Timeline for compact quotient
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 6, 2012 at 18:26 | comment | added | Nik Weaver | ("at least one point of which has no countable neighborhood" can be replaced by "which is uncountable") | |
| Aug 6, 2012 at 1:53 | comment | added | Nik Weaver | Here's a general version of the $D^2$ idea. Replace $D^2$ with any first countable compact Hausdorff space $Y$ at least one point of which has no countable neighborhood, and let $X$ be the disjoint union of all subsets of $X$ of the form: a convergent sequence together with its limit. | |
| Aug 3, 2012 at 19:11 | comment | added | Trevor Wilson | @Nik: Oh, I see. Thanks for the explanation. | |
| Aug 3, 2012 at 18:51 | comment | added | Nik Weaver | @Trevor: the hard part is getting the natural map onto $D^2$ to be a quotient map. Since there are countably many $K_\alpha$, you can find a sequence of values $(\theta_n, r_n)$ that converges to $0$ and eventually avoids every $K_\alpha$. Then $D^2$ minus this set (but including $0$) will be open in the quotient topology, so you get the wrong topology. | |
| Aug 3, 2012 at 16:11 | comment | added | Trevor Wilson | To get a separable counterexample, can't you just let $X$ be the disjoint union of a countable family of thin subspaces that cover $D^2$, for example, $K_\alpha$ for rational angles $\alpha$ where $K_\alpha = \{(\theta, r) : |\theta - \alpha| \le r\}$? | |
| Aug 3, 2012 at 15:11 | comment | added | Nik Weaver | Nice! I'm deleting my faulty answer. | |
| Aug 3, 2012 at 14:30 | history | answered | André Henriques | CC BY-SA 3.0 |