A problem of nets in topology Let $G$ be a topological group with neutral element $e$. Let $\pi \colon G \to B(E)$ a (non-continuous) representation of $G$ on a Banach space $E$ by bounded linear operators. 
Let $T$ an element of the intersection $\bigcap_{V \in \mathcal{V}_e} \overline{\{\pi(t)\ :\ t \in V\}}$ where $\mathcal{V}_e$ is the set of neighbourhoods of $e$ and where the closure is for the weak operator topology.
How prove that there exists a net $(t_i)_{i \in I}$ of $G$ converging to $e$ such that $(\pi(t_i))_{i \in I}$ converge to $T$ in the weak operator topology?
ps: The weak operator topology on $B(E)$ is defined by the seminorms $p_{x,x^*}(T)=|\langle T(x),x^* \rangle|$ for $x \in E$ and $x^* \in E^*$.
 A: You need the result that the topological closure of a set $A$ is the set of limits points of nets in $A$.
To show it let $x$ be an element of $\overline{A}$, the set $\mathcal V_x$ of neighbourhoods of $x$ is a directed set and each neighbourhood of $x$ intersects $A$. So take as a net the map $\mathcal V_x \to A$ where each neighbourhood $V$ is sent to an element that lies in $V\cap A$. This construction converges trivially to $x$ and every element lies in $A$.
Now let $x$ be a limit point of some net $I \to A$ (write the elements as $x_i$ for $i\in I$). Since $x_i \to x$ every neighbourhood of $x$ must meet the set $\{x_i\}_{i\in I}$, but then (since $x_i \in A$) every neighbourhood of $x$ must meet $A$ and then $x$ lies in the closure of $A$.
With this the result is easy. Let $T$ be in the closure
$$\bigcap_{V \in \mathcal{V}_e} \overline{\{\pi(t)\ :\ t \in V\}}=\overline{\bigcap_{V \in \mathcal{V}_e} \{\pi(t)\ :\ t \in V\}}$$
Then there exists a net $I \to B(E)$ (write the elements as $T_i$ for $i \in I$), so that $T_i \in \bigcap_{V \in \mathcal{V}_e} {\{\pi(t)\ :\ t \in V\}}$ for all $i$ and $T_i \to T$.
Now let $t_i$ be any pre-image of $T_i$, then $t_i \in  \bigcap_{V \in \mathcal{V}_e} V$ for all $i$. So every element of the net lies in every neighbourhood of $e$ and $t_i$ must converge to $e$.
