Given the following condition, show that $|\operatorname{trcl}$(y)$| < \kappa$

Question

Suppose $\kappa$ is regular, if we assume that $y \subseteq H(\kappa)$ and $|y| < \kappa$

Show that |$\operatorname{trcl}(y)$| < $\kappa$.

$H(\kappa)$ is the {$x\ :\ |\operatorname{trcl}(x)| < \kappa$}

$\operatorname{trcl}(x)$ is just the transitive closure of $x$.

How do I show this ?

From the assumption all I can conclude is that $\operatorname{trcl}(y)$ $\subseteq H(\kappa)$. Where do I make of the regularity of $\kappa$ ?

Any help is appreciated.

Cheers.

Answer 1

As $y \subseteq H(\kappa)$, $\vert \operatorname{trcl}(x) \vert < \kappa$ for every element $x \in y$. $\operatorname{trcl}(y) = y \cup \bigcup_{x \in y} \operatorname{trcl}(x)$, thus follow by regularity of $\kappa$ that $\vert \operatorname{trcl}(y) \vert = \vert y \vert + \sup_{x \in y} \vert \operatorname{trcl}(x) \vert < \kappa$.