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](https://en.wikipedia.org/wiki/Almost_disjoint_sets), are uncountable, and it is consistent that every clique in $G$ has cardinality $< 2^{\aleph\_0}$.


Is it consistent that $\chi(G) < 2^{\aleph\_0}$?



Coloring almost-disjointness
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.


