Equality of two expressions describing a filter Let $U$, $W$ be boolean lattices with order $\sqsupseteq$, and $U \supseteq W$. The top element of $U$ is the same as the top element of $W$. The bottom element of $U$ is the same as the bottom element of $W$.
Let $f$ be a filter on $W$.
Are the following two expressions equal?


*

*$\bigcup_{X \in f} \left\{ Y \in U \mid Y
  \sqsupseteq X \right\}$;

*$\left\{ L \in U \mid \forall X \in W : (X
  \sqsupseteq L \Rightarrow X \in f) \right\}$.
Are they both filters on $U$?
 A: Here's a counterexample, using the corrected version of formula 2. Let $U$ be the Boolean algebra of subsets of a 2-element set, say $\{0,1\}$.  In this 4-element Boolean algebra, let $W$ be the 2-element subalgebra containing just $\varnothing$ and $\{0,1\}$.  Let $f$ be the unique proper filter in $W$; it consists of just $\{0,1\}$.  Then your expression 1 gives just $f$ itself.  But your expression 2 gives the family $\{\{0\},\{1\},\{0,1\}\}$.  For example, it contains $\{0\}$ because the only element of $W$ above $\{0\}$ is $\{0,1\}$, which is in $f$.  
So expressions 1 and 2 are not in general equal.  Expression 1 always gives a filter in $U$, but, as the example shows, expression 2 need not give a filter.
A: $A=\{L\in U:(\exists X\in f)(X\sqsubseteq L)\}$
$B=\{L\in U:(\forall X\in U)(X\sqsupseteq  L\implies X\in f)\}$
I believe these two sets are equivalent, using the fact that filters are closed upwards.
Suppose $L\in B$. Then $L\in f$. So $L\in A$.
Suppose $L\in A$.  Suppose $X\in U$ with $X\sqsupseteq L$. There exists $X'\in f$ with $X'\sqsubseteq L$. So $L\in f$. So $X\in f$ thus $L\in B$
Now I will try to show they are filters...
