Closure is Closed

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

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

Let $x \in S$.

Then $\operatorname{cl} \left({x}\right)$ is a closed element of $S$ with respect to $\operatorname{cl}$.

Power Set
When the ordering in question is the subset relation on a power set, the result can be expressed as follows:

Proof
By the definition of closure operator, $\operatorname{cl}$ is idempotent.

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

It follows by definition that $\operatorname{cl} \left({x}\right)$ is a closed element.

Also see

 * Topological Closure is Closed