Timeline for Choice sets in hypergraphs with finite edges
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 7, 2022 at 10:33 | vote | accept | Dominic van der Zypen | ||
| Jun 7, 2022 at 9:25 | comment | added | Fedor Petrov | Lines on the usual (affine) real plane also work. Or a rational plane, if you prefer countably many vertices and edges. | |
| Jun 7, 2022 at 9:21 | comment | added | bof | For a counterexample with infinite edges, consider a maximal almost disjoint family of infinite subsets of $\mathbb N$ which contains an infinite collection of pairwise disjoint sets. Alternatively, consider the lines of the real projective plane. | |
| Jun 7, 2022 at 9:20 | comment | added | Fedor Petrov | Ooops. I was thinking on perfect matching instead, sorry | |
| Jun 7, 2022 at 9:10 | answer | added | Fedor Petrov | timeline score: 3 | |
| Jun 7, 2022 at 9:03 | comment | added | bof | @FedorPetrov How is the hypergraph consisting of all infionite subsets of a countably infinite set a counterexample to the original question? If $E_0=\{e_1,e_2,e_1\cup e_2\}$ where $e_1\cap e_2=\varnothing$, then $E_0$ has no choice set. What am I missing? | |
| Jun 7, 2022 at 8:10 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
restriction to finite edges
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| Jun 7, 2022 at 8:08 | comment | added | Dominic van der Zypen | Right, Fedor, thanks for your answer. Is it still the same if all the members of $E$ are finite? I will modify the question accordingly. - I was hoping some compactness argument would work, but I didn't manage it. If you write down the compactness argument in an answer (or provide another answer), I'll be more than happy to accept | |
| Jun 6, 2022 at 9:58 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |