Image of Closure Operator Inherits Infima

Theorem
Let $L = \struct {S, \preceq}$ be an ordered set.

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

Then $R = \struct {f \sqbrk S, \precsim}$ inherits infima,

where
 * $\mathord \precsim = \mathord \preceq \cap \paren {f \sqbrk S \times f \sqbrk S}$
 * $f \sqbrk S$ denotes the image of $f$.

Proof
Let $X$ be subset of $f \sqbrk S$ such that
 * $X$ admits an infimum in $L$.

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

By definition of image of mapping:
 * $\inf_L X \in f \sqbrk S$

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