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 Mar26 comment An elementary proof about filters @AndreasBlass: You are right, the first $\bigcap$ on the right side of the equation was intended to be $\bigcup$. I've corrected the question. Dec14 comment About a function ranging filters Oh, I found an easy solution (but only for people who has read my book): $\operatorname{Back}(f;k)$ is a complete funcoid; from this the conjecture follows. Dec5 comment Why “thin groupoids” are not ubiquitous? Oh, I see: Some people use the word "setoid" differently. So my downvote was a mistake. I can't upvote it back now, sorry. Dec5 comment Why “thin groupoids” are not ubiquitous? This your definition of setoids is not equivalent (however, it is equivalent up to equivalence of categories) to the standard definition of setoids, that is a set with an equivalence relation. Equivalence up to equivalence of categories is not enough for my purposes. I still think that I downvoted correctly, Dec5 comment Why “thin groupoids” are not ubiquitous? Setoid is a set with an equivalence relation on it. (And I know it long ago before I've read your answer.) I understand this. I downvote because switching from a groupoid to a setoid leads to information loss, and this makes it not an answer to my question. Dec5 comment Why “thin groupoids” are not ubiquitous? I need to describe a category. A category contains not only objects but also morphisms. As my category happens to be thin, there is a (not necessarily entire defined) function from pairs of objects into a morphism. Having a setoid we cannot define such a function. But I need this function. I need to be able to get the morphisms whenever a pair of objects is specified. Having only a setoid, I cannot do this. Dec5 comment Why “thin groupoids” are not ubiquitous? But having two elements of setoids, I cannot restore particular morphism (such as $f\mapsto f\cap\Gamma$). It is not what I need Dec5 comment Why “thin groupoids” are not ubiquitous? I yet don't understand you and don't see how to express this with setoids. When we switch from a thin groupoid to a setoid, the information about particular morphisms is lost (they are just replaced with a pair of objects), but the whole thing I need is information about what are particular morphisms, depicted as arrows in my diagrams. Dec5 comment Why “thin groupoids” are not ubiquitous? I don't get how this is related with setoids. How to express my diagrams using setoids? I need particular isomorphisms not just the fact that two objects are isomorphic. Dec2 comment Another way to express certain filter See also counter-examples in this thread: groups.google.com/forum/#!topic/sci.math/Plru0S8ePzs Dec1 comment Another way to express certain filter The other direction is surprisingly difficult Dec1 comment Another way to express certain filter In one direction: Let $X \in [\mathcal{A}]_{\mathfrak{B}}$. Then $\exists Y' \in \mathcal{A} : X \geq Y'$. Thus if $Y \geq X$ for $Y \in \mathfrak{A}$ then $Y \geq Y'$ and thus $Y \in \mathcal{A}$. So $\forall Y \in \mathfrak{A}: (Y \geq X \Rightarrow Y \in \mathcal{A})$. Sep24 comment Equality of two expressions describing a filter @AndreasBlass: Thanks, it was my error. Now have been corrected Sep24 comment Equality of two expressions describing a filter @StevenStadnicki: It seems that $T$ does not witness that $V$ can't be in (2). Sep24 comment Equality of two expressions describing a filter @TomCruise: No, I have edited the question, and now 2. means the filter(?) on the boolean lattice $U$ consisting of all elements $L\in U$ such that every $X$ majorating $L$ is an element of the filter $f$ Sep24 comment Equality of two expressions describing a filter @TomCruise: Yes, I was wrong. I will edit the question. Sep24 comment Equality of two expressions describing a filter @TomCruise: I don't understand your question. By $Y\in U$ I mean that $Y$ is an elemetn of the set $U$. Sep24 comment Compositions of filters on finite unions of Cartesian products 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 Sep14 comment Another conjecture about filters and cartesian products In mathematics21.org/binaries/funcoids-are-filters.pdf I have shown that $\Gamma$ in both my questions are the same. So the answer to this question is: yes, it can be proved Sep14 comment A conjecture about filters and finite unions of cartesian products I provided a counter-example as an answer. That counter-example was wrong.