How can I work with the net associated to a filter? I have a question about the net which one can always associate to a filter. First let me write down our definition:

If $\mathfrak{F}$ is a filter on $M$ then we define $$I_\mathfrak{F}=\{(A,p): A\in \mathfrak{F}, p\in A\}$$which directed via $(A,p)\leq (B,q)$ if $B\subseteq A$. Then we have a net $$I_\mathfrak{F}\rightarrow M;\,\,\,(A,p)\mapsto p$$called the net associated to the filter $\mathfrak{F}$.

Now I somehow don't see how to work with this, because normally we always have some index $i\in I$ which is mapped to $a_i$ and this is called a net, but here I'm a bit confused.
So for example I wanted to solve the problem:

Let $\mathfrak{F}$ be a filter with associated net $(p_i)_{i\in I_\mathfrak{F}}$. Show that $p\in M$ is a cluster point of $\mathfrak{F}$ iff $p$ is a cluster point of the net $(p_i)_{i\in \mathfrak{F}}$.

But I haven't found a solution since I don't really see how my index need to look like. I mean I know that to prove $p$ is a cluster point of the net $I_\mathfrak{F}$, we need to show that for each neighborhood $U$ of $p$ and $i_0\in I_\mathfrak{F}$, there exists some $i\geqslant i_0$ such that $p_i\in U$.
I hope you understand what I mean, maybe it would also help if someone could explain me how to work with this index.
Thanks for your help.
 A: It's just about knowing the definitions.
Suppose that $p \in M$ is a cluster point of $\mathfrak{F}$ and we will show that $p$ is a cluster point of $(p_i)_{i \in I}$ where I'll write $I:=I_{\mathfrak{F}}$ for this proof, to simplify the notation.
So we have to show
$$\forall U \text{ open with } p \in U: \forall i \in I:\exists  I \ni i' \ge i : p_{i'} \in U\tag{1}$$
which I hope you agree is the definition for a cluster point. $U$ could also be taken to be any neighbourhood of $p$, that doesn't matter; it depends on the text book which is preferred.
So take $U$ open with $p \in U$. We only know that $p$ is a cluster point of $\mathfrak{F}$ so we know $$\forall A \in \mathfrak{F}: U \cap A \neq \emptyset\tag{2}$$ Next let $i \in I$ (we need to prove a universal statement $(1)$ and we've already picked arbitrary $U$ and now we get $i$, which is, being an "index" in the associated net just a pair $i=(A,a)$ where $A \in \mathfrak{F}$ and $a \in A$. We need a larger index $i'=(A',a')$ whose point $p_{i'}=a' \in A'$, lies in $U$. A larger index means by definition that $A' \subseteq A$.
So take $a' \in U \cap A$ and let $i':=(A,a')$ and this lies in the set $I= I_{\mathfrak{F}}$ and $p_{i'}=a' \in U$ and $A \subseteq A$ so $i' \ge i$. So $(1)$ has been shown.
The converse is a similar definition chasing:
We know $(1)$ ($p$ is a cluster point of the net) and we need that $p$ is a cluster point of the filter so let $U$ be any open neighbourhood of $p$ again, and $A \in \mathfrak{F}$; we need to show that $A \cap U \neq \emptyset$.
In order to apply $(1)$ we make an index $i=(A,a)$ where $a \in A$ is arbitrary (filters have non-empty sets in them so such an $a$ is easily picked). Applying $(1)$ to $U$ and my $i$ we get some $i' \ge i$ so that $p_{i'} \in U$. OK, we know $i'$ is also of the form $i'=(A',a')$ with $A' \in \mathfrak{F}$, $a' \in A'$ and $i' \ge i$ tells us $A' \subseteq A$. So $a'=p_{i'} \in A' \cap U \subseteq A \cap U$ so indeed $A \cap U \neq \emptyset$, as required. So $p$ is a cluster point of the filter.
I hope this illustrates how to work with such a net. Limits are similarly preserved, a nice exercise.
See how allowing all choices in the second component of members of $I$ makes for easy proofs. We can construct larger indices by shrinking the set or merely changing the point etc.
Subnets correspond to superfilters etc. The map going back from nets to filters is via the tail sets, as you probably know: check that the tail filter of $(p_i)_{i \in I}$ is exactly $\mathfrak{F}$ again.
