Characterization of Prime Filter by Finite Suprema
Theorem
Let $L = \struct {S, \vee, \preceq}$ be a join semilattice.
Let $F$ be a filter in $L$.
Then
- $F$ is a prime filter
Proof
Sufficient Condition
Let $F$ be a prime ideal.
Define $\map \PP X: \equiv X \ne \O \land \sup X \in F \implies \exists x \in X: x \in F$
where $X$ is subset of $S$.
Let $A$ be a non-empty finite subset of $S$.
By definition of empty set:
- $\map \PP \O$
We will prove that:
- $\forall x \in A, B \subseteq A: \map \PP B \implies \map \PP {B \cup \set x}$
Let $x \in A, B \subseteq A$ such that:
- $\map \PP B$ (Induction Hypothesis)
Assume that:
- $B \cup \set x \ne \O$ and $\map \sup {B \cup \set x} \in F$
Case $B = \O$:
- $B \cup \set x = \set x$
- $\sup \set x = x$
By definition of singleton:
- $x \in \set x$
Thus
- $\exists a \in B \cup \set x: a \in F$
$\Box$
Case $B \ne \O$:
By Subset of Finite Set is Finite:
- $B$ is finite.
By Existence of Non-Empty Finite Suprema in Join Semilattice:
- $B$ admits a supremum.
- $\set x$ admits a supremum.
\(\ds \map \sup {B \cup \set x}\) | \(=\) | \(\ds \map \sup {\bigcup \set {B, \set x} }\) | Definition of Set Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \sup \set {\sup B, x}\) | Supremum of Suprema: | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\sup B} \vee x\) | Definition of Join (Order Theory) |
By definition of prime filter:
- $\sup B \in F$ or $x \in F$
Case $\sup B \in F$:
By Induction Hypothesis:
- $\exists a \in B: a \in F$
By definition of union:
- $a \in B \cup \set x$
Thus:
- $\exists a \in B \cup \set x: a \in F$
$\Box$
Case $x \in F$:
By definition of union:
- $x \in B \cup \set x$
Thus:
- $\exists a \in B \cup \set x: a \in F$
$\Box$
- $\map \PP A$
Thus the result.
$\Box$
Necessary Condition
Suppose that
Let $x, y \in S$ such that
- $x \vee y \in F$
- $\set {x, y}$ is a finite set.
By definition of unordered tuple:
- $x \in \set {x, y}$
By definition of non-empty set:
- $\set {x, y}$ is a non-empty set.
By definition of join:
- $\sup \set {x, y} = x \vee y$
By assumption:
- $\exists a \in \set {x, y}: a \in F$
Thus by definition of unordered tuple:
- $x \in F$ or $y \in F$
Hence $I$ is prime filter.
$\blacksquare$
Sources
- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices
- Mizar article WAYBEL_7:14