Closure Operator does not Change Infimum of Subset of Image

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

Let $c: S \to S$ be a closure operator on $L$.

Let $X$ be a subset of $c\left[{S}\right]$ such that
 * $X$ admits an infimum,

where $c\left[{S}\right]$ denotes the image of $c$.

Then $\inf X = c\left({\inf X}\right)$

Proof
We will prove that
 * $c\left({\inf X}\right)$ is lower bound for $X$.

Let $x \in X$.

By definition of subset:
 * $x \in c\left[{S}\right]$

By definition of image of mapping:
 * $\exists y \in S: x = c\left({y}\right)$

By definition of closure operator/idempotent:
 * $x = c\left({x}\right)$

By definition of infimum:
 * $\inf X$ is lower bound for $X$.

By definition of lower bound:
 * $\inf X \preceq x$

Thus by definition of closure operator/increasing:
 * $c\left({\inf X}\right) \preceq x$

By definition of infimum:
 * $c\left({\inf X}\right) \preceq \inf X$

By definition of closure operator/inflationary:
 * $c\left({\inf X}\right) \succeq \inf X$

Thus by definition of antisymmetry:
 * $c\left({\inf X}\right) = \inf X$