Sunflowers in maximal almost disjoint families

Question

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 $n$?

Answer 1

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 $k+1$.

Answer 2

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?

Answer 3

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.