# 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.

$\blacksquare$