Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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($\mathbb{N}, +$))?

share|improve this question
    
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 $(\mathbb Q,+,\le)$. –  Emil Jeřábek Oct 26 '12 at 14:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.