Image of Closure Operator Inherits Infima

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

Let $f$ be a closure operator on $L$.

Then $R = \left({f\left[{S}\right], \precsim}\right)$ inherits infima,

where
 * $\mathord\precsim = \mathord\preceq \cap \left({f\left[{S}\right] \times f\left[{S}\right]}\right)$
 * $f\left[{S}\right]$ denotes the image of $f$.

Proof
Let $X$ be subset of $f\left[{S}\right]$ such that
 * $X$ admits an infimum in $L$.

By Closure Operator does not Change Infimum of Subset of Image:
 * $f\left({\inf_L X}\right) = \inf_L X$

By definition of image of mapping:
 * $\inf_L X \in f\left[{S}\right]$

Thus by Infimum in Ordered Subset:
 * $X$ admits an infimum in $R$ and $\inf_R X = \inf_L X$