# A conjecture about filters and finite unions of cartesian products

Let $U$ be some set.

Let $\Gamma$ be the set of all finite unions of cartesian products ($X_1 \times Y_1 \cup \dots \cup X_n \times Y_n$) of sets on $U$.

Obviously, $\Gamma$ is a a distributive lattice.

Let $\mathcal{A}$ and $\mathcal{B}$ be filters on the lattice $\Gamma$. (Note that I consider improper filter to be a filter.)

I denote $\langle K \rangle a = \bigcup \left\{ Y_i \mid i \in \{1,\dots,n\}, X_i \in a \right\}$ where $K = X_1 \times Y_1 \cup \dots \cup X_n \times Y_n$ for every $K \in \Gamma$ and ultrafilter $a$.

Let $\left\{ \langle P \rangle a \mid P \in \mathcal{A} \right\} = \left\{ \langle P \rangle a \mid P \in \mathcal{B} \right\}$ for every ultrafilter $a$ on $U$.

Prove (or give a counter-example) that $\mathcal{A} = \mathcal{B}$.

Consequences of this conjecture (if true) are very important for my research. If you prove it, it is a major step forward in mathematics. Please help.

• It seems that there is a simple solution: Just restrict for the case when $a$ are trivial ultrafilters. No I am checking that this is a correct solution. – porton Aug 23 '14 at 3:38
• No, not that simple. I keep thinking – porton Aug 23 '14 at 3:47
• In the future I'd be cautious about that last paragraph. (1) Nobody should do your research for you; (2) it's quite arrogant to say that anything you do is a major step forward in mathematics, you should leave that to other people to judge and say. – Asaf Karagila Aug 25 '14 at 5:33
• I provided a counter-example as an answer. That counter-example was wrong. – porton Sep 14 '14 at 13:55