# 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 $$n$$?

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 $$l=m.
Let $$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 $$k+1$$.

• Indeed. Thanks. Jun 15, 2021 at 7:20
• Is there any hope to modify the construction to work with an arbitrary almost disjoint family that is maximal with respect to not including 3-element sunflowers? This would lead to an answer to my question at mathoverflow.net/q/394987/1946. Jun 15, 2021 at 8:34
• Hope springs eternal but this particular example is tied very strongly to the geometry of the binary tree and it very much uses that both families have the same szie: $\mathfrak{c}$. But maybe you can take an AD family without $3$-element sunflowers and embed it into a binary tree (each member into its own branch). Jun 15, 2021 at 8:53

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 $$k, promising to add no additional elements of $$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, 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 $$\frak{a}_3\neq\frak{a}_4$$?

• Yes, and opens up the question about m.a.d. families of size larger than $\aleph_1$. Jun 9, 2021 at 7:41