Consider the function $F: \mathscr{P}(\mathbb{N}) \to \mathscr{P}(\mathbb{N})$ given by $$F(A) = \overline{A} = \{k \cdot n : k \in A, n \in \mathbb{N}\} = \bigcup_{n\in \mathbb{N}} nA$$
a) Show that $F$ is a Kuratowsky closure operator and describe the unique topology induced by $F$, that is, describe the open and closed sets of $(\mathbb{N}, \tau)$, where $\tau$ is the topology induced by $F$.
b) Show that a function $f: (\mathbb{N}, \tau) \to (\mathbb{N}, \tau) $ is continuous if and only if $m \vert n \implies f(m) \vert f(n)$.
For a), it's really trivial to show that $F$ satisfies the Kuratowsky closure axioms, so I don't need help with that. I understand that $A \subset \mathbb{N}$ is closed if and only if it contains all of the multiples of elements of $A$, so I think the only closed sets of $\mathbb{N}$ in this topology are $\emptyset, \mathbb{N}, 2\mathbb{N},3\mathbb{N}, 4\mathbb{N},5\mathbb{N}, 6\mathbb{N},7 \mathbb{N}, \cdots$, and so the only open sets would be the complements of those sets, but I have yet to convince myself that I'm not missing any sets. Are these really the only ones or have I forgot something here?
For b), I thought about using that $f$ is continuous $\iff f(\overline{A}) \subset \overline{f(A)}$, but I got nowhere with that unfortunately.
Help?
Some progress: about a), I now know I should say that those along with unions and intersections of them are really the only closed sets, it's a simple argument really:
Let $A = \{a_1, a_2, \cdots\}$ be a closed subset of $\mathbb{N}$ in this topology. Then it's pretty clear (because of how we constructed the closure operator) that $A = a_1 \mathbb{N} \cup a_2 \mathbb{N} \cup \cdots$
As $K:\mathcal{P}(\mathbb{N}) \to \mathcal{P}(\mathbb{N})$, $A \to A\cdot \mathbb{N} = \{ an \mid a \in A, n \in \mathbb{N} \}$ is a closure operator, $K(A)$ is the closure of $A$.
If for all $a,n \in \mathbb{N}$, $a|$n implies $f(a)|f(n)$,
then $f$ is continuous.
This is proven by showing $f[K(A)] \subseteq K(f[A])$:
If $y \in f[K(A)] = f(A\cdot \mathbb{N})$, then
there exist $a \in A$, $n \in \mathbb{N}$ with $y = f(an)$.
As $a|an$, we have that $f(a)|f(an)$. So there exists a $k$ with $f(an) = k\cdot f(a)$.
Whence, $y \in f(a)\cdot \mathbb{N} \subseteq f[A]\cdot \mathbb{N} = K(F[A])$. QED.
If $f$ is continuous and $a|n$, then there exists $k$ with $n = ak$.
So $f(n) \in f(a\cdot \mathbb{N}) = f[K(\{a\}] \subseteq K(\{f(a)\}) = f(a)\cdot \mathbb{N}$
and $f(a)|f(n)$.