True or false: infinite sequence in a compact topological group is dense.

Question

This is from an exercise in Bredon's Topology and Geometry:

Let $G$ be a compact topological group (assumed to be Hausdorff). Let $g\in G$ and define $A=\{g^n:n=0,1,2...\}$. Then show that the closure $\bar{A}$ is a topological subgroup.

Note if the assumption of compactness is dropped, then the statement is false. A counterexample is $\mathbb{N}\subset \mathbb{R}$ as additive sets.

Note: if we add the assumption of $G$ being 1st countable, this question is easy to answer.

So, if the set $A$ is finite, then it is a cyclic group and it is already closed. In the infinite case, I can only think of the example of an infinite subgroup of the circle group, which is dense in compact group. I am not sure how to proceed with the proof of this general fact. Any hint will be appreciated.

Answer 1

Here is a proof which assumes that $G$ is first countable. Since $G$ is compact, there is a subsequence $(g^{n_k})_{k\in\Bbb Z_+}$ of $(g^n)_{n\in\Bbb Z_+}$ which converges to some $a\in G$. Clearly, $a$ is in fact an element of $A$. But then the sequence $(g^{n_k-1}a)_{k\in\Bbb Z_+}$ is a sequence of elements of $A$ which converges to $g^{-1}a$. So, $g^{-1}a\in A$. By the same argument, $(\forall N\in\Bbb N):g^{-N}a\in A$. But then $(g^{-n_k-1}a)_{k\in\Bbb Z_+}$ is a sequence of elements of $A$ which converges to $g^{-1}$ and this proves that $g^{-1}\in A$. By the same argument, $(\forall N\in\Bbb N):g^{-N}\in A$. Can you take it from here?

If you drop the assumption that $G$ is first countable, you can still get the conclusion that you want to get, using nets in the proof, as suggested by Eric Wofsey in the comments.

Answer 2

Multiplication by an element of $G$ is a homeomorphism of $G$, so $g\bar A = \overline{gA}$. Also $\bar A \bar A \subset \bar A$, and all elements of $\bar A$ commute with each other (I think this proof works: Let $G$ be a compact group. If $\{a^n\}_{n \in \mathbb{Z}}$ is dense in $G$, then $G$ is abelian.).

Case 1: $1 \in g\bar A$. Then $g^{-1} \in \bar A$, and we are done.

Case 2: $1 \notin g \bar A$. Then there exists an open set $N \ni 1$ such that $N \cap g\bar A = \emptyset$. Thus $(g\bar A) \bar A \subset g \bar A$, and hence for any $x \in g\bar A$ we have that $x \bar A \cap N = \emptyset$.

Now $\{x N:x \in \bar A\}$ is an open cover for $\bar A$. We show it has no empty subcover. Given elements $x_1,\dots,x_n \in \bar A\setminus\{1\} = g\bar A$, see that for $x_k^{-1} x_1 \cdots x_n = x_1 \cdots x_{k-1} x_{k+1} \cdots x_n \in g\bar A$. Thus $N \cap x_k^{-1} x_1 \cdots x_n \bar A = \emptyset$. Thus $x_1 \cdots x_n \notin x_k N$. Therefore $x_1\cdots x_n \notin N \cup x_1 N \cup \dots \cup x_n N$. Thus Case 2 leads to a contradiction.