**Conjecture** Let $latex a$ and $latex b$ are filters on a set $latex U$. Then

$latex a\cap b = \{U\} \Rightarrow \\ \exists A,B\in\mathcal{P}U: (\forall X\in a: A\subseteq X \wedge \forall Y\in b: B\subseteq Y \wedge A \cup B = U).$ [corrected]

This conjecture can be equivalently reformulated in terms of filter objects:

**Conjecture** Let $latex \mathcal{A}$ and $latex \mathcal{B}$ are filter objects on a set $latex U$. Then

$latex \mathcal{A}\cup^{\mathfrak{F}}\mathcal{B}=U \Rightarrow \exists A,B\in\mathcal{P}U: (A\subseteq\mathcal{A} \wedge B\subseteq\mathcal{B} \wedge A\cup B=U).$

(where $latex \mathfrak{F}$ is the set of filter objects on $latex U$).

It is simple to show (by applying the above conjecture twice) that it is equivalent to a yet simpler conjecture

**Conjecture** Let $latex \mathcal{A}$ and $latex \mathcal{B}$ are filter objects on a set $latex U$. Then

$latex \mathcal{A}\cup^{\mathfrak{F}}\mathcal{B}=U \Rightarrow \exists A\in\mathcal{P}U: (A\subseteq\mathcal{A} \wedge A\cup^{\mathfrak{F}}\mathcal{B}=U).$

Put in yet simpler words, this conjecture can be formulated: the filtrator of filters on a set is with co-separable center.

Looking innocent? Indeed I currently don’t know how to attack this looking simple problem. Maybe you will help?

I solved this problem (positively). I will publish the proof on the Web soon. The solution is concise (simple) and more or less beautiful.

How to exclude the problem from the list of polymath proposals? (at http://en.wordpress.com/tag/polymath-proposals/) Will it work if I’ll delete the “polymath proposals” tag?