Underlying set of structure is $\mathbb{N}$ , our language only have (<) interpreted as usual order. Why is this minimal? (Every definable subset finite or cofinite)
I try to show with contradiction that we have set not of this form (so definable set which is infinite but not cofinite) but I cannot see how to describe this with only this order relation <. I first try to see without quantifiers, which is not hard, but when quantifiers come I begin to have troubles. Please give advise.
First, remark that $\omega \equiv \omega +\zeta$ where $\zeta$ is the order type of the integers $\mathbb{Z}$. The language is always $<$, I omit it in notion. Next since the automorphism group of $\omega+\zeta$ is transitive on $\zeta$ any definable subset of $\omega+\zeta$ must either contain or be disjoint from $\zeta$.
Now assume that $\varphi(x)$ defines an infinite, coinfinite subset of $\omega$ then $$\omega \vDash \forall x\exists y (x<y\wedge \varphi(y))$$ $$\omega \vDash \forall x\exists y (x<y\wedge \neg\varphi(y))$$ and thus $\omega+\zeta$ must satisfy the same formulas. This contradicts the earlier remark about the subset of $\omega+\zeta$ defined by $\varphi$.