I'm approaching to topological groups and I was reading "Introduction to topological groups" by Dikranjan: https://users.dimi.uniud.it/~dikran.dikranjan/ITG.pdf . In demonstrating Theorem 3.1.5, he asserts: "Let $\mathcal{V}$ be a filter on $G$ satisfying all conditions (a), (b) and (c). Let us see first that every $U \in \mathcal{V}$ contains $e_{G}$. In fact, take $W \in \mathcal{V}$ with $W^{2} \subseteq U$ and choose $V \in \mathcal{V}(e_{G})$ with $V \subseteq W$ and $V^{-1} \subseteq W$".
Why does this $V$ exists? And why is it in $\mathcal{V}(e_{G})$? He hasn't yet defined a topology on $G$, so who is $\mathcal{V}(e_{G})$?
Anyone can help me?
It's just a typo: it should say $V\in\mathcal{V}$, not $V\in\mathcal{V}(e_G)$. This is just using properties (a) and (b) that are assumed to hold for the filter $\mathcal{V}$: since $U\in\mathcal{V}$, by (a) there exists $W\in\mathcal{V}$ such that $W\cdot W\subseteq U$ and then by (b) there exists $V'\in\mathcal{V}$ such that $V^{-1}\subseteq W$. If you then take $V=V'\cap W$ it will be an element of $\mathcal{V}$ that satisfies both $V\subseteq W$ and $V^{-1}\subseteq W$.