# MAD family with the choosability property

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 $$M\in {\cal M}$$?

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.

$$\newcommand{\mc}{\mathcal{#1}}$$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 $$\mc M$$.

• In Case 2, the set $M\cup R$ intersects with $M$ by infinitely many elements, so $\mathcal M’$ is not almost disjoint... Jun 7, 2022 at 12:58
• @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.) Jun 7, 2022 at 13:04

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 $$R=\{a,b\}$$.

• 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. Jun 7, 2022 at 12:51
• @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? Jun 7, 2022 at 12:56