Proof of Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders

By Partial and Total Orders on a Set and the Associated Strict Relation §partial, $\le$ is reflexive, antisymmetric and transitive on $a$; these three properties are meant below by reflexivity, antisymmetry and transitivity.

Least-unique. Let $s$ be a subclass of $a$, and let $m,m'\in s$ be such that $m\le v$ and $m'\le v$ for every $v\in s$. Since $m'\in s$, $m\le m'$; since $m\in s$, $m'\le m$. As $m,m'\in s\subseteq a$, antisymmetry gives $m=m'$. If instead $m,m'\in s$ satisfy $v\le m$ and $v\le m'$ for every $v\in s$, then likewise $m'\le m$ and $m\le m'$, so $m=m'$.

Strict-characterization. Let $u$ and $v$ be sets. By Partial and Total Orders on a Set and the Associated Strict Relation §strict and Class Abstraction: the Class of All Sets Satisfying a Predicative Formula §abstraction, $(u,v)\in{<}$ holds if and only if there are sets $u'$ and $v'$ with $(u,v)=(u',v')$, $u'\le v'$ and $u'\neq v'$. By The Characteristic Property of Ordered Pairs and Nested Tuples of Sets §characteristic, $(u,v)=(u',v')$ if and only if $u=u'$ and $v=v'$. Hence $u<v$ if and only if $u\le v$ and $u\neq v$. This characterization is used in the remaining parts without further mention.

Strict-irreflexive. Let $u\in a$. If $u<u$ held, then $u\neq u$, which is false. So $u<u$ does not hold.

Strict-transitive. Let $u,v,w\in a$ with $u<v$ and $v<w$. Then $u\le v$ and $v\le w$, so $u\le w$ by transitivity. Suppose $u=w$. Then $v\le w$ reads $v\le u$, and together with $u\le v$ antisymmetry gives $u=v$, contradicting $u\neq v$ (part of $u<v$). Hence $u\neq w$, and $u<w$.

Weak-strict. Let $u,v\in a$. If $u\le v$, then either $u=v$, or $u\neq v$ and so $u<v$. Conversely, if $u<v$ then $u\le v$; and if $u=v$ then $u\le v$ by reflexivity, since $u\in a$.

Trichotomy. Let $\le$ be total and $u,v\in a$. At least one holds: if $u=v$ we are done. Otherwise $u\neq v$, and by Partial and Total Orders on a Set and the Associated Strict Relation §total $u\le v$ or $v\le u$; in the first case $u<v$, and in the second, since $v\neq u$, $v<u$. At most one holds: $u<v$ and $u=v$ cannot both hold, since $u<v$ requires $u\neq v$; likewise $v<u$ and $u=v$ cannot both hold. If $u<v$ and $v<u$, then $u\le v$ and $v\le u$, so $u=v$ by antisymmetry, contradicting $u\neq v$. Hence exactly one of $u<v$, $u=v$ and $v<u$ holds.

Total-negation. Let $\le$ be total and $u,v\in a$. Suppose $u\le v$ fails. By Partial and Total Orders on a Set and the Associated Strict Relation §total, $v\le u$; and $v\neq u$, since otherwise $u\le v$ would hold by reflexivity. Hence $v<u$. Conversely, suppose $v<u$, so that $v\le u$ and $v\neq u$. If $u\le v$ held, antisymmetry would give $u=v$, a contradiction; so $u\le v$ fails.