A property of compact group $G$ is a compact group with identity $e$. Prove: $$\forall x\in G,   e\in \overline{(x^n)}_{n\in\mathbb{N}}$$
The overline indicates the closure of this sequence. I'm confused because there is no distance or norm defined in the question, I don't know what to do with the condition "compact". Could someone help me? 
Thanks~
 A: First of all if $x$ is of finite order then the statement is trivial since $e=x^n$ for some $n\in\mathbb{N}$. So assume that $x$ is of infinite order.
From the general topology we know that
$$z\in\overline{A}$$
if and only if for any open neighbourhood $U$ of $z$ we have $U\cap A\neq\emptyset$.
So lets take any open neighbourhood $U$ of $e$. Then there exists a symmetric, almost product-closed sub-neighbourhood $V$ of $U$, i.e.
$$V\mbox{ is open}$$
$$e\in V$$
$$V\subseteq U$$
$$V^{-1}=V$$
$$VV\subseteq U.$$
The last property is what I call "almost product-closed" (it's not a standard nomenclature, just a name I came up with just now). So $V$ is pretty close to being a subgroup, except that $VV$ can actually be a bit bigger then $V$. We will use these properties later.
Now since $G$ is a topological group, then $\{gV\}_{g\in G}$ is an open cover of $G$. Furthermore since $G$ is compact then this cover has a finite subcover
$$\{g_1V,\ldots,g_mV\}$$
Now since $x$ is of infinite order, then the pigeonhole principle implies that there are $i>j$ such that $x^i,x^j\in g_kV$ for some $k$. So write
$$x^i=g_ka$$
$$x^j=g_kb$$
for some $a,b\in V$. Then $x^{-j}=b^{-1}g_k^{-1}$ and so
$$x^{i-j}=x^{-j}x^i=b^{-1}g_k^{-1}g_ka=b^{-1}a$$
In particular $x^{i-j}\in V^{-1}V$ and since $V$ is symmetric and almost product-closed, then $x^{i-j}\in U$. Finally $i>j$ implies that $x^{i-j}\in\{x^n\}_{n\in\mathbb{N}}$ and thus we've shown that
$$U\cap \{x^n\}_{n\in\mathbb{N}}\neq\emptyset$$
which completes the proof. $\Box$
