Ordering on Extended Real Numbers is Ordering

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Theorem

Denote with $\le$ the usual ordering on the extended real numbers $\overline \R$.


Then $\le$ is an ordering, and so $\overline \R$ is an ordered set.


Proof

Transitive

Let $a, b, c \in \overline \R$.

Suppose that $a \le b$ and $b \le c$.

If $c = +\infty$, then by the definition of $\le$, $a \le c$.

If $b = +\infty$, then by Positive Infinity is Maximal, $b = c$, so $a \le c$.

If $a = +\infty$, then applying Positive Infinity is Maximal twice yields $a \le c$.

The cases for $a$, $b$, or $c$ being $-\infty$ are similar.

The only remaining case is that $a, b, c \in \R$.

Since $\le$ is transitive on $\R$, $a \le c$.

$\Box$


Antisymmetric

Let $a, b \in \overline{\R}$.

Suppose that $a \le b$ and $b \le a$.

If $a \in \R$ and $b \in \R$, then since $\le$ is antisymmetric on $\R$, $a = b$.

If $a = +\infty$, then since $a \le b$ and Positive Infinity is Maximal, $b = +\infty$.

Thus $a = b$.

If $a = -\infty$, then since $b \le a$ and Negative Infinity is Minimal, $b = -\infty$.

Similarly $b = +\infty \implies a = +\infty$ and $b = -\infty \implies a = -\infty$.

$\Box$


Reflexive

Let $a \in \overline \R$.

If $a \in \R$, then since $\le$ is reflexive on $\R$, $a \le a$.

If $a = +\infty$, then since $\left({+\infty, +\infty}\right) \in \left\{{ \left({ x, +\infty }\right): x \in \overline \R}\right\}$, $a \le a$.

Similarly, if $a = -\infty$, then $a \le a$.

$\blacksquare$