Closed Element of Composite Closure Operator

Theorem
Let $(S, \preceq)$ be an ordered set.

Let $f, g: S \to S$ be closure operators.

Let $h = f \circ g$, where $\circ$ represents composition.

Suppose that $h$ is also a closure operator.

Then an element $x \in S$ is closed with respect to $h$ iff it is closed with respect to $f$ and with respect to $g$.

Reverse Implication
Let $x$ be closed with respect to $f_i$ for each $i \in \N_n$.

Then $x$ is a fixed point of each $f_i$.

Thus by Fixed Point of Mappings is Fixed Point of Composition, $x$ is a fixed point of $g$.

Forward Implication
Let $x$ be closed with respect to $h$.

Then $x$ is a fixed point of $f \circ g$.

Since $f$ and $g$ are closure operators, they are inflationary.

Thus $x$ is a fixed point of $f$ and a fixed point of $g$ by Fixed Point of Composition of Inflationary Mappings.

Thus $x$ is closed with respect to $f$ and with respect to $g$.