Set of Chains is Chain Complete for Inclusion

From ProofWiki
Jump to navigation Jump to search


Let $\left({P, \leq}\right)$ be an ordered set.

Let $\operatorname{ch} \left({P}\right)$ be the set of chains in $P$.

Let $\mathcal C$ be a chain in $\operatorname{ch} \left({P}\right)$ with respect to the subset ordering.

Then the union $\bigcup \mathcal C$ of $\mathcal C$ is also a chain in $P$.


Let $D = \bigcup \mathcal C$.

Observe that any $C \in \mathcal C$ 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 $A, B \in \mathcal C$ such that $a \in A$ and $b \in B$.

Since $\mathcal C$ is a chain in $\left({\operatorname{ch} \left({P}\right), \subseteq}\right)$, 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 $a \leq b$ or $b \leq a$.

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

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