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.

Your title and body seem to ask two different questions. It is not in general true that $A$ is dense in $G$ if $G$ is compact and $g$ has infinite order. — Wojowu, Aug 02 '20 at 18:18
For a simple counterexample of an $A$ which is not dense look at $S^1\times S^1$ — Alessandro Codenotti, Aug 02 '20 at 18:21
I was thinking of using sequential compactness argument, but I noticed that 1st countability assumption is not included in the definition of topological groups. This leads me to condider density. — kzkzkzz, Aug 02 '20 at 18:27
A more general result is that for every subgroup $H$ of a topological group $G$, the closure $\overline{H}$ is a subgroup of $G$. The group $G$ need not be compact and the subgroup $H$ need not be cyclic. See https://proofwiki.org/wiki/Closure_of_Subgroup_is_Group or Prop. 1.4 in http://www.math.wm.edu/~vinroot/PadicGroups/topgroups.pdf. — KCd, Aug 02 '20 at 18:32
The setting here is that $A$ is just a subset, not necessarily a group. — kzkzkzz, Aug 02 '20 at 18:37
Ah, I misread the definition of $A$ as being all integral powers of $g$. — KCd, Aug 02 '20 at 19:12

José Carlos Santos · Accepted Answer · 2020-08-02T19:10:12.997

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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.

edited Aug 02 '20 at 19:10

answered Aug 02 '20 at 18:14

José Carlos Santos

427,504

We can't assume $a^{-1}\in A$, instead what we can easily show is $A$ closed under action of $g^{-1}$. – Berci Aug 02 '20 at 18:20
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Thanks for answering. My confusion is whether sequential compactness is a valid argument, since $G$ might not be first countable. – kzkzkzz Aug 02 '20 at 18:39
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@Berci Sure. I've edited my answer. I hope that everything is correct now. – José Carlos Santos Aug 02 '20 at 18:40
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@kzkzkzz Every compact topological space is sequentially compact. – José Carlos Santos Aug 02 '20 at 18:42
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@José $[0,1]^{[0,1]}$ and $\beta\omega$ are two of the many examples of spaces that are compact but not sequentially compact – Alessandro Codenotti Aug 02 '20 at 18:45
@AlessandroCodenotti Right. I made a mistake. I thought that every compact topological space is sequentially compact. In fact, in general, neither of these properties implies the other one. – José Carlos Santos Aug 02 '20 at 18:49
Maybe in this question we do need to assume 1st countability. – kzkzkzz Aug 02 '20 at 18:52
I don't have a copy of the book at hand right now. Perhaps the the author is assuming that. I don't know. – José Carlos Santos Aug 02 '20 at 18:54
No the author did not assume this. – kzkzkzz Aug 02 '20 at 18:56
The argument works if you just use nets instead of sequences. – Eric Wofsey Aug 02 '20 at 19:04
@EricWofsey Sure! I missed that. I shall edit my answer. Thank you. – José Carlos Santos Aug 02 '20 at 19:07
How do you know that the subnet of $g^n$ doesn't contain the identity element? Possibly infinitely many times? – Stephen Montgomery-Smith Aug 02 '20 at 20:47
I don't know. Is that a problem? – José Carlos Santos Aug 02 '20 at 20:50
You need to know that $n_k - 1 \ge 0$. – Stephen Montgomery-Smith Aug 02 '20 at 20:51
Otherwise you don't know that $g^{n_k-1} a \in A$. – Stephen Montgomery-Smith Aug 02 '20 at 20:52
But $(g^{n_k}){k\in\Bbb Z+}$ is a subsequence of $(g^n){n\in\Bbb Z+}$ and therefore $\lim_{k\to\infty}n_k=\infty$. – José Carlos Santos Aug 02 '20 at 20:58
It's not a subsequence. It is a subnet. – Stephen Montgomery-Smith Aug 02 '20 at 20:59
Right! I will have t tink about that. – José Carlos Santos Aug 02 '20 at 21:01

Stephen Montgomery-Smith · Answer 2 · 2020-08-02T23:08:08.740

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.

But this proof https://math.stackexchange.com/q/452375 is much better in my opinion. — Stephen Montgomery-Smith, Aug 02 '20 at 23:28

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

2 Answers2