2
$\begingroup$

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

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

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Matheus Andrade Jan 28 '19 at 10:10
  • $\begingroup$ @MatheusAndrade no, he just refuses to use MathJax or something like that... $\endgroup$ – Henno Brandsma Jan 28 '19 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.