Closed Element of Composite Closure Operator

Theorem
Let $\left({S, \preceq}\right)$ 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 $g$.

Proof
An element is closed with respect to a closure operator iff it is a fixed point of that operator.

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

Thus the result follows from Fixed Point of Composition of Inflationary Mappings.