# Equivalence of Definitions of Closure Operator

## Theorem

The following definitions of the concept of Closure Operator are equivalent:

### Definition 1

Let $\struct {S, \preceq}$ be an ordered set.

A closure operator on $S$ is a mapping:

$\cl: S \to S$

which satisfies the closure axioms as follows for all elements $x, y \in S$:

 $(\text {cl} 1)$ $:$ $\cl$ is inflationary: $\ds x$ $\ds \preceq$ $\ds \map \cl x$ $(\text {cl} 2)$ $:$ $\cl$ is increasing: $\ds x \preceq y$ $\ds \implies$ $\ds \map \cl x \preceq \map \cl y$ $(\text {cl} 3)$ $:$ $\cl$ is idempotent: $\ds \map \cl {\map \cl x}$ $\ds =$ $\ds \map \cl x$

### Definition 2

Let $\struct {S, \preceq}$ be an ordered set.

A closure operator on $S$ is a mapping:

$\cl: S \to S$

which satisfies the following condition for all elements $x, y \in S$:

$x \preceq \map \cl y \iff \map \cl x \preceq \map \cl y$

## Proof

Let $\struct {S, \preceq}$ be an ordered set.

Let $\cl: S \to S$ be a mapping.

### Definition 1 implies Definition 2

Let $\cl$ be an inflationary, increasing and idempotent mapping.

It is necessary to show that for all $x, y \in S$:

$x \preceq \map \cl y \iff \map \cl x \preceq \map \cl y$

#### Necessary Condition

 $\ds x$ $\preceq$ $\ds \map \cl y$ $\ds \leadsto \ \$ $\ds \map \cl x$ $\preceq$ $\ds \map \cl {\map \cl y}$ $\cl$ is increasing $\ds \leadsto \ \$ $\ds \map \cl x$ $\preceq$ $\ds \map \cl y$ $\cl$ is idempotent

$\Box$

#### Sufficient Condition

Suppose that $\map \cl x \preceq \map \cl y$.

 $\ds x$ $\preceq$ $\ds \map \cl x$ $\cl$ is inflationary $\ds \leadsto \ \$ $\ds x$ $\preceq$ $\ds \map \cl y$ $\preceq$ is transitive by dint of being an ordering

$\Box$

### Definition 2 implies Definition 1

Suppose that:

$x \preceq \map \cl y \iff \map \cl x \preceq \map \cl y$

It is necessary to show that $\cl$ is inflationary, increasing and idempotent.

#### Inflationary

Let $x \in S$.

 $\ds \map \cl x$ $\preceq$ $\ds \map \cl x$ $\preceq$ is reflexive by dint of being an ordering $\ds \leadsto \ \$ $\ds x$ $\preceq$ $\ds \map \cl x$ by hypothesis: $\map \cl x \preceq \map \cl y \implies x \preceq \map \cl y$

That is, $\cl$ is inflationary.

$\Box$

#### Increasing

It has been demonstrated that $\preceq$ is inflationary.

Let $x, y \in S$ such that $x \preceq y$.

Then:

 $\ds y$ $\preceq$ $\ds \map \cl y$ $\cl$ is inflationary $\ds \leadsto \ \$ $\ds x$ $\preceq$ $\ds \map \cl y$ $\preceq$ is transitive by dint of being an ordering $\ds \leadsto \ \$ $\ds \map \cl x$ $\preceq$ $\ds \map \cl y$ by hypothesis: $x \preceq \map \cl y \implies \map \cl x \preceq \map \cl y$

That is, $\cl$ is increasing.

$\Box$

#### Idempotent

It has been demonstrated that $\preceq$ is inflationary.

Let $x \in S$.

Then:

 $\ds \map \cl x$ $\preceq$ $\ds \map \cl x$ $\preceq$ is reflexive by dint of being an ordering $\ds \leadsto \ \$ $\ds \map \cl {\map \cl x}$ $\preceq$ $\ds \map \cl x$ by hypothesis: $x \preceq \map \cl y \implies \map \cl x \preceq \map \cl y$

Then as $\cl$ is inflationary:

$\map \cl x \preceq \map \cl {\map \cl x}$

As $\preceq$ is antisymmetric by dint of being an ordering:

$\map \cl {\map \cl x} = \map \cl x$

That is, $\cl$ is idempotent.

$\Box$

Thus $\cl$ has been shown to be inflationary, increasing and idempotent as required.

$\blacksquare$