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visits member for 3 years, 11 months
seen Nov 15 at 1:57

An amateur general topology researcher.


Sep
24
comment Equality of two expressions describing a filter
@AndreasBlass: Thanks, it was my error. Now have been corrected
Sep
24
comment Equality of two expressions describing a filter
@StevenStadnicki: It seems that $T$ does not witness that $V$ can't be in (2).
Sep
24
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 $f$
Sep
24
comment Equality of two expressions describing a filter
@TomCruise: Yes, I was wrong. I will edit the question.
Sep
24
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 $U$.
Sep
24
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
Sep
14
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
Sep
14
comment A conjecture about filters and finite unions of cartesian products
I provided a counter-example as an answer. That counter-example was wrong.
Sep
11
comment A “rearrangement” of a finite set
For pairwise non-equivalent elements $a_0, \ldots, a_k$ we have $\forall i, j \in \{ 0, \ldots, k \} : (i \neq j \Rightarrow a_i \not\in X_{i, j} \wedge a_j \in X_{i, j})$ where $X_{i, j} \in T$. How to derive that $k \leqslant 2^n$?
Sep
11
comment A “rearrangement” of a finite set
@ThomasAndrews: I've spent more then a hour trying to prove this. I feel I am very near to the solution, but something prevents me to find it. Please explain how to derive the solution. It is not a homework and your help won't harm
Sep
11
comment A “rearrangement” of a finite set
@ThomasAndrews I feel that I am near to the answer. Probably I need to select a canonical representative element from every equivalence class
Sep
11
comment A “rearrangement” of a finite set
@ThomasAndrews I am stalled in the attempt to figure out the property of equivalence relations which corresponds to finiteness. Why would you not give me an answer? It is not a homework
Sep
11
comment A “rearrangement” of a finite set
@ThomasAndrews Maybe this is an eclipse in my mind. Why $S$ has at most $2^n$ elements?
Sep
11
comment A “rearrangement” of a finite set
@Hippalectryon I tried to meditate a half of minute and had no ideas
Sep
10
comment A lattice generated by two particular sublattices of the lattice of binary relations
@JairTaylor: Yes. Maybe the answer is the set of all finite unions of cartesian products?
Sep
10
comment A lattice generated by two particular sublattices of the lattice of binary relations
Maybe, it is the lattice consisting of all finite unions of pairwise non-intersecting cartesian products?
Sep
10
comment A lattice generated by two particular sublattices of the lattice of binary relations
@Keinstein: No, I mean all finite partitions of $U$ and every function $Y$ for a given partition. $Y$ is a mapping from $S$ into $\mathscr{P}U$ not from $\mathscr{P}U$ to $\mathscr{P}U$
Aug
26
comment Directed multigraph with numbered edges
@Casteels: Accordingly my understanding "edge-labeled directed multigraph" may have labels which are not natural numbers, have duplicate labels for two different edges starting in the same vertex, etc. So, I think this is not an answer
Aug
23
comment A conjecture about filters and finite unions of cartesian products
No, not that simple. I keep thinking
Aug
23
comment A conjecture about filters and finite unions of cartesian products
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.