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This question concerns a new cardinal characteristic of the continuum that arose out of issues in my answer to the question, Sunflowers in maximal almost disjoint families.

A family $\cal A$ of infinite subsets of $\omega$ is almost disjoint, if any two members of the family have finite intersection. Such a family is a maximal almost disjoint family if it cannot be extended to a larger almost disjoint family.

A $\Delta$-system, also known as a sunflower, is a family of sets with all pairs having the same pairwise intersection.

In his earlier question, Dominic had asked whether every maximal almost disjoint family must contain an infinite sunflower. In the general case, this seems still to be open, but my answer there shows that under the continuum hypothesis, there is a maximal almost disjoint family containing no sunflowers even of size $3$. Indeed, the construction there shows that there is a maximal almost disjoint family $\langle A_\alpha\mid\alpha<\omega_1\rangle$ such that every $A_\alpha$ has different intersections with every earlier $A_\beta$, for $\beta<\alpha$. This property implies that there can be no sunflower of size $3$ in this family. (But notice by a simple pigeon-hole argument that it will be impossible to extend this stronger property to enumerations beyond $\omega_1$.)

My questions concern the property of almost disjoint families that are maximal with respect to the property of not containing any sunflower of a certain size.

Question 1. If an almost disjoint family of infinite subsets of $\omega$ is maximal amongst almost disjoint families with respect to the property of not containing a sunflower of size $3$, is it a maximal almost disjoint family?

And more generally, I ask the same for sunflowers of any particular size.

The question leads naturally to new cardinal characteristics of the continuum. Namely, let us define almost-disjointness sunflower number, officially denoted $\frak{a}_{\kappa}^\Delta$, but let me immediately drop the superscript and write just $\frak{a}_\kappa$, to be the size of the smallest almost-disjoint family that is maximal among almost-disjoint families with respect to the property of not containing a sunflower of size $\kappa$. (We consider only $\kappa\geq 3$.)

The construction on my other answer shows that $\omega_1\leq\frak{a}_{\kappa}$.

Question 2. Can we separate these various cardinal characteristics $\frak{a}_\kappa$ from each other, and from the almost-disjointness number $\frak{a}$?

For example, is it consistent with ZFC that $\frak{a}_{3}<\frak{a}$? This would be a strong refutation of question 1. Is it consistent that $\frak{a}_{3}\neq\frak{a}_{4}$?

At first I had though it was clear that $\frak{a}_{\kappa}\leq\frak{a}$, the almost-disjointness number, which is the smallest size of any maximal almost disjoint family. But upon reflection, this no longer seems clear to me, since perhaps there could be a small maximal almost disjoint family, but it contains a lot of sunflowers, and the smallest maximal sunflower-free family might be larger. Or strictly smaller, since a maximal sunflower-free family might not be a maximal almost disjoint family. I had similarly expected that if $\kappa<\lambda$, then there should be some trivial provable relation between $\frak{a}_{\kappa}$ and $\frak{a}_{\lambda}$. But unless I am mistaken, this now also doesn't seem to be immediate.

Question 3. What are the provable relations between $\frak{a}_\kappa$, $\frak{a}_\lambda$, and $\frak{a}$ when $\kappa<\lambda$?

Can we say something even about the relation of $\frak{a}_{3}$ and $\frak{a}_{4}$, or their relation to $\frak{a}$?

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    $\begingroup$ These are fun questions. I do not see at the moment even that $\mathfrak a_3\not =\omega_1$. $\endgroup$ Jun 22, 2021 at 22:40

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The answer to Question 1 is NO:

Split $\omega$ into two disjoint infinite sets $X_0$ and $X_1$, fix $\mathcal A_0$ a countable AD family of subsets of $X_0$ such that

  1. $\mathcal A_0$ does not contain a $\Delta$-system of size $3$, yet
  2. for every finite $F\subseteq X_0$ there are $A,B\in \mathcal A_0$ such that $F=A\cap B$.

Added after comments below (To construct $\mathcal A_0$ start with a partition $\{B_n^{i}: n\in\omega, i\in 2\}$ of $\omega$ into infinite pieces and enumerate $[\omega]^{<\omega}$ as $\{F_n: n\in\omega\}$ . We shall recursively construct the family $\mathcal A_0=\{A_n^{i}:n\in\omega\}$ so that

(1) $A_n^{i}\subseteq^* B_n^{i}$,

(2) $F_n=A_n^0\cap A_n^1$ and

(3) $|A_n^{i}\cap A_m^{j}|\not = |A_n^{i'}\cap A_{m´}^{j'}|$ for every $m,m'<n$ and $i,i',j,j'\in 2$.

To accomplish this note that (1) gives you "enough room" to pick arbitrarily large disjoint pieces from the previous $A_m^{j}$'s which do not affect (2) to satisfy (3). Now, by (2) we have that $\mathcal A_0$ has the property that for every finite set $F\subseteq \omega$ there are $A,B\in \mathcal A_0$ such that $F=A\cap B$. The fact that we do not have "accidental" $\Delta$-systems of size $3$ follows directly from (3) as one (or two) of the lower indices of the potential 3-element $\Delta$-system come before the other and (3) guaranties that the corresponding intersections have different sizes.

Having done this just copy $\mathcal A_0$ on $X_0$).

Extend $\mathcal A_0$ to a maximal AD family $\mathcal A$ of subsets of $X_0$ not containing a $\Delta$-system of size $3$.

Claim. $\mathcal A$ is a maximal AD family of subsets of $\omega$ not containing a $\Delta$-system of size $3$ (yet not MAD).

$\mathcal A$ as a family of subsets of $\omega$ is not MAD as all elements of $\mathcal A$ are disjoint from $X_1$.

To show that it is maximal with respect to the property of not containing a $\Delta$-system of size $3$, assume that $B\in [\omega]^\omega$ AD with every element of $\mathcal A$. The proof has two simple cases:

Case 1: $C=B\cap X_0$ is infinite.

Then $C\cap A= B\cap A$ for every $A\in \mathcal A$ as $\mathcal A\subseteq \mathcal P(X_0)$. By maximality of $\mathcal A$ there are $A_0, A_1 \in\mathcal A$ such that $C\cap A_0=C\cap A_1= A_0\cap A_1$. Then, however, $B\cap A_0=B\cap A_1= A_0\cap A_1$.

Case 2: $C=B\cap X_0$ is finite.

Then by the property of $\mathcal A_0$ there are $A_0, A_1 \in\mathcal A$ such that $C= A_0\cap A_1$. Then $C=B\cap A_0=B\cap A_1= A_0\cap A_1$.

In both cases $\{B,A_0,A_1\}$ forms a $\Delta$-system.

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  • $\begingroup$ Of course, a similar argument also works for $\kappa>3$. $\endgroup$ Jun 22, 2021 at 21:40
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    $\begingroup$ Thank you for your answer! Shortly after asking my question, I had tried to mount exactly this construction, since I realized that having a family like your ${\cal A}_0$ was key. But I wasn't able to see easily how to produce it, since all my initial obvious attempts led to sunflowers. Can you explain how you produce that initial family? $\endgroup$ Jun 23, 2021 at 7:20
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    $\begingroup$ Thanks for the question, I have added a paragraph to the original answer. $\endgroup$ Jun 23, 2021 at 10:48
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    $\begingroup$ Thanks so much! I see it was your condition (3), which I was missing, that does the trick. (How funny to realize now that I had used this idea in my answer to Dominic's original sunflower question.) $\endgroup$ Jun 23, 2021 at 11:38

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