Let "$\mathrm{Power\ set}$" denote the Power set axiom. I'd like if some of you could tell me if the solution of the following exercise is correct.
I want to prove that $(V_\omega,\in)\models \mathrm{Power\ set}$, namely $(V_\omega,\in)\models \forall x \exists y \forall z (z\subseteq x \rightarrow z \in y)$.
Let $a\in V_\omega$, notice that $P(a)$ itself belongs to $V_\omega$. In fact, since $a\in V_\omega$, then $rk(a)\leq n$, for some $n\in \omega$. If $z\in P(a)$, then $z\subseteq a$, so $rk(z)\leq rk(a)$. It follows that $rk(P(a))\leq n+1$ and so $P(a)\in V_\omega$.
Now, since $P(a)\in V_\omega$ for all $a\in V_\omega$, we have that $\mathrm{Power\ set}$ is satisfied in $(V_\omega,\in)$ because, for all $a \in V_\omega$, $(V_\omega,\in)\models \forall z (z\subseteq a \rightarrow z\in P(a))$. In fact suppose that $a\in V_\omega$ and that for $z\in V_\omega$ $(V_\omega, \in)\models z\subseteq a$. So we have that $(V, \in)\models z\subseteq a$ $(**)$, because the "formula" "$z\subseteq a$" is absolute over $(V_\omega,\in)$ since it is a $\Delta_0$-formula and $V_\omega$ is transitive. Finally, from $(**)$ we have that for all $z \in V_\omega$ $z\in P(a)$.
And so we have that $(V_\omega,\in)\models \mathrm{Power\ set}$.