---

Statement. There is a Sidon set $A\subseteq\mathbb N_{>0}$ meeting every progression $P(a,d)=\{a+td:t\in\mathbb N_0\}$, where $a\ge0$ and $d\ge1$. Consequently $\mathbb N_0\setminus A$ contains no infinite arithmetic progression.

Source. The problem page gives this construction and regards it as implicit in Baumgartner's 1975 work. See the source record for the unresolved historical attribution.

Proof. The pairs $(a,d)\in\mathbb N_0\times\mathbb N_{>0}$ are countable, so enumerate their progressions as $P_1,P_2,\ldots$. Choose a positive $a_1\in P_1$. Once $a_n$ is chosen, the unbounded progression $P_{n+1}$ contains an integer $a_{n+1}>2a_n$; choose its least such element. The set $A=\{a_n:n\ge1\}$ is Sidon by the doubling-gap lemma. It meets $P_n$ at $a_n$ for every $n$, so no enumerated progression lies in its complement. These are all infinite progressions in $\mathbb N_0$. The same argument applies to positive integers. $\square$

Method. Enumeration makes countably many hitting requirements compatible with successively larger gaps. This proves the integer statement; it does not enumerate all progressions in $\mathbb R$ or prove Problem 199's corresponding real-set assertion.

Bears on. Problem 198: the set constructed is a Sidon set whose complement contains no infinite arithmetic progression, a negative answer.