Intersection of Complete Meet Subsemilattices invokes Closure Operator

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Theorem

Let $\struct {S, \preccurlyeq}$ be an ordered set.

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

Let $C_i = \map {f_i} S$ 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 $S$ such that $\inf D \in C_i$.


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


Proof

Lemma

Let $\struct {S, \preccurlyeq}$ 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, $\inf D \in C$.

$\Box$


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

Let $x \in S$.

Then $C \cap x^\succcurlyeq$ has an infimum in $S$ which lies in $C$, where $x^\succcurlyeq$ is the upper closure of $x$.

By the definition of infimum:

$x \preceq \inf \struct {C \cap x^\succcurlyeq}$

so this infimum is in fact the smallest element of $C \cap x^\succcurlyeq$.

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

$\blacksquare$