Timeline for A new generalisation of dimension? part 2
Current License: CC BY-SA 4.0
26 events
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| Oct 17, 2019 at 8:04 | comment | added | Dattier | Yes, I must find the condition (ii) not too stronger, for have theorem 2 and the examples true. | |
| Oct 17, 2019 at 8:02 | history | edited | Dattier | CC BY-SA 4.0 |
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| Oct 16, 2019 at 20:39 | comment | added | Alex Kruckman | Responding to your new edit: I'm confused - doesn't every structure with a dimension have a good dimension? After all, if $F\subseteq E$ and $\dim(F) = n$, then the largest free subset $X\subseteq F$ has cardinality $n$, and also $X\subseteq E$, so the largest free subset of $E$ has cardinality at least $n$, and hence $\dim(F) \leq \dim(E)$. Right? | |
| Oct 16, 2019 at 14:22 | history | edited | Dattier | CC BY-SA 4.0 |
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| Jan 21, 2019 at 21:30 | comment | added | Alex Kruckman | Responding to your edit: Your Theorem 2 is not correct, I've added a counterexample at the bottom of my answer. | |
| Jan 21, 2019 at 15:54 | history | edited | Dattier | CC BY-SA 4.0 |
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| Jan 21, 2019 at 15:10 | history | edited | Dattier | CC BY-SA 4.0 |
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| Jan 21, 2019 at 14:58 | history | edited | Dattier | CC BY-SA 4.0 |
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| Jan 14, 2019 at 23:37 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
Formatting
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| Jan 14, 2019 at 18:41 | answer | added | Alex Kruckman | timeline score: 12 | |
| Jan 14, 2019 at 11:24 | history | edited | Dattier | CC BY-SA 4.0 |
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| Jan 14, 2019 at 10:26 | comment | added | მამუკა ჯიბლაძე | Wikipedia has some information about infinite matroids. | |
| Jan 14, 2019 at 10:05 | comment | added | მამუკა ჯიბლაძე | @Vincent E. g. the set of subspaces of a vector space. | |
| Jan 14, 2019 at 10:03 | history | edited | Wolfgang | CC BY-SA 4.0 |
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| Jan 14, 2019 at 9:53 | comment | added | Vincent | I was wondering about something that sounds a bit like Andreas question above: is there an easy example where the set $\mathcal{T}$ is not the set of closed sets of some topology? | |
| Jan 14, 2019 at 9:49 | comment | added | მამუკა ჯიბლაძე | Is not the linear-algebra tag irrelevant here? | |
| Jan 14, 2019 at 8:39 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| Jan 14, 2019 at 7:22 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
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| Jan 14, 2019 at 2:00 | comment | added | Dattier | We can working in dimension infinite, without change the theory, it's not the case of matroïds. | |
| Jan 14, 2019 at 1:43 | comment | added | j.c. | Note that what you call a "structure" is also known as a "Moore family": en.wikipedia.org/wiki/Closure_operator | |
| Jan 14, 2019 at 1:38 | comment | added | Andreas Blass | It seems that a matroid gives rise to a structure with a dimension, by taking $\mathcal T$ to consist of the closed sets (the flats) of the matroid. In particular, your notion of "having a dimension" seems to correspond to the exchange property of matroids. Is that correct? Do you have examples of structures with dimension that don't come from matroids in this way? | |
| Jan 14, 2019 at 0:55 | history | edited | Dattier | CC BY-SA 4.0 |
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| Jan 14, 2019 at 0:54 | comment | added | Noah Schweber | (Also, it seems that the actually important object here is the closure operator $cl_S: U\mapsto\langle U\rangle_S$; just talking about the pair $(X, cl_S)$ seems much easier than talking about $(X,\mathcal{T})$, and also makes the relationship with matroids etc. clearer.) | |
| Jan 14, 2019 at 0:53 | comment | added | Noah Schweber | But it has nothing to do with any of the questions you mention, as far as I can tell, so why bring it up? (Also, should "$\mathcal{U}$" be "$\mathcal{T}$"?) | |
| Jan 14, 2019 at 0:50 | comment | added | Noah Schweber | As far as I can tell, $O_\mathcal{T}$ plays no role here; why introduce it? | |
| Jan 14, 2019 at 0:44 | history | asked | Dattier | CC BY-SA 4.0 |