Timeline for A new generalization of the dimension?
Current License: CC BY-SA 4.0
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| Dec 20, 2018 at 19:08 | history | edited | Dattier | CC BY-SA 4.0 |
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| Dec 16, 2018 at 8:38 | history | edited | Dattier | CC BY-SA 4.0 |
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| Dec 15, 2018 at 15:58 | history | undeleted | Dattier | ||
| Dec 15, 2018 at 14:54 | history | deleted | Dattier | via Vote | |
| Dec 15, 2018 at 12:14 | history | edited | Dattier | CC BY-SA 4.0 |
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| Dec 15, 2018 at 12:08 | history | edited | Dattier | CC BY-SA 4.0 |
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| Dec 15, 2018 at 10:13 | comment | added | YCor | Definition 3 is quite hard to read, the long sentence is cut somewhat arbitrarily at the middle, and you switch from symbolic notation to English words before the end. Also it's quite helpful, for readability, to denote sets of subsets with mathcal letters. | |
| Dec 15, 2018 at 10:12 | history | edited | YCor | CC BY-SA 4.0 |
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| Dec 15, 2018 at 9:27 | history | edited | Dattier | CC BY-SA 4.0 |
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| Dec 15, 2018 at 9:17 | history | edited | Dattier | CC BY-SA 4.0 |
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| Dec 15, 2018 at 9:09 | history | edited | Dattier | CC BY-SA 4.0 |
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| Dec 14, 2018 at 16:28 | comment | added | Alex Kruckman | To be clear, I'm not saying that proper containments between sets of the same dimension is necessarily a bad thing - I'm just curious if you intended that kind of behavior. I think it's the main thing making your definition different than the notion of a matroid (at least in the case when $X$ is finite). If you strengthen the definition of "has a dimension" to rule out this kind of behavior, I think you'll exactly end up with a definition that's equivalent to the definition of a matroid. | |
| Dec 14, 2018 at 16:25 | comment | added | Alex Kruckman | Oh, I liked the old definition much better... You can easily adjust my example to fit your new definition: $X = \{0,1,*\}$, $T = \{\{*\}, \{0,*\}, \{0,1,*\}\}$. Again $(X,T)$ is a "structure with dimension", with $O_T = \{*\}$, and $\dim(\{0,*\}) = \dim(\{0,1,*\}) = 1$, since the former is the closure of $\{0\}$, and the latter is the closure of $\{1\}$. | |
| Dec 14, 2018 at 16:10 | history | edited | Dattier | CC BY-SA 4.0 |
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| Dec 14, 2018 at 15:33 | comment | added | Alex Kruckman | You didn't include it in your question, but I assume that for $C\in T$, you want to say that the dimension of $C$ is the size of any independent set with closure $C$ (and $\infty$ if there is no such set)? Consider $X = \{0,1\}$ and $T = \{\emptyset, \{0\}, \{0,1\}\}$. Now it's easy to check that $(X,T)$ is a "structure with a dimension". And we have $\text{dim}(\{0\}) = \text{dim}(\{0,1\}) = 1$. So you have have proper containments between sets of the same dimension. Is this a kind of behavior you want to allow? | |
| Dec 14, 2018 at 11:55 | comment | added | Gerald Edgar | Perhaps read about matroids ... en.wikipedia.org/wiki/Matroid | |
| Dec 14, 2018 at 11:34 | history | edited | Dattier |
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| Dec 14, 2018 at 11:18 | history | edited | Dattier |
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| Dec 14, 2018 at 10:55 | history | edited | Dattier |
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| Dec 14, 2018 at 10:47 | history | asked | Dattier | CC BY-SA 4.0 |