How many monotonic functions on sets of naturals are there?

Question

A function $f\colon P(\mathbb{N}) \to P(\mathbb{N})$ is monotonic iff $x \subseteq y \implies f(x) \subseteq f(y)$.

What is the cardinality of the set $F$ of all such functions?

What I have tried:
I know the cardinality is either $2^\mathbb{N}$ or $(2^{\mathbb{N}})^{2^\mathbb{N}} = 2^{\mathbb{N}\cdot 2^\mathbb{N}} = 2^{2^\mathbb{N}}$. I know that if I asked the question about natural numbers and not sets of them, the answer would be the (analogue of the) greater possibility, and if I asked about real numbers, the answer would be the (analogue of the) lesser possibility. (So as far as I know, there is a kind of "precedent" for both.)

I tried proving it is $2^{\mathbb{N}}$ by finding a bijection between $P(\mathbb{N})$ and $F$, but I couldn't come up with anything.

I tried proving it is $2^{2^\mathbb{N}}$ in two ways. I tried finding a bijection between $P(\mathbb{N}) \to P(\mathbb{N})$ and $F$. For functions on naturals, this is easy, (and has been asked before): $$g(f) = n \mapsto \begin{cases} f(0) & n = 0\\ f(n) + f(n-1) & \text{else} \end{cases}$$

I tried to generalize this trick by defining $g'(f)$ similarly, in terms of sets containing one fewer element. However, such a definition would not be well-founded, because the set of sets of natural numbers is not well-ordered (take the family $\{ n \mid n > c \}$ for some $c$).

I also tried a diagonalization proof by contradiction: assume there is a bijection $$g\colon P(\mathbb{N}) \to F\,,$$ then $$x \mapsto P(\mathbb{N}) \setminus g(x)(x)$$ or something similar fails to be a monotonic function not in $g$'s image, because it fails to be (necessarily) monotonic, and I don't think this can be fixed.

Answer 1

Let $$M=\{\,S\subseteq P(\Bbb N)\mid A\subseteq B\subseteq\Bbb N, A\in S\to B\in S\,\}$$ (the set of all upward-closed subsets of $P(\Bbb N)$). For $f\in F$ and $n\in \Bbb N$, the set $h_f(n):=\{\,A\subseteq\Bbb N\mid n\in f(A)\}$ is an element of $M$. As This gives us a map $h\colon F\to M^{\Bbb N}$, $f\mapsto (n\mapsto h_f(n))$. From $f(X)=\{\,n\in \Bbb N\mid X\in h_f(n)\,\}$ we see that $h$ is injective, but by the same we readily verify that $h$ is onto. So $|F|=|M|^{\aleph_0}$.

Remains to find $|M|$. Clearly, $|M|\le 2^{2^{\aleph_0}}$. As $\Bbb N\times\Bbb N\approx \Bbb N$, we may as well consider the set of upward-closed subsets of $\Bbb N\times \Bbb N$ instead. For an arbitrary sequence of non-empty subsets $A_n\subseteq\Bbb N$, we obtain the upward-closed set $X:=\{\,S\in\Bbb N\times\Bbb N\mid\exists n\colon \{n\}\times A_n\subseteq S\,\}$. This map $P(\Bbb N)^{\Bbb N}\to M$ is injective because ee can reconstruct $A_n=\bigcap\{\,A\subset\Bbb N\mid \{n\}\times A\in X\,\}$. It follows that $|M|=2^{2^{\aleph_0}}$.