I am looking for a proof of the fact that if $f:\mathbb{R}\to \mathbb{R}$ is a group automorphism of $(\mathbb{R},+)$ that also preserves order, then there exists a positive real number $c$ s.t. $f(x)=cx$ for all $x\in \mathbb{R}$. If anyone can point to a reference, that will be great.
The only group automorphisms of the additive group of real numbers that are also order isomorphisms are multiplication by positive real numbers
2 Answers
IIRC this has been covered a handful of times on this site; the proof, in any case, is pretty straightforward. Obviously $f(0)=0$ and $f(1)\gt 0$ by the order-preserving property; say $f(1)=c$. Then $f(n)=cn$ for all $n\in\mathbb{N}$ ($f(1+1+\ldots+1) = f(1)+f(1)+\ldots+f(1)$) and $f(q) = cq$ for all $q\in\mathbb{Q}$ (if $q={r\over s}$, then $sf(q) = f(sq) = f(r) = cr$, so $f(q) = cr/s = cq$). Now Cauchy sequences give you the full result.
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0Thanks, that clears it up. – 2011-08-30
I don't have a reference, but here is an outline of how I would do this. Since we can approximate any real number from above and below by elements of $\mathbf Q$ and $f$ preserves the order, we can reduce to the case of rational $x$. Clearly we want $f(1) = c > 0$ to be our constant. Now, if $b$ is a positive integer then $ c = f(1) = f(b/b) = bf(1/b). $ It doesn't take much work to get from this to a proof that $f(x) = cx$ for each $x \in \mathbf{Q}$.
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0Thanks for the answer. I get it now. – 2011-08-30