We begin by showing that every non-discrete subgroup of $\mathbf{R}$ is dense in $\mathbf{R}$. If a subgroup $G$ of $\mathbf{R}$ is not discrete, then for every $\varepsilon > 0$ there is a point $x \neq 0$ in $G$ which belongs to the interval $[-\varepsilon, +\varepsilon]$; since all integer multiples of $x$ belong to $G$, every interval of length $> \varepsilon$ contains an element of $G$, and therefore $G$ is dense in $\mathbf{R}$.

Every closed subgroup of $\mathbf{R}$ other than $\mathbf{R}$ itself is therefore discrete. It remains to show that every discrete subgroup $G$ of $\mathbf{R}$ other than $\{0\}$ is of the form $a.\mathbf{Z}$, where $a > 0$. Now the relation $-G = G$ shows that the set $H$ of elements $> 0$ in $G$ is not empty; if $b \in H$, the intersection of the interval $[0, b]$ and $G$ is compact and discrete, and is therefore finite. Let $a$ be the smallest element of $H$ contained in $[0, b]$, and for every $x \in G$ put $m = [x/a]$, the integer part of $x/a$; then we have $x - ma \in G$ and $0 \leq x - ma < a$. By the definition of $a$ it follows that $x - ma = 0$ and therefore $G = a.\mathbf{Z}$.

Every $x \in \mathbf{R}$ is congruent $(\bmod\,1)$ to some number in the interval $[a, a+1]$, namely to $x - [x-a]$; hence $\mathbf{T}$ is the image of this interval under the canonical mapping $\varphi$ of $\mathbf{R}$ onto $\mathbf{R}/\mathbf{Z}$, and is therefore compact and connected (Chapter I, § 9, no. 4, Theorem 2, and § 11, no. 2, Proposition 4). On the other hand, two distinct elements of the interval $[a, a+1]$ are congruent $(\bmod\,1)$ only if they are the end-points of the interval; from the compactness of $\mathbf{T}$ it therefore follows that $\mathbf{T}$ is homeomorphic to the quotient space of $[a, a+1]$ obtained by identifying the end-points (Chapter I, § 9, no. 4, Theorem 2, Corollary 4, and § 10, no. 4, Proposition 8). Finally, $\mathbf{Z}$ being a discrete subgroup of $\mathbf{R}$, $\mathbf{T} = \mathbf{R}/\mathbf{Z}$ is locally isomorphic to $\mathbf{R}$ (Chapter III, § 2, no. 6, Proposition 19) and in particular is locally connected (this last fact is also a consequence of Chapter I, § 11, no. 6, Proposition 12).

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

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

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

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.

The uniqueness of the extension (if it exists) follows from the preceding remarks, because $I$ generates the group $\mathbf{R}$. We have to establish the existence of such an extension.

If $n$ is an integer $> 0$ and if $x \in I$ and $nx \in I$, we have $f(nx) = (f(x))^n$, by induction on $n$ (since $mx \in I$ for all integers $m$ such that $1 \leq m \leq n$). Put $J = \bigcup_{n \in \mathbf{N}} nI$; $J$ is either the whole line $\mathbf{R}$, or one of the intervals $[0, +\infty[$ or $[-\infty, 0]$, according as $o$ is or is not an interior point of $I$. If $x \in J$ we have $x/n \in I$ whenever $n$ is a sufficiently large integer $> 0$. Let $x \in J$, and let $m, n$ be two integers $> 0$ such that $x/n \in I$ and $x/m \in I$; then $x/mn \in I$, and therefore

in other words, the element $(f(x/n))^n$ of $G$ is the same for all integers $n > 0$ such that $x/n \in I$. Let us denote this element by $f_1(x)$; $f_1$ is a mapping of $J$ into $G$, which agrees with $f$ on $I$ and is therefore continuous at the point $o$ (with respect to $J$). Let $x, y$ be two elements of J and let n be a sufficiently large integer > 0 such that $\frac{x}{n} \in I, \frac{y}{n} \in I$ and $\frac{x+y}{n} \in I$; then

which shows that $f\left( \frac{x}{n} \right)$ and $f\left( \frac{y}{n} \right)$ commute; by the definition of $f_1$, we have therefore $f_1(x+y) = f_1(x) f_1(y)$. If $J = \mathbf{R}$, the proof is complete; if not, say $J = [0, +\infty[$, and for each $x < 0$ define $f_1(x)$ to be $(f_1(-x))^{-1}$. Then the relation $f_1(x+y) = f_1(x) f_1(y)$ remains valid for all $x \in \mathbf{R}$ and all $y \in \mathbf{R}$. This is clear if $x < 0$ and $y < 0$; if $x \geqslant 0$, $y < 0$ and $x + y \geqslant 0$ it follows from $f_1(x) = f_1(x+y) f_1(-y)$; similarly if $x \geqslant 0, y < 0$ and $x + y < 0$, for then we have

analogous proofs for $x < 0$ and $y \geqslant 0$. We see therefore that $f_1$ is a homomorphism of $\mathbf{R}$ into G, so that $f_1(0) = e$, the identity element of G; and since $f_1$ is continuous with respect to J, it has a limit on the right at 0, equal to $e$; since $f_1(-x) = (f_1(x))^{-1}$, $f_1$ also has a limit at 0 on the left, equal to $e$; thus $f_1$ is continuous at 0, and the proof is complete.

Let $\overline{f}$ be the continuous homomorphism of $\mathbf{R}$ into G which coincides with $f$ at all points of an open interval I, which contains 0 and is contained in the set on which $f$ is defined; $\overline{f}(\mathbf{R})$ by hypothesis contains a neighbourhood of the identity element of G, hence (Chapter III, § 2, no. 1, Corollary to Proposition 4) is an open subgroup of G; and $\overline{f}$ is a strict morphism of $\mathbf{R}$ onto $\overline{f}(\mathbf{R})$, by Chapter III, § 2, no. 8, Proposition 24.

For a local isomorphism of $\mathbf{R}$ with G extends to a strict morphism of $\mathbf{R}$ onto an open subgroup of G (Corollary to Proposition 6), hence onto G itself since G is connected. Hence G is isomorphic to a quotient group of $\mathbf{R}$; since G is Hausdorff and does not consist of the identity element alone (because it is locally isomorphic to $\mathbf{R}$), it is isomorphic to either $\mathbf{R}$ or $\mathbf{T}$ by Proposition 3 of no. 2.