That some sequences grow without bound leads us to add two elements to the set of real numbers.
The set of extended real numbers is the union of the set of real numbers with a set containing two elements: one which we call positive infinity and the other we call negative infinity. We call an element of the extended real numbers finite if it is not positive or negative infinity. Otherwise, we call it infinite.
The only two infinite extended real numbers are exactly the two elements positive and negative infinity.
We denote positive infinity by $+\infty$ and
the negative infinity by $-\infty$.
We denote the set of extended real numbers by
$\Rbar$ so
\[
\Rbar = \R \cup \set{-\infty, +\infty}
\]
We extend intervals in the obvious way.
We define
\[
[-\infty, \alpha ) := \Set{x \in \R }{x <\alpha } \cup
\set{-\infty}
\] \[
[-\infty, \alpha ] := \Set{x \in \R }{x \leq \alpha } \cup
\set{-\infty}
\] \[
(\alpha , +\infty] := \Set{x \in \R }{x > \alpha } \cup
\set{+\infty}
\] \[
[\alpha , +\infty] := \Set{x \in \R }{x \geq \alpha } \cup
\set{+\infty}
\] \[
(-\infty, \alpha ) := \Set{x \in \R }{x <\alpha }
\] \[
(-\infty, \alpha ] := \Set{x \in \R }{x \leq \alpha }
\] \[
(\alpha , +\infty) := \Set{x \in \R }{x > \alpha }
\] \[
(\alpha , +\infty) := \Set{x \in \R }{x \geq \alpha }
\] \[
\Rbar_+ := [0, \infty] \quad \text{ and } \quad \Rbar_{++}
:= (0, \infty].
\]
Naturally, we define $-\infty < a <
+\infty$ for every $\alpha \in \R $.
This new extended order can be put to good
use, mostly notationally.
For example, the intervals above can be
rewritten by selecting $x$ from $\Rbar$.
As a particular example, we can write
\[
[-\infty, \alpha ] =\Set{x \in \Rbar}{-\infty \leq x \leq
\alpha }
\]
It is logically consistent to define the
greatest lower bound of the empty set to be
$+\infty$ and likwise the least upper bound of
the empty set to be $-\infty$.
In symbols,
\[
\inf \varnothing := +\infty \quad \text{ and } \quad \sup
\varnothing = -\infty
\]
We extend addition to all but one ordered pair of elements of the new set.
\[ \alpha + \infty = \infty + \alpha = \infty \quad \text{for } -\infty < \alpha \leq \infty \]
\[ \alpha - \infty = -\infty + \alpha = -\infty \quad \text{for } -\infty \leq x < \infty \]
Similarly, we extend multiplication
\[
\alpha \infty = \infty\alpha = \infty, \quad \alpha (-\infty)
= (-\infty)\alpha = -\infty \quad \text{for } 0 < \alpha
\leq \infty,
\] \[
\alpha \infty = \infty\alpha = -\infty, \quad \alpha (-\infty)
= (-\infty)\alpha = \infty \quad \text{for } -\infty \leq <
\alpha < 0,
\] \[
0\infty = \infty0 = 0 = 0(-\infty) = (-\infty)0, \quad
-(-\infty) = \infty
\]
Under these rules, the familiar laws remain commutative, associative, and multiplication distributes over addition. This is provided, of course, that we never form the sum $\infty - \infty$ or $-\infty + \infty$. These can be verified by testing all possible finte combinations for the values in the laws. Avoiding a forbidden sum requires some cautious attention, like dividing by zero, and in practice one or the other of the infinities tends to be excluded by hypothesis.