Separating Milliken-Taylor systems in $\mathbb{Q}$

A finite sequence $\vec a= \langle a_i \rangle_{i=1}^k$ in $\mathbb{Q} \setminus \{0 \}$ is compressed provided $a_i \neq a_{i+1}$ for $i<k$. Given a compressed sequence $\vec a= \langle a_i \rangle_{i=1}^k$ in $\mathbb{Q} \setminus \{0 \}$ and given a sequence $ \langle x_n\rangle_{n=1}^\infty$ in a commutative group $(G,+)$, the Milliken-Taylor system generated by $\vec a$ and $ \langle x_n\rangle_{n=1}^\infty$ is $MT( \vec a, \langle x_n\rangle_{n=1}^\infty)= \{ \sum_{i=1}^k a_i \cdot \sum_{n \in F_i} \,x_n:F_1,F_2, \ldots,F_k$ are finite nonempty subsets of $\mathbb{N}$ with $\max F_i< \min F_{i+1}$ for $i<k \}$. It is an easy consequence of the Milliken-Taylor Theorem that Milliken-Taylor systems are partition regular in the strong sense that if $ \langle y_n\rangle_{n=1}^\infty$ is any sequence in $G$, and $MT( \vec a, \langle y_n\rangle_{n=1}^\infty)$ is partitioned into finitely many cells, there is a sequence $ \langle x_n\rangle_{n=1}^\infty$ such that $MT( \vec a, \langle x_n\rangle_{n=1}^\infty)$ is contained in one of those cells.

It is known that if $\vec a$ and $\vec b$ are compressed sequences in $\mathbb{Z} \setminus \{0 \}$ which are not rational multiples of each other, then there is a partition of $\mathbb{Z} \setminus \{0 \}$ into two cells, neither of which contains $MT( \vec a, \langle x_n\rangle_{n=1}^\infty) \cup MT( \vec b, \langle y_n\rangle_{n=1}^\infty)$ for any sequences $ \langle x_n\rangle_{n=1}^\infty$ and $ \langle y_n\rangle_{n=1}^\infty$. In this paper we establish the corresponding statement for Milliken-Taylor systems in $\mathbb{Q}$. (In fact, the entries of $\vec a$ and $\vec b$ are allowed to come from $\mathbb{Q} \setminus \{0 \}$.)

Keywords

Milliken-Taylor systems, partition regular

2010 Mathematics Subject Classification

Primary 05D10. Secondary 22A15, 54D35.

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Published 29 October 2014