# Equivalence of Definitions of Closure Operator

## Theorem

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

### Definition 1

Let $\left({S, \preceq}\right)$ be an ordered set.

A closure operator on $S$ is a mapping:

$\operatorname{cl}: S \to S$

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

 $\operatorname{cl}$ is inflationary $\displaystyle x$ $\displaystyle \preceq$ $\displaystyle \operatorname{cl} \left({x}\right)$ $\operatorname{cl}$ is increasing $\displaystyle x \preceq y$ $\displaystyle \implies$ $\displaystyle \operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$ $\operatorname{cl}$ is idempotent $\displaystyle \operatorname{cl} \left({\operatorname{cl} \left({x}\right)}\right)$ $\displaystyle =$ $\displaystyle \operatorname{cl} \left({x}\right)$

### Definition 2

Let $\left({S, \preceq}\right)$ be an ordered set.

A closure operator on $S$ is a mapping:

$\operatorname{cl}: S \to S$

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

$x \preceq \operatorname{cl} \left({y}\right) \iff \operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$

## Proof

Let $\left({S, \preceq}\right)$ be an ordered set.

Let $\operatorname{cl}: S \to S$ be a mapping.

### Definition 1 implies Definition 2

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

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

$x \preceq \operatorname{cl} \left({y}\right) \iff \operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$

#### Necessary Condition

 $\displaystyle x$ $\preceq$ $\displaystyle \operatorname{cl} \left({y}\right)$ $\displaystyle \implies \ \$ $\displaystyle \operatorname{cl} \left({x}\right)$ $\preceq$ $\displaystyle \operatorname{cl}(\operatorname{cl} \left({y}\right))$ $\operatorname{cl}$ is increasing $\displaystyle \implies \ \$ $\displaystyle \operatorname{cl} \left({x}\right)$ $\preceq$ $\displaystyle \operatorname{cl} \left({y}\right)$ $\operatorname{cl}$ is idempotent

$\Box$

#### Sufficient Condition

Suppose that $\operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$.

 $\displaystyle x$ $\preceq$ $\displaystyle \operatorname{cl} \left({x}\right)$ $\operatorname{cl}$ is inflationary $\displaystyle \implies \ \$ $\displaystyle x$ $\preceq$ $\displaystyle \operatorname{cl} \left({y}\right)$ $\preceq$ is transitive by dint of being an ordering

$\Box$

### Definition 2 implies Definition 1

Suppose that:

$x \preceq \operatorname{cl} \left({y}\right) \iff \operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$

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

#### Inflationary

Let $x \in S$.

 $\displaystyle \operatorname{cl} \left({x}\right)$ $\preceq$ $\displaystyle \operatorname{cl} \left({x}\right)$ $\preceq$ is reflexive by dint of being an ordering $\displaystyle \implies \ \$ $\displaystyle x$ $\preceq$ $\displaystyle \operatorname{cl} \left({x}\right)$ by hypothesis: $\operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right) \implies x \preceq \operatorname{cl} \left({y}\right)$

That is, $\operatorname{cl}$ is inflationary.

$\Box$

#### Increasing

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

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

Then:

 $\displaystyle y$ $\preceq$ $\displaystyle \operatorname{cl} \left({y}\right)$ $\operatorname{cl}$ is inflationary $\displaystyle \implies \ \$ $\displaystyle x$ $\preceq$ $\displaystyle \operatorname{cl} \left({y}\right)$ $\preceq$ is transitive by dint of being an ordering $\displaystyle \implies \ \$ $\displaystyle \operatorname{cl} \left({x}\right)$ $\preceq$ $\displaystyle \operatorname{cl} \left({y}\right)$ by hypothesis: $x \preceq \operatorname{cl} \left({y}\right) \implies \operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$

That is, $\operatorname{cl}$ is increasing.

$\Box$

#### Idempotent

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

Let $x \in S$.

Then:

 $\displaystyle \operatorname{cl} \left({x}\right)$ $\preceq$ $\displaystyle \operatorname{cl} \left({x}\right)$ $\preceq$ is reflexive by dint of being an ordering $\displaystyle \implies \ \$ $\displaystyle \operatorname{cl} \left({\operatorname{cl} \left({x}\right)}\right)$ $\preceq$ $\displaystyle \operatorname{cl} \left({x}\right)$ by hypothesis: $x \preceq \operatorname{cl} \left({y}\right) \implies \operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({y}\right)$

Then as $\operatorname{cl}$ is inflationary:

$\operatorname{cl} \left({x}\right) \preceq \operatorname{cl} \left({\operatorname{cl} \left({x}\right)}\right)$

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

$\operatorname{cl} \left({\operatorname{cl} \left({x}\right)}\right) = \operatorname{cl} \left({x}\right)$

That is, $\operatorname{cl}$ is idempotent.

$\Box$

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

$\blacksquare$