# Are there insane families in $L$?

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.)
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I would be glad to hear constructive remarks, in additional to the less-constructive downvotes! –  Asaf Karagila Sep 9 '13 at 17:13
+1: seems like you are having too much fun with your math, how non serious is that, tsk tsk... :-) –  Suvrit Sep 9 '13 at 17:32
@survit: Well, MAD families is a common term (which I can't, in good conscience, claim as my own) and insane families are just... madder than usual, because that tower thingie is not at all obviously definable from every mad family. :-) –  Asaf Karagila Sep 9 '13 at 17:37
Someone downvoted this? Weird. +1 from me, anyways. –  Noah S Sep 9 '13 at 18:51
If only you could have found a way to replace "L" with "the membrane" ... –  Yemon Choi Sep 9 '13 at 23:13

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.
Oh. I suspected as much. Drats. This means that I am going to have so much more work... Oh well. Thanks. I suppose this will be the same argument if we require $\leq$ and $>$ instead... –  Asaf Karagila Sep 9 '13 at 20:33
In the case of $P(\kappa)$, if you use an ideal $I$ for which $P(\kappa)/I$ is a complete Boolean algebra, then you can realize any maximal antichain as the difference antichain of a tower in the Boolean algebra. But you've got to work modulo $I$ instead of modulo finite. –  Joel David Hamkins Sep 9 '13 at 21:53
This question reminds me of your previous question about the countable completeness of $\mathcal{P}(\omega)$/fin. If an insane family exists, then $A_{\omega}$ (for example) is a least upper bound for the countable chain $\langle A_n: n < \omega \rangle$. For if $A_{\omega}' \subsetneq_* A_{\omega}$ was a strictly smaller upper bound, the difference set $X=A_{\omega} \setminus A_{\omega}'$ would be almost disjoint from all the $B_{\beta}$, contradicting maximality. But countable chains never have least upper bounds in $\mathcal{P}(\omega)$/fin, so it must be that there is no insane family. –  Garrett Ervin Sep 10 '13 at 4:14