Problem of General topology (Schaum) Let $\tau=\{U \subseteq \mathbb{Z}_+ :$ if $n \in U \implies$ all divisor of $n $ is in $U \}$.
Let $f: (\mathbb{Z}_+,\tau) \to (\mathbb{Z}_+,\tau)$. Prove: $f$ is continuous $\leftrightarrow$ if $m|n $ then  $ f(m) | f(n)$.
$(\leftarrow)$ Let $n \in U$ (open) and let $m \in \mathbb{Z}_+$ such that $m|n$. 
I want to prove that $m \in f^{-1} (U)$. We know that  $f(n) \in U$ and $f(m)|f(n)$. Since  $U$ is open , then $f(m) \in U$.  Hence $m \in f^{-1} (U)$
$(\rightarrow)$  ???  

Assume that there exists an $n $ such that the implication $m\mid n\implies f (m)\mid f (n) $ does not hold. Then we find an $m $ such that $m\mid n $ but $f (m)\not\mid f (n) $. Now choose $U\in\tau $ to contain only the divisors of $f (n) $. Then $n\in f^{-1}(U) $ but $m\notin f^{-1}(U) $, so $f^{-1}(U)\notin\tau$, completing the proof of the desired implication.
