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Updated Oct 7

Problem 881

claims/: The 1 claim page of Problem 881, one per claimant's result; the problem's standing derives from them.

Statement. Let $A\subset\mathbb{N}$ be an additive basis of order $k$ which is minimal, in the sense that if $B\subset A$ is any infinite set then $A\backslash B$ is not a basis of order $k$ .

Must there exist an infinite $B\subset A$ such that $A\backslash B$ is a basis of order $k+1$ ?

Status. Claimed: a pending full claim answers the question no; the site's label is OPEN and the site lists no proof claim. A manuscript posted in the site's thread on 2026-05-03 claims a complete answer, no for every $k\ge2$ , and is the pending claim Svyable; a reader's AI check reports that it counts sums of exactly $k$ elements where the site's definition allows at most $k$ , and finds fatal defects in its proof.

Source. erdosproblems.com/881, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #881, https://www.erdosproblems.com/881.

Formalization. Statement in formal-conjectures.

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Linked from (10)

Svyable's counterexamples for every order at least two problems/additive_bases/E0881/claims Additive Bases and Sidon Sets additive_bases/svyable_2026_infinite_deletions_strongly_minimal_additive_bases Lemma 8 (p. 5): private witnesses at order k+1 that are covered at order k with the booster 1 Lemma 9 (p. 7): booster witnesses that are sums of k elements only with the element 1 Main Theorem 6 (p. 3): the deletion question has answer yes for k = 1 and no for every k >= 2, with thin counterexamples Proposition 12 (p. 15): no asymptotic basis of order 2 is minimal as an asymptotic basis of order 3 Proposition 7 (p. 4): removing a suitable infinite set from an asymptotic basis of order 1 leaves one of order 2 Theorem 11 (p. 14): a basis of order k with no infinite deletion of order k+1 is a minimal basis of order k+1 plus a finite set

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