# first-order theory

• Jun 6th 2010, 01:32 PM
nngktr
first-order theory
Recall that the ordering on Q resp. on R is archimedean, i.e. for every x in Q resp. x in R there is some n in N with -n<x<n. Use the compactness theorem to prove that archimedeanity is not a first-order property.

Could anyone please give me some hints how to handle this problem? Any help is appreciated!
• Jun 7th 2010, 04:29 AM
clic-clac
Hi

Basically, to show that archimedeanity is not a first order property, you want to obtain two equivalent structures, one archimedean and the other not.

Let's work with $\displaystyle \mathbb{R},$ we want to find for instance $\displaystyle A\equiv\mathbb{R}$ in the language $\displaystyle \{+,\times ,0,1,\leq\}$ with an element $\displaystyle a\in A$ such that for all $\displaystyle n\in\mathbb{N},$ $\displaystyle n<a.$

So consider the theory $\displaystyle T=Th(\mathbb{R})\cup\{c\ \text{is strictly greater than any natural integer}\}$ where $\displaystyle c$ is a new constant symbol ( $\displaystyle T$ is therefore a $\displaystyle \{+,\times ,0,1,\leq,c\}$-theory )

1. Write $\displaystyle \{c\ \text{is strictly greater than any natural integer}\}$ as a set of first-order formulae.

2. Use compactness to prove that $\displaystyle T$ is consistent.

3.Conclude.