Question

In a Hausdorff topological group, how can I show that every infinite topological group has a countable open subgroup?

Answer 1

I believe the question the poster is trying to ask is, "Why is theorem 3.6 of the article http://link.springer.com/article/10.1023%2FA%3A1010466924961#page-1 true?" Certainly this seems to be something the OP cares about, so I'll address it. (I think this question is borderline for MO, since the answer seems to be rather trivial - feel free to downvote this answer if you think the question is definitely inappropriate, although please say that's why you're downvoting.)

Definition (1.1 in the cited paper): If $G$ is a group, a $T$-sequence $\alpha=\langle a_n\rangle_{n\in\omega}$ is a sequence of elements of $G$ which converges to 0 (the authors say "vanishes;" I presume that's what this means) in some non-discrete topology on $G$. A topology $\tau$ on $G$ is determined by $\alpha$ if $\tau$ is a maximal topology in which $\alpha$ converges to 0.

The theorem the OP is asking about is:

Theorem (3.6): If $\tau_1$ is a topology on an infinite group $G$ determined by some $T$-sequence,then $\tau_1$ is complemented by some topology $\tau_2$ also determined by a $T$-sequence.

The proof of this theorem, in its entirety, is:

Note that $(G, \tau_1)$ has a countable open subgroup. Now, apply Theorems 1.6 and 3.5 and Lemma 2.3.

The part the OP seems to be asking about is the first sentence. The key is that in the theorem's hypothesis the topological group $(G, \tau_1)$ is assumed to be generated by some $T$-sequence $\alpha$. The reason this matters is that if $(G, \tau_1)$ is determined by $\alpha$, then clearly $A\cup\lbrace 0\rbrace$, where $A$ is the underlying set of $\alpha$, must be open - since $\tau_1$ is maximal among the topologies in which $\alpha$ converges to 0. Now the desired countable open subgroup is just the group generated by $A$.