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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

Conventions (pp. 111--112, as on the Theorem 1 page). On a finite set SS of mm elements, a set-mapping of type kk and order l+1l+1 sends each kk-element X⊆SX\subseteq S to a set of at most ll points outside XX; a set P⊆SP\subseteq S is free when f(X)∩P=∅f(X)\cap P=\varnothing for every kk-element X⊆PX\subseteq P. In the paper p,m,l,kp,m,l,k are integers here (pp. 114, 129).

Theorem 12 (p. 129, quoted). "Let p(m,l,k)p(m, l, k) denote the greatest integer pp for which (m,l+1,k)→p(m, l+1, k)\to p is true. Then c1mk+1<p(m,l,k)<c2mlog⁡mkc_1\sqrt[k+1]{m}<p(m, l, k)<c_2\sqrt[k]{m\log m} where the numbers c1c_1 and c2c_2 depend on kk and ll but they do not depend on mm, and c1>0c_1>0."

So every set-mapping of type kk and order l+1l+1 on an mm-element set has a free set of more than c1m1/(k+1)c_1m^{1/(k+1)} elements, and some such set-mapping has no free set of c2(mlog⁡m)1/kc_2(m\log m)^{1/k} or more elements; Section 3 (p. 114) calls c1c_1 and c2c_2 positive real numbers.

Problem 4 (p. 114, quoted). "What is the exact order of magnitude of p(m,l,k)p(m, l, k)?"

Source. P. Erdős and A. Hajnal, On the structure of set-mappings, Acta Math. Acad. Sci. Hungar. 9 (1958), 111--131: Theorem 12 on p. 129, proof pp. 130--131, announced with Problem 4 on p. 114. The edition is the one identified on the source card.

Read depth. Claims checked: the statement, Problem 4 and the definitions they use were read clause by clause on the printed pages. The proof was not checked.

Proof pointer

Lower bound (p. 130): if no free set has pp elements, every pp-set contains a non-free (k+1)(k+1)-set, one consisting of a kk-set and a point of its value; counting such (k+1)(k+1)-sets, of which there are at most l(mk)l\binom{m}{k}, against the pp-sets, each of which must contain one, shows that any such pp is at least c m1/(k+1)c\,m^{1/(k+1)} for some c>0c>0. Upper bound (pp. 130--131): a uniformly random set-mapping of type kk and order l+1l+1 has, with positive probability, no free set of pp elements once p≥c2(mlog⁡m)1/kp\ge c_2(m\log m)^{1/k}, by a union bound over the pp-sets.

Dependencies

None beyond the counting and probability estimates of the proof.

Bears on

  • Problem 1025: the problem's g(n)g(n) is the largest independent set guaranteed for maps sending each pair of {1,…,n}\{1,\ldots,n\} to a point outside it. That is p(n,1,2)p(n,1,2) (an observation of this page: type 2 and order 2, a map whose values are empty being replaced by one with point values, which only removes free sets), so Theorem 12 with k=2k=2, l=1l=1 gives n1/3≪g(n)≪(nlog⁡n)1/2n^{1/3}\ll g(n)\ll(n\log n)^{1/2}. The problem's question, the order of g(n)g(n), is the case k=2k=2, l=1l=1 of the paper's Problem 4. The bounds are recorded as a claim on Erdős and Hajnal 1958.