Proof of Order and Suprema in the Extended Real Line

Write $E=[-\infty,\infty]$. By The Extended Real Line, Its Order, and [0,∞] (i), $-\infty\notin\mathbb{R}$, $\infty\notin\mathbb{R}$ and $-\infty\neq\infty$; by The Extended Real Line, Its Order, and [0,∞] (ii), every element of $E$ is real or equal to $-\infty$ or to $\infty$.

Step 1: four facts. For all $x,y\in E$: (a) $-\infty\le y$ and $x\le\infty$; (b) $x\le-\infty$ if and only if $x=-\infty$; (c) $\infty\le y$ if and only if $y=\infty$; (d) if $x,y\in\mathbb{R}$, then $x\le y$ if and only if $x\le y$ in $\mathbb{R}$. Fact (a) is The Extended Real Line, Its Order, and [0,∞] (iii). For (b), if $x\le-\infty$, then by The Extended Real Line, Its Order, and [0,∞] (iii) $x,-\infty\in\mathbb{R}$ (impossible, as $-\infty\notin\mathbb{R}$), or $x=-\infty$, or $-\infty=\infty$ (impossible); so $x=-\infty$, and the converse is (a). Fact (c) is proved in the same way, using $\infty\notin\mathbb{R}$ and $\infty\neq-\infty$. Fact (d) is stated in The Extended Real Line, Its Order, and [0,∞] (iii).

Proof of (i). By The Extended Real Line, Its Order, and [0,∞] (iii), $\le$ is a subset of $E\times E$, a relation on $E$. Reflexive: for $x\in\mathbb{R}$ by (d) and reflexivity of the order of $\mathbb{R}$; for $x\in\{-\infty,\infty\}$ by (a). Antisymmetric: let $x\le y$ and $y\le x$. If $x=-\infty$, then $y\le-\infty$ gives $y=-\infty$ by (b); if $x=\infty$, then $\infty\le y$ gives $y=\infty$ by (c). If $x\in\mathbb{R}$, then $y\neq-\infty$, since otherwise $x\le-\infty$ and $x=-\infty$ by (b); and $y\neq\infty$, since otherwise $\infty\le x$ and $x=\infty$ by (c); so $y\in\mathbb{R}$, and $x=y$ by (d) and antisymmetry in $\mathbb{R}$. Transitive: let $x\le y$ and $y\le z$. If $x=-\infty$ or $z=\infty$, then $x\le z$ by (a). Otherwise $z\neq-\infty$, since $z=-\infty$ would give $y=-\infty$ and then $x=-\infty$ by (b); so $z\in\mathbb{R}$. Likewise $x\neq\infty$, since $x=\infty$ would give $y=\infty$ and then $z=\infty$ by (c); so $x\in\mathbb{R}$. And $y\neq-\infty$ (else $x=-\infty$ by (b)) and $y\neq\infty$ (else $z=\infty$ by (c)), so $y\in\mathbb{R}$. Then $x\le z$ by (d) and transitivity in $\mathbb{R}$. Total: for $x,y\in\mathbb{R}$, $x\le y$ or $y\le x$ by (d), the order of $\mathbb{R}$ being total by The Real Numbers (i); if one of $x,y$ is $-\infty$ or $\infty$, they are comparable by (a). Hence $\le$ is a partial order on $E$ that is total. The statement on $x\le y$ for real $x,y$ is (d); for $<$ it follows, as $x<y$ means $x\le y$ and $x\neq y$ both in $E$ (The Extended Real Line, Its Order, and [0,∞] (iii)) and in $\mathbb{R}$ (Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (ii)).

Proof of (ii). By (a), $-\infty\in E$ lies below and $\infty\in E$ above every element, so they are the least and the greatest element of $E$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iii); the last assertion is (b) and (c).

Proof of the suprema in (iii), and of (iv) and (v). Let $A\subseteq E$ and $B=A\cap\mathbb{R}$, and let $\operatorname{Ub}(A)$ and $\operatorname{Lb}(A)$ be as in Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (i), formed in $E$. Exactly one of the following four cases occurs.

Case (i): $\infty\in A$. Then $\infty\in\operatorname{Ub}(A)$ by (a), and every $b\in\operatorname{Ub}(A)$ satisfies $\infty\le b$, so $b=\infty$ by (c). Thus $\operatorname{Ub}(A)=\{\infty\}$, whose least element is $\infty$, and $\sup A=\infty$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv).

Case (ii): $\infty\notin A$ and $B=\emptyset$. Then every element of $A$ equals $-\infty$, so every $b\in E$ is an upper bound of $A$ by (a): $\operatorname{Ub}(A)=E$, whose least element is $-\infty$ by (ii). So $\sup A=-\infty$; in particular $\sup\emptyset=-\infty$.

Case (iii): $\infty\notin A$, and $B$ is nonempty and bounded above in $\mathbb{R}$. Then $s=\sup_{\mathbb{R}}B$ exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vii). It is an upper bound of $A$ in $E$: each $x\in A$ is $-\infty\le s$ by (a), or lies in $B$, so $x\le s$ in $\mathbb{R}$ and hence in $E$ by (d). Let $b\in\operatorname{Ub}(A)$, and pick $x\in B$; then $x\le b$, so $b\neq-\infty$ by (b), as $x\in\mathbb{R}$. If $b=\infty$, then $s\le b$ by (a). If $b\in\mathbb{R}$, then by (d) $b$ is an upper bound of $B$ in $\mathbb{R}$, so $s\le b$ in $\mathbb{R}$, hence in $E$. Thus $s$ is the least element of $\operatorname{Ub}(A)$, and $\sup A=s$.

Case (iv): $\infty\notin A$, and $B$ is not bounded above in $\mathbb{R}$. Then $B\neq\emptyset$, as every real number is an upper bound of $\emptyset$; pick $x\in B$. By (a), $\infty\in\operatorname{Ub}(A)$. No $b\in\mathbb{R}$ lies in $\operatorname{Ub}(A)$, since by (d) it would be an upper bound of $B$ in $\mathbb{R}$; and $-\infty\notin\operatorname{Ub}(A)$, since $x\le-\infty$ would give $x=-\infty$ by (b). So $\operatorname{Ub}(A)=\{\infty\}$ and $\sup A=\infty$.

Hence $\sup A$ exists in every case. Case (iii) gives (iv), and Cases (i) and (iv) give (v); for $A\subseteq\mathbb{R}$ nonempty and bounded above in $\mathbb{R}$, we have $\infty\notin A$ and $B=A$, so Case (iii) applies.

Proof of the infima in (iii). Let $A\subseteq E$, and let $m=\sup\operatorname{Lb}(A)$, which exists by the above, as $\operatorname{Lb}(A)\subseteq E$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (i). Every $a\in A$ is an upper bound of $\operatorname{Lb}(A)$, so $m\le a$, as $m$ is the least upper bound; thus $m\in\operatorname{Lb}(A)$. Every $b\in\operatorname{Lb}(A)$ satisfies $b\le m$, as $m$ is an upper bound of $\operatorname{Lb}(A)$. So $m$ is the greatest element of $\operatorname{Lb}(A)$, and $\inf A=m$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv). For $A=\emptyset$, every element of $E$ is a lower bound, so $\operatorname{Lb}(\emptyset)=E$, whose greatest element is $\infty$ by (ii); hence $\inf\emptyset=\infty$. The notation $\sup_{n}u_{n}$ is a definition and needs no proof.

Proof of (vi) and (vii). Let $A\subseteq E$ and $B=A\cap\mathbb{R}$. If $-\infty\in A$, then $-\infty\in\operatorname{Lb}(A)$ by (a), and every $b\in\operatorname{Lb}(A)$ satisfies $b\le-\infty$, so $b=-\infty$ by (b); so $\operatorname{Lb}(A)=\{-\infty\}$ and $\inf A=-\infty$. Let now $-\infty\notin A$. If $B$ is nonempty and bounded below in $\mathbb{R}$, then $t=\inf_{\mathbb{R}}B$ exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (viii). Each $x\in A$ is $\infty\ge t$ by (a), or lies in $B$, so $t\le x$ in $\mathbb{R}$ and in $E$ by (d); thus $t\in\operatorname{Lb}(A)$. Let $b\in\operatorname{Lb}(A)$ and pick $x\in B$; then $b\le x$, so $b\neq\infty$ by (c). If $b=-\infty$, then $b\le t$ by (a); if $b\in\mathbb{R}$, it is a lower bound of $B$ in $\mathbb{R}$ by (d), so $b\le t$. Hence $\inf A=t$. If $B$ is not bounded below in $\mathbb{R}$, then $B\neq\emptyset$; pick $x\in B$. Then $-\infty\in\operatorname{Lb}(A)$ by (a), no $b\in\mathbb{R}$ lies in $\operatorname{Lb}(A)$ by (d), and $\infty\notin\operatorname{Lb}(A)$, as $\infty\le x$ would give $x=\infty$ by (c). So $\operatorname{Lb}(A)=\{-\infty\}$ and $\inf A=-\infty$.

Proof of (viii). If $x\in[0,\infty]$, then either $x\in\mathbb{R}$ with $0\le x$ in $\mathbb{R}$, so $0\le x$ by (d), or $x=\infty$ and $0\le x$ by (a). Conversely let $x\in E$ with $0\le x$. Then $x\neq-\infty$, since $0\le-\infty$ would give $0=-\infty$ by (b), while $0\in\mathbb{R}$. If $x=\infty$, then $x\in[0,\infty]$; if $x\in\mathbb{R}$, then $0\le x$ in $\mathbb{R}$ by (d), and $x\in[0,\infty]$ by The Extended Real Line, Its Order, and [0,∞] (ii).

Proof of (ix). Let $A\subseteq[0,\infty]$. Then $0$ is a lower bound of $A$, so $0\le\inf A$, and $\inf A\in[0,\infty]$. If $A\neq\emptyset$, pick $a\in A$; then $0\le a\le\sup A$, so $0\le\sup A$ by (i), and $\sup A\in[0,\infty]$.

Proof of (x). Let $t<\sup A$, and suppose that no $a\in A$ satisfies $t<a$. Then $a\le t$ for every $a\in A$, by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vii) and (i); so $t\in\operatorname{Ub}(A)$ and $\sup A\le t$. Together with $t\le\sup A$ this gives $t=\sup A$ by antisymmetry, contradicting $t\neq\sup A$. Likewise, let $\inf A<t$ and suppose that no $a\in A$ satisfies $a<t$. Then $t\le a$ for every $a\in A$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders (vii), so $t\in\operatorname{Lb}(A)$ and $t\le\inf A$; with $\inf A\le t$ this gives $t=\inf A$, a contradiction.