What does $\text{cl}(\text{im}(e_\mathbb{N}))\subseteq 2^{2^{\mathbb{N}}}$ look like? This is a follow-up question to that question.
Let $2 = \{0,1\}$ be endowed with the discrete topology. My hunch was the following:
Let $e_\mathbb{N}: \mathbb{N} \to 2^{2^{\mathbb{N}}}$ be the "evaluation map", that is, it is given by $$e_\mathbb{N}(n): f\in 2^\mathbb{N}\mapsto f(n).$$
Can $\text{cl}(\text{im}(e_\mathbb{N}))\subseteq 2^{2^{\mathbb{N}}}$ be described explicitly? (I know this is a bit of an unexact and "hand-wavy" question -- I apologise.)
 A: For each $n \in \mathbb N$ let $\mathfrak e_n : 2^{\mathbb N} \to 2$ be the evaluation map at $n$; i.e., $\mathfrak e_n (f) = f(n)$ for each $f \in 2^{\mathbb{N}}$. Letting $A = \{ \mathfrak{e}_n : n \in \mathbb{N} \}$, wwe are looking for $\overline A$ in $2^{2^{\mathbb N}}$.
Given any $\mathfrak f \in 2^{2^{\mathbb N}}$, a neighborhood basis for $\mathfrak f$ is described as follows. For $f_1, \ldots , f_k \in 2^{\mathbb N}$, define $$U_{\mathfrak f;f_1,\ldots,f_k} = \{ \mathfrak g \in 2^{2^{\mathbb N}} : ( \forall i \leq k ) ( \mathfrak g (f_i) = \mathfrak f (f_i) ) \}.$$ Then $\{ U_{\mathfrak f;f_1,\ldots,f_k} : f_1 , \ldots , f_k \in 2^{\mathbb N} \}$ constitutes a neighborhood basis for $\mathfrak f$.
Now $\mathfrak f \in \overline A$ iff $A \cap U_{\mathfrak f;f_1,\ldots,f_k} \neq \emptyset$ for all $f_1 , \ldots , f_k \in 2^{\mathbb N}$. That is, given any $f_1 , \ldots , f_k \in 2^{\mathbb N}$ there is an $n \in \mathbb N$ such that $\mathfrak e_n \in U_{\mathfrak f;f_1,\ldots,f_k}$, or, given $f_1 , \ldots , f_k \in 2^{\mathbb N}$ there is an $n \in \mathbb N$ such that $\mathfrak f (f_i) = \mathfrak e_n(f_i) = f_i(n)$ for all $i \leq k$. Putting this into words, $\mathfrak f \in \overline A$ iff $\mathfrak f$ is "everywhere locally an evaluation map at some $n \in \mathbb{N}$".
This description is probably not as simple as you would like, but we can get a bit out of this. 


*

*Neither constant function belongs to $\overline A$.  (Given constant $\mathfrak{f} : f \mapsto j$, consider the constant function $f : n \mapsto 1-j$.)

*Given $f \in 2^{\mathbb N}$, let $\hat f$ denote the complement of $f$: $\hat f(n) = 1 - f(n)$ for all $n \in \mathbb N$. If $\mathfrak f \in \overline A$, then $\mathfrak f(\hat f) = 1 - \mathfrak f(f)$ for all $f \in 2^{\mathbb N}$. (Given $f$ there is an $n$ such that $\mathfrak f(\hat f) = \hat f(n)$ and $\mathfrak f(f) = f(n)$. Then $\mathfrak f(\hat f) = \hat f(n) = 1 - f(n) = 1 - \mathfrak f(f)$.)

*In a bit more generality than the above point, suppose $f_1, \ldots , f_k$ is a "partition of unity", i.e., for each $n \in \mathbb N$ exactly one of $f_1(n), \ldots , f_k(n)$ is $1$. If $\mathfrak f \in \overline A$, then $\mathfrak f(f_i) = 1$ for exactly one $i \leq k$.

*Define a partial ordering $\preceq$ on $2^{\mathbb N}$ by $f \preceq g$ iff $f(n) \leq g(n)$ for all $n \in \mathbb N$. If $\mathfrak f \in \overline A$, then $\mathfrak f$ is non-decreasing with respect to $\preceq$. (Given $f \preceq g$ there is an $n$ such that $\mathfrak f(f) = f(n)$ and $\mathfrak f(g) = g(n)$. Then $\mathfrak f(f) = f(n) \leq g(n) = \mathfrak f(g)$.)

A: The closure is exactly the ultrafilters on $\mathbb{N}$.  To be precise, if $U$ is an ultrafilter on $\mathbb{N}$, let $e(U)\in 2^{2^\mathbb{N}}$ be its characteristic function (note that $e_\mathbb{N}(n)$ is just $e(U_n)$ where $U_n$ is the principal ultrafilter at $n$).  Then I claim that the limit of $e_\mathbb{N}(n)$ along $U$ is $e(U)$.  Indeed, for each $S\in 2^\mathbb{N}$, $\{n:e_\mathbb{N}(n)(S)=1\}\in U$ iff $S\in U$ (since that set is exactly $S$).  Since every point in the closure of the image of $e_\mathbb{N}$ must be a limit of some ultrafilter, this is the entire closure.
(From the perspective of Stone duality, this closure is the Stone space of the Boolean algebra $2^\mathbb{N}$, and its embedding in $2^{2^{\mathbb{N}}}$ is the canonical embedding.)
