Hausdorff group topologies on finitely generated groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:57:00Z http://mathoverflow.net/feeds/question/114816 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114816/hausdorff-group-topologies-on-finitely-generated-groups Hausdorff group topologies on finitely generated groups Jeremy Brazas 2012-11-28T22:03:00Z 2012-11-28T23:08:45Z <p>Suppose \$G\$ is a finitely generated Hausdorff topological group. Must \$G\$ be first countable (or perhaps a sequential space)? What if we restrict to the abelian case?</p> <p>I wonder if this is even true for the additive group of integers \$\mathbb{Z}\$. There certainly are non-discrete, Hausdorff group topologies on \$\mathbb{Z}\$ where a basis at \$0\$ consists of subgroups (such as that used in Furstenberg's proof of the infinitude of primes). On the other hand, determining if there is a Hausdorff group topology that makes a given sequence converge to \$0\$ is non-trivial. For instance, it is known that the sequence of squares \$n^2\$ can't converge to \$0\$ in any Hausdorff group topology and that if there is a Hausdorff group topology on \$\mathbb{Z}\$ such that the sequence of primes \$2,3,5,...,p,..\$ converges to \$0\$, then the twin prime conjecture is false.</p> http://mathoverflow.net/questions/114816/hausdorff-group-topologies-on-finitely-generated-groups/114821#114821 Answer by Ramiro de la Vega for Hausdorff group topologies on finitely generated groups Ramiro de la Vega 2012-11-28T22:52:43Z 2012-11-28T23:08:45Z <p>No. The Bohr topology on \$\mathbb{Z}\$ is not first countable, in fact the least size of a local base at \$0\$ is \$2^{\aleph_0}\$. It is also known that this topology is not sequential (because there are no non-trivial convergent sequences).</p>