Meet-Continuous iff Ideal Supremum is Meet Preserving

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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 $\mathscr S$


Then

$\mathscr S$ is meet-continuous

if and only if

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


Proof

Sufficient Condition

Let $\mathscr S$ be meet-continuous.

We will prove that:

for every directed subsets $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$

Let $D_1, D_2$ be directed subsets of $S$.

we will prove as sublemma that:

for every an element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x = \sup \set {x \wedge d: d \in D}$

Let $x \in S$, $D$ be a directed subset of $S$ such that:

$x \preceq \sup D$

Thus:

\(\ds x\) \(=\) \(\ds x \wedge \sup D\) Preceding iff Meet equals Less Operand
\(\ds \) \(=\) \(\ds \sup \set {x \wedge d: d \in D}\) Definition of Meet-Continuous Lattice


By definition of reflexivity:

for every element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x \preceq \sup \set {x \wedge d: d \in D}$

Define a mapping $g: S \times S \to S$:

$\forall \tuple {s, t} \in S \times S: \map g {s, t} = s \wedge t$

By Meet is Directed Suprema Preserving:

$g$ preserves directed suprema as a mapping from the simple order product $\struct {S \times S, \precsim}$ of $\mathscr S$ and $\mathscr S$ into $\mathscr S$.

Thus by Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets:

$\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$


By exemplification:

for every ideals $I_1, I_2$ of $S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {d_1 \wedge d_2: d_1 \in I_1, d_2 \in I_2}$

Thus by Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving:

$f$ preserves meet.

$\Box$


Necessary Condition

Let $f$ be such that it preserves meet.

Thus:

$\mathscr S$ is up-complete

We will prove that:

for every ideals $I_1, I_2$ in $\mathscr S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}$

Let $I_1, I_2$ be ideals in $\mathscr S$.

Thus:

\(\ds \paren {\sup I_1} \wedge \paren {\sup I_2}\) \(=\) \(\ds \map f {I_1} \wedge \map f {I_2}\) Definition of $f$
\(\ds \) \(=\) \(\ds \map f {I_1 \wedge I_2}\) Definition of Mapping Preserves Infimum
\(\ds \) \(=\) \(\ds \map \sup {I_1 \wedge I_2}\) Definition of $f$
\(\ds \) \(=\) \(\ds \sup \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\) Meet in Set of Ideals

We will prove that:

for every directed subsets $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$

Let $D_1, D_2$ be directed subsets of $S$.

By definition of up-complete:

$D_1$ and $D_2$ admit suprema

By Supremum of Lower Closure of Set:

$D_1^\preceq$ and $D_2^\preceq$ admit suprema

Thus:

\(\ds \paren {\sup D_1} \wedge \paren {\sup D_2}\) \(=\) \(\ds \paren {\sup D_1^\preceq} \wedge \paren {\sup D_2^\preceq}\) Supremum of Lower Closure of Set
\(\ds \) \(=\) \(\ds \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\) $D_1^\preceq$ is ideal in $\mathscr S$
\(\ds \) \(=\) \(\ds \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}^\preceq\) Supremum of Lower Closure of Set
\(\ds \) \(=\) \(\ds \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}^\preceq\) Lower Closure of Meet of Lower Closures
\(\ds \) \(=\) \(\ds \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\) Supremum of Lower Closure of Set

It remains to prove (MC) of definition of meet-continuous.

Let $x$ be an element of $S$.

Let $D$ be a directed subset of $S$.

By Singleton is Directed and Filtered Subset:

$\set x$ is directed.

Thus:

\(\ds x \wedge \sup D\) \(=\) \(\ds \paren {\sup \set x} \wedge \sup D\) Supremum of Singleton
\(\ds \) \(=\) \(\ds \sup \set {d_1 \wedge d_2: d_1 \in \set x, d_2 \in D}\)
\(\ds \) \(=\) \(\ds \sup \set {x \wedge d: d \in D}\) Definition of Singleton

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Meet-Continuous_iff_Ideal_Supremum_is_Meet_Preserving&oldid=664926"