Negative Infinity is Minimal

Theorem
Let $\struct {\overline \R, \le}$ be the extended real numbers with the usual ordering.

Then $-\infty$ is a minimal element of $\overline \R$.

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


 * ${\le} = {\le_\R} \cup \set {\tuple {x, +\infty}: x \in \overline \R} \cup \set {\tuple {-\infty, x}: x \in \overline \R}$

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

That is:
 * $\tuple {x, -\infty} \in {\le}$

By the definition of union, $\tuple {x, -\infty}$ must lie in one of the three sets whose union forms $\le$.

Since ${\le_\R} \subseteq \R \times \R$ and $-\infty \notin \R$:


 * $\tuple {x, -\infty} \notin {\le_\R}$

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


 * $\tuple {x, -\infty} \notin \set {\tuple {x, +\infty}: x \in \overline \R}$

Therefore:


 * $\tuple {x, -\infty} \in \set {\tuple {-\infty, x}: x \in \overline \R}$

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

That is, $-\infty$ is a minimal element of $\overline \R$.

Also see

 * Positive Infinity is Maximal