# Positive Infinity is Maximal

## Theorem

Let $\left({\overline \R, \le}\right)$ be the extended real numbers with the usual ordering.

Then $+\infty$ is a maximal element of $\overline \R$.

## Proof

By the definition of the usual ordering on the extended real numbers:

${\le} = {\le_\R} \cup \left\{{ \left({ x, +\infty }\right): x \in \overline \R}\right\} \cup \left\{{ \left({ -\infty, x }\right): x \in \overline \R}\right\}$

Suppose $x \in \overline \R$ and $+\infty \le x$.

That is:

$\left({+\infty, x}\right) \in {\le}$

By the definition of union, $\left({+\infty, x}\right)$ must lie in one of the three sets whose union forms $\le$.

Since $\le_\R \subseteq \R \times \R$ and $+\infty \notin \R$:

$\left({+\infty, x}\right) \notin \le_\R$

Since $+\infty \ne -\infty$ by the definition of the extended real numbers:

$\left({+\infty, x}\right) \notin \left\{{ \left({ -\infty, x }\right): x \in \overline \R}\right\}$

Therefore:

$\left({+\infty, x}\right) \in \left\{{ \left({ x, +\infty }\right): x \in \overline \R}\right\}$

and we conclude that $x = +\infty$.

That is, $+\infty$ is a maximal element of $\overline \R$.

$\blacksquare$