TheoremBase

If a were ∞, the defining condition of ∞ ≤ b would fail in each of its three alternatives, since ∞ is neither real nor −∞ and b is real.

Proof

We prove (i). By The Extended Real Line, Its Order, and [0,∞] (ii), either a∈Ra\in\mathbb{R}, and we are done, or a=∞a=\infty. Suppose, for a contradiction, that a=∞a=\infty. By the definition of the order in The Extended Real Line, Its Order, and [0,∞] (iii), ∞≤b\infty\le b means that ∞,b∈R\infty,b\in\mathbb{R} and ∞≤b\infty\le b in R\mathbb{R}, or ∞=−∞\infty=-\infty, or b=∞b=\infty. By The Extended Real Line, Its Order, and [0,∞] (i), ∞∉R\infty\notin\mathbb{R} and −∞≠∞-\infty\neq\infty, so the first two alternatives fail; and the third would give ∞=b∈R\infty=b\in\mathbb{R}, again contradicting ∞∉R\infty\notin\mathbb{R}. So the case a=∞a=\infty cannot occur, and therefore a∈Ra\in\mathbb{R}.

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