I have a question where working in ZF I need to build a transitive model of Z containing $Ord$ (the class of all ordinals) s.t it does not model Zermelo definition of an infinite set:
$$\exists x\left(\emptyset\in x\wedge \forall z\in x, \{z\}\in x\right)$$
I know (from a previous question) that defining
$$X_0 = \omega,\ X_{n+1}=\mathcal{P}(X_n),\ X=\bigcup_{n<\omega} X_n$$
then $X$ is a transitive model of Z without a Zermelo Infinite set (not containing $Ord$ of course). I thought about using the above but taking
$$X_0 = Ord,\ X_{n+1}=\mathcal{P}(X_n),\ X=\bigcup_{n<\omega} X_n$$
this gives me that $X$ the desired model. What I'm having a problem with is taking the power set of the class $Ord$, $\omega$ many times.
Can we use a "power class" as we would use a power set?
Yes, we can define the idea of a "power class", of a class $A$, but then you need to clarify if you mean:
The second-order (or rather, third-order) object of all the subclasses of $A$, or
The collection of all sets which are subclasses of $A$.
The latter is a first-order object, if $A$ is defined by $\varphi$, then $\mathcal P(A)$ is given by $\{x\mid\forall y(y\in x\to\varphi(y))\}$. You want to iterate this power set, and then it's obviously a question whether or not we can even consider $\bigcup_{n<\omega}\mathcal P^n(A)$. The answer is again yes, since we are quantifying on an internal $\omega$, rather than writing the scheme of $\mathcal P^n(A)$ explicitly, we use the recursion theorem.
But there's a clearer way to see why we can do it: for $\alpha$, let $A_\alpha=A\cap V_\alpha$, then $A_\alpha$ is a set. So we can talk about $\mathcal P^\omega(A_\alpha)=\bigcup_{n<\omega}\mathcal P^n(A_\alpha)$, and we can prove that $\mathcal P^\omega(A)=\bigcup_{\alpha\in\mathrm{Ord}}\mathcal P^\omega(A_\alpha)$.