Timeline for Compact Lie group action on non-Hausdorff (but CGWH) space with Hausdorff quotient
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22 events
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| Mar 25, 2023 at 21:27 | comment | added | PatrickR | Taras's argument shows more generally (1) WH implies US (unique sequential limits), (2) US + first-countable imply Hausdorff. Also see topology.pi-base.org/properties/P000099 for example. | |
| Nov 21, 2017 at 9:14 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 22, 2017 at 9:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
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| Aug 24, 2017 at 16:57 | comment | added | Taras Banakh | @AlexanderKörschgen To prove that a first-countable WH-space is Hausdorff, assume that two distinct points $x,y$ cannot be separated by open neighborhoods and choose decreasing countable neighborhood bases $(U_n)_{n\in N}$ and $(V_n)_{n\in N}$ at $x,y$, respectively. For every $n\in N$ choose a point $x_n\in U_n\cap V_n\setminus\{x,y\}$. Then the sequence $(x_n)_{n\in N}$ converges to $x$ and $y$ simultaneoulsy. Since the set $K=\{x\}\cup\{x_n\}_{n\in N}$ is a continuous image of the compact Hausdorff space $\{0\}\cup\{2^{-n}:n\in N\}$, it is closed in the WH-space, a contradition. | |
| Aug 24, 2017 at 16:38 | comment | added | Alexander Körschgen | @TarasBanakh Unfortunately, I don't know the answer for finite groups. If $G$ is finite, any orbit subspace of $X$ is discrete, but this does not really help to separate two points of the orbit. | |
| Aug 24, 2017 at 16:36 | comment | added | Alexander Körschgen | @TarasBanakh why does WH + first-countable imply Hausdorff? I wasn't able to find this statement, maybe you're confusing this with the fact that every first-countable space is compactly generated (see Strickland's notes, for example). | |
| Aug 23, 2017 at 10:21 | comment | added | Taras Banakh | It seems that a weakly Haudorff first-countable space is Hausdorff, so the answer is affirmative for first-countable spaces $X$. By the way, what is the answer for finite acting group $G$? | |
| Aug 23, 2017 at 8:28 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 24, 2017 at 8:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 24, 2017 at 7:23 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 25, 2017 at 6:27 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 25, 2017 at 6:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 26, 2017 at 5:12 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 24, 2017 at 4:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 25, 2017 at 3:36 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 5, 2017 at 0:08 | comment | added | Alexander Körschgen | @GabrielC.Drummond-Cole: Thanks, I had not realized that the weak Hausdorff property on $X$ is sufficient to have Hausdorff orbits until now. I agree that this makes a potential example even less likely to be found. | |
| Dec 26, 2016 at 16:38 | comment | added | Gabriel C. Drummond-Cole | One reason it may be hard to think of an example by hand is because you assume the quotient is Hausdorff, and then compactness of $G$ implies that each orbit subspace must be Hausdorff. With a Hausdorff "base" and Hausdorff "fibers," non-Hausdorffness in the "total space" has to come from something weird. | |
| Dec 26, 2016 at 3:16 | comment | added | YCor | $X$ weakly Hausdorff: all images of compact Hausdorff spaces in $X$ are closed (implies $T_1$: points are closed). Compactly generated: en.wikipedia.org/wiki/Compactly_generated_space | |
| Dec 26, 2016 at 3:11 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 26, 2016 at 3:06 | answer | added | Jeff Strom | timeline score: 1 | |
| Nov 26, 2016 at 1:20 | history | asked | Alexander Körschgen | CC BY-SA 3.0 |