Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Claim. The question of Problem 880
has the answer yes for and no for every . Write for
the set of sums of pairwise distinct elements of and for
the largest asymptotic gap of a set
. Theorem 1 of N. Hegyvári, F. Hennecart and A. Plagne,
Answer to a question by Burr and Erdős on restricted addition, and related
results, Combin. Probab. Comput. 16 (2007), no. 5, 747--756, states: (i) if
contains all sufficiently large integers, then
, and if does, then ;
(ii) for every there is a set such that contains
all sufficiently large integers and
, and a set with
containing all sufficiently large integers and .
A set is a basis of order in the problem's sense, every large integer
a sum of or fewer elements, exactly when contains all
large integers, and the problem's is $A\cup 2\times A\cup\cdots\cup k\times
A$, so part (i) gives for all large when , and part
(ii) gives a basis of each order with unbounded. The
claim value is disproved: the question asks whether the gaps are bounded
for every basis of every order , and part (ii) refutes that for each
; the case , where the assertion holds, is the part of the
question that survives. The site labels the problem PROVED, although its
remarks report the answer no for every ; the claim value follows the
question as stated, which asks about every order .
Argument. The case is the parity observation the site's remarks also make: an odd element of is a sum of two distinct elements, so when every large integer lies in , every large odd integer lies in , the problem's , whose gaps are therefore at most (when alone contains every large integer, itself contains every large odd integer). For the paper gives an explicit set: with and , let and ; sums of elements of cover a long initial interval, which makes a basis of order , while sums of distinct elements miss arbitrarily long stretches. The same example gives the paper's Theorem 3: its , the largest over all with cofinite of the least with , satisfies , realized by this example, which is why the answer fails for every . The source card is hegyvari_2007_answer_question_burr_erdos_restricted_addition; the theorem numbering used here is that of the authors' preprint. This outline is a reading aid, not proof coverage.
Acceptance. The refereed evidence is the journal publication cited
above (Combinatorics, Probability and Computing; Crossref gives the issue
date 2007-09-01, which is the page's date). The reviewed evidence is the
documented acceptance by the catalog erdosproblems.com: its curator, Thomas
Bloom, credits both answers to this paper in the problem's remarks, and the
page carries the label PROVED (accessed 2026-10-07). The proof is not compiled
or reviewed here.
Formalization. A third party formalized both answers: not_erdos_880 in
src/latest/ErdosProblems/Erdos880.lean of Boris Alexeev's repository
https://github.com/plby/lean-proofs, pinned above at the commit of 2026-08-31
that last touched the file (the file was added on 2026-08-17). Its header
names Hegyvári, Hennecart and Plagne as informal authors and Codex and
GPT-5.6 Sol as formal authors, and its module comment states their theorem, so
it is a formalization of this result and a link on this page, not an
independent claim. Its theorem is a conjunction: every infinite set whose
least order as an asymptotic basis is has sums of at most two distinct
elements with bounded gaps, and for every there is an infinite set of
least order whose sums of at most distinct elements have unbounded
gaps; the file imports Mathlib, contains no sorry and prints the theorem's
axioms. This corpus has not built the file or audited its definitions against
the problem, so the claim carries no formalized evidence.
Depends on. No page of this wiki.