I'm reading the proof following lemma in Horváth's book.
Lemma: Let $E$ be a topological vector space and $A$ a dense subset of $E$. If every Cauchy filter on $A$ converges to a point in $E$, then $E$ is complete.
Proof: Let $\mathfrak{F}$ be a Cauchy filter on $E$. The collection of all sets $x+V$, where $X \in \mathfrak{F}$ and $V$ is a balanced neighborhood of $0$ in $E$ is a basis of a filter $\mathfrak{G}$ on $E$ since if $X \in \mathfrak{F}$, $Y \in \mathfrak{F}$, and $V, W$ are balanced neighborhoods of $0$, then $(X+V)\cap(Y+W) \supset (X\cap Y)+(V \cap W)$. Furhtermore, $\mathfrak{G}$ is a Cauchy filter on $E$ since for any balanced neighborhood $V$ of $0$ exists $X \in \mathfrak{F}$ such that $X-X \subset V$ and thus $$(X+V)-(X+V) \subset V+V+V.$$ The sets $(X+V)\cap A$ are not empty since $A$ is dense in $E$. Thus $\mathfrak{G}$ induces a Cauchy filter $\mathfrak{G}_A$ on $A$, and by hypothesis $\mathfrak{G}_A$ converges to a point $x \in E$. But $x$ adheres to $\mathfrak{G}$ and since $\mathfrak{G}$ is a Cauchy filter , by Proposition 1 it converges to $x$. Consequently, $\mathfrak{F}$ which is a finer than $\mathfrak{G}$ also converges to $x$.
I think that $\mathfrak{G}_A$ is the filter in $A$ generated by the basis $\{(X+V)\cap A: X \in \mathfrak{F} \hbox{ and } V \hbox{ is a balanced neighborhood of 0}\}$, that is, $\mathfrak{G}_A=\{U \subset A: U \supset (X+V)\cap A \hbox{ for some } X \in \mathfrak{F} \hbox{ and } V \hbox{ a balanced neighborhood of 0}\}$
My question: What is the definition of: A filter on $A$ converges to a point of $E$?
According to definition in page 124, a filter $\mathscr{F}$ on $X$ converges to a point of $X$ if $\mathscr{F}$ is finer than $\mathfrak{B}(x)$, where $\mathfrak{B}(x)$ is the filter of all neighborhoods of $x$.
However, with this definition a filter on $A$ converges to a point of $E$ if $\mathfrak{B}(X) \subset \mathscr{F}$, which is strange because the elements of $\mathscr{F}$ are subsets of $A \subsetneq E$ and for example $E$ is a neighborhood of $x$ and consequently $E \in \mathscr{F}$ without that $E \subset A$.