Consider the class of sets of the form $X \cap Y$ where $X \subseteq \mathbb{N}^d$ is defined in FO($\mathbb{N}, +$) and $Y \subseteq \mathbb{Q}^d$ is defined in FO($\mathbb{Q}, +, \leq$). Clearly, this class includes the semilinear sets, since they can be described as $X \cap \mathbb{Q}^d$. Is the converse true? Do all sets of this class are semilinear (i.e. definable in FO($\mathbb{N}, +$))?
1
-
$\begingroup$ Sure. If $Y\subseteq\mathbb Q^d$ is definable in $(\mathbb Q,+,\le)$, then $Y\cap\mathbb N^d$ is definable in $(\mathbb N,+)$ by quantifier elimination of $(\mathbb Q,+,\le)$. $\endgroup$– Emil JeřábekCommented Oct 26, 2012 at 14:57
Add a comment
|