Timeline for Totally bounded group topologies on $\Bbb Q$ with trivial intersection of two neighborhoods
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 9, 2014 at 0:43 | comment | added | Minimus Heximus | (at)AdreasThom: seems correct. | |
| Oct 5, 2014 at 9:27 | comment | added | Andreas Thom | I thought $\mathcal S$ and $\mathcal T$ correspond to compactifications of $\mathbb Q$. Then, $A \times B$ would be a neighborhood of $0$ in the product topology and $A \cap B$ would be the intersection of $A \times B$ with the diagonal embedding of $\mathbb Q$ into the product. If $A \cap B = \{0\}$, then the induced topology follows to be discrete - contradicting the compactness of the product. | |
| Oct 4, 2014 at 22:01 | comment | added | Ramiro de la Vega | @AndreasThom: I don´t understand your comment, what does the product topology have to do here? | |
| Oct 4, 2014 at 18:00 | vote | accept | Minimus Heximus | ||
| Oct 4, 2014 at 16:21 | comment | added | Andreas Thom | I see, isn't this quit obviously impossible -- basically since the product of two compact spaces is compact. How can $\mathbb Q$ be discrete in the product? Maybe I am missing something. | |
| Oct 4, 2014 at 15:52 | comment | added | Ramiro de la Vega | @AndreasThom: The topologies must be different since they are totally bounded. | |
| Oct 4, 2014 at 15:50 | answer | added | Ramiro de la Vega | timeline score: 2 | |
| Oct 4, 2014 at 12:54 | comment | added | Andreas Thom | You mean with $\mathcal S \neq \mathcal T$? | |
| Oct 4, 2014 at 10:27 | history | asked | Minimus Heximus | CC BY-SA 3.0 |