| or ∣bi∣|b_i| is assumed.

Proof pointer

The paper gives no proof. It refers the reader (p. 6) to its reference [3], B. Bollobás, Combinatorics: Set systems, hypergraphs, families of vectors, and combinatorial probability, Cambridge University Press, 1986.

Dependencies

External to the paper. Inside it, Theorem 3 is the tool behind Theorem 4 (through the claim (5), p. 7) and Theorem 7 (through the claim (7), p. 9).

Read depth. Claims checked: the statement and its equality clause were read clause by clause on p. 6. The paper contains no proof to check.

Bears on

  • Problem 7: background only. The theorem is about finite sets and says nothing about congruences.