Generated group by a compact set is finitely generated Let $G$ be a locally compact group, and let $K\subseteq G$ be a compact set containing the identity. I want to make sure if the following statements are correct:
(i) Is it true that  $\langle K\rangle= \underset{m \in \mathbb{Z}}{\cup}K^m$?
(ii) Is it true that $\langle K\rangle$ is finitely generated?
(iii) Is $\langle K \rangle$ closed?
I have a suspicion that all three are true but am not sure.
The question stems from me trying to prove that a locally compact group has an open subgroup which is locally compact and $\sigma$-compact. I suspect $\langle K\rangle$ is the desired subpgroup.
These three points are questions which I am not sure on since I am a bit weak on group theory. I'm pretty sure the first one is true. The second and third are a 'hail mary', which I have not seen somewhere but suspect them to be true.
 A: All three of these are incorrect, as it turns out. Here are counterexamples. For all three, take $G = \mathbb{R}$. 


*

*Let $K = \{0,\frac{1}{2}, \frac{1}{3}\}$. Note that $\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\in\langle K\rangle$, but since


$$K^m = \left\{\frac{a}{2}+\frac{b}{3}: a,b\in\mathbb{Z},~\operatorname{sgn}(a) = \operatorname{sgn}(b), ~|a+b|\le m\right\}$$
there is no $m$ such that $\frac{1}{6}\in K^m$.


*Let $K = [0,1]$. Since any real number is some integer multiple of some $k\in K$, we have $\langle K\rangle = G$. However, $G$ is not finitely generated.

*[See Yanko's answer for a good counterexample here.]
A: I find some flaws with the existing answer (Now all are corrected) so I post my own:
The answer to $(i)$ is a little complicated because you didn't define $K^n$ properly so I'll answer it below. The answer to $(ii),(iii)$ is no, even if $G$ is compact!
Counter-examples
(ii). Take $G=S^1$ (i.e. the unit circle, $S^1=\{z\in\mathbb{C}:|z|=1\}$ with complex multiplication, and $K=G$. 
(iii). Take $G=S^1$ and $K = \{1,\alpha\}$ where $\alpha\in S^1$ any element that is not a root of unity (i.e. there's no $n\in\mathbb{N}$ such that $\alpha^n=1$). Then $\left<K\right>$ is dense and not closed.
For $(i)$ it really depends on how you define $K^n$.
Case 1: (corrected) it is false! (check out santana's answer)
Case 2: $K^n = \{k^n:k\in K\}$ in this case the answer is negative. Take $G=S^1$ and $K=\{1,e^{2\pi i\alpha},e^{2\pi i \beta}\}$ where $1,\alpha,\beta$ are independent over $\mathbb{Q}$. In this case the element $e^{2 \pi i (\alpha+\beta)}$ does not exist in $\bigcup_{n\in\mathbb{Z}} K^n$ as it's not a power of any element or an inverse of an element in $\{1,e^{2\pi i \alpha},e^{2\pi i \beta}\}$.
ADDED

The question stems from me trying to prove that a locally compact group has an open subgroup which is locally compact and σ-compact. I suspect ⟨K⟩ is the desired subpgroup.

This is true if you choose $K$ correctly. If the set $K$ is the closure of a symmetric open set then the group $\left<K\right>$ is $\sigma$-compact (by $(i)$ if $K$ is symmetric) it is open (because it contains an open set, so $K=\bigcup_{k\in K} kU$) and hence it is also closed (any open subgroup is also closed, as it is the complement of it's non-trivial co-sets).
In other words if you choose a compact set $K$ which contains a neighborhood of the identity $1\in U \subseteq K$ (you can do that, because $G$ is locally compact) then $\left<K\right>$ is a closed $\sigma$-compact subgroup of $G$. If you understand my answer to your other questions above you can see that everything fails if $K$ does not contain an open neighborhood.
