Correct definition of filter convergence?

Question

In Stefan Waldmann's Topology An Introduction (2014) on page 66 the convergence of a filter is defined as

How should this be interpreted?

A) $\forall M,\mathcal{M},\mathfrak{F},p: \left[(M,\mathcal{M})\text{ is a topological space and }\mathfrak{F}\text{ is a filter on }M\right] \Longrightarrow \left[\mathfrak{F}\text{ converges to }p \iff \left(p\in M \text{ and }\mathfrak{F}\text{ is finer than the neighbourhood filter of }p \right)\right]$

B) $\forall M,\mathcal{M},\mathfrak{F},p: \left[(M,\mathcal{M})\text{ is a topological space and }\mathfrak{F}\text{ is a filter on }M\text{ and }p\in M \right] \Longrightarrow \left[\mathfrak{F}\text{ converges to }p \iff \left(\mathfrak{F}\text{ is finer than the neighbourhood filter of }p \right)\right]$

So basically does the convergence imply that $p\in M$ or is it a requirement before we could even resolve the definition?

Answer 1

We are working in the context of a given topological space, so a filter converging to $p$ means that $p \in M$, otherwise we cannot even talk about the neighbourhood filter of $p$. So if you want to be formal, your second statement best approximates, what I just said.

Answer 2

I could see how one could interpret B as being "more correct", but really the two are equivalent. We can't discuss the convergence of a filter on $M$ to a point $p$ unless $p\in M$, so this does imply $p\in M$ if it hasn't been explicitly stated elsewhere.

This isn't unique to filter convergence either. Some authors may write something like: "Let $(x_n)_n$ be a sequence in a topological space $X$ such that $x_n\to x$." In this case it is understood that $x$ must be an element of $X$.