1
$\begingroup$

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").

$\endgroup$
2
  • $\begingroup$ 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 $f:S\to S$. $\endgroup$ Commented 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 $e_1=e_2$. $\endgroup$ Commented Jun 7, 2020 at 16:32

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.