Proposition 5

Tag 0BV8, permanent. Cite it as [Bourbaki, Tag 0BV8].

General Topology, chapter V, § 1. Subgroups and quotient groups of R

Every continuous homomorphism $f$ of the topological group $\mathbf{R}$ into itself is of the form $x \to ax$ , where $a \in \mathbf{R}$ ; it is an automorphism of $\mathbf{R}$ if $a \neq 0$ .

For every $x \in \mathbf{R}$ and every integer $p \in \mathbf{Z}$ , we have $f(px) = pf(x)$ ; replacing $x$ by $(1/p)x$ , it follows that

f\left(\frac{1}{p}x\right) = \frac{1}{p} f(x) \quad \text{if } p \neq 0;

hence, for all integers $p$ and $q \neq 0$ , we have

f\left(\frac{p}{q}x\right) = \frac{p}{q} f(x).

In other words, $f(rx) = rf(x)$ for all rational numbers $r$ . If now $t$ is any real number, by reason of the continuity of $f$ we have

f(tx) = \lim_{r \to t, r \in \mathbf{Q}} f(rx) = \lim_{r \to t, r \in \mathbf{Q}} rf(x) = \left( \lim_{r \to t, r \in \mathbf{Q}} r \right) f(x) = tf(x).

In particular, if $a = f(1)$ we have $f(t) = at$ , and the proposition is proved.

The group of automorphisms of the topological group $\mathbf{R}$ is therefore isomorphic to the multiplicative group $\mathbf{R}^*$ of non-zero real numbers.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Theorem 1, § 4

Cites

Nothing.