Lemma IV (4.9) (Kunen book). Let $M$ be a countable transitive model of $ZFC$ and fix $\mathbb{Q}\in M$ such that $(|\mathbb{Q}| \leq \aleph_{0})^{M}$. Let $G$ be $\mathbb{Q}$-generic over $M$. Then there is no $h \in \omega^{\omega}\cap M[G]$ such that $f\leq^{*} g$ for all $f$ in $\omega^{\omega}\cap M$.
Proof: Suppose that were such an $h$. Fix $\dot{h} \in M^{\mathbb{Q}}$ such that $\dot{h}_{G}=h$. Let $W=\omega^{\omega}\cap M$. Then $M[G] \models \forall{x \in \check{W}}[x \leq^{*} \dot{h}] $, so fix $p \in G$ such that $p\Vdash \dot{h}:\omega \to \omega \wedge \forall{x \in \check{W}}[x \leq^{*} \dot{h}]$. Now, working in $M$, list $p\downarrow$ as $\{r_j:j \in \omega \}$. For each $n$, let $E_n=\{l \in \omega: \exists{j<n}[r_j \Vdash \check{l} \leq \dot{h}(n)]\}$.
Note that $E_n$ is finite.
Why that $E_n$ is finite.?
A suggestion to show this fact.
If $E_n$ was infinite then there would be some $j<n$ such that $r_j\Vdash \check{l}\leq \dot{h}(n)$ holds for infinitely many $l$. But this contradicts the fact that $r_j\Vdash \dot{h}(n)<\check{\omega}$ (since $p\Vdash \dot{h}(n)<\check{\omega}$ and $r_j\leq p$).