Am doing an exercise on the cardinality of Borel sets where we have and arbitrary set $X$ and the set of countable ordinals $\Omega$. Given $\mathcal{E}\subset \mathcal{P}(X) $ with $\emptyset\in \mathcal{E}$ we define a collection of subsets of $\mathcal{P}(X)$ as follows:
Define $\mathcal{E}^c:=\{E^c:E\in\mathcal{E}\}$ and $\mathcal{E}_{\sigma}:=\{\cup_{n=1}^{\infty}E_n:(E_n)\subset\mathcal{E}\}$. Define $\mathcal{F}_1=\mathcal{E}\cup \mathcal{E}^c$. If $x\in\Omega$ and if $x$ has an immediate predecessor $y$ then we set $\mathcal{F}_x:=(\mathcal{F}_y)_{\sigma}\cup ((\mathcal{F}_y)_{\sigma})^c$, and if $x$ has no immediate predecessor then we set $\mathcal{F}_x:=\cup_{y<x}\mathcal{F_y}$.
Am trying to understand why this definition makes sense. I believe it is an application of the principle of recursive definition that appears on page 70 of Munkres' book:
Theorem. Let $J$ be a well ordered set and let $C$ be a set. Let $\mathcal{G}$ be the set of all functions mapping sections of $J$ into $C$. Given a function $\rho:\mathcal{G}\to C$, there exists a unique function $h:J\to C$ such that $h(\alpha)=\rho(h\mid S_{\alpha})$ for each $\alpha\in J$.
If I let $J=\Omega$, $C=\mathcal{P}(\mathcal{P}(X))$ and given $f:S_x\to \mathcal{P}(\mathcal{P}(X))$ with $S_x=\{z\in \Omega:z<x\} $ define
$$ \rho(f) = \begin{cases} (f(y))_{\sigma}\cup ((f(y))_{\sigma})^c & \text{if $x$ has immediate predecesor $y$} \in S_x \\ \cup_{y<x} f(y) & \text{if $x$ has no immediate predecesor and $x\neq \min \Omega$} \\ \mathcal{E}\cup \mathcal{E}^c & \text{if $x= \min \Omega$} \end{cases} $$
then I can invoke the above Theorem to obtain a function $h:\Omega\to \mathcal{P}(\mathcal{P}(X))$ with $\mathcal{F}_x:=h(x)$ that satisfies
$h(1)=\rho(h\mid S_{1})=\rho(\emptyset)=\rho(S_1)=\mathcal{E}\cup \mathcal{E}^c$,
$h(x)=\rho(h\mid S_{x})=(\mathcal{F}_y)_{\sigma}\cup ((\mathcal{F}_y)_{\sigma})^c$ if $x$ has no immediate predecesor, and
$h(x)=\rho(h\mid S_{x})=\cup_{y<x} \mathcal{F}_y$ if $x$ has no immediate predecesor and $x\neq \min \Omega$.
Is this correct? Any feedback is very apprciated.