Elements Well Inside Form Ideal

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Theorem

Let $L = \struct{S, \vee, \wedge, \preceq}$ be a distributive lattice with greatest element $\top$ and smallest element $\bot$.

Let $\eqslantless$ denote the well inside relation on $L$.


Then:

$\forall a \in S : \set{b \in S: b \eqslantless a}$ is a a lattice ideal


Proof

Let $a \in S$.

Let $I = \set{b \in S: b \eqslantless a}$.

$I$ is an Lower Section

Let $b \in I$ and $c \in S : c \preceq b$.


By Ordering Axiom $(3)$: Antisymmetry:

$a \preceq a$

Hence we have:

$c \preceq b \eqslantless a \preceq a$


From Well Inside Relation Extends to Predecessor and Successor:

$c \eqslantless a$

Hence:

$c \in I$


It follows that $I$ is a lower section by definition.

$\Box$

$I$ is a Join Subsemilattice

Let $b, c \in I$.

By definition of well inside relation:

$\exists x, y \in S : b \wedge x = \bot, a \vee x = \top, c \wedge y = \bot, a \vee y = \top$


We have:

\(\ds \paren{b \vee c} \wedge \paren{x \wedge y}\) \(=\) \(\ds \paren{b \wedge \paren{x \wedge y} } \vee \paren{c \wedge \paren{x \wedge y} }\) Definition of Distributive Lattice
\(\ds \) \(=\) \(\ds \paren{b \wedge \paren{x \wedge y} } \vee \paren{c \wedge \paren{y \wedge x} }\) Meet is Commutative
\(\ds \) \(=\) \(\ds \paren{\paren{b \wedge x} \wedge y} \vee \paren{\paren{c \wedge y} \wedge x}\) Meet is Associative
\(\ds \) \(=\) \(\ds \paren{\bot \wedge y} \vee \paren{\bot \wedge x}\) as $\paren{b \wedge x} = \bot, \paren{c \wedge y} = \bot$
\(\ds \) \(=\) \(\ds \bot \vee \bot\) Predecessor is Infimum
\(\ds \) \(=\) \(\ds \bot\) Join is Idempotent


Also we have:

\(\ds a \vee \paren{x \wedge y}\) \(=\) \(\ds \paren{a \vee x} \wedge \paren{a \vee y}\) Definition of Distributive Lattice
\(\ds \) \(=\) \(\ds \top \wedge \top\) as $\paren{a \vee x} = \top, \paren{a \vee y} = \top$
\(\ds \) \(=\) \(\ds \top\) Meet is Idempotent


By definition of well inside relation:

$b \vee c \eqslantless a$

By definition of $I$:

$b \vee c \in I$


It follows that $I$ is an join subsemilattice by definition.

$\Box$


It follows that $I$ is a lattice ideal by definition.


The result follows.

$\blacksquare$


Sources

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