Union of Chain of Orderings is Ordering

Theorem
Let $S$ be a set, and let $C$ be a nonempty nest of orderings on $S$.

Then $\bigcup C$ is an ordering on $S$.

Proof
Let $\preceq$ be an arbitrary element of $C$, and let $\sim = \bigcup C$.

Then $\sim$ is an ordering:


 * For any $a,\,b \in S$, if $a \sim b$ and $b \sim a$, then $a = b$:
 * Since $a \sim b$ and $b \sim a$, there exist $\preceq_1,\,\preceq_2 \in C$ such that $a \preceq_1 b$ and $b \preceq_2 a$.


 * Since $C$ is a chain,


 * $\preceq_1 \subset \preceq_2$ or $\preceq_2 \subset \preceq_1$.


 * Suppose without loss of generality that $\preceq_1 \subset \preceq_2$.


 * Then it must hold that $a \preceq_2 b$.


 * Since $a \preceq_2 b$, $b \preceq_2 a$, and $\preceq_2 \in T$ is an ordering,


 * $a = b$.


 * For any $a \in S$, $a \sim a$:
 * $a \preceq a$ because $\preceq$ is an ordering.
 * Since $\preceq \subseteq \sim$, $a \sim a$.


 * For any $a,\,b,\,c \in S$, if $a \sim b$ and $b \sim c$ then $a \sim c$:
 * Since $a \sim b$ and $b \sim c$,


 * there are elements $\preceq_1$ and $\preceq_2$ of $C$ such that $a \preceq_1 b$ and $b \preceq_2 c$.


 * Since $C$ is a chain, it must hold (as above) that $a \preceq_2 b$ or $b \preceq_1 c$.


 * Suppose without loss of generality, as above, that $a \preceq_2 b$.


 * Then since $a \preceq_2 b$, $b \preceq_2 c$, and $\preceq_2 \in T$ is an ordering,


 * $a \preceq_2 c$.


 * Since $\preceq_2 \subseteq \sim$,


 * $a \sim c$.