Timeline for nonhausdorff dimension
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 22, 2013 at 15:13 | answer | added | John Rognes | timeline score: 5 | |
| Apr 14, 2012 at 21:34 | answer | added | Douglas Somerset | timeline score: 0 | |
| Jan 9, 2010 at 15:06 | answer | added | Joel David Hamkins | timeline score: 11 | |
| Jan 9, 2010 at 0:23 | comment | added | Pete L. Clark | BTW, could you please capitalize the first letter in each sentence? This would make your posts easier to read. | |
| Jan 9, 2010 at 0:21 | answer | added | Pete L. Clark | timeline score: 5 | |
| Jan 8, 2010 at 22:50 | comment | added | Mariano Suárez-Álvarez | Take transitive closures at each step just as you did to define $\sim$. | |
| Jan 8, 2010 at 22:50 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
deleted 48 characters in body; added 91 characters in body
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| Jan 8, 2010 at 22:49 | comment | added | Martin Brandenburg | @mariano: are you sure? this does not look transitive. | |
| Jan 8, 2010 at 22:40 | comment | added | Mariano Suárez-Álvarez | I guess you want something more than the following recursive description: $x\sim_{\alpha+1}y$ iff given $U$ and $V$ open neighborhoods of $x$ and $y$, there exists $x'\in U$ and $y'\in V$ such that $x'\sim_{\alpha} y'$. | |
| Jan 8, 2010 at 22:19 | comment | added | Jonas Meyer | My mistake, I should have read more carefully. The parenthetical remark threw me off, because the generating relation was being defined where I thought the equivalence relation was intended. But yes, it should have been clear. Thanks. | |
| Jan 8, 2010 at 21:57 | comment | added | Pete L. Clark | @JM: He didn't claim it was an equivalence relation; he said "generated by", i.e., the smallest equivalence relation containing the given relation. | |
| Jan 8, 2010 at 21:51 | comment | added | Jonas Meyer | Furthermore I don't think what you describe is an equivalence relation. E.g. in the topology on {a,b,c,d,e} with subbasis {{a,d,e},{b,d},{c,e}}, a~b and a~c but ¬(b~c). | |
| Jan 8, 2010 at 21:23 | comment | added | Pete L. Clark | "Topological indistinguishability" usually refers to a different, finer, equivalence relation: $x \sim y$ iff $x$ and $y$ have exactly the same neighborhoods. The quotient by this relation is called the Kolmogorov quotient and is the universal $T_0$-space mapped to by the given space. Thus your terminology is potentially confusing to readers, and I recommend that you adjust it. | |
| Jan 8, 2010 at 21:16 | history | asked | Martin Brandenburg | CC BY-SA 2.5 |