Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. We say ${\cal A}\subseteq [\omega]^\omega$ is *almost disjoint* if $A \cap B$ is finite whenever $A\neq B \in {\cal A}$. Zorn's Lemma implies that every almost disjoint family is contained in a maximal one. Moreover, a diagonalization argument shows that every maximal almost disjoint (MAD) family is uncountable.


A *sunflower* is a set ${\cal X}$ of sets such that ${\cal X} \neq \emptyset$ and there is $K\subseteq \bigcup{\cal X}$ such that $X\cap Y = K$ whenever $X\neq Y\in {\cal X}$. (We allow for $K=\emptyset$.)


Does every MAD family ${\cal A}\subseteq [\omega]^\omega$ contain an infinite sunflower? If not, is it true that given $n\in \omega$, every MAD family contains a sunflower of cardinality $n$?



Sunflowers in maximal almost disjoint families
The following is a ZFC example, ***due to Michael Hrušák***, of a MAD family without sunflowers of
cardinality $3$.


Start with the standard AD family $\mathcal{B}=\{B\_f:f\in{}^\omega2\}$ of
branches through the binary tree $2^{<\omega}$, so $B\_f=\{f|n:n\in\omega\}$.
Extend $\mathcal{B}$ to a MAD family by adding a family $\mathcal{C}$ that
consists of antichains in the tree with the additional property
that each $C\in\mathcal{C}$ converges to point $b\_C$ in ${}^\omega2$ in the
sense that for every $n$ the set $\{c\in C:b\_C|n\subseteq c\}$ is cofinite
in $C$.
Every infinite subset of the tree that is almost disjoint from all members
of $\mathcal{B}$ contains such a set, so this yields a MAD family.


Next enumerate $\mathcal{C}$ as $\{C\_f:f\in{}^\omega2\}$ in a one-to-one fashion
and in such a way that $f\neq b\_{C\_f}$; we write $b\_f$ for $b\_{C\_f}$.
Define $A\_f=B\_f\cup D\_f$, where $D\_f$ is a co-finite subset of $C\_f$
specified as follows: let $k=\min\{n:f(n)\neq b\_f(n)\}$, then
$D\_f=\{c\in C\_f:\operatorname{dom}c\ge k+2$ and $c(k)\neq f(k)\}$.


The family $\{A\_f:f\in{}^\omega2\}$ is a MAD family without $3$-element sunflowers.
Let $f,g,h\in{}^\omega2$ and assume without loss of generality that
$k=\min\{n:f(n)\neq g(n)\}$ is larger than or equal to
$l=\min\{n:f(n)\neq h(n)\}$ and
$m=\min\{n:g(n)\neq h(n)\}$.
It follows easily that then in fact one has $l=m<k$.  

Let $s$ be the point in $B\_f\cap B\_g$ whose domain is $l+1$.
Then $s$ is not in~$A\_h$: it is not in~$B\_h$ because $s(l)\neq h(l)$,
it is also not in $D\_h$, because its direct predecessor is in~$B\_h$
and none of the points in $D\_h$ have their direct predecessor in $B\_h$.
It follows that $s\in (A\_f\cap A\_g)\setminus A\_h$, so $\{A\_f,A\_g,A\_h\}$
is not a sunflower.


If one uses the tree $k^{<\omega}$ instead of the binary tree then one create a MAD family with many sunflowers of cardinality $k$ but none of cardinality $k+1$.


