Ideals are Continuous Lattice Subframe of Power Set

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.

Let $I = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an inclusion ordered set

where

$\mathit{Ids}\left({L}\right)$ denotes the set of all ideals in $L$,
$\mathord\precsim = \mathord\subseteq \cap \left({\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right)}\right)$

Let $P = \struct {\mathcal P\left({S}\right), \precsim'}$ be an inclusion ordered set

where

$\mathcal P\left({S}\right)$ denotes the power set of $S$,
$\mathord\precsim' = \mathord\subseteq \cap \left({\mathcal{P}\left({S}\right) \times \mathcal{P}\left({S}\right)}\right)$


Then $I$ is continuous lattice subframe of $P$.


Proof

By definition of subset:

$\mathit{Ids}\left({L}\right) \subseteq \mathcal P\left({S}\right)$

Then

$\mathord\precsim = \mathord\precsim' \cap \left({\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right)}\right)$

Hence $I$ is ordered subset of $P$.

Infima Inheriting

Let $A$ be a subset of $\mathit{Ids}\left({L}\right)$ such that

$A$ admits an infimum in $P$.

By the proof of Power Set is Complete Lattice:

$\inf_P A = \bigcap A$

By Intersection of Semilattice Ideals is Ideal/Set of Sets:

$\inf_P A \in \mathit{Ids}\left({L}\right)$

Thus by Infimum in Ordered Subset:

$A$ admits an infimum in $I$ and $\inf_I A = \inf_P A$

Hence $I$ inherits infima.

$\Box$

Directed Suprema Inheriting

Let $D$ be a directed subset of $\mathit{Ids}\left({L}\right)$ such that

$D$ admits a supremum in $P$.

By the proof of Power Set is Complete Lattice:

$\sup_P D = \bigcup D$

We will prove that

$\bigcup D$ is an ideal in $L$.

Directed

Let $x, y \in \bigcup D$.

By definition of union:

$\exists I_1 \in D: x \in I_1$

and

$\exists I_2 \in D: y \in I_2$

By definition of directed:

$\exists I \in D: I_1 \precsim I \land I_2 \precsim I$

By definition of $\precsim$:

$I_1 \subseteq I$ and $I_2 \subseteq I$

By definition of subset:

$x, y \in I$

By definition of directed:

$\exists z \in I: x \preceq z \land y \preceq z$

Thus by definition of union:

$\exists z \in \bigcup D: x \preceq z \land y \preceq z$

$\Box$

Lower Set

Let $x \in \bigcup D$, $y \in S$ such that

$y \preceq x$

By definition of union:

$\exists I \in D: x \in I$

By definition of lower set:

$y \in I$

Thus by definition of union:

$y \in \bigcup D$

$\Box$

Non-Empty Set

By definition of directed:

$D$ is non-empty and $\forall I \in D: I$ is non-empty.

Thus by definitions of non-empty set and union:

$\bigcup D$ is non-empty.

$\Box$

By definition of $\mathit{Ids}$:

$\bigcup D \in \mathit{Ids}\left({L}\right)$

Hence $I$ inherits directed suprema.

$\blacksquare$


Sources