Non-Archimedan Groups 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? 
 A: 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.
A: Unless I have misunderstood your question badly, you don't really need compactness for this, or anything nearly as complicated.
${\bf Z}^2$ with lexicographic order fits the description nicely: no multiple of $(0,1)$ is larger than $(1,0)$.
Actually, any product of (nontrivial) ordered groups ordered lexicographically will exhibit the same behaviour.
