# Finite limit to size of Intersecting Family with no Smaller Intersecting Set?

I'm interested in intersecting families of sets of size $r$, but where no set smaller than $r$ intersects the entire family. Meaning:

1. For every $A,B \in F$ we have $A \subset \{0,...,n-1\}$, $|A| = r$, $A \cap B \neq \emptyset$.
2. If $|C| < r$ there exists $D \in F$ such that $D \cap C = \emptyset$.

3. I also ask that we "use" every element in $\{0,...,n-1\}$, i.e for all $i < n$ there exists $A \in F$ such that $i \in A$.

Question: Given $r$, is there a finite upper limit to $n$?

Motivation:

For two simple cases, the answer is yes:

1. For $r=1$ trivially the maximal $n$ is $1$.
2. For $r=2$, I claim the maximal $n$ is $3$. For suppose we have in $F$ w.l.o.g sets $A = \{1,2\}$ and $B= \{1,3\}$. To "use" $4$, while intersecting $A$, we must have a set $\{1,4\}$ or $\{2,4\}$. However $\{2,4\}$ obviously doesn't intersect $B$, so we're left with $\{1,3\}$. So every "new" element $x$ we want to use after $3$ must add set $\{1,x\}$ to $F$. However this means $\{x\}$ intersects all the sets in $F$, violating requirement (2). Therefore for $r =2$ we have $n \le 3$.

But I have no idea how to generalize this result, if it's even true for greater $r$. The Fano plane shows that for $r=3$ the maximal $n \ge 7$.

The straightforward way of creating an intersecting family is just to pick a fixed element $x$, and take all the subsets of $\{1,...,n\}$ that include $x$. However, that means $\{x\}$, a set of size $1$, intersects all the other sets in the family, violating requirement (2) for $r > 1$.

Fedor Petrov provided an affirmative answer - for each $k$ there is a finite limit to the size $F$ (and hence also to $n$).