MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

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.

share|cite|improve this question
up vote 10 down vote accepted

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).

share|cite|improve this answer
Ok, I now know what the Bohr topology is. Is there a standard text for reading up on this (that includes the failure to be sequential)? – Jeremy Brazas Nov 28 '12 at 23:38
Jeremy, I don´t know of a standard text but a good paper is "The maximal totally bounded group topology on G and the biggest minimal G-space, for abelian groups G" of E.K. van Douwen. It includes a proof of the fact that an infinite abelian group with its Bohr topology contains no non-trivial convergent sequence. – Ramiro de la Vega Nov 29 '12 at 12:44
Thanks, this is an excellent reference. – Jeremy Brazas Nov 30 '12 at 14:19

It is not the answer but an important point, too long for a comment. There is no Hausdorff group topology on Z so that the sequence of prime numbers $2, 3, 5 ...$ converges to zero. If there were such topology, there would be finitely many pairs of primes of each given gap $k$. But that contradicts Zhang's Gap Theorem

share|cite|improve this answer
This was short enough for a comment :) – YCor Sep 30 '14 at 13:51
prime numbers... corrected – Bugs Bunny Sep 30 '14 at 13:56
and Hausdorff as well – Bugs Bunny Sep 30 '14 at 14:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.