Let $A,B\subseteq\omega$. We write $A\subseteq^\*B$ if $A\setminus B$ is finite, if additionally $B\setminus A$ is infinite then we write $A\subsetneq^\*B$, otherwise we write $A=^\*B$.


We say that a $\cal A\subseteq P(\omega)$ is almost disjoint if for every two distinct $A,B\in\cal A$ we have $A\cap B=^\*\varnothing$. We say that $\cal A$ is maximal almost disjoint, or MAD, if there is no $\cal B$ strictly containing $\cal A$ which is almost disjoint.


At the other end of the spectrum we say that $\cal A\subseteq P(\omega)$ is a tower if $\cal A$ is well-ordered by $\subsetneq^\*$.


Finally, we define $\mathcal B=\{B\_\alpha\mid\alpha<\kappa\}$ to be **insane** if it is MAD, and there exists a tower $\mathcal A=\{A\_\alpha\mid\alpha<\kappa\}$ with the following property: $$\beta<\alpha\implies B\_\beta\subseteq^\*A\_\alpha\\ \beta\geq\alpha\implies B\_\beta\cap A\_\alpha=^\*\varnothing.$$
In that case we say that $\cal A$ is an associated tower for $\cal B$.


Note, for example, that if $\cal B$ is insane and $\cal A$ is an associated tower then $A\_{\alpha+1}\setminus A\_\alpha=^\*B\_\alpha$.




---



> 
> **Questions.**
> 
> 
> 1. Is the existence of insane families consistent with $\sf ZFC$?
> 2. If the answer is yes to the previous question, is there an insane family in $L$?
> 3. If the answer is yes to the previous question, can this notion be extended to every regular cardinal $\kappa$? (replacing "finite" by ${<}\kappa$ in the definition of $\subseteq^\*$ and so on.)
> 
> 
> 



Are there insane families in $L$?
Your requirements are inconsistent; there is no insane family.


Suppose towards contradiction that we have an insane family
$\mathcal{B}=\{B\_\alpha\mid\alpha\lt\kappa\}$, witnessed by tower
$\langle A\_\alpha\mid\alpha\lt\kappa\rangle$. For finite $k$, let $b\_k$ be any
element in $[(A\_\omega-A\_k)\cap B\_k]-\bigcup\_{j\lt k}B\_j$. There are such elements, since $B\_k$ is almost disjoint from $A\_k$ and from the earlier $B\_j$ for $j\lt k$, and $B\_k$ is almost contained in $A\_\omega$. Note that the $b\_k$ are
distinct, and so $B=\{b\_k\mid k\lt\omega\}$ is infinite. By
maximality, $B$ must have infinite intersection with some
$B\_\beta$. Note that $B$ has exactly one element from each $B\_k$
for $k\lt\omega$. So it must be that $\beta\geq\omega$. But in
this case, since $B\subset A\_\omega$, we have infinitely many
elements in $B\_\beta\cap A\_\omega$, which violates the second insanity clause. 


So there is no insane family.


