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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.

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up vote 7 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).

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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
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