be an automorphism
. By considering s
(1) we see that s(n)
for all integer
; since s
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.