Timeline for Chromatic number of a topological space
Current License: CC BY-SA 3.0
12 events
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| Feb 5, 2018 at 21:21 | comment | added | Taras Banakh | @SimonHenry Yes. This follows from the fact that each zero-dimensional set is contained in a zero-dimensional $G_\delta$-set (and this fact follows from Lavrentev Theorem). So, the Urysohn decomposition theorem implies that each metrizable separable space $X$ of dimension $n$ can be covered by $n+1$ zero-dimensional $G_\delta$-sets. | |
| Feb 5, 2018 at 21:14 | comment | added | Simon Henry | That is impressive ! And does it also answer the question in the case of partition by borelian ? (i.e. does the decomposition theorem of Urysohn can produce decomposition in borelian subsets ?) | |
| Feb 5, 2018 at 20:08 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Feb 5, 2018 at 19:47 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Feb 5, 2018 at 19:41 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Feb 5, 2018 at 19:38 | vote | accept | N. de Rancourt | ||
| Feb 5, 2018 at 19:23 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Feb 5, 2018 at 19:06 | comment | added | Taras Banakh | I suspect that Theorem 1 uses some weak forms of choice (like countable or dependent choice), but this acceptable (I hope). | |
| Feb 5, 2018 at 18:21 | comment | added | Asaf Karagila♦ | Just a remark about choice (to the OP and to those interested). The proof of Theorem 1 is without choice, modulo the blackbox theorems used (which I suspect one can get in a fairly choice free in the cases of interest); the second proof certainly utilizes some choice, since $\omega_1$ need not be regular. | |
| Feb 5, 2018 at 17:27 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Feb 5, 2018 at 17:01 | history | edited | Taras Banakh | CC BY-SA 3.0 |
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| Feb 5, 2018 at 16:52 | history | answered | Taras Banakh | CC BY-SA 3.0 |