Characterization of precompact sets 
Definition: We say that a set $A$ of a Hausdorff topological vector space is precompact if every ultrafilter on $A$ is a Cauchy filter.

I'm reading the proof of the following characterization, from John Horváth's book, of precompact sets.

Theorem 2: Let $E$ be a Hausdorff topological vector space and $A$ a subset of $E$. The following conditions are equivalent:
a) $A$ is precompact
b) If $\hat{E}$ is a completion of $E$ and $j:E \rightarrow \hat{E}$ the isomorphism onto the dense subspace $E_0$ of $\hat{E}$, then the closure of $j(A)$ is a compact subset of $\hat{E}$
c) For every neighbohood $V$ of $0$ in $E$ there exists a finite family $(x_i)_{1 \leq i \leq n}$ of points of $A$ such that $A \subset \bigcup_{i=1}^{n} (x_i+A)$.

I'm having trouble understanding the proof of $a) \Rightarrow b)$
Proof: $a) \Rightarrow b)$ Suppose that $B=\overline{j(A)}$ is not compact in $\hat{E}$ and let us prove that $A$ is not precompact. There exists on $B$ a filter $\mathfrak{F}$ which has no adherent point. The collection of all sets $X+V$, where $X \in \mathfrak{F}$ and $V$ is a balanced neightborhood of $0$ in $\hat{E}$, is the basis of a filter $\mathfrak{O}$ on $\hat{E}$. Since every set $X+V$ meets $A$, the trace $\mathfrak{O}_A$ of $\mathfrak{O}$ on $A$ is a filter on $A$. Let $\mathfrak{U}$ be an ultrafilter on $A$ finer than $\mathfrak{O}$. If $\mathfrak{U}$ were a Cauchy filter, it would generate a Cauchy filter on $B$ which would converge to some point $x \in B$. But then $x$ wold be clearly an adherent point of $\mathfrak{F}$, in contradiction to our original assumption.
More precisely, I was unable to justify the following passage:

If $\mathfrak{U}$ were a Cauchy filter, it would generate a Cauchy filter on $B$ which would converge to some point $x \in B$. But then $x$ wold be clearly an adherent point of $\mathfrak{F}$, in contradiction to our original assumption.

The author is identifying $A$ with $j(A)$, and this is confusing me. To arrive at a contradiction, we would have to find an ultrafilter in $A$ that is not a Cauchy filter. The natural candidate would be $j^{-1}(\mathfrak{U})$, but I couldn't understand what is the relationship of this ultrafilter with $\mathfrak{F}$.
 A: If you avoid identifying $A$ with $j(A)$, then the filter which the proof calls $\mathfrak{O}_A$ is $\{j^{-1}(S)\cap A:S\in\mathfrak{O}\}$.  So, $\mathfrak{U}$ is just some ultrafilter on $A$ containing this filter $\mathfrak{O}_A$.  Now suppose $\mathfrak{U}$ is Cauchy.  Then the pushforward filter $j_*(\mathfrak{U})=\{S\subseteq B:j^{-1}(S)\in\mathfrak{U}\}$ is also a Cauchy filter on $B$, and thus converges to some point $x\in B$.
The key claim is now that this point $x$ is an adherent point of the original filter $\mathfrak{F}$ (giving a contradiction since $\mathfrak{F}$ was supposed to not have an adherent point).  To prove this, we unravel the definitions.  Let $U$ be a neighborhood of $x$ in $B$.  Since $j_*(\mathfrak{U})$ converges to $x$, $U\in j_*(\mathfrak{U})$.  By definition, this means $j^{-1}(U)\in\mathfrak{U}$.  Since $\mathfrak{O}_A\subseteq\mathfrak{U}$, this means that $j^{-1}(U)$ has nonempty intersection with every element of $\mathfrak{O}_A$.  By definition of $\mathfrak{O}_A$, this means that for any $S\in\mathfrak{O}$, $(j^{-1}(S)\cap A)\cap j^{-1}(U)\neq \emptyset$, and therefore $S\cap U\neq \emptyset$.  By definition of $\mathfrak{O}$, this means that for any balanced neighborhood $V$ of $0$ in $\hat{E}$ and any $X\in\mathfrak{F}$, $(X+V)\cap U\neq \emptyset$, or equivalently $X\cap (U+V)\neq \emptyset$.  Since $U$ here is an arbitrary neighborhood of $x$ in $B$ and $V$ is an arbitrary neighborhood of $0$, $U+V$ can be an arbitrarily small neighborhood of $x$ in $B$.  Thus every $X\in\mathfrak{F}$ has nonempty intersection with every neighborhood of $x$, which means exactly that $x$ is an adherent point of $\mathfrak{F}$.
