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Statement

Lemma 1 (p. 92). Let s≥1s\ge1 and t≥0t\ge0, and let S=⋃i=1sSiS=\bigcup_{i=1}^{s}S_i and T=⋃k=1tTkT=\bigcup_{k=1}^{t}T_k be sets such that

  • (1a) ∣Si∣∈{1,2}\lvert S_i\rvert\in\{1,2\} and ∣Tk∣∈{1,2}\lvert T_k\rvert\in\{1,2\} for all i≤si\le s and k≤tk\le t;
  • (1b) the SiS_i are pairwise disjoint and the TkT_k are pairwise disjoint;
  • (1c) Si≠TkS_i\ne T_k for all ii and kk.

Let Φ(S,T)\Phi(S,T) be the number of sets X⊆SX\subseteq S with (1d) ∣X∣=s\lvert X\rvert=s, (1e) ∣X∩Si∣=1\lvert X\cap S_i\rvert=1 for every ii, and (1f) X∩Tk≠∅X\cap T_k\ne\emptyset for every kk. Then Φ(S,T)≤2s(3/4)t\Phi(S,T)\le2^s(3/4)^t, and the paper calls the estimate best possible.

The extremal example (pp. 94--95): when s≥2ts\ge2t, take 2s2s distinct elements a1,…,as,b1,…,bsa_1,\ldots,a_s,b_1,\ldots,b_s, Si={ai,bi}S_i=\{a_i,b_i\} and Tk={ak,at+k}T_k=\{a_k,a_{t+k}\}; then Φ(S,T)=3t2s−2t=2s(3/4)t\Phi(S,T)=3^t2^{s-2t}=2^s(3/4)^t.

Remark (p. 95). The paper calls it an open combinatorial problem to find good estimates for Φ(S,T)\Phi(S,T) when (1a) is replaced by 1≤∣Si∣≤h1\le\lvert S_i\rvert\le h and 1≤∣Tk∣≤h1\le\lvert T_k\rvert\le h for h≥3h\ge3.

Proof pointer

Pp. 92--95. Induction on tt, the case t=0t=0 giving 2s2^s. A TkT_k missing SS gives Φ=0\Phi=0; a TkT_k meeting SS in one point forces that point and removes one SiS_i and one TkT_k. Otherwise every TkT_k is a pair inside SS, and the two cases S=TS=T and S≠TS\ne T each remove a few SiS_i and TkT_k at a time, the recursion closing because 2⋅2−2(3/4)−2≤12\cdot2^{-2}(3/4)^{-2}\le1 and 2−2(3/4)−1+2−1(3/4)−1=12^{-2}(3/4)^{-1}+2^{-1}(3/4)^{-1}=1.

Read depth

Claims checked: the statement, the extremal example and the remark were read clause by clause on the page images of the print, and the induction was followed. Nothing here is independently reviewed.

Dependencies

None.

Source. P. Erdős and M. B. Nathanson, Systems of distinct representatives and minimal bases in additive number theory, in: Number Theory, Carbondale 1979, Lecture Notes in Math. 751, Springer, Berlin, 1979, pp. 89--107 (MR 81k:10089); the edition read is named on the source card.

Bears on

The lemma bears on no problem directly; through Lemma 2 it is the counting step behind Theorem 1 and the threshold c>1/log⁡(4/3)c>1/\log(4/3) there. The remark on p. 95 calls the analogous estimate for sets of size up to hh, h≥3h\ge3, an open combinatorial problem; order h≥3h\ge3 is the setting of Problem 870, and the paper proves nothing for h≥3h\ge3.