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By $[\omega]^\omega$ we denote the collection of infinite subsets of $\omega$. Two sets $A,B\in[\omega]^\omega$ are said to be almost disjoint if $A\cap B$ is finite. An almost disjoint family is a set ${\cal A}\subseteq [\omega]^\omega$ in which every two distinct members are almost disjoint. A standard application of Zorn's Lemma shows that any almost disjoint family is contained in a maximal almost disjoint (MAD) family (maximal with respect to $\subseteq$).

A "pathological" MAD family is $\{E, \omega\setminus E\}$ where $E = \{2n:n\in \omega\}$. We will consider infinite MAD families only. (A diagonalisation argument shows that every infinite MAD family is uncountable.)

Question. Is there an infinite MAD family ${\cal M}\subseteq [\omega]^\omega$ with $\bigcap {\cal M} = \emptyset$ and a set $R\subseteq \omega$ such that $|R\cap M| = 1$ for all $M\in {\cal M}$?

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2 Answers 2

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This answer only deals with the case that $R$ is infinite. I thought that I would be able to modify it to the finite case - thanks to Ilya Bogdanov for spotting the mistake in my argument. (His answer shows that for finite $R$ such family indeed exists.) And thanks to bof for explaining in a comment that my original argument was unnecessarily complicated.


$\newcommand{\mc}[1]{\mathcal{#1}}$Let us assume that $R$ and $\mc M$ fulfill the conditions given in the question. Moreover, let $R$ be an infinite set.

Then the system $\{R\}\cup\mc M$ is an almost disjoint family. (For every $M\in \mc M$, the intersection $R\cap M$ is finite.) Thus from maximality we get $R\in\mc M$.

But now $|R\cap R|=|R|\ne 1$, contradicting "choosability".


The above argument can be shortly summarized as follows: If an infinite sets has a finite intersection with each element of a MAD family $\mc M$, then this set belongs to $\mc M$.

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  • $\begingroup$ In Case 2, the set $M\cup R$ intersects with $M$ by infinitely many elements, so $\mathcal M’$ is not almost disjoint... $\endgroup$ Jun 7, 2022 at 12:58
  • $\begingroup$ @IlyaBogdanov Thanks for the clarification - I have to admit that this was quite an embarrassing mistake on my part. (This clarifies the comment I posted under your answer too - my objection was clearly wrong.) $\endgroup$ Jun 7, 2022 at 13:04
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Take any MAD family on $\omega\setminus\{a,b\}$ whose intersection is $\varnothing$. Then add $a$ to some of its elements and $b$ to all other elements. Then you can choose $R=\{a,b\}$.

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  • $\begingroup$ If we have the family $\mathcal M$ described in your answer, doesn't adding $\{a,b\}$ to each member of the family create a strictly larger almost disjoint family (thus contradicting the maximality)? In the other words, I am not sure that the family you've described is maximal - of course, I might simply have missed something. $\endgroup$ Jun 7, 2022 at 12:51
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    $\begingroup$ @MartinSleziak As far as I understand, we are not allowed to modify the members of the family, just to add new sets. So I javelin a family some of whose elements contain only $a$ (as I’ve added only $a$ to them), the others contain only $b$. This family is maximal, as you cannot add an infinite set to it, so what is the trouble? $\endgroup$ Jun 7, 2022 at 12:56

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