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").



Chromatic self-maps for almost disjoint families
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.


