Timeline for Topologizing free abelian groups
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 30, 2010 at 21:26 | comment | added | HenrikRüping | Sry, I messed things up. | |
| Mar 30, 2010 at 16:11 | comment | added | François G. Dorais | With the additional maps I suggested, this always gives a positive answer to Henrik's question if (ii) is weakened to require that addition is separately continuous. Addition will be jointly continuous iff this topology agrees with the Markov topology. So I think that Henrik's question has a positive answer precisely when the Markov topology agrees with the final topology for the affine combinations of $S\to\mathbb{Z}[S]$. | |
| Mar 30, 2010 at 15:36 | comment | added | François G. Dorais | @Henrik: I get the discrete topology. The final topology is the finest topology which makes all the maps continuous. | |
| Mar 30, 2010 at 15:36 | comment | added | Keenan Kidwell | @Henrik: I'm not understanding what you're saying. If $S$ is discrete then the final topology on $\mathbb{Z}[S]$ is also discrete. | |
| Mar 30, 2010 at 14:52 | comment | added | HenrikRüping | @François: If one considers $S=\{a\}$ then the resulting final topology w.r.t. all those maps is just the cofinite topology. Hence this also doesn't suffice to make the map to $\mathbb{Z}$ equipped with the discrete topology continuous. | |
| Mar 30, 2010 at 14:44 | comment | added | François G. Dorais | In addition to $S \to \mathbb{Z}[S]$, I would take all of its multiples and translates. (You probably need more, but the single map $S \to \mathbb{Z}[S]$ will clearly not do.) | |
| Mar 30, 2010 at 14:38 | comment | added | HenrikRüping | If one considers a discrete set $S$, the topology on $\mathbb{Z}[S]$ shoudl also be discrete. But the map $\mathbb{Z}[S]\rightarrow \mathbb{Z}[S]$ is not continuous, where the source is equipped with the final topology and the target is equipped with the discrete topology. | |
| Mar 30, 2010 at 14:24 | history | answered | Keenan Kidwell | CC BY-SA 2.5 |