User:Dfeuer/Complete Totally Ordered Group is Archimedean

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Theorem

Let $G$ be a Dedekind-complete totally ordered group.

Then $G$ is Archimedean.

Proof

Let $e$ be the identity of $G$.

Let $e < x < y$.

Suppose for the sake of contradiction that $x^n < y$ for all $n \in \N_{>0}$.

Let $P = \{ x^n: n \in \N_{>0} \}$.

Then $P$ is bounded above by $y$.

Since $G$ is Dedekind-complete, $P$ has a supremum $m$.

Then $mx^{-1} < m$, so $mx^{-1}$ is not an upper bound of $P$, so there is an $n \in \N_{>0}$ such that $mx^{-1}\le x^n$.

But then $m \le x^{n+1} < x^{n+2}$, contradicting the fact that $m$ is an upper bound of $P$.

Thus $x^n \ge y$ for some $n \in \N_{>0}$.

$\blacksquare$