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An amateur general topology researcher.


Sep
11
revised A conjecture about filters and finite unions of cartesian products
joins -> unions
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
revised A lattice generated by two particular sublattices of the lattice of binary relations
added 55 characters in body
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 $\mathscr{P}U$
Sep
9
asked A lattice generated by two particular sublattices of the lattice of binary relations
Sep
9
revised Another conjecture about filters and cartesian products
typo
Sep
9
revised Another conjecture about filters and cartesian products
minor correction
Sep
9
revised Another conjecture about filters and cartesian products
clarified non-counterexample
Sep
9
revised Another conjecture about filters and cartesian products
Forgotten definition of \Delta
Sep
9
asked Another conjecture about filters and cartesian products
Sep
6
revised For every pair of vertexes there is at most one path
a counter-example for an answer
Sep
6
asked For every pair of vertexes there is at most one path
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
26
asked Directed multigraph with numbered edges
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.
Aug
23
revised A conjecture about filters and finite unions of cartesian products
added 33 characters in body
Aug
23
revised A conjecture about filters and finite unions of cartesian products
added 12 characters in body
Aug
23
asked A conjecture about filters and finite unions of cartesian products