Timeline for Hall's Marriage Theorem and intervals
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2021 at 21:32 | comment | added | Tri | Are you certain the members of $G\setminus B$ are "out of luck"? | |
| Apr 18, 2019 at 9:15 | comment | added | Fedor Petrov | This is of course too late and probably does not differ from your argument, but for what it worth. We may reformulate the Hall condition in terms of grooms: the obstruction is provided by a set $T$ of grooms such that $|b:A_b\subset T|>|T|$. Then it is clear that if $T$ is an obstruction, then one of its "connected components" also is an obstruction. | |
| Apr 11, 2013 at 14:03 | comment | added | Vidit Nanda | I haven't seen this generalization before, but you should make sure it doesn't follow obviously from something like exercise III.4.6 in Bourbaki's set theory (I don't think it does, but it's too early in the morning). | |
| Apr 11, 2013 at 11:16 | comment | added | Allen Knutson | Thanks Wolfgang, yes I did mean that and have edited. | |
| Apr 11, 2013 at 11:15 | history | edited | Allen Knutson | CC BY-SA 3.0 |
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| Apr 11, 2013 at 10:46 | comment | added | Wolfgang | Also, I guess you could further generalize it by replacing "intervals in $\mathbb N$" with "sets in $\mathbb N^m$ that are convex hulls" or "convex sets in $\mathbb N^m$". (not sure if the latter works, as intersections may not be convex) | |
| Apr 11, 2013 at 10:27 | comment | added | Wolfgang | Do you mean in your last sentence "it restricts the form of $S$"? | |
| Apr 11, 2013 at 3:46 | history | asked | Allen Knutson | CC BY-SA 3.0 |