---

Claim. There is a comeager set $G\subset(2,\infty)$ such that for every $x\in G$ the set $S_x=\{\lfloor x^n\rfloor : n\ge 1\}$ is Sidon and meets every infinite arithmetic progression; the complements of these uncountably many sets contain no infinite arithmetic progression, so the answer is no.

For $x\ge 2$ the terms of $S_x$ at least double, so $S_x$ is Sidon. For $d\ge 1$, $0\le r<d$ and $N\ge 1$, let $U_{d,r,N}$ be the set of $x$ with $m<x^n<m+1$ for some $n\ge N$ and some $m\equiv r\pmod d$, so that $\lfloor x^n\rfloor=m$; it is open, as a union of the intervals $(m^{1/n},(m+1)^{1/n})$, and dense, since $b^n-a^n$ grows without bound for fixed $b>a$. The Baire category theorem makes the intersection of the $U_{d,r,N}$ over all $d$, $r$ and $N$ comeager, and for $x$ in it $S_x$ meets every residue class modulo every $d$ infinitely often, hence every infinite progression. The comment states the argument on $(1,\infty)$ with $S_x$ Sidon for $x\ge 2$; the library's construction page writes the density, tail and distinctness steps out on $(2,\infty)$.

Postings. Sayan Dutta's comment of 2025-09-02 in the site's discussion thread; the site's page records Dutta's observation that these sets give examples for uncountably many $x$ and thanks Dutta. The page's record of the remark appears in a web archive capture of 2026-01-28 and not in one of 2025-12-06.

Acceptance. Reviewed: the site's curator, Thomas Bloom, records the construction on the problem page as a further example and credits it to Dutta, a documented acceptance outside this project and independent of the claimant. No refereed publication exists, and no formalization of this argument is posted.

Depends on. No wiki page; the claim rests on the argument stated above.