Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving

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Theorem

Let $\mathscr S = \struct {S, \wedge, \preceq}$ be an up-complete meet semilattice.

Let $f: \map {\it Ids} {\mathscr S} \to S$ be a mapping such that:

$\forall I \in \map {\it Ids} {\mathscr S}: \map f I = \sup_{\mathscr S} I$

where

$\map {\it Ids} {\mathscr S}$ denotes the set of all ideals in $\mathscr S$

Let

$\forall I_1, I_2 \in \map {\it Ids} {\mathscr S}: \paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {i \wedge j: i \in I_1, j \in I_2}$


Then: $f$ preserves meet as a mapping from $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$ into $\mathscr S$


Proof

Let $I, J \in \map {\it Ids} {\mathscr S}$ such that

$\set {I, J}$ admits an infimum in $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$.

By definition of image of set:

$\map {f^\to} {\set {I, J} } = \set {\map f I, \map f J}$

Thus by definition of meet semilattice:

$\map {f^\to} {\set {I, J} }$ admits an infimum in $\mathscr S$.

Thus:

\(\ds \map \inf {\map {f^\to} {\set {I, J} } }\) \(=\) \(\ds \map f I \wedge \map f J\) Definition of Meet (Order Theory)
\(\ds \) \(=\) \(\ds \paren {\sup I} \wedge \paren {\sup J}\) Definition of $f$
\(\ds \) \(=\) \(\ds \sup \set {x \wedge y: x \in I, y \in J}\) Assumption
\(\ds \) \(=\) \(\ds \map \sup {I \wedge J}\) Meet in Set of Ideals
\(\ds \) \(=\) \(\ds \map f {I \wedge J}\) definition of $f$
\(\ds \) \(=\) \(\ds \map f {\inf \set {I, J} }\) Definition of Meet (Order Theory)

$\blacksquare$


Sources

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