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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 paper's notation and the set S(k,l,m)S(k,l,m) of systems are those of Theorem 1; S(2,2r,4r)S(2,2r,4r) consists of the systems of subsets of [0,4r)[0,4r), each of at most 2r2r elements, no member containing another, any two members sharing at least two elements.

Example (p. 319). For r>0r>0, the 2r2r-subsets aa of [0,4r)[0,4r) with ∣a∩[0,2r)∣>r\lvert a\cap[0,2r)\rvert>r form a system in S(2,2r,4r)S(2,2r,4r) with

n=12(4r2r)−12(2rr)2n=\frac12\binom{4r}{2r}-\frac12\binom{2r}{r}^2

members. The paper notes that this exceeds (4r−22r−2)\binom{4r-2}{2r-2}, the bound of Theorem 2 (b), for every large rr, possibly for every r>2r>2. The paper introduces it as a more general example than S. H. Min's example of p. 318: the 4-subsets aa of [0,8)[0,8) with ∣a∩[0,4)∣=3\lvert a\cap[0,4)\rvert=3, sixteen sets forming a system in S(2,4,8)S(2,4,8), against (62)=15\binom62=15.

Conjecture (p. 319). The authors conjecture that for these values of k,l,mk,l,m the example is a case of largest nn: if r>0r>0 and (a0,…,an−1)∈S(2,2r,4r)(a_0,\ldots,a_{n-1})\in S(2,2r,4r), then

n≤12(4r2r)−12(2rr)2.n\le\frac12\binom{4r}{2r}-\frac12\binom{2r}{r}^2.

The conjecture covers systems whose members may have fewer than 2r2r elements, provided no member contains another; the example has all members of size 2r2r.

Source. P. Erdős, Chao Ko and R. Rado, Intersection theorems for systems of finite sets, Quart. J. Math. Oxford Ser. (2) 12 (1961), 313–320, as identified on the source card: concluding remark (i), pp. 318–319, with the conjecture on p. 319.

Read depth. Claims checked: the example, its count and the conjecture were read clause by clause on the print. Nothing here is independently reviewed.

Proof pointer

A conjecture; the paper proves only the count of the example, by summing (2rλ)(2r2r−λ)\binom{2r}\lambda\binom{2r}{2r-\lambda} over r<λ≤2rr<\lambda\le2r and using the symmetry λ↔2r−λ\lambda\leftrightarrow2r-\lambda (p. 319).

Dependencies

None.

Bears on

  • Problem 83: the problem's statement, with its nn the paper's rr, is the conjecture restricted to systems all of whose members have exactly 2r2r elements, for which the incomparability condition is automatic. The paper poses the conjecture and gives the example showing the bound would be attained. For r≥2r\ge2 it proves no bound of that size: Theorem 2 (a) applies to this case, but its bound is larger (that comparison is arithmetic, not stated in the paper).