Timeline for Non-archimedean group over the reals

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Feb 15, 2020 at 18:10	comment	added	YCor		I guess the question was intended with abelian groups. Anyway, let me provide the argument for arbitrary left-ordered groups $G$. Namely, the claim is that if the topology on the order on $G$ is connected, then $G$ is archimedean. By contradiction, suppose $G$ non-archimedean: there exists $x>1$ such that $X=\{x^n:n\ge 0\}$ is bounded above, hence by connectedness it has a supremum $y$. Hence $y$ is also the sup of $xX$, which by left-invariance implies $y=xy$, so $x=1$, contradiction. (In general this shows that in a left-ordered group, $\{x^n:n\ge 0\}$ has no upper bound for every $x>1$.)
May 29, 2012 at 8:45	vote	accept	chros
May 28, 2012 at 22:08	history	answered	Andreas Blass	CC BY-SA 3.0