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Statement. For every vector space $V$ over $\mathbb Q$ and every fixed positive integer $k$, there exists a set $X_k\subseteq V$ meeting every one-sided infinite arithmetic progression in $V$ such that

$$\left|X_k\cap\{a+jb:0\le j<k\}\right|\le2 \quad\text{for every }a,b\in V\text{ with }b\ne0.$$

The choice of $X_k$ may depend on $k$. For $k=1,2$ the upper bound is automatic. For $k=3$ it is the exclusion in the main theorem; larger $k$ give the stated additional restriction. The source does not assert one set working simultaneously for every $k$.

The print states it as follows (p. 233), introducing it as "the following still stronger theorem": "Let V be a vector space over the rationals and let k be a fixed positive integer. Then there is a set X_k ⊆ V such that X_k meets every infinite arithmetic progression in V but X_k intersects every k-element arithmetic progression in at most two points."

Source and proof scope. J. E. Baumgartner, Partitioning vector spaces, J. Combin. Theory Ser. A 18 (1975), 231–233: the unnumbered final theorem on p. 233, read 2026-09-06 and again 2026-10-08. The author says that a slight modification of the preceding proof yields this result but supplies no modified argument. This page records the source-stated result with an omitted proof; it is not a complete proof reconstruction. The preceding theorem and complete main proof are compiled separately, including their external basis and choice dependency.

Bears on. Problem 199 through its $k=3$ instance; the extra fixed-length exclusion strengthens that instance.