How many subsets are produced? (a transfinite induction argument)

Question

Suppose that $\varkappa$ is the least ordinal of infinite cardinality $|\varkappa|.$ A lot of standard constructions produce by the transitive induction a sequence of subsets $I_r \subseteq \varkappa,$ where $r \in \varkappa,$ by acting as follows. Set $I_0=\varnothing.$ In the case when $r$ is a limit ordinal $I_r$ is the union of all the preceding subsets $I_s,$ and if $r=s+1,$ then $I_r$ is chosen as the minimal initial segment of $\varkappa$ which contains $I_s$ and some nonempty finite subset $U_s$ of $\varkappa$ such that $$ I_s \cap U_s=\varnothing \text{ and } s \in U_s. $$ Now is it true that the cardinality of the sequence $I_r$ is equal to $\varkappa?$ It should but the proof escapes me.

Answer 1

With the given information, there is no definite answer.

Take for example $\kappa = \omega_1$ and $U_s = s$. Then $(I_s \mid s < \kappa) = (s \mid s < \kappa)$ and hence $$ \{ I_s \mid s < \kappa \} $$ has size $\kappa$.

On the other hand, let $U_s = \kappa$ for all $s < \kappa$. Then $$ \{ I_s \mid s < \kappa \} = \{ \kappa \} $$ has size $1$.