Compact Element iff Principal Ideal

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Theorem

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

Let $P = \struct {\map {\operatorname {Ids} } L, \precsim}$ be an inclusion ordered set

where

$\map {\operatorname {Ids} } L$ denotes the set of all ideals in $L$,
$\mathord\precsim = \mathord\subseteq \cap \paren {\map {\operatorname {Ids} } L \times \map {\operatorname {Ids} } L}$

Let $x \in \map {\operatorname {Ids} } L$


Then $x$ is compact element if and only if $x$ is principal ideal in $L$.


Proof

By Ideals are Continuous Lattice Subframe of Power Set:

$P$ is continuous lattice subframe of $\struct {\powerset S, \subseteq'}$

where

$\powerset S$ denotes the power set of $S$,
$\mathord\subseteq' = \mathord\subseteq \cap \struct {\powerset S \times \powerset S}$


Sufficient Condition

Assume that

$x$ is compact element.

By Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset:

$\exists F \in \map {\operatorname {Fin} } S: x = \bigcap \set {I \in \map {\operatorname {Ids} } L: F \subseteq I} \land F \subseteq x$

where $\map {\operatorname {Fin} } S$ denotes the set of all finite subsets of $S$.

We will prove that

$\exists y \in x: y$ is upper bound for $x$.

Define $y = \sup_L F$.

By Directed in Join Semilattice with Finite Suprema:

$F \ne \O \implies y \in x$

By Supremum of Empty Set is Smallest Element:

$F = \O \implies y = \bot_L$

where $\bot_L$ is the smallest element in $L$.

By Bottom in Ideal:

$F = \O \implies y \in x$

Thus $y \in x$

Let $z \in x$.

We will prove that

$F \subseteq y^\preceq$

Let $u \in F$.

By definitions of supremum and upper bound:

$u \preceq y$

Thus by definition of lower closure of element:

$u \in y^\preceq$

$\Box$

By Lower Closure of Element is Ideal:

$y^\preceq$ is ideal in $L$.

Then

$y^\preceq \in \set {I \in \map {\operatorname {Ids} } L: F \subseteq I}$

By definition of intersection:

$z \in y^\preceq$

Thus by definition of lower closure of element:

$z \preceq y$

$\Box$

Hence $x$ is principal ideal.

$\Box$


Necessary Condition

Assume that $x$ is principal ideal.

By definition of principal ideal:

$\exists y \in x: y$ is upper bound for $x$.

We will prove that

$\exists F \in \map {\operatorname {Fin} } S: F \subseteq x \land x = \bigcap \set {I \in \map {\operatorname {Ids} } L: F \subseteq I}$

Define $F = \set y$

By Singleton is Finite:

$F$ is finite.

Thus by definition of $\operatorname {Fin}$:

$F \in \map {\operatorname {Fin} } S$

Thus by definitions of subset and singleton:

$F \subseteq x$

We will prove that

$\forall z: z \in x \iff z \in \bigcap \set {I \in \map {\operatorname {Ids} } L: F \subseteq I}$

Thus by definition of set equality:

$x = \bigcap \set {I \in \map {\operatorname {Ids} } L: F \subseteq I}$

$\Box$

Thus by Compact Element iff Existence of Finite Subset that Element equals Intersection and Includes Subset:

$x$ is compact element.

$\blacksquare$


Sources

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