I recently proved the following conjecture (now a theorem):

Theorem $A\cap^{\mathfrak{F}}\bigcup{}^{\mathfrak{F}}S = \bigcup{}^{\mathfrak{F}} \{ A\cap^{\mathfrak{F}} X | X\in S \}$ for every set $A\in\mathscr{P}\mho$ and every $S\in\mathscr{P}\mathfrak{F}$ where $\mathfrak{F}$ is the set of filter objects on some set $\mho$.

This theorem is a direct consequence of the following lemma:

Lemma $A\cap^{\mathfrak{F}}$ is a lower adjoint of $(\mho\setminus A)\cup^{\mathfrak{F}}$ for every set $A\in\mathscr{P}\mho$.

Proof That $A\cap^{\mathfrak{F}}$ and $(\mho\setminus A)\cup^{\mathfrak{F}}$ are monotone is obvious.

We need to prove (for every $x,y\in\mathfrak{F}$) that

$x \subseteq (\mho\setminus A)\cup^{\mathfrak{F}}(A\cap^{\mathfrak{F}}x)$ and $A\cap^{\mathfrak{F}}((\mho\setminus A)\cup^{\mathfrak{F}}y) \subseteq y$.

Really, $(\mho \setminus A) \cup^{\mathfrak{F}} (A \cap^{\mathfrak{F}} x) =\\ ((\mho \setminus A) \cup^{\mathfrak{F}} A) \cap^{\mathfrak{F}} ((\mho\setminus A) \cup^{\mathfrak{F}} x) =\\ \mho \cap^{\mathfrak{F}} ((\mho\setminus A) \cup^{\mathfrak{F}} x) =\\ (\mho \setminus A) \cup^{\mathfrak{F}} x\supseteq x$
and
$A \cap^{\mathfrak{F}} ((\mho\setminus A)\cup^{\mathfrak{F}} y) =\\ (A \cap^{\mathfrak{F}} (\mho \setminus A))\cup^{\mathfrak{F}} (A \cap^{\mathfrak{F}} y) =\\ \emptyset \cup^{\mathfrak{F}}(A \cap^{\mathfrak{F}} y) =\\ A \cap^{\mathfrak{F}} y \subseteq y.$