$f \in cl(P)$ iff $\{ A \subset \mathbb{N} : f(A) = 1 \}$ is an ultrafilter I have the following definitions:
$p_n : \mathbb{P}(\mathbb{N}) \rightarrow \{0,1\}$ are functions defined as: $p_n(A) = 1$ if $n \in A$ or $p_n(A) = 0$ if not. Now, $P = \{p_n : n \in \mathbb{N} \} \subset \{0,1\}^{\mathbb{P}(\mathbb{N})}$, where the last one is the set of all functions from $\mathbb{P}(\mathbb{N})$ to $\{0,1\}$.
So, what I have to show is: $x \in \bar{P}$ iff $\{ A \subset \mathbb{N} : x(A) = 1 \}$ is an ultrafilter over $\mathbb{N}$. 
Up until now, I haven't been able to prove any of the directions, because I struggle to work with $\{0,1\}^{\mathbb{P}(\mathbb{N})}$ as a topologic space (which are its opens? basis?). Also, I know that $\{0,1\}$ comes with the discrete topology. \
I've tried to prove it using the topologic definition of closure (every nbhd. of x in $\{0,1\}^{\mathbb{P}(\mathbb{N})}$ intersects $P$, but I can't see how can I write a full proof of both directions.
 A: The standard base for $\{0,1\}^I$, for any indexing set $I$, is the family of all sets $$\langle I_0; I_1 \rangle = \{f \in \{0,1\}^I\mid \forall i \in I_0: f(i)=0; \forall i \in I_1: f(i)=1\}$$ where $I_0, I_1$ are finite disjoint subsets of $I$.
These are finite intersections of the subbasic sets $\pi_i^{-1}[\{0\}]$ and $\pi_i^{-1}[\{1\}]$ which form the generating set for the product topology in this case (as $\{0\}, \{1\}$ are all the non-trivial open sets of $\{0,1\}$.
In this case, where $I= \mathscr{P}(\Bbb N)$, each point of $\{0,1\}^I$ is (i.e. can be seen as) a family of subsets of $\Bbb N$ via the identification $f(A) = 1$ iff $A \subseteq \Bbb N$ is a member of that family. So the $p_n$ are actually the so-called fixed ultrafilters on $\Bbb N$. So using this way of viewing points, the basic open sets are really of the form
$$\langle \mathcal{F}_0; \mathcal{F}_1\rangle = \{\mathcal{F} \subseteq \mathscr{P}(\Bbb N)\mid \mathcal{F}_0 \subseteq \mathcal{F}; \mathcal{F}_1 \cap \mathcal{F} = \emptyset\}$$
where $\mathcal{F}_0, \mathcal{F}_1$ are finitely many families of subsets of $\Bbb N$.
Hopefully this will gave you a bit of a handle on the topology. I think the view of $\overline{P}$ as the set of all $\mathcal{F}$ such that every basic open set contains a point of $P$ will be enough to show the required properties for $\mathcal{F}$ to be an ultrafilter on $\Bbb N$.
E.g if $\emptyset \in \mathcal{F}$, then $\langle \{\emptyset\}; \emptyset\rangle$ is a basic neighbourhood of $\mathcal{F}$ that contains no $p_n$ (as these never contain $\emptyset$ as a member) and so $\mathcal{F} \notin \overline{P}$ by the definition of closure, and so no family in $\overline{P}$ contains $\emptyset$, one of the properties of a filter.
Similarly for other properties that (ultra)filters should have: if we have two non-empty subsets $A \subseteq B$, then $\langle \{A\}; \{B\}\rangle$ (the set of all families that contain $A$ and do not contain $B$) is basic open and contains no member of $P$, so no family that is in the closure of $P$ can be contained in it. This implies that all $\mathcal{F} \in \overline{P}$ obey the property that with any subset they contain all supersets.
That all such $\mathcal{F}$ are closed under intersections is clear from considering basic open sets like $\langle \{A,B\}; \{A \cap B\}\rangle$ and the final ultrafilter property follows from considering $\langle \emptyset;\{A, \Bbb N\setminus A\} \rangle$ for any subset $A$ of $\Bbb N$... All families that fail a certain ultrafilter property form open sets disjoint from $P$, that’s the recurring theme. The precise $P$ doesn't matter for this, only that it's a set of ultrafilters.
You do need the precise form of $P$ to show that all ultrafilters on $\Bbb N$ are in $\overline{P}$: let $\mathcal{F}$ be an ultrafilter and $\langle \mathcal{F}_0; \mathcal{F}_1\rangle$ be a basic open neighbourhood of $\mathcal{F}$ (recall that this means that $\mathcal{F}_0$ is a finite collection of subsets of $\Bbb N$ that are all in $\mathcal{F}$, and $\mathcal{F}_1$ is a finite collection of subsets of $\Bbb N$ that are not in $\mathcal{F}$ (so their complements are in $\mathcal{F}$, being an ultrafilter). Now pick (this is possible as the right hand set is in $\mathcal{F}$, so a non-empty set in particular) $$m \in \bigcap \{A \mid A \in \mathcal{F}_0\} \cap \bigcap \{\Bbb N\setminus A \mid A \in \mathcal{F}_1\}$$ (with the convention that an intersection over an empty family is just $\Bbb N$ to avoid trivialities). Then $p_m \in \langle \mathcal{F}_0; \mathcal{F}_1\rangle \cap P$ shows that indeed $\mathcal{F} \in \overline{P}$.
