Let $\mathscr{C}_E$ be the collection of all Cauchy filters on the TVS $E$ and define the following relation in $\mathscr{C}_E$:
$\mathscr{F} \sim_R \mathscr{G} \Leftrightarrow$ for all neighborhood $U$ of the origin in $E$, there exists $A \in \mathscr{F}$, $B \in \mathscr{G}$ such that $A − B ⊂ U$.
Consider the following operations in $\hat{E}=\mathscr{C}_E/\sim_R$, $$\bar{\mathscr{F}}+\bar{\mathscr{G}}=\overline{(\mathscr{F}+\mathscr{G})},$$ where $(\mathscr{F}+\mathscr{G})$ is the filter generated by the basis $(\mathscr{F}+\mathscr{G})_0=\{A+B: A \in \mathscr{F}, B \in \mathscr{G}\}$ and $$\lambda \bar{\mathscr{F}}=\overline{\langle \lambda \mathscr{F}\rangle }$$ where $\langle \lambda \mathscr{F} \rangle$ is the filter generated by the basis $\{\lambda A:A \in \mathscr{F}\}$.
My question: Let $i:E \rightarrow \hat{E}$ given by $i(x)=\{\mathscr{F} \in \mathscr{C}_E:\mathscr{F}\rightarrow x\}$, how to prove that $i(x+y)=i(x)+i(y)$ and $i(\lambda x)=\lambda i(x)$.
Let $\mathscr{F} \in i(x+y)$ I was trying to split the filter $\mathscr{F}$ into a sum of the type $(\mathscr{F}-\mathscr{F}(y))+\mathscr{F}(y)$, since $\mathscr{F}(y) \rightarrow y$, but I couldn't prove that $\mathscr{F}=(\mathscr{F}-\mathscr{F}(y))+\mathscr{F}(y)$.