The Use of Ordinals and Cardinals in Recursive Proofs

I'm reading something on ordinal numbers and I have a probably silly doubt.

Well, let $Z:=\{x_i: \, i \in I\}$ be a subset of a given set $X$. Define in $X$ a sequence $Y_\alpha=\Gamma(Y_{\alpha-1})$ when $\alpha$ is non limit and $Y_\alpha=\bigcup\limits_{\beta <\alpha} \Gamma(Y_\beta)$ (here $\Gamma$ is a given extensive function on the powerset of $X$). In this way, I obtain an increasing sequence $Y_0 \subseteq Y_1 \subseteq \dots \subseteq Y_{\omega} \subseteq ...$ Denote by $Y$ the corresponding union and suppose that $Z \subseteq Y$. By Bergman - An Invitation to General Algebra [prop. 5.5.4 (ii)] "The union of any set of ordinals is an ordinal". Thus, there exists an ordinal $\beta$ such that $Z \subseteq Y_\beta$.

Now, in order to make more concrete my request, I will consider the following situation: suppose that the elements $z,w$ are pairs of subsets of a given ground set $\Omega$ and $X=\mathcal{P}(\Omega) \times \mathcal{P}(\Omega)$. This means that the subsets $Y_\alpha$ are collections of pairs of subsets of $\Omega$. I assume that if $z=(A,B)$ and $w=(C,B)$, then the pair $(A \cup C,B)$ is the result of an operation $\times$ defined between $z$ and $w$.

At this point, suppose that if $z,w \in Y_\gamma$, then the result of $z \times w \in \Gamma(Y_\gamma)=Y_{\gamma+1}$. . My question is: if the members of $Z$ are pairs $(A_z,B)$ with the second component given by the same subset $B$, when I apply the operation on all the elements of $Z$ I get a pair $(\bigcup\limits_{z \in Z} A_z, B)$. Does the resulting pair belong to $\bigcup Y_\alpha$?

Here is a simplified version of your question, which I think should settle your confusion.

Let $(Y_\gamma)_{\gamma\in\mathrm{Ord}}$ be a continuous increasing sequence of collections of subsets of $\Omega$, indexed by the ordinals, and let $(A_z)_{z\in Z}$ be a family of subsets of $\Omega$.

Assume that (1) for all $z\in Z$, there is some $\gamma$ such that $A_z\in Y_\gamma$, and (2) if $A_z$ and $A_{z'}$ are in $Y_\gamma$, then $A_z\cup A_{z'}\in Y_{\gamma+1}$.

Does it follow that $\bigcup_{z\in Z} A_z\in \bigcup_{\gamma\in \mathrm{Ord}}Y_\gamma$?

The answer is no: For example, take $\Omega = \mathbb{N}$, $Y_\gamma = \mathcal{P}_{\mathrm{fin}}(\mathbb{N}) = \{B\subseteq \mathbb{N}\mid B\text{ is finite}\}$ for all $\gamma$, so $(Y_\gamma)_{\gamma\in \mathrm{Ord}}$ is the constant sequence and $\bigcup_{\gamma\in \mathrm{Ord}} Y_\gamma = \mathcal{P}_{\mathrm{fin}}(\mathbb{N})$. Let $Z = \mathbb{N}$, and define $A_n = \{n\}$ for all $n$. Then (1) and (2) are satisfied, since $A_n\in Y_\gamma$ and $A_n\cup A_m\in Y_\gamma$ for all $n,m\in \mathbb{N}$ and all $\gamma$. But $\bigcup_{n\in \mathbb{N}} A_n = \mathbb{N}\notin \mathcal{P}_{\mathrm{fin}}(\mathbb{N})$.

The point is that the closure condition (2) ensures that any finite union of the $A_z$ is in some $Y_\gamma$, but there's no reason to expect that this extends to infinite unions.