Equality of two expressions describing a filter

Question

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?

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

2. $\left\{ L \in U \mid \forall X \in W : (X \sqsupseteq L \Rightarrow X \in f) \right\}$.

Are they both filters on $U$?

Answer 1

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.

Answer 2

$A=\{L\in U:(\exists X\in f)(X\sqsubseteq L)\}$

$B=\{L\in U:(\forall X\in U)(X\sqsupseteq L\implies X\in f)\}$

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...