# Exercise on topology on the natural numbers induced by a Kuratowski closure operator

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 $$A = a_1 \mathbb{N} \cup a_2 \mathbb{N} \cup \cdots$$

• Get the definition of F and your set builder notation right: F(A) = { nk : n in N, k in A } – William Elliot Jan 28 '19 at 1:39
• @WilliamElliot fixed it, thanks. – Matheus Andrade Jan 28 '19 at 1:44
• What is the closure of {3,5}? – William Elliot Jan 28 '19 at 1:44
• Your set notation is still confusing. Write { x : x is even } with a colon. – William Elliot Jan 28 '19 at 1:52
• @MatheusAndrade What is this: $\overline{A} = \{k \cdot n, k \in \mathbb{N}, n \in \mathbb{N}\}$ supposed to mean? – feynhat Jan 28 '19 at 2:11

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 $$f(a)|f(n)$$.

• I figure you're probably on mobile or something and that's the reason for the formatting, so I don't mind that. With a little bit of effort, I was able to fully understand your answer, so thanks a lot! – Matheus Andrade Jan 28 '19 at 10:10
• @MatheusAndrade no, he just refuses to use MathJax or something like that... – Henno Brandsma Jan 28 '19 at 15:07