Timeline for Minimum number of elements needed to represent a lattice with a union-closed family of sets
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jul 24 at 12:10 | vote | accept | Fabius Wiesner | ||
| Jul 24 at 12:10 | answer | added | Fabius Wiesner | timeline score: 0 | |
| Jul 17 at 15:50 | comment | added | Richard Stanley | @FabiusWiesner: in your diagrams, the white elements are the meet-irreducibles (they are covered by exactly one element), and every element is a meet of meet-irreducibles. If you want to generate the lattice by taking joins, then the unique minimal generating set is the set of join-irreducibles. | |
| Jul 15 at 18:48 | comment | added | Fabius Wiesner | @RichardStanley I am not doing the opposite, the join is above in the diagram. I don't understand if your comment is aimed at simplifying my argument or if it gives the proof required in the very last question in my post. | |
| Jul 15 at 18:00 | comment | added | Richard Stanley | @FabiusWiesner: I'm just going by your diagrams. Traditionally the join of two elements is written above the two elements. Are you doing the opposite? | |
| Jul 14 at 14:30 | comment | added | Fabius Wiesner | @RichardStanley excuse me, but with a union closed family aren't we generating $L$ with the join operation rather than the meet operation? | |
| Jul 14 at 13:59 | comment | added | Richard Stanley | @FabiusWiesner: we can't express a meet-irreducible element as a meet of other elements, so it must appear in any set of elements that generate $L$ using the meet operation. | |
| Jul 14 at 8:51 | history | edited | Fabius Wiesner | CC BY-SA 4.0 |
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| Jul 14 at 8:44 | comment | added | Fabius Wiesner | @SamHopkins thank you for your comments. I think that I managed to do the easy part, and can be satisfied by having shown that we can restrict the elements to only meet-irreducible ones. However, proving or disproving that this is a representation with a minimum number of elements seems more difficult. | |
| Jul 14 at 8:39 | history | edited | Fabius Wiesner | CC BY-SA 4.0 |
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| Jul 11 at 21:15 | comment | added | Sam Hopkins | See for example my answer at the other question: mathoverflow.net/a/469052/25028 | |
| Jul 11 at 21:09 | comment | added | Sam Hopkins | "Maybe the only needed elements are those with out-degree exactly 1?": in lattice theory these are called the meet-irreducible elements, and I think that yes this should be easy to show. | |
| Jul 11 at 21:07 | history | edited | Fabius Wiesner | CC BY-SA 4.0 |
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| Jul 11 at 12:09 | history | edited | Fabius Wiesner | CC BY-SA 4.0 |
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| Jul 11 at 11:32 | history | edited | Fabius Wiesner | CC BY-SA 4.0 |
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| Jul 11 at 10:49 | history | asked | Fabius Wiesner | CC BY-SA 4.0 |