实数域自同构

被问到，刚查的，水很深。
证明居然不长。

(idea)

by Alexander Pushkin

Tue Sep 26 2000 at 12:31:02

Let s:R->R be an automorphism. By considering s(1+1+...+1)=s(1)+s(1)+...+s(1) we see that s(n)=n for all integers n; since s preserves division, s(r)=r for any rational number r. So much is standard. Now, if we could prove that s is continuous, we would be done (since it is the identity function on a dense set). But what properties does s have that we could use? NONE! Instead, we'll use some properties of R. Call a number x "nonnegative" if x=y2 for some y, in R. Note that this corresponds exactly with our notion of this concept. Note also that s(x)=s(y)2, so s preserves nonnegativity. But this means that s preserves the order on R (since a<b iff a+c=b for some nonnegative c). Now, to every real number corresponds a unique "section" of the rationals (a partition of the rationals into those "smaller than" the real number and those "greater than" (or equal, if it's rational) it. But s maps all elements of the section to themselves, and preserves order. So the section of every real number x is transformed into the section of s(x) (since s preserves order), and yet remains constant (since s is the identity on the rationals). Thus it must be that s(x)=x.

但如果是复数域就有很多不同的自同构，连续的，不连续的都有。