Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Ideals

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Then

$\mathscr S$ is meet-continuous if and only if
for every ideals $I, J$ in $\mathscr S$: $\paren {\sup I} \wedge \paren {\sup J} = \sup \set {i \wedge j: i \in I, j \in J}$


Proof

Sufficient Condition

Let $\mathscr S$ be meet-continuous.

Define $\II$, the set of all ideals in $\mathscr S$

Define a mapping $f: \II \to S$ such that

$\forall I \in \II: \map f I = \sup I$

By Meet-Continuous iff Ideal Supremum is Meet Preserving:

$f$ preserves meet.

Let $I, J \in \II$.

By definition of mapping preserves meet:

$f$ preserves the infimum of $\set {I, J}$

Thus

\(\ds \paren {\sup I} \wedge \paren {\sup J}\) \(=\) \(\ds \map f I \wedge \map f J\) Definition of $f$
\(\ds \) \(=\) \(\ds \inf \set {\map f I, \map f J}\) Definition of Meet (Order Theory)
\(\ds \) \(=\) \(\ds \map \inf {\map {f^\to} {\set {I, J} } }\) Definition of Image of Subset under Mapping
\(\ds \) \(=\) \(\ds \map f {\inf \set {I, J} }\) Definition of Mapping Preserves Infimum
\(\ds \) \(=\) \(\ds \map f {I \wedge J}\) Definition of Meet (Order Theory)
\(\ds \) \(=\) \(\ds \map \sup {I \wedge J}\) Definition of $f$
\(\ds \) \(=\) \(\ds \sup \set {i \wedge j: i \in I, j \in J}\) Meet in Set of Ideals

$\Box$


Necessary Condition

Assume that for every ideals $I, J$ in $\mathscr S$:

$\paren {\sup I} \wedge \paren {\sup J} = \sup \set {i \wedge j: i \in I, j \in J}$

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

$f$ preserves meet.

Thus by Meet-Continuous iff Ideal Supremum is Meet Preserving:

$\mathscr S$ is meet-continuous.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Meet-Continuous_iff_Meet_of_Suprema_equals_Supremum_of_Meet_of_Ideals&oldid=664925"