the closure of a set in a topological group

Question

This question is a exercise from Folland's Real analysis (chapter 11, question 1).

If $G$ is a topological group and $E\subset G$, then the closure $\bar{E}=\cap\{EV: V$ is a neighborhood of $ e\}$.

Any help would be apreciated!

Answer 1

I assume that $G$ is Hausdorff and second countable (to avoid nets). Let $F(E)=\cap \{VE$, $V$ is a neighborhood of the identity$\}$. Firstly, we remark that $F$ is closed. Suppose that $x$ is not in $F(E)$, there exists a neighborhood $V$ such that $x$ is not in $EV$. There exists an open subset $U$ such that $U=U^{-1}$ and $UU^{-1}\subset V$, see the reference. $x$ is also not in $EU$, as $EU \subseteq EV$.

We claim that $EU\cap xU=\emptyset$: if not, let $y=eu=xu', u,u'\in U, e\in E$, we have $x=eu{u'}^{-1}$, which implies that $x\in EU$ since $UU^{-1}\subset V$. This implies that the complement of $F$ is open and $F(E)$ is closed.

Let $L$ be any closed subset, we claim that $F(L)=L$. Let $x\in F(L)$, and $U_n$ a sequence of neighborhoods of $e$ such that $\cap_nU_n=\{e\}$. $x=u_nf_n, u_n\in U_n, f_n\in F$. This implies that $f_n=xu_n^{-1}$ and converges towards $x$, we deduce that $x\in F$.

Let $E$ be any subset, and $L$ a closed subset which contains $E$, for every neighborhood $V$ of the identity, $EV\subset LV$ implies that $F(E)\subset F(L)=L$, this implies that $F(E)$ is closed subset contained in every closed subset which conatains $E$ so it is the adherence of $E$.

Every neighborhood of identity in a topological group contains the product of a symmetric neighborhood of identity.