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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.

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So $\lambda$-saturation means that if we have $\lambda$ positive sets at least two intersect positively? Also, in $(c)$, what is $\mathbb F$? – Asaf Karagila Jul 14 '12 at 20:18
@Asaf: $\Bbb F$ is a typo for $\Bbb I$. – Brian M. Scott Jul 14 '12 at 20:27
Ah, thanks @Brian. Pavel: The second one is pretty obvious, if there is no family of size $\lambda$ there is certainly no bigger family! – Asaf Karagila Jul 14 '12 at 20:28
up vote 2 down vote accepted

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.

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More trivial than I thought... Thank you! – Pavel Jul 14 '12 at 21:08
@Pavel: I don’t quite understand exactly what you’re asking. Can you phrase the question more clearly? – Brian M. Scott Jul 20 '12 at 20:27
Oh, sorry, thought I will not be answered here, so I opened another question topic at…. I'll be very glad, if you could help me there. – Pavel Jul 20 '12 at 20:36

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