Timeline for Generalized geometries
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jul 16, 2015 at 13:16 | vote | accept | Dominic van der Zypen | ||
| Jul 16, 2015 at 2:07 | answer | added | Will Sawin | timeline score: 2 | |
| Jul 16, 2015 at 1:02 | history | edited | Andrés E. Caicedo |
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| Jul 15, 2015 at 21:43 | comment | added | Gerhard Paseman | It seems to me that one should be able to construct an injection into "halves" of $\omega$ (infinite sets with infinite complements) from the set of (n-1) subsets of $\omega$ so that each finite subset is a subset of its image and conditions 2) and 3) are held. Gerhard "Is This Ramsey In Disguise?" Paseman, 2015.07.15 | |
| Jul 15, 2015 at 19:11 | comment | added | მამუკა ჯიბლაძე | I have doubts about 3 - ($n-1$)-planes in $\mathbb P^n$ almost never satisfy this, so most likely it is a very severe restriction. | |
| S Jul 15, 2015 at 15:59 | history | suggested | Anurag |
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| Jul 15, 2015 at 15:39 | review | Suggested edits | |||
| S Jul 15, 2015 at 15:59 | |||||
| Jul 15, 2015 at 15:07 | history | edited | Dominic van der Zypen |
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| Jul 15, 2015 at 9:45 | comment | added | Dominic van der Zypen | I agree - beautiful argument! | |
| Jul 15, 2015 at 8:57 | comment | added | Tom De Medts | If I'm not mistaken, the previous argument also shows that, as soon as $n \geq 2$, all members of $\mathfrak P$ have the same number of elements. | |
| Jul 15, 2015 at 8:56 | comment | added | Tom De Medts | More to the point: the non-existence of a geometry of type $m$ would imply the non-existence of such geometries of type $n$ for all $n>m$. Indeed, if a geometry of type $n$ exists, then one can form the "derived geometry" w.r.t. an element $s \in S$ obtained by considering $S' = S \setminus \{s\}$, and $\mathfrak P' = \{ B \setminus \{s\} \mid B \in \mathfrak{P}, s \in B \}$, which is then a geometry of type $n-1$. | |
| Jul 15, 2015 at 8:55 | history | edited | Dominic van der Zypen | CC BY-SA 3.0 |
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| Jul 15, 2015 at 8:53 | comment | added | Tom De Medts | You should require $|\mathfrak P| \geq 2$ to avoid $\mathfrak P = \{ S \}$. Condition 4 is superfluous as soon as $|S| \geq n+2$. | |
| Jul 15, 2015 at 7:55 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |