Equivalence of Definitions of Closure Operator

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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.

$\blacksquare$