# If $B$ is an infinite complete Boolean algebra, then its saturation is a regular uncountable cardinal

I am trying to understand the proof of the statement (Jech 7.15)

If $B$ is an infinite complete Boolean algebra, then $\operatorname{sat}(B)$ is a regular uncountable cardinal. I understand the following:

If we let $\kappa = \operatorname{sat}(B)$, then it must be uncountable. Assume $\kappa$ is singular, and we hope to get a contradiction. We define $B_u:= \{v \in B \mid v \le u \}$ and let $\operatorname{sat}(u)$ denote $\operatorname{sat}(B_u)$. An element $u \in B$ is stable if $\operatorname{sat}(v) = \operatorname{sat}(u)$ whenever $v \le u$. The set $S$ forms a dense subalgebra of B, and if we construct a maximal set of pairwise disjoint elements of $S$ (via Zorn's Lemma), then this in fact forms a partition of $B$, with $|T| < \kappa$.

Now here is the part I don't understand

"First we show that $\operatorname{sup}\{ sat(u) | u \in T \} = \kappa$. For every regular $\lambda < \kappa$ such that $\lambda > |T|$, consider a partition $W$ of $B$ of size $\lambda$. Then at least one $u \in T$ is partitioned by $W$ into $\lambda$ pieces."

Now why does only considering regular cardinals $\lambda$ give any information about what the supremum of that set is? And why is the last sentence true?

Any help would be appreciated.

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Since $\kappa$ is singular then it is not a successor. So let say $\kappa=sup\{\theta_\xi: \xi<cof(\kappa)\}$ then $\kappa=sup\{\theta^+_\xi: \xi< cof(\kappa)\}$. Thus, it suffices to consider the regular cardinals.
On the other hand, the partition $W$ induces a partition on the elements of $T$ if every element $u\in T$ is partitioned by $W$ in less than $\lambda$ pieces then by the regularity of $\lambda$ and $\lambda > |T|$ would imply that $W$ has cardinality strictly less than $\lambda$.