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 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 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>