Timeline for A set for which it is hard to determine whether or not it is countable.
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Oct 7, 2011 at 14:42 | comment | added | Junyan Xu | By the way, unlike exotic R^4's, there are at most countably many distinct exotic 4-spheres, according to Wikipedia. | |
| Mar 24, 2011 at 6:35 | comment | added | Kimball | @Tim: ah yes, I suppose it was easier than I was thinking. Is it less clear if you ask about the number of subgroups of $PSL_2(\mathbb Z)$? | |
| Mar 22, 2011 at 5:23 | comment | added | Tim O | @Kimball : That's uncountable. It's even uncountable if you look at cyclic subgroups up to conjugacy, which are classified by the absolute value of the trace of a generator. | |
| Mar 22, 2011 at 4:04 | comment | added | Kimball | A variant for $d=2$: the number of discrete subgroups of $PSL_2(\mathbb R)$ up to conjugacy. | |
| Mar 22, 2011 at 0:28 | comment | added | villemoes | A quick google search yields deepblue.lib.umich.edu/bitstream/2027.42/32762/1/0000133.pdf , but I don't know what the "canonical" reference is. | |
| Mar 22, 2011 at 0:24 | comment | added | Mariano Suárez-Álvarez | Work of Kerekjarto and Richards implies there are at least as many non-compact separable surfaces as there are triples $(X,Y,Z)$ with $Z\subseteq Y\subseteq X$ a chain of of compact, separable, totally disconnected spaces. There are too many of these for surfaces to be countable, no? | |
| Mar 22, 2011 at 0:21 | comment | added | David Cohen | Surely there are uncountably many. Just remove countably many tame knots from $\mathbb{R}^{3}$, there are uncountably many ways to do this. | |
| Mar 21, 2011 at 23:53 | history | answered | Matthew Kahle | CC BY-SA 2.5 |