# Chromatic self-maps for almost disjoint families

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").

• 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 $f:S\to S$. – bof Jun 7 '20 at 10:49
• 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 $e_1=e_2$. – KP Hart Jun 7 '20 at 16:32

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.