# Kunen chapter II exercise $62$ about $\kappa$-complete, $\lambda$-saturated ideals

The property $$S(\kappa,\lambda, I)$$ states that $$I$$ is a $$\kappa$$-complete ideal on $$\kappa$$ which contains all the singletons, and there is no family $$\{X_{\alpha}: \alpha<\lambda\}\subseteq P(\kappa)$$ such that $$X_{\alpha}\notin I$$ for all $$\alpha<\kappa$$, but $$X_{\alpha}\cap X_{\beta}\in I$$ for all $$\alpha\ne\beta<\kappa$$.

Problem: Assume $$S(\kappa,\lambda, I)$$ where $$2^{<\lambda}<\kappa$$. Show that there is an atom, i.e a set $$A\subseteq\kappa$$ such that $$A\notin I$$ and for all $$X\subseteq A$$ we have $$X\in I$$ or $$A\setminus X\in I$$.

Attempt: I managed to show this in the case $$\lambda=\omega$$, which was the previous exercise. In this case, suppose there are no atoms. Let $$X=\kappa\notin I$$. Since $$X$$ is not an atom we can write it as a disjoint union $$X=X_0\cup X_1$$ where $$X_0, X_1\notin I$$. Now, $$X_0, X_1$$ are also not atoms and so we can write them as disjoint unions $$X_0=X_{00}\cup X_{01}, \ \ X_1=X_{10}\cup X_{11}$$ of sets which are not in $$I$$. Similarly we write $$X_{00}=X_{000}\cup X_{001}$$, and so on. We continue by induction, and then the sets:

$$X_1, X_{01}, X_{001}, X_{0001}, X_{00001}$$,...

Are a family of $$\omega$$ pairwise disjoint sets which don't belong to $$I$$. But the existence of such a family contradicts $$S(\kappa,\omega, I)$$.

So now I tried to do something similar for a general $$\lambda$$. Again, I assumed there are no atoms and started to break sets like before: $$X=X_0\cup X_1, \ X_0=X_{00}\cup X_{01}$$ and so on. But now in the induction process we also have to deal with limit steps. The natural thing to do at a limit step is to intersect the sets in every branch of the binary tree, which will give sets like $$X_0\cap X_{00}\cap X_{000}\cap X_{0000}\cap...$$. However, there is a problem with that idea, because we can't be sure the intersection will not be in $$I$$. So I'm stuck here.

Obviously, I have to use the assumption $$2^{<\lambda}<\kappa$$, but I don't see how. Any ideas?

• Sorry for my previous comment. I think one way to deal with this situation is to prove that there always exists at least one branch at every limit level of the tree with intersection not in $I$ and now you may continue the construction from that branch onwards. Here you will use $\kappa$-completeness and $2^{<\lambda}<\kappa$. Commented Jun 26, 2021 at 21:46
• Yes, I just thought about that a few minutes ago. Indeed, there are at most $2^{<\lambda}$ branches, so one of them will not be in $I$. Then when we seek a contradiction we will take the sets from a different branch. I believe it is very annoying to describe this process formally, but it is what it is. Thanks for your help.
– Mark
Commented Jun 26, 2021 at 21:52
• No problem. (That's exactly why I decided to write a comment rather than a complete answer describing the process at 2:30 a.m. :D) Commented Jun 26, 2021 at 21:57

You're on the right track. The next key idea is to ignore those bad sets, the sets in $$I$$, that you encountered at limit steps, and just keep splitting the remaining (good, $$\notin I$$) sets through $$\lambda$$ levels. Notice that the complete binary tree with $$\lambda$$ levels would have cardinality $$2^{<\lambda}$$, so you have at most $$2^{<\lambda}<\kappa$$ bad sets. Their union is therefore in $$I$$, by $$\kappa$$-completeness. In particular, you can fix some $$x\in X$$ that is not in any of your bad sets. The construction of your tree ensures that this $$x$$ is in exactly one set at each level $$\alpha<\lambda$$. Thus, $$x$$ determines a path of length $$\lambda$$ through the tree, which doesn't get stuck at any bad node. Now use this path the same way that you used the all-zeros path in the case $$\kappa=\omega$$. That is, consider the nodes that are just one step off the path; they are $$\lambda$$ pairwise disjoint sets, none of which are in $$I$$.