I'm trying to come up with the largest family of sets that obeys the following properties:


Consider $X = \{1,\dots,n\}$ and take $\mathcal{F} \subset 2^X$ such that for any three subsets $A,B,C \in \mathcal{F}$ we have that (at least) two of the numbers $|A \cap B|, |B \cap C|, |A \cap C|$ are the same size.


There is a trivial example here, $\{1\}, \{1,2\}, \{1,2,3\},\dots$ which gives a family $\mathcal{F}$ of size $n$, but I don't see how to do better (I've tried using modular arithmetic, which doesn't seem to help all that much.)


In this problem it doesn't particularly matter (to me at least) what $n$ is, eg, for the modular arithmetic way I was trying I set $n = L^2$ for some integer.


So, my question is as follows, is there a better choice for $\mathcal{F}$, eg, one that is larger than the trivial?



Common sizes of intersections
Consider $\ \binom A2\ $ for an arbitrary finite set $\ A.\ $ Then you get a proper family with $\ \binom n2\ $ members for $\ n:=|A|.\ $ You get an improvement for every $\ n>3$.


**EDIT:**   Let me coin the name *"Three Is a Crowd"* (or *TIsC* for short) for the families introduced in the ***Question*** by **James Kilbane**. Now thanks to the comments let me record certain improvements:


* **The Masked Avenger** has proposed to add the whole set $\ A,\ $ and the empty set too--let's be greedy. Thus the enlarged Three Is a Crowd family looks like this:


$$ \binom A2\cup\{A\,\ \emptyset\} $$


for a new record $\ \binom n2+2\ $ for $\ n=|A|$.


* **Tony Huynh** has successfully added the singletons (and the empty set). This gives the following increased Three Is a Crowd family:


$$\binom{A}{2} \cup \binom A 1 \cup \binom A 0$$


for a new record $\ \binom {n+1}2$.


* The semi-dual example is just as good:


$$\binom A{n-2}\cup\binom A{n-1}\cup\{\emptyset\}$$


for $\ n:=|A|.\ $ We get $\ \binom {n+1}2\ $ different members again.



> 
> It'd be nice to find the **maximum**.
> 
> 
> 


