I have been reading a proof for the following proposition
Proposition: Let $X$ be a Urysohn space. If $A$ is a subset of $X$, then $|[A]_\theta|\leq |A|^{\chi(X)}$
Here, $[A]_\theta$ denotes the $\theta-$closed hull of , which is the smallest $\theta-$closed set that contains $A$, and a $\theta-$closed set $C$ is such that $x\in C$ if and only if for every closed neighborhood of $x$ intersects $C$. Also, $\chi$ denotes the character of the space. Well, there is a step in this proof that I do not quite understand, and here it is:
Let $\kappa=|A|$, $A_0=A$ and by transfinite induction, let us define, for any $\alpha\in \kappa^+$ sets $A_\alpha$, in such a way that $A_\alpha=Cl_\theta (\bigcup_{\beta \in \alpha }A_\beta)$. It is easy to see that $$ \bigcup_{\alpha \in \kappa^+} A_\alpha \subset [A]_\theta$$
And I do not really get this, I hope someone can explain why this is true, any comment is appreciated, thanks in advance.
It suffices to see that $A_\alpha\subset [A]_\theta$ for each $\alpha\in\kappa^+$. We can easily see that by transfinite induction. Namely, we have $A_0\subset [A]_\theta$. Assume that $\alpha\in\kappa^+$ and we have $A_\beta\subset [A]_\theta$ for each $\beta \in \alpha$. Thus $\bigcup_{\beta \in \alpha }A_\beta\subset [A]_\theta$. Since the set $[A]_\theta$ is $\theta$-closed, $A_\alpha=Cl_\theta (\bigcup_{\beta \in \alpha }A_\beta)\subset [A]_\theta$.