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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

Setting (p. 345). Let S\mathcal S be a set and ff a set function assigning to each finite subset AA of S\mathcal S an element f(A)∈S−Af(A)\in\mathcal S-A. For S1⊆S\mathcal S_1\subseteq\mathcal S put

F(S1)=⋃A⊆S1f(A),F(\mathcal S_1)=\bigcup_{A\subseteq\mathcal S_1}f(A),

the union over all finite subsets AA of S1\mathcal S_1. A set S1\mathcal S_1 is independent when S1∩F(S1)\mathcal S_1\cap F(\mathcal S_1) is empty. For ∣S∣=n<ℵ0\lvert\mathcal S\rvert=n<\aleph_0:

  • h(n)h(n) is the largest number such that for every ff the set S\mathcal S has an independent subset S1\mathcal S_1 with ∣S1∣≥h(n)\lvert\mathcal S_1\rvert\ge h(n);
  • H(n)H(n) is the least number for which some ff has F(S2)=SF(\mathcal S_2)=\mathcal S for every S2⊆S\mathcal S_2\subseteq\mathcal S with ∣S2∣≥H(n)\lvert\mathcal S_2\rvert\ge H(n).

Theorem (Tétel, p. 345, unnumbered). Let ε>0\varepsilon>0 and n>n0(ε)n>n_0(\varepsilon). Then

log⁡nlog⁡2<H(n)<log⁡nlog⁡2+(3+ε)log⁡log⁡nlog⁡2.\frac{\log n}{\log 2}<H(n)<\frac{\log n}{\log 2} +\frac{(3+\varepsilon)\log\log n}{\log 2}.

The English summary (p. 348) states the result as log⁡n/log⁡2<H(n)<log⁡n/log⁡2+3log⁡log⁡n/log⁡2+o(log⁡log⁡n)\log n/\log 2<H(n)<\log n/\log 2+3\log\log n/\log 2+o(\log\log n). It defines ff on every subset AA of an nn-element set SS, with f(A)∈S−Af(A)\in S-A, and F(A)F(A) as the union of f(B)f(B) over the subsets BB of AA.

Remarks printed with the theorem.

  • Improved constant (p. 347, no proof given). The authors say a small change of their method gives H(n)<log⁡n/log⁡2+(2+ε)log⁡log⁡n/log⁡2H(n)<\log n/\log 2+(2+\varepsilon)\log\log n/\log 2 for n>n0(ε)n>n_0(\varepsilon), and that they do not yet see how to determine H(n)H(n) to within o(log⁡log⁡n)o(\log\log n).
  • The function h(n)h(n) (pp. 345 and 347). The authors have only very weak bounds for h(n)h(n). They say the theorem easily gives h(n)<(log⁡n+3log⁡log⁡n)/log⁡2+o(log⁡log⁡n)h(n)<(\log n+3\log\log n)/\log 2+o(\log\log n), and that h(n)>ckh(n)>ck, where kk is the least number with log⁡kn<1\log_k n<1 and log⁡k\log_k is the kk-fold iterated logarithm. They think both bounds are very far from the truth.
  • Reformulation (p. 348). The authors restate the problem of determining H(n)H(n) as finding the least tt for which the subsets of at most tt elements of an nn-element set can be split into nn classes so that the subsets of every S1⊆S\mathcal S_1\subseteq\mathcal S with ∣S1∣=2t\lvert\mathcal S_1\rvert=2t meet all nn classes. They cite their paper On a property of families of sets (the paper's reference [3]) for this reformulation.

Proof pointer

Lower bound (p. 345). A set S2\mathcal S_2 has at most 2∣S2∣2^{\lvert\mathcal S_2\rvert} subsets, so ∣F(S2)∣≤2∣S2∣\lvert F(\mathcal S_2)\rvert\le2^{\lvert\mathcal S_2\rvert}, which gives H(n)≥log⁡n/log⁡2H(n)\ge\log n/\log 2. For strictness, (n2)>n\binom n2>n when n>3n>3, so two 2-element sets A≠BA\ne B have f(A)=f(B)f(A)=f(B). A set S2\mathcal S_2 containing both then has ∣F(S2)∣<2∣S2∣\lvert F(\mathcal S_2)\rvert<2^{\lvert\mathcal S_2\rvert}.

Upper bound (pp. 346-347). The paper counts functions. It restricts ff to the tt-element subsets, which allows (n−t)(nt)(n-t)^{\binom nt} functions, and lets F′(S1)F'(\mathcal S_1) be the union of f(A)f(A) over the tt-element A⊆S1A\subseteq\mathcal S_1. Counts (1) and (2) give the number of functions whose F′(S2)F'(\mathcal S_2) misses a given point xx, for a given 2t2t-element S2\mathcal S_2, in the cases x∉S2x\notin\mathcal S_2 and x∈S2x\in\mathcal S_2. Bound (3) sums them over xx, and (4) sums over all 2t2t-element sets. Inequality (5) shows the result is fewer than all (n−t)(nt)(n-t)^{\binom nt} functions for n>n0n>n_0. So some ff has F′(S2)=SF'(\mathcal S_2)=\mathcal S for every 2t2t-element S2\mathcal S_2, and then H(n)≤2tH(n)\le2t. The last step reduces (5) to 4t>4nt2log⁡n4^t>4nt^2\log n. The value of tt is printed in two forms: on p. 346 it is t=⌊log⁡n/log⁡2+(3+ε)log⁡log⁡n/(2log⁡2)⌋t=\bigl\lfloor\log n/\log2+(3+\varepsilon)\log\log n/(2\log2)\bigr\rfloor, and on p. 347 it is t=⌊log⁡n/log⁡2+(3+ε)log⁡log⁡n/log⁡2⌋t=\bigl\lfloor\log n/\log2+(3+\varepsilon)\log\log n/\log2\bigr\rfloor, with square brackets for the integer part. (Observation of this page, not of the paper.) Neither printed value fits the theorem. The bound H(n)≤2tH(n)\le2t and the final inequality both fit t=⌊(log⁡n+(3+ε)log⁡log⁡n)/(2log⁡2)⌋t=\bigl\lfloor(\log n+(3+\varepsilon)\log\log n)/(2\log2)\bigr\rfloor: then 4t4^t is about n(log⁡n)3+εn(\log n)^{3+\varepsilon}, and 2t2t is at most the theorem's upper bound.

Source. P. Erdős and A. Hajnal, Egy kombinatorikus problémáról (On a combinatorial problem), Matematikai Lapok 19 (1968), 345-348; MR 39 #5378. The edition read is identified on the source card.

Read depth. Claims checked: the definitions, the theorem, the English summary and the remarks were read clause by clause on the page images of the print. The proof on pp. 345-347 was followed but not checked line by line. The reading of tt above is this page's own. Nothing here is independently reviewed.

Dependencies

None in the corpus. The paper cites its references [1] (On the structure of set mappings, 1958) and [2] (On a problem of B. Jónsson, 1965) for the infinite case, and [3] for the reformulation on p. 348.

Bears on

  • Problem 624: the theorem places H(n)−log⁡n/log⁡2H(n)-\log n/\log2 strictly between 00 and (3+ε)log⁡log⁡n/log⁡2(3+\varepsilon)\log\log n/\log2 for n>n0(ε)n>n_0(\varepsilon). It neither proves nor refutes that this difference tends to infinity, which the problem asks; see the conjecture on p. 346. The paper requires f(A)∈S−Af(A)\in\mathcal S-A, while the problem's statement lets f(A)f(A) be any element of XX.