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 $m \in f^{-1} (U)$
$(\rightarrow)$  ???  

Suppose $f$ is continuous and $m|n$.  Let $U$ be the set of $a$ such that $a|f(n)$.  Then $U$ is open and $f(n)\in U$.  Thus $n\in f^{-1}(U)$ which is also open.  Thus since $m|n$, $m\in f^{-1}(U)$. Thus $f(m)\in U$.  Thus $f(m)|f(n)$.
