Another way to express certain filter Let $\mathfrak{A}$, $\mathfrak{B}$ be posets, $\mathfrak{A} \subseteq
\mathfrak{B}$ (and $\mathfrak{A}$ is the induced order of $\mathfrak{B}$). Let $\mathcal{A}$ be a filter on $\mathfrak{A}$.
Help me to prove (or disprove):
$$ \left\{ L \in \mathfrak{B} \mid \forall Y \in
   \mathfrak{A}: (Y \geq L \Rightarrow Y \in \mathcal{A}) \right\} =
   [\mathcal{A}]_{\mathfrak{B}} $$
where $[\mathcal{A}]_{\mathfrak{B}}$ is the filter on $\mathfrak{B}$ generated
by the set $\mathcal{A}$.
If it does not hold for arbitrary posets $\mathfrak{A}$, $\mathfrak{B}$, does
it holds for $\mathfrak{A}$, $\mathfrak{B}$ being booleans lattices of subsets
of some sets?
By the way, is $$ \left\{ L \in \mathfrak{B} \mid \forall Y \in
   \mathfrak{A}: (Y \geq L \Rightarrow Y \in \mathcal{A}) \right\} $$
necessarily a filter on $\mathfrak{B}$?
I am also interested in the following special case:


*

*$\mathfrak{B}$ is the boolean lattice $\mathscr{P}(U\times U)$ of all subsets of a set $U\times U$;

*$\mathfrak{A}$ is the boolean lattice $\Gamma(U;U)$ of all finite unions of cartesian products of two subsets of the set $U$.



My main question is answered.
However, it is yet an open question whether $\left\{ L \in \mathfrak{B} \mid \forall Y \in \mathfrak{A}: (Y \geq L \Rightarrow Y \in
\mathcal{A}) \right\}$ is always a filter.
 A: Counterexample
  $\mathfrak{B}= \mathscr{P} (U \times U)$, $\mathfrak{A}= \Gamma (U ; U)$,
  $\mathcal{A} = \left\{ X \in \Gamma (U ; U) \mid X
  \supseteq {(\mathord{=})} |_U \right\}$ for $U=\mathbb{R}$. (It is easy to generalize for every infinite set $U$.)
Proof Let $L \in \mathscr{P} (U \times U)$.
Let's prove
$$
  \forall Y \in \mathfrak{A}: (Y \geq L \Rightarrow Y \in \mathcal{A})
  \Leftrightarrow L \supseteq {(\mathord{=})} |_U
$$
  that is
$$ \forall Y \in \Gamma (U ; U) : \left( Y \geq L \Rightarrow Y \supseteq
     {(\mathord{=})} |_U \right) \Leftrightarrow L \supseteq {(\mathord{=})}
     |_U .
$$
$L \supseteq {(\mathord{=})} |_U \Rightarrow \forall Y \in \Gamma (U ; U) :
  \left( Y \geq L \Rightarrow Y \supseteq {(\mathord{=})} |_U \right)$ is
obvious.
Let now $L \nsupseteq {(\mathord{=})} |_U$ that is $(x ; x) \not\ni L$. Then
  there exists $Y \in \Gamma (U ; U)$ such that $Y \geq L$ but $Y \nsupseteq
  {(\mathord{=})} |_U$ (take $Y = (\mathbb{R} \times \mathbb{R}) \setminus
  \{ (x ; x) \}$).
Thus we have proved the above equivalence.
Let $L_0 = \bigcup_{i \in \mathbb{Z}} ([i ; i + 1 [\times [i ; i + 1 [)$.
$L_0 \supseteq {(\mathord{=})} |_U$ but $L_0 \not\in
  [\mathcal{A}]_{\mathfrak{B}}$ because otherwise there would be $P \in \Gamma
  (U ; U)$ such that both $P \supseteq {(\mathord{=})} |_U$ and $P \subseteq
  L_0$. This is clearly impossible because $P$ must have a point in every $[i
  ; i + 1 [\times [i ; i + 1 [$ for $i \in \mathbb{Z}$ and thus isn't
  representable as a finite union of cartesian products.
Thus $\forall Y \in \mathfrak{A}: (Y \geq L_0 \Rightarrow Y \in \mathcal{A})
  \nLeftrightarrow L_0 \in [\mathcal{A}]_{\mathfrak{B}}$.
It is yet an open question whether $\left\{ L \in \mathfrak{B} \mid \forall Y \in \mathfrak{A}: (Y \geq L \Rightarrow Y \in
\mathcal{A}) \right\}$ is always a filter.
