If $X\neq\emptyset$ is a set, then ${\cal S}\subseteq {\cal P}(X)$ with ${\cal S}\neq \emptyset$ is said to be a *sunflower* if there is $K\subseteq X$ such that whenever $A\neq B\in{\cal S}$ ten $A\cap B = K$. ($K$ is sometimes said to be the *kernel* of ${\cal S}$, and it is allowed that $K = \emptyset$, in which case ${\cal S}$ is a collection of pairwise disjoint sets).


Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. There are examples of countable collections ${\cal K} \subseteq [\omega]^\omega$ such that for every sunflower ${\cal S}\subseteq {\cal K}$ we have $|{\cal S}| \leq 2$. (At the end of this post I construct such a ${\cal K}$.)


**Question.** Is there an uncountable collection ${\cal L}\subseteq [\omega]^\omega$ such that every sunflower ${\cal S}\subseteq {\cal L}$ has cardinality at most $2$?


**Construction** for ${\cal K}$. For every $n\in \omega$, let $p\_n$ be the $n$th prime, so $p\_0=2, p\_1=3$, etc. For $n\in\omega$ set $$W\_n = \{0, \ldots, p\_n\} \cup \{p^n: n\in \omega\},$$ and let ${\cal K}=\{W\_n: n\in \omega\}$.



Sunflowers in $\omega$ consisting of infinite sets
There exist uncountable chains in the poset $([\omega]^\omega,\subseteq)$ (e.g. by bijecting $\omega$ with $\mathbb Q$ and taking all downwards-closed sets). Let $L$ be such a chain. The only sunflowers in $L$ have two elements, since if $A,B,C\in L$ and $A\subsetneq B\subseteq C$, then $A\cap B=A\neq B=B\cap C$.


