Timeline for Are all Hawaiian Earrings homeomorphic?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| May 12, 2017 at 0:14 | vote | accept | john mangual | ||
| May 11, 2017 at 13:18 | history | edited | Todd Trimble | CC BY-SA 3.0 |
the coproduct topology rules out the type of example proposed by John Mangual
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| Aug 20, 2010 at 3:52 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jan 5, 2010 at 23:35 | comment | added | Mariano Suárez-Álvarez | That sequence does not converge to zero---I have not made any claim about the resulting space. | |
| Jan 5, 2010 at 23:26 | comment | added | john mangual | @Mariano I might buy this in the monotonic case. However, if we let the radii be 1 + 1/n the points (0, 2 + 2/n) (antipodal to (0,0) converge to (0,2) which does not lie in any circle. So, this space is also not compact. | |
| Jan 5, 2010 at 21:03 | comment | added | Mariano Suárez-Álvarez | @Anweshi: the special point in an infinite wedge of circles does not have a countable basis of neighborhoods, so in particular that wedge is not metrizable and, as a consequence, it is not a subspace of $\mathbb R^2$. | |
| Jan 5, 2010 at 21:01 | comment | added | Anweshi | @Mariano. Supposing a_n goes to infinity. Then we should get the wedge. If a_n goes to zero, then it should be the Hawaiian. No? | |
| Jan 5, 2010 at 21:00 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
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| Jan 5, 2010 at 20:59 | comment | added | Anweshi | Oops, I screwed by deleting my comment and re-posting in a different form.. The one of mine above, should have been the first. Sorry. | |
| Jan 5, 2010 at 20:57 | comment | added | Anweshi | That is the wedge of infinitely many circles. Isn't that non-homeomorphic to the Hawaiian? | |
| Jan 5, 2010 at 20:57 | comment | added | Mariano Suárez-Álvarez | In the wedge of infinitely many circles, you can construct a sequence (whose points are the antipodal points in each circle, of the joining point) which does not have a convergent subsequence: the wedge is, therefore, not compact. | |
| Jan 5, 2010 at 20:57 | comment | added | john mangual | See Example 1.25 in "Algebraic Topology" by Allen Hatcher. The Hawaiian Earring has a different fundamental group than the infinite wedge of circles. | |
| Jan 5, 2010 at 20:50 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |