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Let $\Gamma$ be the lattice of all finite unions of Cartesian products $A\times B$ of two arbitrary sets $A,B\subseteq U$ for some set $U$. See this note for other equivalent ways to describe the set $\Gamma$: http://www.mathematics21.org/binaries/funcoids-are-filters.pdf

It is obvious that $G\circ F\in\Gamma$ for every binary relations $F,G\in\Gamma$.

Thus for every filters $f$ and $g$ on $\Gamma$ we can define $g\circ f$ as the filter on $\Gamma$ defined by the base $\{ G\circ F \mid F\in f,G\in g \}$.

I need to prove (or disprove) that if $K\in g\circ f$ then there exist $F\in f$ and $G\in g$ such that $K\supseteq G\circ F$ (for all filters $f$, $g$ on $\Gamma$).


It seems that it's enough to prove this conjecture only for the case if $f$ or $g$ is a principal filter.

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Let $f$, $g$ be filters on $Γ$.

"define $g∘f$ as the filter on $Γ$ defined by the base $\{G∘F∣F∈f,G∈g\}$."

So as claimed, $g∘f$ is the filter generated by (=smallest filter containing) the base $\{G∘F∣F∈f,G∈g\}$.

This means that

$\{G∘F∣F∈f,G∈g\}$ is a base for the filter $g\circ f$.

Which, by definition of a base for a filter, implies:

For each $K$, if $K∈g∘f$ then there exist $F∈f$ and $G∈g$ such that $K⊇G∘F$.

So there remains nothing to prove.

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    $\begingroup$ I thought (without writing a detailed proof), that this my question is equivalent to an other open problem I work about. Now I see my problem does not follow trivially from this question. Because my open problem is more hard, I thought this question is also hard and stupidly overlooked a trivial solution. It is one of my biggest mistakes. Thanks anyway and get my bounty $\endgroup$ – porton Sep 24 '14 at 14:28
  • $\begingroup$ I wonder why no-one answered in last 5 or 6 days. $\endgroup$ – user795571 Sep 24 '14 at 14:31

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