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Statement. There is a comeager set $G\subset(2,\infty)$ such that for every $x\in G$, the set $S_x=\{\lfloor x^n\rfloor:n\ge1\}$ is Sidon and meets every infinite arithmetic progression in $\mathbb N_0$. The resulting family contains uncountably many distinct sets.

Source. Sayan Dutta's comment of 2025-09-02 in the Problem 198 discussion, read in the dated source record. The reconstruction spells out the density, tail, and distinctness steps. This is a different supplied argument from the factorial construction, not a new problem-solving claim.

External input. The Baire category theorem: in the open interval $(2,\infty)$, a countable intersection of open dense sets is dense. This locally compact Hausdorff space is a Baire space. The comment invokes that theorem; its general proof is not included here.

Proof. For integers $d\ge1$, $0\le r<d$, and $N\ge1$, define

$$U_{d,r,N}=(2,\infty)\cap \bigcup_{n\ge N} \bigcup_{\substack{m\ge0\\m\equiv r\pmod d}} \left(m^{1/n},(m+1)^{1/n}\right).$$

This set is open. To prove density, take $2<a<b$. For arbitrarily large $n\ge N$, the length $b^n-a^n$ exceeds $d+1$. An integer $m\equiv r\pmod d$ can then be chosen with $a^n<m<m+1<b^n$. Its displayed interval lies inside $(a,b)$, so $U_{d,r,N}$ meets every nonempty open subinterval of $(2,\infty)$.

By Baire category,

$$G=\bigcap_{d\ge1}\ \bigcap_{r=0}^{d-1}\ \bigcap_{N\ge1}U_{d,r,N}$$

is comeager and dense. For $x\in G$, each residue class modulo each $d$ is attained by $\lfloor x^n\rfloor$ for arbitrarily large $n$. These values tend to infinity, so they eventually exceed the initial term of any specified progression $P(a,d)$. Thus $S_x$ meets that progression, not merely its residue class below $a$.

For $x>2$, every term is positive and

$$\lfloor x^{n+1}\rfloor\ge\lfloor2x^n\rfloor\ge2\lfloor x^n\rfloor.$$

The doubling-gap lemma shows that $S_x$ is Sidon. Finally, $G$ is uncountable: a countable subset of an interval is meager, whereas this comeager dense subset cannot also be meager in a nonempty Baire space. If $S_x=S_y$, their strictly increasing enumerations agree. Since $\lfloor x^n\rfloor^{1/n}\to x$, taking $n$th roots gives $x=y$. Therefore different parameters give different sets. $\square$

Method and limit. Baire category satisfies countably many modular tail conditions simultaneously; it supplies a residual parameter set rather than a particular computable parameter. Countability remains essential to this proof.

Bears on. Problem 198: each $S_x$ with $x\in G$ is a Sidon set whose complement contains no infinite arithmetic progression, a negative answer, and there are uncountably many such sets.