[Originally posted at [math.stackexchange](https://math.stackexchange.com/q/4626210/573047) without answer]


Consider a family $\mathcal{F}$ of $n=|\mathcal{F}|$ sets, $\emptyset \not\in \mathcal{F}$ and an universe $U(\mathcal{F})$ of $q=|U(\mathcal{F})|$ elements.


It is known that at least a fraction $r\binom{n}2$, $0 \lt r \lt 1$, of the unordered couples of sets of $\mathcal{F}$, have at least one element in common, i.e. $|\{\{A\_1,A\_2\}: A\_1,A\_2 \in \mathcal{F} \land A\_1 \not= A\_2 \land A\_1 \cap A\_2 \not= \emptyset \}| \ge r\binom{n}2$.


If we find the lowest $m$ such that:


$$\binom{m}2 \ge \frac{r\binom{n}2}q$$


we can then say that there exists an element belonging to at least $m$ sets of $\mathcal{F}$.


However since those $m$ sets cannot be made of only one element, I think the bound can be improved, i.e. we can say that there is an element in at least $m'$ sets:


$$m' = f(r,n,q) \gt m$$


Any idea for doing that?



Improving a lower bound for the minimum of the maximum frequency of an element in a family of sets
I do not think the bound can be improved without further assumptions. For example, consider the family $\mathcal{F}$ consisting of all singleton subsets of $\{1, \dots, n-1\}$ together with $\{1, \dots, n-1\}$. Here $q=n-1$, and $r\binom{n}{2}=n-1$. Thus, the smallest $m$ for which $$\binom{m}{2} \geq \frac{r\binom{n}{2}}{q}$$ is $m=2$. However, there is no element which is contained in $3$ sets of $\mathcal{F}$, so the bound is tight.


