Closure of a certain subset in a compact topological group Suppose that $G$ is a compact Hausdorff topological group and that $g\in G$. Consider the set $A=\{g^n : n=0,1,2,\ldots\}$ and let $\bar{A}$ denote the closure of $A$ in $G$.
Is it true that $\mathbf{\bar{A}}$ is a subgroup of $\mathbf{G}$?
From continuity of multiplication and the fact that $A\cdot A\subseteq A$ it is clear that $\bar{A}\cdot\bar{A}\subseteq\bar{A}$. therefore, $a,b\in \bar{A}$ yields $a\cdot b\in \bar{A}$. However, I am having trouble showing that inverses of elements in $\bar{A}$ are also in $\bar{A}$.

I really like Fischer's solution. I would like to post an answer based on Fischer's answer in the "filling the blank" spirit - certainly helpful for beginners like me.
We need to get help from the two theorems below (refer to section 1.15 in Bredon's Topology and Geometry):


*

*In a topological group $G$ with unity element $1$, the symmetric neighborhoods of $e$ form a neighborhood basis at $1$.

*If $G$ is a topological group and $U$ is any neighborhood of $1$ and $n$ is any positive integer, then there exists a symmetric neighborhood $V$ of $1$ such that $V^n\subset N$.


Since $A.A\subset A$, by the same argument using the continuity of the multiplication map to prove the closure of a subgroup is a subgroup, we have $\bar{A}.\bar{A}\subset\bar{A}$. Therefore, if $g^{-1}\in\bar{A}$, then $g^{-n}\in\bar{A}$ for any $n\in\textbf{N}$, and so the group $\langle g\rangle\subset\bar{A}$. Taking bar of $A\subset\langle g\rangle\subset\bar{A}$ yields $\bar{A}=\bar{\langle g\rangle}$, which means $\bar{A}$ is a subgroup due to being the closure of a subgroup.
We now show that $g^{-1}\in \bar{A}$. The case $g$ has finite order is trivial, so let $g$ has infinite order. Suppose $g^{-1}\notin\bar{A}$. Then by definition, there is a neighborhood $U'$ of $1$ that, by the homeomorphic translation, $g^{-1}U'\cap A=\emptyset$. Since the symmetric neighborhoods form a neighborhood basis of $1$, we can choose a symmetric neighborhood $U$ of $1$ such that $g^{-1}U\cap A=\emptyset$.
Since $A$ is an infinite set in a compact space, $A$ has a limit point $p$. We can choose a symmetric neighborhood $V$ of $1$ such that $V^2\subset U$. Then, $pV$, being a neighborhood of a limit point of $A$, must contain at least two distinct points $g^m$, $g^n$ ($m, n\in \textbf{N}$), i.e. $g^m=pa$, $g^n=pb$ for some $a,b\in V$. Due to $V$ being symmetric and $V^2\subset U$, $b^{-1}a=u$ for some $u\in U$, and we have $g^m=pa=pb.u\in g^n U\cap A$
Now, if $m>n$, then $g^{m-n-1}\in g^{-1}U\cap A$ (contradict with our definition of $U$). If $n>m$, then since $U^{-1}=U$, $g^m\in g^n U^{-1}$, and so $g^n\in g^m U\cap A$ (also contradict with our definition of $U$). Therefore $g^{-1}\in \bar{A}$.
