Assume $\sf GCH$.

Let $\kappa$ be a regular cardinal, we say that $\{A_\alpha\mid\alpha<\kappa^+\}\subseteq\mathcal P(\kappa)$ is an almost disjoint family, if whenever $\alpha\neq\beta$, $A_\alpha\cap A_\beta$ is bounded in $\kappa$.

A family is MAD if it is a maximal almost disjoint family.

Given an almost disjoint family, we say that it is a broken family if whenever $A\subseteq\kappa^+$ is bounded, there is a refinement $B_\alpha\subseteq A_\alpha$ (and $A_\alpha\setminus B_\alpha$ bounded) for $\alpha\in A$, such that $\{B_\alpha\mid\alpha\in A\}$ are a pairwise disjoint family with the property that $A_\xi\subseteq^*\bigcup_{\alpha\in A}B_\alpha$ if and only if $\xi\in A$.

We can produce broken families from towers, as the pointwise difference. But those are not maximal.

  1. Is it provable, or at least consistent (with $\sf GCH$), that there are almost disjoint families which are not broken?
  2. How about MAD families? Can we prove that every MAD family is not broken, or at least consistently obtain broken MAD families?

(If the general case is a bit too hard, I'd be interested in the case for $\kappa=\omega$ as a particular case.)


1 Answer 1


To answer your Question 1, under GCH there is always an almost disjoint family which is not broken. To construct such a family, begin with a partition $\{A_\alpha \mid \alpha < \kappa\}$ of $\kappa$ into pairwise disjoint, unbounded subsets. Enumerate each $A_\alpha$ in increasing order as $\{\eta^\alpha_\beta \mid \beta < \kappa\}$. By GCH, fix a sequence of functions $\vec{f} = \langle f_\gamma \mid \gamma < \kappa^+ \rangle$ that is increasing and cofinal in ${^\kappa}\kappa$ modulo the bounded ideal. Now, for $\gamma \in [\kappa, \kappa^+)$, let $A_\gamma = \{\eta^\alpha_{f_\gamma(\alpha)} \mid \alpha < \kappa\}$. Since $\vec{f}$ is increasing modulo the bounded ideal, the family $\{A_\alpha \mid \alpha < \kappa^+\}$ is almost disjoint. Suppose for sake of contradiction that it is broken. Then there is a function $g:\kappa \rightarrow \kappa$ such that, setting $B_\alpha = \{\eta^\alpha_\beta \mid g(\alpha) \leq \beta < \kappa\}$, we have that $\{B_\alpha \mid \alpha < \kappa\}$ witnesses brokenness for the set $A = \kappa$. But then, if $\gamma \in [\kappa, \kappa^+)$ is such that $g <^* f_\gamma$, we have $A_\gamma \subseteq^* \bigcup_{\alpha < \kappa} B_\alpha$.

This seems like sort of a cheap example, though, since, as far as I can tell, the family might be made broken simply by removing the first $\kappa$ elements. Perhaps a more involved version of this construction could produce a more robust failure of brokenness.

  • $\begingroup$ I think that you've used $A_\alpha$ too many times. Nice construction! At least this reaffirms the idea that you can't immediately argue that a MAD family is broken. $\endgroup$
    – Asaf Karagila
    Aug 25, 2016 at 17:06
  • 1
    $\begingroup$ The $A_\alpha$'s for $\alpha < \kappa$ are the original $A_\alpha$'s that partition $\kappa$. I only used $\vec{f}$ to define $A_\alpha$ for $\alpha \in [\kappa, \kappa^+)$, so I think the notation is at least correct, if possibly confusing. $\endgroup$ Aug 25, 2016 at 17:11
  • $\begingroup$ Ohh, now I understand the idea entirely (I thought you redefined a new family entirely. I missed the $\gamma\in[\kappa,\kappa^+)$ part). Thanks! $\endgroup$
    – Asaf Karagila
    Aug 25, 2016 at 17:15

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