Suppose that $V$ is a model of $\sf ZFC$, and for concreteness I should point that at this point I am interested in $V=L$ as a ground model.

Suppose that $V[c]$ is a Cohen extension of $V$ where $c$ is a real number, and $\{A_n\mid n\in\omega\}$ is an almost disjoint family in $V[c]$ of subsets of $\omega$. Can we always find an almost disjoint family $\{A'_n\mid n\in\omega\}$ such that $A_n\subseteq A'_n$?

What about larger cardinals? Suppose that $\{A_\alpha\mid\alpha<\kappa\}$ is an almost disjoint family of subsets of a regular $\kappa$ (almost disjoint means that the intersection of any distinct two is of size ${<}\kappa$). Can we find in $V$ a separation family?

While this is not the case I care about, I am curious about the case where the almost disjoint family is maximal, or at least of size $\kappa^+$. Can we still find such family?

  • $\begingroup$ Interestingly enough, both a positive and a negative answer would be useful. Although in different ways. $\endgroup$
    – Asaf Karagila
    Sep 14, 2013 at 21:58

2 Answers 2


Perhaps I am misunderstanding your question, but I think you intend that $A_n'\in V$ and also that the family $\{A_n'\mid n\in\omega\}$ is in $V$. Is that right?

In this case, the answer is no. Let $A_0$ consist of every other element of $c$, and $A_1$ consist of every other element of the remaining elements of $c$ and so on with $A_n$. These are fully disjoint. Now, suppose that we have $A_n\subset A_n'\in V$ and the family $\{A_n'\mid n\in\omega\}$ is in $V$ and almost disjoint. So $A_0'$ is a specific infinite co-infinite set in $V$. Let $p$ force that $\dot A_0\subset \check A_0'$, using a name $\dot A_0$ for this definition of $A_0$. But this is impossible, since we can extend $p$ so as to control the later elements of $c$ in such a way that would violate $A_0\subset A_0'$, by inserting the next two elements into $c$ from outside of $A_0'$. So this is an almost disjoint family in $V[c]$ that cannot be covered by a ground model family $A_n'$ in the way you described.

The same argument seems to work with higher cardinals and larger families. One simply extends the every-other-element set $A_0$ to a maximal almost disjoint family of any kind, and runs the same argument. Basically, the obstacle is that there is no ground model co-infinite set $A$ that you can force to cover of the every-other-element-set of the generic set.


Here's a possibly simpler example than Joel's, in the case of $\omega$. Let $A_0$ be $c$ (where I think of the Cohen real as a subset of $\omega$), and let the rest of the $A_n$'s be any partition of $\omega-c$. (In fact, you could let $A_1=\omega-c$ and let all the later $A_n$'s be empty, if you're willing to let the empty set into your almost disjoint family.) An easy genericity argument shows that the only ground-model subsets $A_0'$ of $\omega$ that satisfy $c\subseteq A_0'$ are cofinite. So $A_0'$ would have to meet $A_1'$ infinitely.

  • 1
    $\begingroup$ Yes, this is the right way to do it. I don't know why I started with every-other-element of $c$ instead of just $c$ itself; but otherwise, this is the same idea. (I also had initially posted about $c$ and $\omega-c$, but deleted it...) $\endgroup$ Sep 15, 2013 at 0:46
  • $\begingroup$ Thank you Andreas, and @Joel. The situation I have in mind is much closer to Joel's answer than to this. But both answers have been very helpful, thank you very much! $\endgroup$
    – Asaf Karagila
    Sep 15, 2013 at 5:17

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