I'm trying to think of an explicit example of a non-Archimedian group.
The definition of Archimiedean is s.t. if for all $x$ and $y$, there is some $m$ a Natural number s.t. $mx = \underbrace {x + x + ... + x}_{m \text { times}}> y$.
An example I thought was playing around with the files, but not understanding why that would do it. Also, how can we use the compactness theorem to prove that one does exist?
Here's an explicit example of a non-Archimedean ordered field that should be accessible to almost anyone who has taken real analysis, or maybe even just freshman calculus.
Consider the field of rational functions, that is, functions $f(x)$ that can be written as:
$$f(x) = \frac{p(x)}{q(x)}$$
Where $p$ and $q$ are polynomials. Define addition as $(f+g)(x) = f(x) + g(x)$ and multiplication as $(f\cdot g)(x) = f(x) \cdot g(x)$.
It is easy to verify that this set forms a field. The ordering is a bit more tricky to define, but we can manage it by defining $f<g$ if $g$ eventually exceeds $f$: $$f < g \iff \exists r \in \mathbb R \forall x > r: f(x) < g(x)$$
The only tricky part of the proof that $<$ makes the field an ordered field is proving that for all rational functions $f \ne g$, $f<g$ or $g < f$. For a hint, prove the simpler statement $f < 0 \lor 0 < f$ by taking a maximal root of a polynomial and using Bolzano's theorem.
But it should be clear that this field is not Archimedean. Taking the multiplicative identity $1(x) = 1$, it should be clear that for any fixed $n$, $n\cdot 1 < I$ where $I(x) = x$.
This is an example of a non-Archimedean ordered field $( X , + , \cdot , <)$. So a fortiori, the set of rational functions, under addition and the order $<$ form a non-Archimedean group.