Set of Chains is Chain Complete for Inclusion

Theorem
Let $\struct {P, \le}$ be an ordered set.

Let $\map {\operatorname {ch} } P$ be the set of chains in $P$.

Let $\CC$ be a chain in $\map {\operatorname {ch} } P$ with respect to the subset ordering.

Then the union $\ds \bigcup \CC$ of $\CC$ is also a chain in $P$.

Proof
Let $\ds D = \bigcup \CC$.

Observe that any $C \in \CC$ is a chain in $P$, hence $C \subseteq P$.

Therefore, $D \subseteq P$, by Set Union Preserves Subsets.

Now suppose that $a, b \in D$.

By definition of union, there exist $A, B \in \CC$ such that $a \in A$ and $b \in B$.

Since $\CC$ is a chain in $\struct {\map {\operatorname {ch} } P, \subseteq}$, either $A \subseteq B$ or $B \subseteq A$.

Hence either $a \in B$ or $b \in A$.

Both $A$ and $B$ are chains in $P$, and thence $a \le b$ or $b \le a$.

Therefore, $D$ is totally ordered by $\le$.

That is, it is a chain in $P$.