For $n\in\mathbb{N}$, suppose $p_n$ denote the set of all prime divisors of $n$. Define $f:\mathbb{N}\to \mathcal{P}\{\mathbb{N}\}$ by $f(n)=p_n$ for all $n\in\mathbb{N}$.
Show that $f$ is order preserving. Here $(\mathbb{N},\le)$ and $(\mathcal{P}\{\mathbb{N}\},\subseteq)$ are ordered sets.
So all I have to prove is that for $m,n\in\mathbb{N}, m\le n\implies f(m)\subseteq f(n)$.
Let $m=5, n=7$. Then $5\le 7$ and $f(5)=p_5=\{1,5\}, f(7)=p_7=\{1,7\}$ but $f(5)\not\subseteq f(7)$. So how is $f$ order-preserving?
EDIT: From the comment of J.G, $1$ is not a rime. So $f(5)=p_5=\{5\}, f(7)=p_7=\{7\}$