# Countable open subgroup

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

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The real numbers under addition is a Hausdorff group with no countable open subgroup. What actually did you want to ask? – Robin Chapman Jun 29 '10 at 19:00
Indeed, no nondiscrete locally compact group has a countable open subgroup, as follows from the regularity properties of a Haar measure. – Pete L. Clark Jun 29 '10 at 19:10
Please give author, journal and year (or even two out of three) for this article. The definition you quote is rather hard to understand in isolation. Also, the comments above show that not every Hausdorff abelian group has a countable open subgroup, so I think you need to consider much more carefully what the article is actually saying. – Yemon Choi Jun 29 '10 at 23:40
(I'm not sure if low-rep users can edit their own questions. If not, please write something in an answer and someone else can paste it into your original question as an edit. But as it stands the question is just wrong.) – Yemon Choi Jun 29 '10 at 23:42
Voting to close since the question still has not been edited to make sense – Yemon Choi Jul 12 '10 at 7:39

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.

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.
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 $A$.