Every subset of $\overline{\mathbb{R}}$ has a least upper bound and a greatest lower bound in $\overline{\mathbb{R}}$, agreeing with the real supremum and infimum on nonempty sets bounded in $\mathbb{R}$

Case S1 for the supremum: $+\infty \in A$.

Case S2 for the supremum: $+\infty \notin A$ and $A_{\mathbb{R}} = \emptyset$, so that every element of $A$ equals $-\infty$.

Case S3 for the supremum: $+\infty \notin A$, $A_{\mathbb{R}} \ne \emptyset$, and $A_{\mathbb{R}}$ is bounded above in $\mathbb{R}$.

Case S4 for the supremum: $+\infty \notin A$, $A_{\mathbb{R}} \ne \emptyset$, and $A_{\mathbb{R}}$ is not bounded above in $\mathbb{R}$.

Case I1 for the infimum: $-\infty \in A$.

Case I2 for the infimum: $-\infty \notin A$ and $A_{\mathbb{R}} = \emptyset$, so that every element of $A$ equals $+\infty$.

Case I3 for the infimum: $-\infty \notin A$, $A_{\mathbb{R}} \ne \emptyset$, and $A_{\mathbb{R}}$ is bounded below in $\mathbb{R}$.

Case I4 for the infimum: $-\infty \notin A$, $A_{\mathbb{R}} \ne \emptyset$, and $A_{\mathbb{R}}$ is not bounded below in $\mathbb{R}$.

In case S1 the element $+\infty$ is an upper bound of $A$, being the greatest element of $\overline{\mathbb{R}}$; and if $v$ is any upper bound of $A$ then $+\infty \in A$ gives $+\infty \le v$, whence $v = +\infty$ by antisymmetry. So $+\infty$ is the least upper bound of $A$.

In case S2 every element of $A$ equals $-\infty$, so $-\infty$ is an upper bound of $A$ by reflexivity; and $-\infty \le v$ for every $v \in \overline{\mathbb{R}}$, being the least element. So $-\infty$ is the least upper bound of $A$.

In case S3 the real number $\sigma := \sup A_{\mathbb{R}}$ exists, and it is an upper bound of $A$ in $\overline{\mathbb{R}}$: an element of $A$ is either real, hence lies in $A_{\mathbb{R}}$ and satisfies $a \le \sigma$ in $\mathbb{R}$ and so in $\overline{\mathbb{R}}$, or equals $-\infty$, which is $\le \sigma$; the value $+\infty$ does not occur in $A$ in this case.

In case S4 the element $+\infty$ is an upper bound of $A$; and if $v$ is an upper bound then $v \ne -\infty$, because fixing $a \in A_{\mathbb{R}}$, which is possible in this case, gives $a \le v$ with $a$ real and $-\infty$ is below no real, while $v$ real would make $v$ a real upper bound of $A_{\mathbb{R}}$ and contradict the case hypothesis. So $v = +\infty$, and $+\infty$ is the least upper bound of $A$.

In case I1 the element $-\infty$ is a lower bound of $A$, being least; and any lower bound $w$ satisfies $w \le -\infty$ because $-\infty \in A$, whence $w = -\infty$ by antisymmetry. So $-\infty$ is the greatest lower bound of $A$.

In case I2 every element of $A$ equals $+\infty$, so $+\infty$ is a lower bound of $A$ by reflexivity, and $w \le +\infty$ for every $w$. So $+\infty$ is the greatest lower bound of $A$.

In case I3 the real number $\iota := \inf A_{\mathbb{R}}$ exists and is a lower bound of $A$ in $\overline{\mathbb{R}}$: an element of $A$ is either real, hence in $A_{\mathbb{R}}$ and $\ge \iota$, or equals $+\infty \ge \iota$; the value $-\infty$ does not occur in $A$ in this case.

In case I4 the element $-\infty$ is a lower bound of $A$; any lower bound $w$ satisfies $w \ne +\infty$, because fixing $a \in A_{\mathbb{R}}$ gives $w \le a$ with $a$ real and $+\infty$ is above no real, while $w$ real would be a real lower bound of $A_{\mathbb{R}}$ and contradict the case hypothesis. So $w = -\infty$ is the greatest lower bound of $A$.

In case S3 let $v$ be any upper bound of $A$ and fix $a \in A_{\mathbb{R}}$, which is possible since $A_{\mathbb{R}} \ne \emptyset$. From $a \le v$ with $a$ real we get $v \ne -\infty$, since $-\infty$ is below no real. If $v = +\infty$ then $\sigma \le v$ because $+\infty$ is greatest. Otherwise $v$ is real, and it bounds $A_{\mathbb{R}}$ above in $\mathbb{R}$, so $\sigma \le v$ by leastness of the real supremum. Hence $\sigma$ is the least upper bound of $A$.

In case I3 let $w$ be a lower bound of $A$ and fix $a \in A_{\mathbb{R}}$. From $w \le a$ with $a$ real we get $w \ne +\infty$. If $w = -\infty$ then $w \le \iota$; otherwise $w$ is real and bounds $A_{\mathbb{R}}$ below in $\mathbb{R}$, so $w \le \iota$. Hence $\iota$ is the greatest lower bound of $A$.

The four supremum cases are exhaustive and mutually exclusive: either $+\infty \in A$, which is S1, or not, and then either $A_{\mathbb{R}} = \emptyset$, which is S2, or $A_{\mathbb{R}} \ne \emptyset$ and it is bounded above in $\mathbb{R}$, which is S3, or it is not, which is S4. In each case a least upper bound was produced, and it is unique. The same four alternatives with $-\infty$, $+\infty$ and "below" in place of $+\infty$, $-\infty$ and "above" are I1 to I4, and in each a greatest lower bound was produced.

The agreement clause follows: a nonempty $A \subseteq \mathbb{R}$ bounded above in $\mathbb{R}$ satisfies $+\infty \notin A$ and $A_{\mathbb{R}} = A$, so case S3 applies and $\sup A = \sup A_{\mathbb{R}}$ is the real supremum; a nonempty $A \subseteq \mathbb{R}$ bounded below satisfies case I3 and $\inf A$ is the real infimum; and $A = \emptyset$ falls under S2 and I2, giving $\sup \emptyset = -\infty$ and $\inf \emptyset = +\infty$.