Timeline for Which topological spaces are (topological) groups?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Apr 29, 2013 at 16:41 | answer | added | Włodzimierz Holsztyński | timeline score: 3 | |
| Feb 24, 2013 at 20:29 | comment | added | Włodzimierz Holsztyński | I feel that the very definition of the topological group notion is already the needed characterization (tautological as it is). One may characterize a specific topological group, that's different. But in the case of the whole class, it is characterized by its strong homogeneity (homogeneousness) properties which virtually amount to the definition itself or something very similar. | |
| Feb 24, 2013 at 7:46 | answer | added | Włodzimierz Holsztyński | timeline score: 4 | |
| Feb 24, 2013 at 4:19 | comment | added | David White | @Unnikrishnan: I think you're using the "accept" checkmark all wrong. You should reserve it till you have an answer you're really happy with, or until a long time goes by and it seems you're not getting any new answers. Don't just shift it to whatever the newest answer is | |
| Feb 24, 2013 at 2:50 | comment | added | Pete L. Clark | I asked this question on our sister website a while back: math.stackexchange.com/questions/16829/…. | |
| Feb 23, 2013 at 20:36 | comment | added | Gerald Edgar | Any completely regular Hausdorff space $X$ is homeomorphic to a subset of a cartesian product $\mathbb R^Y$ in a standard way, namely $Y = C(X,\mathbb R)$, the set of continuous functions from $X$ to $\mathbb R$. | |
| Feb 23, 2013 at 19:10 | vote | accept | N Unnikrishnan | ||
| Feb 24, 2013 at 5:58 | |||||
| Feb 23, 2013 at 18:42 | comment | added | Emil Jeřábek | @N Unnikrishnan: As for subsets, every $T_{3\frac12}$ space is homeomorphic to a subspace of $\mathbb R^\kappa$ (or $(S^1)^\kappa$ if you prefer a compact group) for some $\kappa$, no homogeneity is needed. I believe that likewise every completely regular space can be embedded into an appropriate abelian topological group whose Kolmogorov quotient is, say, $\mathbb R^\kappa$. | |
| Feb 23, 2013 at 18:39 | vote | accept | N Unnikrishnan | ||
| Feb 23, 2013 at 19:09 | |||||
| Feb 23, 2013 at 18:30 | comment | added | N Unnikrishnan | By the separation axiom complete regularity, I mean it in the weaker sense of Kelly and Willard - without the T0 axiom. It is this property which is equivalent to uniformisability (in the class of all topological spaces). This, I hope, every topological group (which is not assumed T0, I emphasise) must satisfy, as does the indiscrete topology. But then I regret having written metrisability and not pseudo-metrisability in my question. | |
| Feb 23, 2013 at 18:30 | answer | added | Ramiro de la Vega | timeline score: 5 | |
| Feb 23, 2013 at 18:27 | comment | added | N Unnikrishnan | With Gerald Edgar's suggestion, can we expect every (strongly?)homogeneous uniformisable topological space to be (homeomorphic to) a subset of a topological group? | |
| Feb 23, 2013 at 17:56 | history | edited | N Unnikrishnan | CC BY-SA 3.0 |
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| Feb 23, 2013 at 17:36 | history | edited | N Unnikrishnan | CC BY-SA 3.0 |
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| Feb 23, 2013 at 17:35 | vote | accept | N Unnikrishnan | ||
| Feb 23, 2013 at 18:39 | |||||
| Feb 23, 2013 at 14:39 | comment | added | Gerald Edgar | Much easier is: which spaces are homeomorphic to a subset of a topological group. | |
| Feb 23, 2013 at 14:39 | history | edited | Lee Mosher | CC BY-SA 3.0 |
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| Feb 23, 2013 at 12:13 | comment | added | Tom Goodwillie | By "uniform" I think you mean that the space can be given a uniform structure, which is necessary as you say. Metrizability or any kind of separation axiom is not necessary; the indiscrete topology on any group gives a topological group. | |
| Feb 23, 2013 at 11:46 | answer | added | Cleft | timeline score: 10 | |
| Feb 23, 2013 at 11:43 | answer | added | Ralph | timeline score: 14 | |
| Feb 23, 2013 at 10:54 | history | asked | N Unnikrishnan | CC BY-SA 3.0 |