Closure of identity element in a topological group is the intersection of its neighborhood system Consider a topological group $G$. I want to prove that 

if $\cal{U}$ is the set of all neighborhoods of $e$ (the identity element), then $\overline{\{e\}}=\cap_{U\in \cal{U}}U$.

What I thought was:
i) If $x\in \overline{\{e\}}$, then every neighborhood $V$ of $x$ intersects $\{e\}$, hence every $V$ is a neighborhood of $e$ as well, i.e. $V\in \cal{U}$.
ii) On the other hand, if $x\in \cap_{U\in \cal{U}}U$, then $x\in U$ for all $U \in \cal{U}$, meaning every $U$ is a neighborhood of $x$ intersecting $\{e\}$.
Now, I think that the missing part is an argument justifying that there is no neighborhood of $e$ that is not a neighborhood of $x$ (part i)), and that there is no neighborhood of $x$ that is not a neighborhood of $e$ (part ii)). My idea for this is to argue that there is a homeomorphism taking $e$ to $x$, therefore the neighborhood system of one cannot have more elements that the neighborhood system of the other.
Is this correct? If not, is there a way I can fix it?
 A: You have the right idea, but the details need a bit of work. I’ll take you through it by a slightly roundabout path that corresponds to how you might actually discover the argument.
The group operation is continuous, so the map 
$$h_x:G\to G:y\mapsto xy$$
is continuous, and $h_x(e)=x$. It’s easy to check that $h_{x^{-1}}$ is its inverse, so $h_x$ is in fact a surjective homeomorphism. Since $\mathscr{U}$ is the nbhd system at $e$, this means that $\{h_x[U]:U\in\mathscr{U}\}$ is the nbhd system at $x$. 
Suppose that $x\in\left(\bigcap\mathscr{U}\right)\setminus\operatorname{cl}\{e\}$; then there must be a $U\in\mathscr{U}$ such that $e\notin h_x[U]$. This in turn implies that $x^{-1}=h_{x^{-1}}(e)=h_x^{-1}(e)\notin U$. If we could infer from this that $x\notin U$, we’d have our contradiction and could conclude that $\bigcap\mathscr{U}\subseteq\operatorname{cl}\{e\}$, thereby completing the proof. Unfortunately, that inference isn’t legitimate.
It would be legitimate, however, if $U$ were a symmetric nbhd of $e$, meaning that for each $y\in G$, $y\in U$ iff $y^{-1}\in U$. There’s no guarantee that our original $U$ is symmetric, but I claim that it contains a symmetric nbhd of $e$. Specifically, if $U^{-1}=\{u^{-1}:u\in U\}$, use continuity of the inverse operation to verify that $U^{-1}\in\mathscr{U}$. Then let $V=U\cap U^{-1}$, and verify that $V$ is a symmetric nbhd of $e$ contained in $U$. Then you can replace $U$ in the preceding paragraph by $V$ and get the desired contradiction.
