Intersection of Complete Meet Subsemilattices induces Closure Operator

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

Let $f_i$ be a closure operator on $S$ for each $i \in I$.

Let $C_i = f_i(S)$ be the set of closed elements with respect to $f_i$ for each $i \in I$.

Suppose that for each $i \in I$, $C_i$ is a complete meet subsemilattice of $S$ in the following sense:
 * For each $D \subseteq C_i$, $D$ has an infimum in $S$ and $\inf D \in C_i$.

Then $C = \displaystyle \bigcap_{i \mathop \in I} C_i$ induces a closure operator on $S$.

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

Let $C_i$ be a complete meet subsemilattice of $S$.

Then $C = \displaystyle \bigcap_{i \mathop \in I} C_i$ is also a complete meet subsemilattice.

Proof
Let $D \subseteq C$.

By Intersection is Largest Subset, $D \subseteq C_i$ for each $i \in I$.

Thus $D$ has an infimum in $S$ and $\inf D \in C_i$ for each $i \in I$.

By the definition of intersection, $\inf D \in C$.

By the lemma, $C$ is a complete meet semilattice.

Let $x \in S$.

Then $C \cap {\bar\uparrow} x$ has an infimum in $S$ which lies in $C$, where ${\bar\uparrow} x$ is the upper closure of $x$.

By the definition of infimum, $x \preceq \inf (C \cap {\bar\uparrow} x)$ so this infimum is in fact the smallest element of $C \cap {\bar\uparrow} x$.

Thus $C$ induces a closure operator on $S$ by Closure Operator from Closed Elements.