Proposition 7

The function $x + y$ can be extended by continuity to each of the sets $A' \times A'$ and $A'' \times A''$, according to the formulae

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: