Characterization of Compact Element in Frame or Locale
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Theorem
Let $L = \struct{S, \preceq}$ be a frame or locale.
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 $: a \preceq \sup F$
- $(4)\quad \forall A \subseteq S : a = \sup A \implies \exists F \subseteq A : F$ is finite $: a = \sup F$
Proof
Recall, a frame or locale is a complete lattice satisfying the infinite join distributive law:
\(\ds \forall a \in L, S \subseteq L:\) | \(\ds a \wedge \bigvee S = \bigvee \set {a \wedge s : S \in S} \) |
where $\bigvee S$ denotes the supremum $\sup S$.
From Characterization of Compact Element in Complete Lattice:
- Statements $(1)$, $(2)$ and $(3)$ are equivalent.
Statement $(3)$ implies Statement $(4)$
Let $a$ satisfies:
- $\forall A \subseteq S : a \preceq \sup A \implies \exists F \subseteq A : F$ is finite $: a \preceq \sup F$
Let $A \subseteq S : a = \sup A$.
We have by hypothesis:
- $\exists F \subseteq A : F$ is finite $: a \preceq \sup F$
From Supremum of Subset:
- $\sup F \preceq \sup A = a$
By Ordering Axiom $(2)$: Transitivity:
- $a = \sup F$
The result follows.
$\Box$
Statement $(4)$ implies Statement $(3)$
Let $a$ satisfies:
- $\forall A \subseteq S : a = \sup A \implies \exists F \subseteq A : F$ is finite $: a = \sup F$
Let $A \subseteq S : a \preceq \sup A$.
From Predecessor is Infimum:
- $a = a \wedge \sup A$
By the infinite join distributive law:
- $a = \sup \set{a \wedge x : x \in A}$
We have by hypothesis:
- $\exists F \subseteq A : F$ is finite $: a = \sup \set{a \wedge x : x \in F}$
By the infinite join distributive law:
- $a = a \wedge \sup F$
From Predecessor is Infimum:
- $a \preceq \sup F$
The result follows.
$\blacksquare$
Sources
- 1982: Peter T. Johnstone: Stone Spaces: Chapter $\text {II}$: Introduction to Locales, $\S3.1$