Supremum of Ideals is Increasing
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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 $L$.
- $\sup I \preceq \sup J$
Thus by definition of $f$:
- $f\left({I}\right) \preceq f\left({J}\right)$
Hence $f$ is an increasing mapping.
$\blacksquare$
Sources
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article YELLOW_2:51