Timeline for Are all Hawaiian Earrings homeomorphic?
Current License: CC BY-SA 2.5
10 events
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| Jan 6, 2010 at 1:42 | comment | added | Joel David Hamkins | I think the last sentence in my answer is not quite exactly the right information, since for example, one could have 0 being a limit of limit points, or only a limit of isolated points, even for countable sets that are homeomorphic. I think the right way to say it is that the homeomorphism type of the earring determined by a countable nonempty set A is determined by the homeomorphism type of A union {0}, with 0 as a distinguished point. | |
| Jan 6, 2010 at 1:13 | comment | added | Joel David Hamkins | Yes, I think that's right. | |
| Jan 6, 2010 at 1:12 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Jan 6, 2010 at 0:53 | comment | added | Pete L. Clark | Wait -- no, you're right again. The sequences $a_n = 2 - \frac{1}{n}$ and $a_n = n$ give rise to homeomorphic earrings. | |
| Jan 6, 2010 at 0:50 | comment | added | Pete L. Clark | But the homeomorphism type of a sequence converging to infinity is the same as the homeomorphism type of a sequence converging to a finite number (but not including that number). | |
| Jan 6, 2010 at 0:45 | comment | added | Joel David Hamkins | Wait a minute. I think the case of converging to infinity is the same as having any nonconvergent sequence, no? I don't think infinity is special in the way that 0 is special in this construction. | |
| Jan 6, 2010 at 0:31 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Jan 6, 2010 at 0:29 | comment | added | Joel David Hamkins | Well, that one isn't compact, so it isn't the same as the converging to 0 case, right? But you're right, this info will affect the homemorphism type. | |
| Jan 6, 2010 at 0:05 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Jan 5, 2010 at 23:53 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |