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

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

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

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

so $x = +\infty$.

Also see

 * Negative Infinity is Minimal