# Building a model using the power set of $Ord$

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:
1. The second-order (or rather, third-order) object of all the subclasses of $$A$$, or
2. 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 $$\mathcal P^\omega(A)=\bigcup_{\alpha\in\mathrm{Ord}}\mathcal P^\omega(A_\alpha)$$.