Closure in a topological group Let $G$ be a topological group with identity element $e$. If $A,B$ are subsets of $G$ we define
$A^{-1} := \{ a^{-1} \mid a \in A\} \quad \quad AB:=\{ab \mid a \in A , b \in B\}$
Let $A \subseteq G$, then $\overline{A} = \bigcap \{AU | U \text{ neighborhood of } e\} = \bigcap \{AU^{-1} \mid U \text{ neighborhood of } e\} $
Proof:


*

*If $U$ is a neighborhood of $e$ also $U^{-1}$ is a neighborhood of $e$:


If $U$ is a neighborhood of $e$ then there is $A \subseteq U$ open s.t. $e \in A$. Then $e= e^{-1} \in A^{-1} \subseteq U^{-1}$ and $A^{-1}$ is open because the inverse is an homeomorphism. Then $U^{-1}$ is a neighborhood of $e$.


*Let $x \in G$, then every neighborhood of $x$ is in the form $xU$ where $U$ is some neighborhood of $e$:


Let $V$ be a neighborhood of $x$, then $U:=x^{-1}V$ is a neighborhood of $e$ and $xU = V$.


*Thesis:


Let $x \in \bigcap \{AU \mid U \text{ neighborhood of } e\}$. By contradiction suppose exists a neighborhood of $x$, call it $V$, s.t. $V \cap A = \emptyset$. By 2. exists $W$ neighborhood of $e$ s.t. $V=xW$. Then $e \notin A(xW)^{-1} = AW^{-1}x^{-1}$ and then $x \notin AW^{-1}$ which is a contradiction. Then $V \cap A \ne \emptyset$ for every $V$ neighborhood of $x$, i.e. $x \in \overline{A}$ and then $\bigcap \{AU \mid U \text{ neighborhood of } e\} \subseteq \overline{A}$.
Let $x \in \overline{A}$, i.e. $\forall \, V$ neighborood of $x$ we have $V \cap A \ne \emptyset$, then $\forall \, U$ neighborood of $e$ we have $(xU) \cap A \ne \emptyset$, then $\forall \, U$ neighborood of $e$ we have $e \in AU^{-1}x^{-1}$ i.e. $\forall \, U$ neighborood of $e$ we have $x \in AU^{-1}$. Then $x \in \bigcap \{AU \mid U \text{ neighborhood of } e\}$ and then $\overline{A} \subseteq \bigcap \{AU \mid U \text{ neighborhood of } e\}$.
Then the first equality is proved. The second one follows from 1.
Is it correct?
 A: Looks good to me. You can shorten the proof of 1: taking the inverse is a homeomorphism, so $U^{-1}$ is a neighborhood of $e^{-1}=e$ because it's the inverse image under “taking the inverse” of the neighborhood $U$ of $e$ (continuity suffices).
Part 2 is good.
For the sake of brevity, let me denote by $\mathscr{U}_e$ the set of neighborhoods of $e$.
In part 3 you can avoid most notation and also contradiction.
Suppose $x\in\bigcap\{AU\mid U\in\mathscr{U}_e\}$ and $x\notin\overline{A}$ and consider an arbitrary neighborhood $xW$ of $x$ (for some $W\in\mathscr{U}_e$). By assumption and 2, $x\in AW^{-1}$, which means that, for some $a\in A$ and $w\in W$, $x=aw^{-1}$. Hence $xw=a\in A$ and $xW\cap A\ne\emptyset$. Therefore $\bigcap\{AU\mid U\in\mathscr{U}_e\}\subseteq\overline{A}$.
Conversely, suppose $x\in\overline{A}$ and let $W\in\mathscr{U}_e$. Then $xW^{-1}\cap A\ne\emptyset$ so we can write $xw^{-1}=a$ for some $w\in W$ and $a\in A$. Hence $x=aw\in AW$. As $W$ is arbitrary, we conclude $\overline{A}\subseteq\bigcap\{AU\mid U\in\mathscr{U}_e\}$
