Consider first a bounded open interval $I = ]a, b[$ ($a < b$). For each $x \in I$ put $f(x) = -\left( \frac{1}{x-a} + \frac{1}{x-b} \right)$. This function is continuous and strictly increasing on $I$, for we have seen that $\frac{1}{x-b}$ is strictly decreasing in $]-, b[$ and $\frac{1}{x-a}$ strictly decreasing in $]a, \to[$. It follows that $f$ is a homeomorphism of $I$ onto an interval $f(I)$ of $\mathbf{R}$ (§ 2, no. 6, Theorem 5). $f(I)$ is neither bounded above nor below; if, for example, we had $f(x) \leq c$ for all $x \in I$, it would follow that $1 - \frac{b-x}{x-a} \leq c(b-x)$, since $b-x > 0$; and this leads to a contradiction when $x$ is sufficiently near $b$ (by virtue of the continuity of the two sides of the inequality, which are rational functions, at the point $b$). Hence $f(I) = \mathbf{R}$, and therefore every bounded open interval is homeomorphic to $\mathbf{R}$. Let $g$ be the inverse of $f$: it maps every unbounded open interval of $\mathbf{R}$ onto an interval $J$ contained in $I$ and open in $I$. Since $I$ is open in $\mathbf{R}$, $J$ is also open in $\mathbf{R}$; since it is bounded, it is homeomorphic to $\mathbf{R}$; and thus we have proved that every unbounded open interval is homeomorphic to $\mathbf{R}$.

Let $f$ be an increasing homeomorphism. If we extend $f$ to $\mathbf{R}$ by putting $\overline{f}(-\infty) = a$ and $\overline{f}(+\infty) = b$, it is obvious that $\overline{f}$ is a strictly increasing mapping (and therefore a bijection) of $\overline{\mathbf{R}}$ onto $[a, b]$. Hence $\overline{f}$ maps every open interval of $\overline{\mathbf{R}}$ onto an interval which is open with respect to $[a, b]$, and is therefore a homeomorphism of $\mathbf{R}$ onto $[a, b]$ by reason of Definition 1 and Proposition 5 of § 1, no. 4.

If $f$ is decreasing, we apply what has been proved to the increasing homeomorphism $x \to -f(x)$ of $\mathbf{R}$ onto ]$-b, -a[$.

All the properties of the interval $[a, b]$ obtained in § 2, which involve only the order structure and the topology of the interval, can therefore be transported to $\overline{\mathbf{R}}$; hence the following propositions:

Hence (Chapter II, § 4, no. 1, Theorem 1) there is a unique uniformity on $\overline{\mathbf{R}}$ compatible with its topology; this uniformity is isomorphic with the uniformity induced on $[a, b]$ by the additive uniformity of $\mathbf{R}$. But it should be remarked that the uniformity induced on $\mathbf{R}$ by that of $\overline{\mathbf{R}}$ is not the additive uniformity of $\mathbf{R}$ (although it is compatible with the topology of the real line); for $\mathbf{R}$ is a complete space with respect to its additive uniformity, but is not a complete subspace of $\overline{\mathbf{R}}$, since it is not closed in $\overline{\mathbf{R}}$.

The least upper bound (resp. greatest lower bound) of a non-empty subset $A$ of $\mathbf{R}$ is denoted by $\sup A$ (resp. $\inf A$). Clearly

(1)

If $A \subset B$, then $\sup A \leq \sup B$ and $\inf A \geq \inf B$ (Set Theory, Chapter III, § 1, no. 9, Proposition 4).

Finally, the functions $\sup(x, y)$ and $\inf(x, y)$ are continuous on $\overline{\mathbf{R}} \times \overline{\mathbf{R}}$.

Let us show, for example, that as $(x, y)$ tends to the point $(a, +\infty)$ ($a \neq -\infty$) while remaining in $\mathbf{R} \times \mathbf{R}$, $x + y$ tends to $+\infty$. There exists a finite number $b < a$, and the interval $]b, +\infty[$ is a neighbourhood of $a$ in $\mathbf{R}$; given any finite $c$, the relations $x > b$ and $y > c - b$ imply $x + y > c$, and this shows that $x + y$ is as near as we please to $+\infty$ whenever $(x, y)$ is near enough to $(a, +\infty)$. The argument is similar in the other cases.

On the contrary, $x + y$ has no limit at the points $(-\infty, +\infty)$ and $(+\infty, -\infty)$ of $\overline{\mathbf{R}} \times \overline{\mathbf{R}}$. For if $x + y$ had limit $k$ (finite or infinite) as $(x, y)$ tends to $(+\infty, -\infty)$ while remaining in $\mathbf{R} \times \mathbf{R}$, it would follow that, for each finite $a$, the function $(x + a) - x$ would tend to $k$ as $x$ tends to $+\infty$ while remaining in $\mathbf{R}$; and this is absurd, since $(x + a) - x = a$ and $a$ is arbitrary.

The function $x + y$ maps $A' \times A'$ (resp. $A'' \times A''$) into $A'$ (resp. $A''$). It is therefore a law of composition on $A'$ (resp. $A''$) which extends the law of addition on $\mathbf{R}$. By the principle of extension of identities (Chapter I, , no. 1, Corollary 1 to Proposition 2) this law is commutative and associative; o is the identity element for this law; and the only non-regular element (Algebra, Chapter I, § 2, no. 2) of $A'$ is $+ \infty$, by formulae (2)

If $x, y, z, t$ are points of $\overline{\mathbf{R}}$ such that $x \leq y$ and $z \leq t$, then $x + z \leq y + t$ whenever both sides of this inequality are defined.

Note that in $\overline{\mathbf{R}}$ the relation $x < y$ implies $x + z < y + z$ only if $z$ is finite, by formulae (2); it is easily verified that the relations $x < y$ and $z < t$ imply $x + z < y + t$ whenever both sides of this inequality are defined.

In $\overline{\mathbf{R}}$ we put $x^+ = \sup (x, 0), x^- = \sup (-x, 0), |x| = \sup (x, -x)$; thus $(+\infty)^+ = (-\infty)^- = +\infty$ and $(+\infty)^- = (-\infty)^+ = 0$, and $|+\infty| = |- \infty| = +\infty$. The sums $x^+ - x^-$ and $x^+ + x^-$ are defined for all $x \in \overline{\mathbf{R}}$ and are therefore equal to $x$ and $|x|$ respectively by the principle of extension of identities. Also, whenever the sum $x + y$ is defined, we have $|x + y| \leq |x| + |y|$.

Note on the contrary that formulae (6) and (7) of § 1, no. 1 may no longer make sense for certain values of $x$ and $y$ in $\overline{\mathbf{R}}$; for example, if $x = -\infty$ and $y = 0$ we have $\sup (x, y) = 0$, but the sum $x + (y - x)^+$ is not defined, because $(y - x)^+ = +\infty$.

Let $\overline{\mathbf{R}}^*$ denote the complement of o in $\overline{\mathbf{R}}$. Then the analogue of Proposition 7 for multiplication runs as follows:

(3)

We leave the proof to the reader; it is analogous to the proof of Proposition 7.

Likewise, we see that $xy$ has no limit at the points $(0, +\infty), (+\infty, 0), (0, -\infty), (-\infty, 0)$ of $\overline{\mathbf{R}} \times \overline{\mathbf{R}}$.

The function $xy$ is a law of composition on $\overline{\mathbf{R}}^*$ which extends the law of multiplication on $\mathbf{R}$; this law is associative and commutative (principle of extension of identities); it has 1 as identity element; and the non-regular elements in $\overline{\mathbf{R}}^*$ are $+\infty$ and $-\infty$.

If $x \leq y$ and $z > 0$ we have $xz \leq yz$ whenever both sides of this inequality are defined. If the product $xy$ is defined, then so is $|x| \cdot |y|$ and we have $|xy| = |x| \cdot |y|$.

Finally, the distributivity formula

is still valid, by virtue of the principle of extension of identities, whenever all the operations which figure on either side are defined.

Note that it can happen that the left-hand side of (4) is defined while the right-hand side is not: for example, consider the case where $x = +\infty$, $y = 2$ and $z = -1$. The distributivity formula should therefore be used with caution in $\overline{\mathbf{R}}$.