12
$\begingroup$

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.)
$\endgroup$
5
  • 1
    $\begingroup$ I would be glad to hear constructive remarks, in additional to the less-constructive downvotes! $\endgroup$
    – Asaf Karagila
    Sep 9, 2013 at 17:13
  • 1
    $\begingroup$ +1: seems like you are having too much fun with your math, how non serious is that, tsk tsk... :-) $\endgroup$
    – Suvrit
    Sep 9, 2013 at 17:32
  • 2
    $\begingroup$ @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. :-) $\endgroup$
    – Asaf Karagila
    Sep 9, 2013 at 17:37
  • 1
    $\begingroup$ Someone downvoted this? Weird. +1 from me, anyways. $\endgroup$ Sep 9, 2013 at 18:51
  • 5
    $\begingroup$ If only you could have found a way to replace "L" with "the membrane" ... $\endgroup$
    – Yemon Choi
    Sep 9, 2013 at 23:13

1 Answer 1

12
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ 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... $\endgroup$
    – Asaf Karagila
    Sep 9, 2013 at 20:33
  • $\begingroup$ I find the issue related to the connection between maximal antichains and difference antichains in a complete Boolean algebra. See mathoverflow.net/a/139743/1946 and a few other places. $\endgroup$ Sep 9, 2013 at 20:37
  • $\begingroup$ Hmmm, yeah. Well, I guess this means that tomorrow there's a lot of work cut out for me. Thank you very much! $\endgroup$
    – Asaf Karagila
    Sep 9, 2013 at 21:34
  • 3
    $\begingroup$ 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. $\endgroup$ Sep 10, 2013 at 4:14
  • 1
    $\begingroup$ I agree with Garrett. The issue is that $P(\omega)/\text{Fin}$ is not a complete Boolean algebra. One can take $A_\omega'$ in Garrett's argument as $A_\omega-B$ in my construction. $\endgroup$ Sep 10, 2013 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.