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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.

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  • $\begingroup$ 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. $\endgroup$
    – porton
    Aug 23, 2014 at 3:38
  • $\begingroup$ No, not that simple. I keep thinking $\endgroup$
    – porton
    Aug 23, 2014 at 3:47
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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Aug 25, 2014 at 5:33
  • $\begingroup$ I provided a counter-example as an answer. That counter-example was wrong. $\endgroup$
    – porton
    Sep 14, 2014 at 13:55

1 Answer 1

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I proved it with theory of funcoids in http://www.mathematics21.org/binaries/funcoids-are-filters.pdf

I cannot present the proof here, because theory of funcoids is not in common knowledge.

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