Decreasing sequences and MAD Families I'm trying to understand something M. Hrusák wrote on "Almost Disjoint Families and Topology".
Given an Almost Disjoint Family $\mathcal A$, we define $\mathcal I(\mathcal A)=\{X\subset \omega: \exists \mathcal B \in [\mathcal A]^{<\omega}(X \subset^*\bigcup \mathcal B)\}$. And $\mathcal I^+(\mathcal A)=\mathcal P(\omega)\setminus \mathcal I(\mathcal A)$.
Now suppose $\{X_n: n \in \omega\}$ is a decreasing sequence of elements of $\mathcal I^+(\mathcal A)$. The author states there we may extend $\mathcal A$ to a MAD family $\mathcal B$ so that each $X_n$ is in $\mathcal I^+(\mathcal B)$. Question is: How?
 A: (This is almost certainly much too late to be of use to the OP, but it may still be of use to someone. The paper in question by Hrušák can be seen here [PDF], and the OP’s question is about a step in the proof of its Proposition $2$.)
It’s a straightforward Zorn’s lemma argument to show that $\mathscr{A}$ can be extended to an AD family $\mathscr{B}$ that is maximal among AD families containing $\mathscr{A}$ and having the property that $\{X_n:n\in\omega\}\subseteq\mathscr{I}^+(\mathscr{B})$.
Suppose that $\mathscr{B}$ is not a MAD family; then there is a $Y\in[\omega]^\omega\setminus\mathscr{B}$ such that $\mathscr{B}'=\mathscr{B}\cup\{Y\}$ is AD, and the choice of $\mathscr{B}$ ensures that there is an $m\in\omega$ such that $X_m\notin\mathscr{I}^+(\mathscr{B}')$, i.e., such that $X_m\in\mathscr{I}(\mathscr{B}')$. Clearly $Y\cap X_m\notin\mathscr{I}(\mathscr{B})$, since $X_m\in\mathscr{I}(\mathscr{B}')\setminus\mathscr{I}(\mathscr{B})$, and it’s immediate that $Y\cap X_n\notin\mathscr{I}(\mathscr{B})$ for $n<m$ as well.
Let $n\ge m$. Then $X_n\subseteq X_m$, so $X_n\in\mathscr{I}(\mathscr{B}')\setminus\mathscr{I}(\mathscr{B})$, and it follows that $Y\cap X_n\notin\mathscr{I}(\mathscr{B})$ and hence that $(Y\cap X_n)\setminus\bigcup\mathscr{F}$ is infinite for each $\mathscr{F}\in[\mathscr{B}]^{<\omega}$. Thus, we can recursively choose $x_n,y_n\in Y\cap X_n$ for $n\in\omega$ so that the $x_n$ and $y_n$ are all distinct. Let $C=\{x_n:n\in\omega\}$ and $D=Y\setminus C$; $C$ and $D$ are almost disjoint from each member of $\mathscr{B}$, and each has infinite intersection with each $X_n$.
For each $n\in\omega$ let $C_n=C\cap X_n$ and $D_n=D\cap X_n$; $C_n\cup D_n=Y\cap X_n\notin\mathscr{I}(\mathscr{B})$, so at least one of $C_n$ and $D_n$ is not in $\mathscr{I}(\mathscr{B})$. The sequences $\langle C_n:n\in\omega\rangle$ and $\langle D_n:n\in\omega\rangle$ are decreasing, and $\mathscr{I}(\mathscr{B})$ is an ideal, so at least one of the sequences must lie entirely outside $\mathscr{I}(\mathscr{B})$; without loss of generality assume that $D_n\notin\mathscr{I}(\mathscr{B})$ for each $n\in\omega$. Let $\mathscr{C}=\mathscr{B}\cup\{C\}$; $\mathscr{C}$ is an AD family properly extending $\mathscr{B}$.
Suppose that some $X_n\in\mathscr{I}(\mathscr{C})$, so that there is an $\mathscr{F}\in[\mathscr{B}]^{<\omega}$ such that $X_n\subseteq^*C\cup\bigcup\mathscr{F}$. Then $D_n\subseteq X_n\subseteq^*C\cup\bigcup\mathscr{F}$, and $D_n\cap C=\varnothing$, so $D_n\subseteq^*\bigcup\mathscr{F}$, which is impossible, since $D_n\notin\mathscr{I}(\mathscr{B})$. Thus, $X_n\in\mathscr{I}^+(\mathscr{B})$ for all $n\in\omega$, contradicting the choice of $\mathscr{B}$, and it follows that $\mathscr{B}$ must be a MAD family after all.
