Suppose $A$ is a finite set, $B$ is a collection of subsets of $A$, satisfying the following condition:
$$\forall a, b \in A, a \neq b: \exists C \in B: (a \in C) \land (b \notin C)$$
What is the least possible size of $B$.
Currently, I know that the minimal size of $B$ is not less than $\lceil \log_2 |A| \rceil$ (by pigeonhole principle), and it does not exceed $2\lceil \log_2 |A| \rceil$ (an example of that size can trivially be constructed). However, I do not know the exact answer to the question.
The answer by @FabioSomenzi is not just an upperbound - it is actually tight.
Define $S(a) = \{C \in B: a \in C\}$ for all $a \in A$. Note that $S(a) \subset B$.
The main condition $\forall a\neq b: \exists C \in B: a \in C \land b \notin C$ becomes $\forall a\neq b: S(a) \not\subset S(b)$
Therefore the sets $S(a)_{a\in A}$ form a Sperner family of size $|A|$.
The result now follows from Sperner's theorem, that the maximum sized Sperner family formed by subsets of $B$ has size ${|B| \choose |B|/2}$
UPDATE 2019-05-29: Lets see if we can get an estimate on the coefficient. Using Stirling's approximation $n! \sim \sqrt{2 \pi n} ({n \over e})^n$, we have
$${n \choose n/2} = {n! \over (n/2)! (n/2)!} \sim {\sqrt{2 \pi n} ({n \over e})^n \over \sqrt{\pi n} ({n/2 \over e})^{n/2} \sqrt{\pi n} ({n/2 \over e})^{n/2}} = \sqrt{2 \over \pi n} 2^n > |A|$$
So asymptotically we have $n \sim \log_2 |A|$, i.e. the coefficient is $1$, i.e. the pigeonhole-based lower bound is pretty tight.