Wiki
Wiki

Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

Updated


Source. The unnumbered result on pp. 133--135 of P. Erdős, Set-theoretic, measure-theoretic, combinatorial, and number-theoretic problems concerning point sets in Euclidean space, Real Anal. Exchange 4 (1978/79), no. 2, 113--138, doi:10.2307/44151159, as identified on the source card. Pages are those of the journal print.

Read depth. Claims checked: the statement (p. 133) was read clause by clause, the proof (pp. 133--135) for its structure. Nothing here is independently reviewed.

Statement

Assume c>ℵ1\mathfrak c>\aleph_1. For n=1,2,…n=1,2,\ldots let SnS_n be a set of real numbers in which all sums x+yx+y with x,y∈Snx,y\in S_n are distinct, which the paper glosses as all distances between points of SnS_n being distinct. Then there are ℵ1\aleph_1 rationally independent reals bαb_\alpha, 1≤α<ω11\le\alpha<\omega_1, and a real tt such that every number t+∑βrβbβt+\sum_\beta r_\beta b_\beta, with rational rβr_\beta and finitely many terms, lies outside ⋃n=1∞Sn\bigcup_{n=1}^\infty S_n (p. 133). So the complement of the union contains a translate of an ℵ1\aleph_1-dimensional linear subspace of the reals over the rationals.

The paper asks (p. 135) whether the translate can be dropped, the complement containing the rational span itself, and whether ℵ1\aleph_1 can be replaced by ℵ2\aleph_2, saying it does not think the latter likely. It adds that Baumgartner proved, answering an earlier question of Erdős, that the complement of a single set of reals with all sums x+yx+y distinct contains an infinite arithmetic progression, and that the proof here borrows from Baumgartner's unpublished proof (p. 135).

Proof pointer

Pp. 133--135. The argument of the Hajnal-Erdős lemma of Theorem 2 gives a set AA of ℵ2\aleph_2 reals such that rara lies outside the union for every rational rr and every a∈Aa\in A. Against a fixed set BB of ℵ1\aleph_1 rationally independent reals, call a∈Aa\in A bad when ℵ1\aleph_1 numbers ra+∑cβbβra+\sum c_\beta b_\beta lie in the union. Counting choices shows at most ℵ1\aleph_1 elements of AA are bad, since ℵ2\aleph_2 bad ones would put four numbers u+vu+v, u+v′u+v', u′+vu'+v, u′+v′u'+v' into one SnS_n, against distinct sums. A good aa then serves after discarding countably many elements of BB. The printed proof writes the overlined union, the complement, at places where the union itself is meant (pp. 133--135).

Dependencies

The lemma of Hajnal and Erdős recorded on the Theorem 2 page.

Bears on

No Erdős problem page of the corpus cites this result.