Timeline for Is every subgroup closed in this complete, nondiscrete topological group?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 13 at 9:55 | vote | accept | Nick Belane | ||
| Nov 13 at 9:22 | answer | added | KP Hart | timeline score: 2 | |
| Nov 5 at 12:45 | comment | added | Nick Belane | I meant all the functions, not only the finitely supported ones. However, now I see the point. If any of you post the answer, I’ll accept it. | |
| Nov 5 at 12:18 | comment | added | YCor | And in any case, $F^{(X)}$ is not complete then (it's a dense proper subgroup of the compact group $F^X$). | |
| Nov 5 at 12:17 | comment | added | YCor | However, this countable group ($F^{(X)}$, $F$ nontrivial finite, $X$ infinite countable, with induced topology from the product topology on $F^X$) has a non-closed abelian subgroup. First if $F$ is abelian, the homomorphism $F^{(X)}\to F$, $f\mapsto \sum_x f(x)$, has a dense kernel. In general, let $C$ be a nontrivial cyclic subgroup of $F$, and then $C^{(X)}\subset F^{(X)}$ has a non-closed subgroup. | |
| Nov 5 at 12:12 | comment | added | YCor | When you write the power $F^X$, it means all functions $X\to F$, not only finitely supported ones. One notation for the restricted power (elements of the power with finite support) is $F^{(X)}$. | |
| Nov 5 at 10:28 | comment | added | Alessandro Codenotti | As per Moishe's comment on the linked answer, $\bigoplus_\Bbb Z G'$ is a dense, proper subgroup. More generally your group is uncountable and Polish, so for every $2<\alpha<\omega_1$ it will have a Borel subgroup which is $\mathbf{\Pi}^0_\alpha$ but not of lower complexity, by Farah, Ilijas, and Sławomir Solecki. "Borel subgroups of Polish groups." Advances in Mathematics 199.2 (2006): 499-541. | |
| Nov 5 at 10:17 | history | asked | Nick Belane | CC BY-SA 4.0 |