Characterization of Compact Element in Complete Lattice

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Theorem

Let $L = \struct{S, \preceq}$ be a complete lattice.

Let $a \in S$.


The following statements are equivalent::

$(1)\quad a$ is a compact element
$(2)\quad \forall I \subseteq S : I$ is an ideal $: a \preceq \sup I \implies a \in I$
$(3)\quad \forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$


Proof

Recall, $a$ is a compact element if and only if:

for every directed subset $D$ of $S$ such that $a \preceq \sup D$
$\exists d \in D: a \preceq d$

Statement (1) implies Statement (3)

Let $a$ be a compact element.


Let $A \subseteq S : a \preceq \sup A$.


Let $D = \leftset{b \in A : \exists F \subseteq A : F}$ is finite $\rightset{: b = \sup F}$

$D$ is a Directed Subset

We show that $D$ is a directed subset.


Let $x, y \in D$.

By definition of $D$:

$\exists F, G \subseteq A: F, G$ are finite $: x = \sup F, y = \sup G$


Let $H = F \cup G$.

From Union of Finite Sets is Finite:

$H$ is finite of $A$

Hence:

$z = \sup H \in D$.


Furthermore:

$z = \sup \set{x, y}$

By definition of supremum:

$x, y \preceq z$


Since $x,y$ were arbitrary, then $D$ is a directed subset by definition.

$\Box$

Supremum of $D$ Equals Supremum of $A$

We show that $\sup D = \sup A$.


Let $x \in D$.

By definition of $D$:

$\exists F \subseteq A: F$ is finite $: x = \sup F$

From Supremum of Subset:

$\sup F \preceq \sup A$


Since $x$ was arbitrary, it follows that:

$\sup A$ is an upper bound for $D$

By definition of supremum:

$\sup D \preceq \sup A$


Note that for all $b \in A$:

$b = \sup \set b \in D$

Hence:

$\sup D$ is an upper bound for $A$

By definition of supremum:

$\sup A \preceq \sup D$


From Ordering Axiom $(3)$: Antisymmetry:

$\sup A = \sup D$

$\Box$


By definition of compact element:

$\exists d \in D : a \preceq d$

By definition of $D$:

$\exists F \subseteq A : \sup F = d$

Hence:

$\exists F \subseteq A : F$ is finite $: a \preceq \sup F$


Since $A$ was arbitrary:

$\forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$

$\Box$

Statement (3) implies Statement (2)

Let $a$ satisfy:

$\forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$


Let:

$I \subseteq S : I$ is an ideal $: a \preceq \sup I$

We have by hypothesis:

$\exists F \subseteq I : F$ is finite $: a \preceq \sup F$

By Join Semilattice Ideal Axiom $\paren{\text {JSI} 2 }$: Subsemilattice of Join Semilattice:

$\sup F \in I$

By Join Semilattice Ideal Axiom $\paren{\text {JSI} 1 }$: Lower Section of Join Semilattice:

$a \in I$


Since $I$ was arbitrary:

$\forall I \subseteq S : I$ is an ideal $: a \preceq \sup I \implies a \in I$

$\Box$

Statement (2) implies Statement (1)

Let $a$ satisfy:

$\forall I \subseteq S : I$ is an ideal $: a \preceq \sup I \implies a \in I$


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

$a \preceq \sup D$

Let $I = {b \in S : \exists d \in D : b \preceq d}$.

$I$ is an Ideal of $S$

We will show that $I$ satisfies the join semilattice ideal axioms.

$I$ is a Lower Section

Let $x \in I$.

Let $y \preceq x$.


By definition of $I$:

$\exists d \in D : x \preceq d$

By Ordering Axiom $(2)$: Transitivity:

$y \preceq d$

Hence:

$y \in I$

It follows that $I$ is a lower section of $S$.

$\Box$

$I$ is a Join Subsemilattice

Let $x, y \in I$.

By definition of $I$:

$\exists d_1, d_2 \in D : x \preceq d, y \preceq e$

By definition of a directed subset:

$\exists d \in D : d_1, d_2 \preceq d$

By Ordering Axiom $(2)$: Transitivity:

$x, y \preceq d$

By definition of join:

$x \vee y \preceq d$

Hence:

$x \vee y \in I$

$\Box$


This proves $I$ satisfies the join semilattice ideal axioms.

It follows that $I$ is an ideal by definition.

$\Box$


Supremum of $I$ Equals Supremum of $D$

We show that $\sup I = \sup D$.


Let $x \in I$.

By definition of $I$:

$\exists d \in D: x \preceq d$

By definition of supremum:

$x \preceq \sup D$


Since $x$ was arbitrary, it follows that:

$\sup D$ is an upper bound for $I$

By definition of supremum:

$\sup I \preceq \sup D$


Note that for all $d \in D$:

$d \in I$

Hence:

$\sup I$ is an upper bound for $D$

By definition of supremum:

$\sup D \preceq \sup I$


From Ordering Axiom $(3)$: Antisymmetry:

$\sup I = \sup D$

$\Box$


We have by hypothesis:

$a \in I$

By definition of $I$:

$\exists d \in D : a \preceq d$


Since $D$ was arbitrary:

for every directed subset $D$ of $S$ such that $a \preceq \sup D$
$\exists d \in D: a \preceq d$

Hence $a$ is a compact element by definition.

$\blacksquare$

Sources

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