Proof of Greatest Elements as Suprema, Suprema and Infima of Nested Sets in [−∞,∞], and a Scaling Criterion for Order in [0,∞]

Greatest elements. Let $P$, $\preceq$, $A$ and $g$ be as in (i). As $v\preceq g$ for every $v\in A$ and $g\in A\subseteq P$, $g$ is an upper bound of $A$, so $g\in\operatorname{Ub}(A)$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (i)). If $b\in\operatorname{Ub}(A)$, then $g\preceq b$ because $g\in A$. Thus $g$ is a least element of $\operatorname{Ub}(A)$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iii)), that is, a supremum of $A$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)). By Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv), $A$ has at most one supremum and $\sup A$ denotes it; hence $\sup A=g$.

In the remaining clauses, $\le$ is a partial order on $[-\infty,\infty]$, by Order and Suprema in the Extended Real Line (i), so Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order applies with $[-\infty,\infty]$ and $\le$.

Nested sets. Let $A\subseteq B\subseteq[-\infty,\infty]$. The supremum $\sup B$ and the infimum $\inf B$ exist by Order and Suprema in the Extended Real Line (iii), as do $\sup A$ and $\inf A$. By Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv) and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (i), $v\le\sup B$ and $\inf B\le v$ for every $v\in B$, hence for every $v\in A$; so $\sup B$ is an upper bound and $\inf B$ a lower bound of $A$. As $\sup A$ is the least element of $\operatorname{Ub}(A)$ and $\inf A$ the greatest element of $\operatorname{Lb}(A)$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv), Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iii)), $\sup A\le\sup B$ and $\inf B\le\inf A$.

Scaling. Let $a,\alpha\in[0,\infty]$ with $c\cdot a\le\alpha$ for every $c\in\mathbb{R}$ with $0<c<1$. Each such $c$ lies in $[0,\infty]$ by The Extended Real Line, Its Order, and [0,∞] (ii), so $c\cdot a$ is defined by Addition and Multiplication on [0,∞] (ii). By The Extended Real Line, Its Order, and [0,∞] (ii), each of $a$ and $\alpha$ is either real and nonnegative or equal to $\infty$.

If $\alpha=\infty$, then $a\le\alpha$ by The Extended Real Line, Its Order, and [0,∞] (iii). Let now $\alpha\in\mathbb{R}$.

Suppose $a=\infty$. Take $c=1/2$, so $0<c<1$. Then $c\ne0$, $a\ne0$ and $a\notin\mathbb{R}$ (The Extended Real Line, Its Order, and [0,∞] (i)), so $c\cdot a=\infty$ by Addition and Multiplication on [0,∞] (ii). Hence $\infty\le\alpha$, so $\alpha=\infty$ by Order and Suprema in the Extended Real Line (ii), contrary to $\alpha\in\mathbb{R}$ and $\infty\notin\mathbb{R}$ (The Extended Real Line, Its Order, and [0,∞] (i)). So $a\in\mathbb{R}$.

Thus $a,\alpha\in\mathbb{R}$ with $0\le a$ and $0\le\alpha$, and for real numbers the order of $[-\infty,\infty]$ and its strict relation agree with those of $\mathbb{R}$, by Order and Suprema in the Extended Real Line (i). Suppose $a\le\alpha$ fails. As $\le$ is total on $\mathbb{R}$, $\alpha<a$, so $0<a$ and $0<\alpha+a$. Let $c=(\alpha+a)/(2a)\in\mathbb{R}$. Then $0<c$, and $c<1$ because $\alpha+a<2a$. As $c$ and $a$ are real, $c\cdot a$ is their real product by Addition and Multiplication on [0,∞] (ii), so $c\cdot a=(\alpha+a)/2$, and $\alpha<(\alpha+a)/2$ because $\alpha<a$. This contradicts $c\cdot a\le\alpha$. Hence $a\le\alpha$.