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 $M\in {\cal M}$?



MAD family with the choosability property
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 $\mc M$.


