User:Dfeuer/Archimedean Totally Ordered Group is Isomorphic to Subgroup of Real Numbers

Theorem
Let $(G,+,\le)$ be an Archimedean totally ordered group.

Then $G$ is isomorphic (as an ordered group) to some subgroup of $\R$.

Proof
In this proof, we will take $\Z \subseteq \Q \subseteq \R$.

We will denote the set of strictly positive integers by $\Z_+$.

We will denote the set of lower sets in $\Q$ that are bounded above and have no greatest elements by $R$, and we will define the group operations and ordering on $R$ in the usual fashion.

If $G$ has a smallest strictly positive element, $\epsilon$, then by User:Dfeuer/Archimedean Totally Ordered Group is Abelian/Lemma 1, $(G,+,\le)$ is the infinite cyclic group generated by $\epsilon$, with the natural ordering, and is thus isomorphic to $\Z$.

In the remainder of the proof, we will assume that $G$ has no smallest strictly positive element. We will construct a strictly order-preserving homomorphism from $G$ into $R$.

Let $0_G$ be the identity of $G$, let $1_G$ be an arbitrary strictly positive element of $G$, and let $Z_G$ be the subgroup generated by $1_G$.

Define $\psi:Z_G \to \Z$ by letting $\psi(n \cdot 1_G) = n$ for each $n$.

Define $\phi: G \to \mathcal P(\Q)$ thus:
 * $\phi(x) = \left\{ \frac k n: k \in \Z, n\in \Z_+, \psi^{-1}(k) < n \cdot x \right\}$

We must show that $\psi(x)\in R$ for each $x\in G$.

Since $G$ is Archimedean, there is an $m\in \Z$ such that $x < \psi^{-1}(m)$.

Suppose $\frac k n > m$.

Then $k > nm$, so $\psi^{-1}(k) > \psi^{-1}(nm) = n \cdot \psi^{-1}(m) > n \cdot x$ (explain more), so $\frac k n \notin \phi(x)$. Thus $\phi(x)$ is bounded above by $m$.

Suppose for the sake of contradiction that $\phi(x)$ has a greatest element .............

Let $k\in\Z$, $n\in\Z_+$, and $psi^{-1}(k)<n\cdot x$, so that $\frac k n\in \phi(x)$.

Let $j \in \Z$, $m \in \Z_+$, and $\frac j m < \frac k n$.

Then $nj < mk$, so $\psi^{-1}(nj) < \psi^{-1}(mk)$.

Thus $n \cdot \psi^{-1}(j) < m \cdot \psi^{-1}(k) < mn \cdot x$, so $\psi^{-1}(j) < m \cdot x$ (explain this better), so $\frac j m \in \phi(x)$.

Thus $\phi(x)$ is a lower set.