$\kappa$-complete, $\lambda$-saturated ideal properties Kunen, II.56. Having trouble proving the properties of the following:
The definition: $S(\kappa,\lambda,\mathbb{I})$ is the statement that $\kappa > \omega$ and $\mathbb{I}$ is a $\kappa$-complete ideal on $\kappa$ which contains each singleton and which is $\lambda$-saturated, meaning: there is no family $\{X_\alpha : \alpha < \lambda\}$, such that each $X_\alpha \notin \mathbb{I}$ but $\alpha \ne \beta \rightarrow (X_\alpha \cap X_\beta) \in \mathbb{I}$. Need to show that:
a) $\exists\lambda\exists\mathbb{I} S(\kappa,\lambda,\mathbb{I}) \rightarrow \kappa$ is regular.
b) $S(\kappa,\lambda,\mathbb{I}) \land \lambda < \lambda' \rightarrow S(\kappa,\lambda',\mathbb{I})$
c) $\exists\mathbb{I}S(\kappa,\kappa,\mathbb{I}) \rightarrow \kappa$ is weakly inaccesible.
I think, the main problem for me here, is not knowing what sets are in or out of $\mathbb{I}$. 
I'll appreciate any help.
Thanks in advance.
 A: For (a), suppose that $\mathscr{I}$ is a $\kappa$-complete ideal on $\kappa$ that contains the singletons. Then if $\mathscr{A}\subseteq\mathscr{I}$, and $|\mathscr{A}|<\kappa$, $\bigcup\mathscr{A}\in\mathscr{I}$. Thus, $\mathscr{I}$ contains every subset of $\kappa$ of cardinality less than $\kappa$. Suppose that $\operatorname{cf}\kappa=\mu<\kappa$. Then $\kappa$ is the union of $\mu$ sets of cardinality less than $\kappa$, each of which is in $\mathscr{I}$, and $\mathscr{I}$ is $\kappa$-complete, so $\kappa\in\mathscr{I}$, contradicting Definition 6.2(a).
Part (b) is pretty trivial, as Asaf notes in the comments.
For (c), use Theorem 6.11 to conclude that if $\kappa$ were a successor cardinal, then there would be pairwise disjoint sets $X_\alpha\subseteq\kappa$ for $\alpha<\kappa$ such that each $X_\alpha\notin\mathscr{I}$. $\{X_\alpha:\alpha<\kappa\}$ would then be a witness to $\kappa$ not being $\kappa$-saturated, since $X_\alpha\cap X_\beta=\varnothing\in\mathscr{I}$ whenever $\alpha<\beta<\kappa$. Thus, $\kappa$ must be a limit cardinal, and by (a) it must be regular.
