Let $X= \{ 2,3,\dots\} \subset \mathbb N$. We define the divisor topology on $X$ by establishing a basis $\mathcal B = \{S_n \}_{n \in X}$ where $$S_n = \{p \in X : p|n\}$$
($p$ divides $n$).\
Under this topology we want to study the continuity of the function $f:X \to X$ given by the expression $f(n)=n^2, \forall n \in \mathbb N$.
We know that this function is continuous because there exists a criterion under this topology that states that a function $g:X \to X$ is continuous iff $n |m$ implies that $g(n)|g(m)$, and that is the case for this function.
However, I'd like to prove it rather by definition (checking whether $f^{-1}(S_n)$ is an open set, but I'm quite lost on how to approach it. Thanks in advance.
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Simply observe that if $n = k^2m$ where $m$ is square-free, then $$f^{-1}(S_n) = \{p : p^2|n\} = \{p : p|k\} = S_k.$$
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$\begingroup$ But how could I guarantee n can be divisible by a square-free integer? $\endgroup$Perch– Perch2025-11-26 13:09:45 +00:00Commented Nov 26 at 13:09
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$\begingroup$ @Perch Why do you need it to be divisible by a square-free integer? (and $1$ is square-free so all naturals are divisible by some square-free integer) $\endgroup$Jakobian– Jakobian2025-11-26 13:18:04 +00:00Commented Nov 26 at 13:18
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$\begingroup$ Well, yes, but the divisor 1 doesn't provide k unless n is a perfect square. I'm guessing k must be the biggest number such that k^2 divides n. In that case one must prove $S_k=f^{-1}(S_n)$, which I haven't been able to accomplish $\endgroup$Perch– Perch2025-11-26 14:31:05 +00:00Commented Nov 26 at 14:31
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$\begingroup$ @Perch any natural number $n$ can be written as a product of powers of distinct primes $p_1, ..., p_m$, say $n = p_1^{\alpha_1} ... p_m^{\alpha_m}$. Then write $\alpha_i = 2\beta_i + r_i$ where $r_i\in \{0, 1\}$. Then you can take $k = p_1^{\beta_1}...p^{\beta_m}$. Then $m = p/k^2 = p_1^{r_1}...p_m^{r_m}$ will be square-free. $\endgroup$Jakobian– Jakobian2025-11-26 14:39:53 +00:00Commented Nov 26 at 14:39
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$\begingroup$ And for that given $k$ is it true that $f^{-1}(S_n)=S_k$? $\endgroup$Perch– Perch2025-11-26 15:39:05 +00:00Commented Nov 26 at 15:39