Question

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?

Answer 1

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.

Answer 2

Consistently, you can get even more, as noted by Hrusak and Ramos-Garcia in this paper: http://www.matmor.unam.mx/~michael/reprints_files/precompact-groups.pdf

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 $x$.