Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. We say that ${\cal A}\subseteq [\omega]^\omega$ is an almost disjoint family if $A \neq B \in {\cal A}$ implies $|A\cap B|< \aleph_0$.

Let $X\neq\varnothing$ be a set and let ${\cal E}\subseteq {\cal P}(X)\setminus\{\varnothing\}$ be a collection of non-empty subsets. We say that a map $f: {\cal E}\to X$ is a chromatic self-map if

  1. $f(e) \in e$ for all $e\in {\cal E}$, and

  2. if $e_1\neq e_2 \in {\cal E}$ and $e_1\cap e_2 \neq \varnothing$, then $f(e_1)\neq f(e_2)$.

Question. Does every almost disjoint family ${\cal A}\subseteq [\omega]^\omega$ have a chromatic self-map?

Remark. It suffices to answer the question for maximum almost disjoint families ("MAD families").

  • $\begingroup$ I find your use of the term "self-map" a bit confusing. I guess you mean functions with the property $f(e)\in e$? I'm used to seeing those called choice functions, or selectors. And a "self-map" of a set $S$ is a map $f:S\to S$. $\endgroup$
    – bof
    Jun 7, 2020 at 10:49
  • 1
    $\begingroup$ The map would be injective: if $f(e_1)=f(e_2)=n$ then $n\in e_1\cap e_2$, so by condition 2 you'd get $e_1=e_2$. $\endgroup$
    – KP Hart
    Jun 7, 2020 at 16:32

1 Answer 1


If such an $f$ exists then $\mathcal A_n=\{e\in\mathcal A:f(e)=n\}$ is a collection of pairwise disjoint subsets of $\omega$, so $\mathcal A_n$ is countable, so $\mathcal A=\bigcup_n\mathcal A_n$ is countable. So the answer is "no" if $\mathcal A$ is uncountable. Of course it is "yes" if $\mathcal A$ is countable.


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