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Claim. Desmond Weisenberg, in a post of 17 August 2025 in the discussion thread of Problem 949, claims the answer no to the erdosproblems.com wording of that time, which asked for a set $A\subseteq\mathbb R\setminus S$ of cardinality continuum with $A+A\subseteq A$. The counterexample is $S=\bigcup_{n\in\mathbb Z}[3n+1,3n+2)$, which is sum-free; every $a\notin S$ that is not a multiple of $3$ has a multiple in $S$, so the subsets of $\mathbb R\setminus S$ closed under addition are sub-semigroups of $3\mathbb Z$, all countable. A later edit of the post says that the corrected problem asks for $A+A\subseteq\mathbb R\setminus S$, which the construction does not resolve.

Submission note. Posted to the erdosproblems.com forum by Desmond Weisenberg (account DesmondWeisenberg) on 17 August 2025:

The answer is no. Let $S$ be the set of all intervals of the form $[3n+1, 3n

• 2)$, with $n \in \mathbb{Z}$. Then $S$ is sum-free. Furthermore, if $a \in \mathbb{R}\backslash S$ and $a$ is not a multiple of $3$, then $a$ has a multiple in $S$. As such, the only subsets of $\mathbb{R}\backslash S$ that are closed under addition are the sub-semigroups of $3\mathbb{Z}$, so $\mathbb{R}\backslash S$ certainly has no uncountable subsets that are closed under addition.

The website mentions the variant where we instead require that $S$ is Sidon. That's probably a lot harder, and my construction is very much not Sidon.

EDIT: Vjeko pointed out that the problem should actually ask about $A$ such that $A + A \subseteq \mathbb{R}\backslash S$, not $A + A \subseteq A$. So my construction actually doesn't resolve the corrected problem.

Why it is rejected. It answers a variant, not the Statement. The construction answers erdosproblems.com's earlier wording, which asked for $A$ closed under addition. Problem 949 asks, as Erdős did and as the erdosproblems.com wording corrected on 17 and 18 August 2025 does, for $A\subseteq\mathbb R\setminus S$ of cardinality continuum with $A+A\subseteq\mathbb R\setminus S$, and its Formulation sets the earlier wording aside. For the construction's $S$ that question has the answer yes, with $A=[0,1/2)$, so the counterexample settles no instance of it; its mathematics answers the variant and does not settle this problem. The value disproved is its answer to that variant.

Standing. Rejected, as answering a variant. The thread's next comment pointed out the mismatch with Erdős's wording and gave a set $A$ that works for this $S$, and the author replied that no counterexample to the corrected formulation was at hand. This corpus has not checked the construction beyond what the problem page records.

Depends on. No page of this wiki.