Convergence of a filter defined in a subspace I am reading about convergence of filters in topological spaces and cluster points. My doubt is the following: let $X$ be a topological space, $A\subset X$ and $\mathcal F$ a filter on $A$. Given $x \in \overline A$, what is the meaning of $\mathcal F \longrightarrow x$?
Bourbaki defines the convergence of a filter (page 68 - topology 1) which is defined in the whole space $X$. Of course that is not the case of $\mathcal F$ if $A\ne X$. But in Proposition 6 of page 70 they use this kind of convergence. I am assuming that this means that  $\mathcal F$ is finer than $\mathcal B(x)_A$, which is the trace of the filter of the neighborhoods of $x$. Is that correct? I found it very strange since Bourbaki's book (as far as I know) does not make this kind of imprecision (to work with a not given definition).
In Dugundji's book only the definition of convergence of a filterbase is given (Def. 2.1 of Chapter 10). So his Theorem 4.1 (that is the cited proposition of Bourbaki's book) is coherent.
 A: $\mathscr F\to x$ means that $\mathcal{N}_x\subset\mathscr F$, where $\mathcal N_x$ is the filter of neighbourhoods of $x$.
The fact that $x\in \overline A$ can be characterized in terms of convergence of filters. Indeed, $x\in \overline{A}$ iff there is some filter $\mathscr{F}\to x$ such that $A\in \mathscr{F}$.
$\Rightarrow$) Let $x\in \overline{A}$. For every $N\in\mathcal N_x$, $N\cap A\neq \emptyset$, so that $\{N\cap A:N\in\mathcal N_x\}$ is a filter base. Let $\mathscr F$ be the filter generated by it. Obviously, $A\in\mathscr F$.
$\Leftarrow$) Let $N\in\mathcal N_x$. Then $N\cap A\in\mathscr F$, so that $N\cap A\neq\emptyset$. This says that $x\in \overline A$.
A: If $\mathcal{F}$ is a filter on $X$ that contains $A \subseteq X$, then $\mathcal{F}_A:= \{B \subseteq A\mid B \in \mathcal{F}\}=\mathscr{P}(A) \cap \mathcal{F}$ is a filter on $A$ (it's clear that $\emptyset \notin \mathcal{F}_A \subseteq \mathcal{F}$, and $\mathcal{F}_A$ is also clearly closed under intersections and enlargements "within $A$"). And clearly any filter on $A$ contains $A$ and is a filterbase for a filter $\mathcal{F}^\uparrow$ (still containing $A$) on $X$ that obeys $\mathcal{F}^\uparrow_A = \mathcal{F}$ again.
So there is a natural correspondence between them.
It's also easy to see that for such a filter $\mathcal{F}$ on $X$ that contains $A$, and any $x \in A$, $\mathcal{F} \to x$ in $X$ iff $\mathcal{F}_A \to x$ in the subspace topology on $A$.
If $\mathcal{F}$ is a filter on $A$ then $\mathcal{F} \to x$ is defined as the convergence of a filterbase. So as $\mathcal{F}^\uparrow \to x$ if you want to restrict yourself to convergence of (full) filters on $X$. But IIRC Bourbaki defines convergence for filter bases in general anyway.
"Every point of $x$ of $\overline{A}$ is a limit point of a filter on $A$" is true as we can take that filter to be $\mathcal{F}:=\{N \cap A: N \in \mathcal{N}_x\}$, which is a filter on $A$ and a filterbase on $X$ for which $\mathcal F \to x$ is true and makes sense.
A: I agree with you that a filter on $A$ is a filter of subsets of $A$. However, if we denote by $\Bbb F(Y)$ the set of filters on a set $Y$, then $\Bbb F(A)=\{\mathscr{F}\cap\wp(A):A\in\mathscr{F}\in\Bbb F(X)\}$, so there is a natural correspondence between filters on $A$ and filters on $X$ that contain $A$.
But you don’t actually need to make this explicit for Prop. $6$. The first part of the proposition is clear: if $\mathfrak{B}$ is a filter base on $A$, then it is automatically a filter base on $X$ for a filter that contains $A$, so by the remark following Definition $2$ every cluster point of $\mathfrak{B}$ is in $\overline{A}$. Conversely, if $x\in\overline{A}$, then every nbhd of $x$ in $X$ meets $A$, so the trace on $A$ of the nbhd filter of $x$ in $X$ is a filter on $A$ and therefore a filter base on $X$; clearly every nbhd of $x$ in $X$ contains a member of this filter base, so by Prop. $1$ of $7.1$ this filter base converges to $x$ in $X$.
