TITLE: Are there insane families in $L$?
QUESTION [12 upvotes]: 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.

Is the existence of insane families consistent with $\sf ZFC$?
If the answer is yes to the previous question, is there an insane family in $L$?
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.)

REPLY [12 votes]: 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.