Proof of The Maximum and Minimum of Two Elements of a Total Order

Each result cited below is universally quantified over the data in its own statement and is applied to the data named where it is cited. Since $\le$ is a partial order, it is reflexive and transitive, and since it is total, $a\le b$ or $b\le a$. As $a$ and $b$ are sets, $\{a,b\}$ is a set, the unordered pair of $a$ and $b$; as $a,b\in X$, it is a subset of $X$, and its greatest and least elements are as in Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §least.

Exists. Suppose $a\le b$. Then $b\in\{a,b\}$, and $a\le b$ and $b\le b$ by reflexivity, so $b$ is a greatest element of $\{a,b\}$; likewise $a\in\{a,b\}$ with $a\le a$ and $a\le b$, so $a$ is a least element. If $b\le a$, the same argument with $a$ and $b$ exchanged shows that $a$ is a greatest and $b$ a least element. In either case $\{a,b\}$ has a greatest and a least element, so $\max\{a,b\}$ and $\min\{a,b\}$ are defined.

Values. By Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §least-unique, $\{a,b\}$ has at most one greatest and at most one least element, and $\max\{a,b\}$ and $\min\{a,b\}$ denote them. So the elements found in the proof of the clause exists are $\max\{a,b\}$ and $\min\{a,b\}$: if $a\le b$, then $\max\{a,b\}=b$ and $\min\{a,b\}=a$, and if $b\le a$, then $\max\{a,b\}=a$ and $\min\{a,b\}=b$.

Bounds. $\max\{a,b\}$ is a greatest element of $\{a,b\}$, so $a\le\max\{a,b\}$ and $b\le\max\{a,b\}$; $\min\{a,b\}$ is a least element, so $\min\{a,b\}\le a$ and $\min\{a,b\}\le b$.

Least-upper. If $\max\{a,b\}\le c$, then $a\le\max\{a,b\}\le c$ and $b\le\max\{a,b\}\le c$ by the bounds clause, so $a\le c$ and $b\le c$ by transitivity. Conversely, let $a\le c$ and $b\le c$. Since $\max\{a,b\}\in\{a,b\}$, it equals $a$ or $b$, and in both cases $\max\{a,b\}\le c$.

Greatest-lower. If $c\le\min\{a,b\}$, then $c\le a$ and $c\le b$ by the bounds clause and transitivity. Conversely, if $c\le a$ and $c\le b$, then $c\le\min\{a,b\}$, since $\min\{a,b\}\in\{a,b\}$ equals $a$ or $b$.

Strict. All of $a,b,c,\max\{a,b\},\min\{a,b\}$ lie in $X$. By Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, $c<\max\{a,b\}$ holds if and only if $\max\{a,b\}\le c$ fails. By the least-upper clause, this holds if and only if $a\le c$ fails or $b\le c$ fails, that is, again by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, if and only if $c<a$ or $c<b$. Likewise, $\min\{a,b\}<c$ holds if and only if $c\le\min\{a,b\}$ fails, which by the greatest-lower clause holds if and only if $c\le a$ fails or $c\le b$ fails, that is, if and only if $a<c$ or $b<c$.