Timeline for Generalizing König's Lemma
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Dec 31, 2019 at 5:03 | comment | added | Péter Komjáth | Actually the result is in Aharoni, Korman: Greene-Kleitman theorem for infinite posets, Order, 9(1992), 245-253. (Sorry for having claimed that it is new.) | |
| Sep 1, 2019 at 6:21 | comment | added | Péter Komjáth | The names of Konigs: this is a strange thing. In old Hungary, quite some jews were named Konig, they used double strokes as in the name of Erdos. When they published in German (and foreign language publications were almost exclusively German those days) father and son used umlaut. Denes Konig (the son) was a teacher of Erdos. He commited suicide in October 1944 when a Nazi party (the Arrow Cross Party) took power in Hungary and murdering Jews was just very general.. | |
| Sep 1, 2019 at 6:11 | comment | added | Péter Komjáth | I think that the proof from the book uses Rado's selection lemma or some form of it, as suggested below by Pace Nielsen. This is a special case of a conjecture of Ron Aharoni (I have to check), so Pace should publish it. | |
| Aug 30, 2019 at 5:07 | comment | added | Gerhard Paseman | Combine that with something that changes your typing right after you type it. Not to mention the occasional delays between entry and display. (I'm sure it's possible to add Di a critical marks, but see what happens after I type normally.) Gerhard "Is Using The Wrong Tool" Paseman, 2019.08.29. | |
| Aug 30, 2019 at 4:49 | comment | added | David Roberts♦ | @GerhardPaseman no hold-and-select-choice-of-accent option? | |
| Aug 29, 2019 at 22:53 | comment | added | Gerhard Paseman | I'm using a cell phone with a virtual keyboard and a demon inside it called spellcheck. It's a wonder I get as much done on this forum as I do. Gerhard "Double Quotes Easier To Type" Paseman, 2019.08.29. | |
| Aug 29, 2019 at 19:54 | comment | added | Asaf Karagila♦ |
@GerhardPaseman: I don't know what operating system you're using, but on Linux you can define a Compose key. Then Compose+"+o (in sequence) produces ö, and Compose+=+o (in sequence) produces ő. No more cutting and pasting!
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| Aug 29, 2019 at 19:51 | comment | added | Asaf Karagila♦ | @Monroe: Ancient Sumerian, yes. | |
| Aug 29, 2019 at 19:47 | comment | added | Monroe Eskew | @AsafKaragila There’s still only one good language. | |
| Aug 29, 2019 at 18:34 | comment | added | Gerhard Paseman | Koenig the father also tried to disprove the continuum hypothesis, and used the first name Julius as well in publications. The lemma was named to honor the son, who likely was involved in the discovery or invention. As far as I know, the lemma had nothing at all to do with checkers. Gerhard "Not Doing Cut And Paste" Paseman, 2019.08.29. | |
| Aug 29, 2019 at 18:03 | comment | added | Asaf Karagila♦ | @Monroe: Yes, it's a German word, for Hungarian mathematicians. Just because you're now in Vienna doesn't mean that everyone and everywhere and everything should be German. | |
| Aug 29, 2019 at 18:02 | comment | added | Andreas Blass | @PéterKomjáth Doesn't part 2 of your proof also cover the case where there is a highest nonempty level? (The $\phi$ in 2 will be the level after the highest, the only important $\alpha$ will be the highest level, and $\beta$ will range over the levels below the highest. So $U$ will concentrate on level $\alpha$, it will therefore contain a top-level singleton $\{x\}$, and the branch you get from $U$ in case 2 will be the same as what you got from $x$ in case 1.) | |
| Aug 29, 2019 at 17:55 | comment | added | Andreas Blass | @MonroeEskew A German word can be a Hungarian name (or can become one by changing the shape of the accent). | |
| Aug 29, 2019 at 16:46 | comment | added | Monroe Eskew | @AsafKaragila what are you talking about? It's a German word. | |
| Aug 29, 2019 at 11:22 | comment | added | Asaf Karagila♦ | König the father wrote his name with ö; but Kőnig the son wrote his name with ő. | |
| Aug 29, 2019 at 9:20 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added Google Books link
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| Aug 29, 2019 at 8:41 | answer | added | Pace Nielsen | timeline score: 2 | |
| Aug 21, 2019 at 15:07 | comment | added | Péter Komjáth | Another argument: 1. If there is a highest level $S(\phi)$, then for each $x\in S(\phi)$ the set $\{y:y\leq x\}$ is a branch intersecting all levels. 2. If there is no highest level, then let $U$ be an ultrafilter such that $\bigcup\{S(\beta):\beta<\alpha\}\notin U$`` for any $\alpha<\phi$, where $\phi$ is least that $S(\phi)=\emptyset$. For each level $S(\beta)$ there is a unique element $x_\beta\in S(\beta)$ such that $U(x_\beta)=\{z:z\geq x_\beta\}\in U$ and one can easily see that $B=\{x_\beta:\beta<\phi\}$ is a branch. | |
| Aug 20, 2019 at 17:29 | comment | added | Andrés E. Caicedo | You probably want to take a look at this: blog.math.toronto.edu/GraduateBlog/files/2014/03/… and the references in there. | |
| Aug 20, 2019 at 17:12 | history | asked | Pace Nielsen | CC BY-SA 4.0 |