[Originally posted at math.stackexchange 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?

EDIT - Restrictions 1

After the example by @Tony Huynh that shows that the bound is tight in the general case, I add the following restrictions:

1. $r =2/3$;
2. $q \le n/4$.

[Originally posted at math.stackexchange 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?

[Originally posted at math.stackexchange 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?

[Originally posted at math.stackexchange 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?

EDIT - Restrictions 1

After the example by @Tony Huynh that shows that the bound is tight in the general case, I add the following restrictions:

1. $r =2/3$;
2. $q \le n/4$.