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Let $\mathcal{F}$ be a family of subsets of $\left\{1, 2, \dots, n\right\}$ with size two or four. For each $X \subset \left\{1, 2,\dots, n \right\}$ with size three, there exists $Y \in \mathcal{F}$ such that $X \subset Y$ or $Y \subset X$. Prove that $\left\vert\mathcal{F}\right\vert \geq \left\lfloor \frac{(n-1)^2}{4} \right\rfloor$ if $n \geq 8$.

My attempt: If $Y \in \mathcal{F}$ has two elements, it is included in $(n-2)$ sets of three-element subsets, and if it has four elements, it is included in $4$ sets of three-element subsets, so I thought that including many sets with two elements would reduce the number of elements in $\mathcal{F}$.

So, I first tried to solve it by assuming that $\mathcal{F}$ has only two-element sets. Since there are $\frac{n(n-1)(n-2)}{6}$ set of three-element subsets, if one two-element set can cover $(n-2)$ of them, we only need $\frac{n(n-1)}{6}$ sets, which is smaller than $\frac{(n-1)^2}{4}$. So, we need to solve the case where a set of three-elements is overlapped by two sets of two-element sets at the same time. I guess this is a problem that can be solved through graph theoretic argument, but I don't know how to approach it. How can I solve this problem?

Dabin Yu
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    When $\mathcal F$ has only two-element sets, it represents a graph whose complement is triangle-free, so you could solve that special case by Mantel's theorem. – Misha Lavrov Mar 18 '25 at 15:26
  • e.g. see https://math.stackexchange.com/questions/3024908/proof-of-mantels-theorem and https://math.stackexchange.com/questions/2489542/mantels-theorem-proof-by-induction?rq=1 – Michael T Mar 19 '25 at 11:57

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