Choosing between incompatible ideals

Suppose \(\mathcal I\) and \(\mathcal J\) are proper ideals on some set \(X\). We say that \(\mathcal I\) and \(\mathcal J\) are incompatible if \(\mathcal I \cup \mathcal J\) does not generate a proper ideal. Equivalently, \(\mathcal I\) and \(\mathcal J\) are incompatible if there is some \(A \subseteq X\) such that \(A \in \mathcal I\) and \(X \setminus A \in \mathcal J\). If some \(B \subseteq X\) is either in \(\mathcal I \setminus \mathcal J\) or in \(\mathcal J \setminus \mathcal I\), then we say that \(B\) chooses between \(\mathcal I\) and \(\mathcal J\). We consider the following Ramsey-theoretic problem: Given several pairs \((\mathcal I_1,\mathcal J_1), (\mathcal I_2,\mathcal J_2), \dots, (\mathcal I_k,\mathcal J_k)\) of incompatible ideals on a set \(X\), find some \(A \subseteq X\) that chooses between as many of these pairs of ideals as possible. The main theorem is that for every \(n \in \mathbb N\), there is some \(I(n) \in \mathbb N\) such that given at least \(I(n)\) pairs of incompatible ideals on any set \(X\), there is some \(A \subseteq X\) choosing between at least \(n\) of them. This theorem is proved in two main steps. The first step is to identify a (purely finitary) problem in extremal combinatorics, and to show that our problem concerning ideals is equivalent to this combinatorial problem. The second step is to analyze the combinatorial problem in order to show that the number \(I(n)\) described above exists, and to put bounds on it. We show \(\textstyle \frac{1}{2}n \log_2 n - O(n) \,