# Characterization of Prime Ideal by Finite Infima

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## Theorem

Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.

Let $I$ be an ideal in $L$.

Then

- $I$ is a prime ideal

## Proof

### Sufficient Condition

Let $I$ be a prime ideal.

Define $\map P X: \equiv X \ne \O \land \inf X \in I \implies \exists x \in X: x \in I$

where $X$ is subset of $S$.

Let $A$ be a non-empty finite subset of $S$.

By definition of empty set:

- $\map P \O$

We will prove that

- $\forall x \in A, B \subseteq A: \map P B \implies \map P {B \cup \set x}$

Let $x \in A, B \subseteq A$ such that

- $\map P B$ (Induction Hypothesis)

Assume that

- $B \cup \set x \ne \O$ and $\map \inf {B \cup \set x} \in I$

Case $B = \O$:

- $B \cup \set x = \set x$

- $\inf \set x = x$

By definition of singleton:

- $x \in \set x$

Thus

- $\exists a \in B \cup \set x: a \in I$

$\Box$

Case $B \ne \O$:

By Subset of Finite Set is Finite:

- $B$ is finite.

By Existence of Non-Empty Finite Infima in Meet Semilattice:

- $B$ admits an infimum.

- $\set x$ admits an infimum.

\(\ds \map \inf {B \cup \set x}\) | \(=\) | \(\ds \map \inf {\bigcup \set {B, \set x} }\) | Definition of Set Union | |||||||||||

\(\ds \) | \(=\) | \(\ds \inf \set {\inf B, x}\) | Infimum of Infima | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {\inf B} \wedge x\) | Definition of Meet |

By Characterization of Prime Ideal:

- $\inf B \in I$ or $x \in I$

Case $\inf B \in I$:

By Induction Hypothesis:

- $\exists a \in B: a \in I$

By definition of union:

- $a \in B \cup \set x$

Thus

- $\exists a \in B \cup \set x: a \in I$

$\Box$

Case $x \in I$:

By definition of union:

- $x \in B \cup \set x$

Thus

- $\exists a \in B \cup \set x: a \in I$

$\Box$

- $\map P A$

Thus the result.

$\Box$

### Necessary Condition

Suppose that

Let $x, y \in S$ such that

- $x \wedge y \in I$

- $\set {x, y}$ 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 meet:

- $\inf \set {x, y} = x \wedge y$

By assumption:

- $\exists a \in \set {x, y}: a \in I$

Thus by definition of unordered tuple:

- $x \in I$ or $y \in I$

Hence by Characterization of Prime Ideal:

- $I$ is prime ideal.

$\blacksquare$

## 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_7:12