# Coloring almost-disjointness

Let $$[\omega]^\omega$$ denote the collection of infinite subsets of $$\omega$$. Let $$E = \big\{\{A,B\}: A,B \in [\omega]^\omega \text{ and } |A\cap B| \text{ is finite}\big\}.$$

We consider the graph $$G=([\omega]^\omega,E)$$. Maximal cliques in $$G$$, also known as maximal almost disjoint families, are uncountable, and it is consistent that every clique in $$G$$ has cardinality $$< 2^{\aleph_0}$$.

Is it consistent that $$\chi(G) < 2^{\aleph_0}$$?

• What is consistent is that $\mathfrak a<2^{\aleph_0}$ but surely there are MAD families of cardinality $2^{\aleph_0}$, here is one: fix a bijection $f:\omega\to \Bbb Q$ and for every $r\in\Bbb R\setminus\Bbb Q$ fix a sequence $(q^r_i)$ of rational numbers converging to $r$. Take preimages through $f$ of those sequences and extend to a maximal family to have a MAD family of cardinality continuum. May 7 at 13:52
• @AlessandroCodenotti Since your correction immediately provides a negative answer to the question, it would be good to convert your comment to an answer, so that the software doesn't think the question is unanswered. May 7 at 17:34

No, $$\chi(G)=\mathfrak c$$, in fact $$G$$ contains a complete subgraph on $$\mathfrak c$$ vertices.
A simple way to construct one is by fixing a bijection $$f\colon\Bbb Q\to\omega$$ and fixing, for every $$r\in\Bbb R\setminus\Bbb Q$$, a sequence $$(q^r_i)_{i\in\omega}$$ of rationals numbers converging to $$r$$. Taking the images through $$f$$ of those sequences gives an almost disjoint family of cardinality $$\mathfrak c$$ (which can be extended to a mad family of the same cardinality by Zorn's lemma but that's not even needed).
The consistency result is that $$\mathfrak a$$ can be strictly between $$\aleph_0$$ and $$\mathfrak c$$, that is there can be some mad family of cardinality less than continuum.