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?
Best Answer
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 $I$.