Product $a\times b$ of filters $a$ and $b$ is defined as the filter (on the set of binary relations) defined by the base $\{ A\times B | A\in a,B\in b \}$.

I will denote the principal filter corresponding to a set $X$ as $\uparrow X$.

Let $a$ and $b$ are filters.

Suppose for some filter $M$ on binary relations we have $M\subseteq \uparrow A \times b$ and $M\subseteq a\times \uparrow B$ for some $A\in a$, $B\in b$.

Does it follow that there exist $X\in a$, $Y\in b$ such that $M\subseteq \uparrow X\times \uparrow Y$?

The above seems to be equivalent to the following conjecture, but formulated in elementary terms:

Conjecture $\left( \mathcal{A} \ltimes \mathcal{B} \right) \cup \left( \mathcal{A} \rtimes \mathcal{B} \right) = \mathcal{A} \times^{\mathsf{RLD}} \mathcal{B}$ for every filter objects $\mathcal{A}$, $\mathcal{B}$. (In words: join of oblique products is the reloidal product.)

See http://www.mathematics21.org/algebraic-general-topology.html for my research.


I'm not entirely sure I understand the notation, because it seems $a\times b$ would be a filter on the set of ordered pairs, not on the set of binary relations (though it would be a filter in the Boolean algebra of binary relations). But, assuming my best guess for what was intended, here's a counterexample. Let both $a$ and $b$ be the filter $\mathcal F$ of cofinite subsets of $\mathbb N$. Let $M$ be the filter on $\mathbb N\times\mathbb N$ generated by all sets of the form $(P\times\mathbb N)\cup(\mathbb N\times Q)$ with $P,Q\in\mathcal F$. Then $M$ is included in both $(\uparrow\mathbb N)\times b$ and $a\times(\uparrow\mathbb N)$. I claim that $M$ is not included in $(\uparrow X)\times(\uparrow Y)$ for any cofinite $X,Y\subseteq\mathbb N$. To see this, consider any such $X$ and $Y$, and let $P\subsetneq X$ and $Q\subsetneq Y$ be slightly smaller but still cofinite sets (for example, remove one element from $X$ and from $Y$). Then $(P\times\mathbb N)\cup(\mathbb N\times Q)$ is in $M$ but not in $(\uparrow X)\times(\uparrow Y)$; i.e., it is not a superset of $X\times Y$. Indeed, if $x\in X-P$ and $y\in Y-Q$ then $(x,y)$ is in $X\times Y$ but not in $(P\times\mathbb N)\cup(\mathbb N\times Q)$. (At the moment, Preview is showing only a part of my answer. I'll post it anyway and hope for the best.)

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    $\begingroup$ I don't know why the question got so many down-votes without being closed, but perhaps that's because I don't know anything about filters. Anyway, this may be the first case I've ever seen where the Reversal badge could be awarded, so here's my +1 towards that goal. $\endgroup$ – David White Aug 29 '11 at 18:37

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