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Completely unaware of the Bohr topology, I recently asked whether or not there was a Hausdorff group topology on the integers $\mathbb{Z}$ which made the group fail to be first countable. For me, this topological group is a bit extreme since there are no non-trivial convergent sequences. I'm very interested to know if there is a sequential example.

If $\mathbb{Z}$ is given a Hausdorff group topology which makes it a sequential space, must it be first countable?

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I believe there are sequential non-Frechet (hence non-first-countable) Hausdorff group topologies on any abelian group. Unfortunately I don´t remember where I remember this from. – Ramiro de la Vega Nov 30 '12 at 18:16
up vote 6 down vote accepted

The answer is no.

It is proved in Topologies on Abelian Groups (E.G. Zelenyuk and I.V. Protasov, Math. USSR Izvestiya, 1991), that on every infinite abelian group there exists a sequential Hausdorff group topology which is not first-countable.

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Thanks very much Ramiro. – Jeremy Brazas Dec 14 '12 at 19:22

Consistently, you can get even more, as noted by Hrusak and Ramos-Garcia in this paper:

There are consistent examples of Fréchet-Urysohn non-first-countable Hausdorff group topologies on $\mathbb{Z}$.

Fréchet-Urysohn means that for every non-closed set $A$ and point $x \in \overline{A} \setminus A$ there is a countable sequence inside $A$ converging to $x$. Fréchet-Urysohn is the same as every subspace is sequential. Malykhin's problem asks for an example of a countable non-metrizable Fréchet-Urysohn topological group. The consistency of a positive answer to it has been known for some time, and recently Hrusak and Ramos-Garcia came up with a proof of the consistency of a negative answer, thus establishing its independence from ZFC.

Take a family of $\omega_1$ many distinct characters on $\mathbb{Z}$ separating the points of $\mathbb{Z}$ and consider the coarsest topology making each of those characters continuous. This is a Hausdorff group topology on $\mathbb{Z}$ with no countable local base and a base of cardinality $\omega_1$. The latter implies that it is Fréchet-Urysohn in any model of ZFC+$\mathfrak{p}>\omega_1$ (see below).

Call a family $\mathcal{F}$ of infinite subsets of a countable set strongly centered if every finite subfamily of $\mathcal{F}$ has infinite intersection. We say that a set $S$ is a pseudointersection of the family $\mathcal{F}$ if $S \setminus F$ is finite for every $F \in \mathcal{F}$. Now $$\mathfrak{p}:=\min \{|\mathcal{F}|: \mathcal{F} \mbox{ is a strongly centered family without an infinite pseudointersection} \}$$

Since $\mathcal{F}$ is a family of subsets of a countable set we have $\mathfrak{p} \leq \mathfrak{c}$. Moreover, you can cook up an infinite pseudointersection of a given countable family of infinite subsets of a countable set by an easy diagonalization, so $\mathfrak{p} \geq \omega_1$. It is known that $\mathfrak{p}>\omega_1$ is consistent. For example, under Martin's Axiom we have $\mathfrak{p}=\mathfrak{c}$, and hence it suffices to take a model of Martin's Axiom plus the negation of the Continuum Hypothesis.

Every countable topological space with a local base of cardinality $<\mathfrak{p}$ at every point is Fréchet-Urysohn.

Proof: Let $A \subset X$ be a non-closed set and $x \in \overline{A} \setminus A$. Let $\{U_\alpha: \alpha < \kappa \}$ enumerate a local base at $x$, where $\kappa < \mathfrak{p}$. Then $\mathcal{F}=\{U_\alpha \cap A: \alpha < \kappa \}$ is a strongly centered family of subsets of the countable set $A$. Since $\mathcal{F}$ has cardinality smaller than $\mathfrak{p}$ we can fix an infinite pseudointersection $S \subset A$ of $\mathcal{F}$. Now $S$ is a sequence inside $A$ which converges to $x$.

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