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.)

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.)
