Timeline for Covering a poset by minimum number of chains and antichains
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2023 at 15:58 | comment | added | Richard Stanley | @AdamP.Goucher, this is exactly the argument I would give. | |
| Apr 30, 2023 at 14:45 | comment | added | Adam P. Goucher | Removing any chain or antichain from the $k$th Richard Stanley poset gives you something that contains an isomorphic copy of the $(k-1)$th Richard Stanley poset. By induction, you need to remove at least $k$ chains/antichains to get down to the empty poset. | |
| Apr 28, 2023 at 6:20 | comment | added | Pritam Majumder | @RichardStanley: Thanks. I believe your claim is that at least $k$ chains/antichains are required to cover the poset you mentioned. Could you please provide an argument to prove this? | |
| Apr 28, 2023 at 2:01 | comment | added | Richard Stanley | Taking a disjoint union of chains of sizes $1,2,\dots,k$ gives the asymptotic lower bound $\sqrt{2|P|}$. | |
| Apr 27, 2023 at 17:35 | history | edited | Pritam Majumder | CC BY-SA 4.0 |
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| Apr 27, 2023 at 14:53 | history | asked | Pritam Majumder | CC BY-SA 4.0 |