Supremum of Ideals is Increasing

Theorem
Let $L = \left({S, \preceq}\right)$ be an up-complete ordered set.

Let $\mathit{Ids}\left({L}\right)$ be the set of all ideals in $L$.

Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an ordered set where $\mathord \precsim = \subseteq\restriction_{\mathit{Ids}\left({L}\right)\times \mathit{Ids}\left({L}\right)}$

Let $f: \mathit{Ids}\left({L}\right) \to S$ be a mapping such that
 * $\forall I \in \mathit{Ids}\left({L}\right): f\left({I}\right) = \sup I$

Then $f$ is an increasing mapping.

Proof
Let $I, J \in \mathit{Ids}\left({L}\right)$ such that
 * $I \precsim J$

By definition of $\precsim$:
 * $I \subseteq J$

By definition of up-complete:
 * $I$ and $J$ admit suprema in $L$.

By Supremum of Subset:
 * $\sup I \preceq \sup J$

Thus by definition of $f$:
 * $f\left({I}\right) \preceq f\left({J}\right)$

Hence $f$ is an increasing mapping.