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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.

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up vote 2 down vote accepted

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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Thanks. However, I don't quite see how to go from rationals to reals without requiring that $f$ commute with limits. –  BMI Aug 30 '11 at 22:38
You can use the sandwich theorem: choose two Cauchy sequences of rationals $p_n$, $q_n$ converging to your value $x$ such that $p_n \lt p_{n+1} \lt x \lt q_{n+1} \lt q_n$ for all $n$; the fact that $f$ is order-preserving then implies that $f(p_n) \equiv cp_n \lt f(x) \lt f(q_n) \equiv cq_n$ for all n. –  Steven Stadnicki Aug 30 '11 at 23:25
Thanks, that clears it up. –  BMI Aug 30 '11 at 23:33
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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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Thanks for the answer. I get it now. –  BMI Aug 30 '11 at 23:34
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