Sunflowers in maximal almost disjoint families 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 $n$?
 A: It is consistent with ZFC that the answer to your question is no. Specifically, I claim, if we assume the continuum hypothesis, then the answer is no, not even for sunflowers of size $n=3$.
Theorem. Assume the continuum hypothesis. Then there is a
maximal almost disjoint family with no sunflower of size 3.
The proof will rely on the following lemma.
Lemma. If $\cal A$ is a countable almost disjoint family and
$B$ is an infinite set almost disjoint from every member of $\cal
A$, then there is an infinite set $A$ having a different finite
intersection with every element of $\cal A$ and containing
infinitely many elements of $B$.
Proof. Enumerate the family ${\cal A}=\{A_0,A_1,A_2,\ldots\}$,
and fix the infinite set $B$ almost disjoint from every $A_n$. We
build the set $A$ in stages. At stage $n$, we will have already
fixed the intersections $A\cap A_k$ for $k<n$, promising to add no
additional elements of $A_k$ to $A$ beyond what has already been
added. Consider $A_n$. By adding some elements of $A_n$ to $A$ not
in any $A_k$ for $k<n$, we can ensure that $A\cap A_n$ is distinct
from the intersections $A\cap A_k$ that we've already fixed. And we
can also add another element of $B$. After doing this, we promise
not to add any more elements from $A_n$. Thus, in countably many
steps, we construct the set $A$ as desired. $\quad\Box$
Proof of theorem. By CH we can well order the infinite subsets
of $\mathbb{N}$ in order type $\omega_1$. We shall now form a
maximal almost disjoint family of sets $\langle
A_\alpha\mid\alpha<\omega_1\rangle$, with the further property that
every $A_\alpha$ has a distinct finite intersection with $A_\beta$
for all $\beta<\alpha$. This property will ensure that the family
has no sunflower of size 3.
At stage $\alpha$, consider the least set $B$ in the well order
that is almost disjoint from every $A_\beta$ for $\beta<\alpha$. By
the lemma, there is a set $A_\alpha$ that has a different finite
intersection with every $A_\beta$ for $\beta<\alpha$ and contains
infinitely many elements of $B$.
By construction, we've made an almost disjoint family containing no
sunflowers of size 3. It is a maximal almost disjoint family, since
if $B$ is a set that is almost disjoint from the family, then at
some stage it would have been the least such set, and then we would
have added a set having infinite intersection with it. So this is a
maximal almost disjoint family with no sunflowers of size 3, as
desired. $\quad\Box$
We might consider an almost disjoint family that is maximal with respect to the property of not containing any 3-sunflower (or $\kappa$-sunflower for any cardinal $\kappa$). This suggests a host of new cardinal characteristics, namely, $\frak{a}_\kappa$ is the size of the smallest almost disjoint family that is maximal with respect to the property of not containing any $\kappa$-sunflower. The argument I give shows that $\omega_1\leq\frak{a}_3$, and I guess it is immediate that $\frak{a}_\kappa\leq\frak{a}$ (Update this is not actually clear). I am less clear on the relation between $\frak{a}_\kappa$ and $\frak{a}_\lambda$ if $\kappa<\lambda$.
Question. Can we separate these cardinal characteristics?
For example, can we find a model where $\frak{a}_3<\frak{a}$ or where $\frak{a}_3\neq\frak{a}_4$?
I have posted a question about this at Can we separate the almost-disjointness sunflower numbers?
A: 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 $k+1$.
A: One remark on KP Hart's answer: if one does the same to a finitely branching tree with unbounded sizes of the splitting sets then one gets a MAD family with no infinite delta-system but delta systems of all finite sizes.
