Minimizing the intersection of three sets

Question

Let the sets $A,B,C$ which are all subsets of a larger set $N$. If $N(A), N(B), N(C), N$ are the populations respectively, then i need to find the minimum value of the population of their intersection i.e. $$N_{min}(A\cap B\cap C)=?$$

In the case of two sets $A,B\subseteq N$, the solution is rather easy to compute, by an almost direct application of the inclusion-exclusion principle: $$ N(A\cup B)=N(A)+N(B)-N(A\cap B)\leq N \Rightarrow \\ \Rightarrow N(A\cap B)\geq N(A)+N(B)-N \Rightarrow \\ \\ \\ \\ \\ \\ \Rightarrow \boxed{N_{min}(A\cap B)= N(A)+N(B)-N} $$ If the above expression is negative, this indicates that the minimum value of the intersection is zero (i.e. the sets can be completely "seperated" while still kept inside $N$).

In the case of three sets $A, B, C\subseteq N$, I have shown that the answer is: $$ \boxed{N_{min}(A\cap B\cap C)= N(A)+N(B)+N(C)-2N} $$ however, I find my argument somewhat intuitive (I have posted it as an answer below) and I would like to see other people's solutions and thoughts.

Finally, can we generalize the above, to produce a general formula for $$ N_{min}\Big(\bigcap_{i=1}^{\kappa}A_i\Big) $$ for the collection $A_{i}$, $i=1,2,...,\kappa$ of $\kappa$ subsets of $N$ ?

Answer 1

Here is my argument: In order to minimize the intersection $A\cap B\cap C$ we have to "achieve" the maximum possible "separation" between $A$, $B$, $C$, while still keeping them inside $N$. This happens when $N(A\cup B \cup C)$ maximizes, thus: $$ N(A\cup B \cup C)=N=N(A)+N(B)+N(C)-N(A\cap B)-N(A\cap C)-N(B\cap C)+N(A\cap B\cap C)\Rightarrow \\ \Rightarrow N(A\cap B\cap C)=N-\big(N(A)+N(B)+N(C)\big)+ \big(N(A\cap B)+N(A\cap C)+N(B\cap C)\big) $$ but $N-\big(N(A)+N(B)+N(C)\big)$ is a constant (provided by the initial data) and the rhs minimizes when the term $N(A\cap B)+N(A\cap C)+N(B\cap C)$ minimizes. Since it is a sum of three positive numbers we have: $$ N_{min}(A\cap B\cap C) = \\ =N-\big(N(A)+N(B)+N(C)\big)+ \big(N(A\cap B)+N(A\cap C)+N(B\cap C)\big)_{min}= \\ = N-\big(N(A)+N(B)+N(C)\big)+ N_{min}(A\cap B)+N_{min}(A\cap C)+N_{min}(B\cap C) $$ Now, substituting in the above, $$ N_{min}(A\cap B)=N(A)+N(B)-N \\ N_{min}(A\cap C)=N(A)+N(C)-N \\ N_{min}(B\cap C)=N(B)+N(C)-N $$ we arrive at: $$ N_{min}(A\cap B\cap C)= N(A)+N(B)+N(C)-2N $$

Answer 2

Your results suggest the generalisation that the minimum is $\min\left(0,\sum_iN(A_i)-(\kappa-1)N\right)$. To see that this is correct, take a $\kappa\times N$ rectangular grid and fill it with the elements of the subsets, from left to right and then from top to bottom. Elements in the same column are the same elements. If $\sum_iN(A_i)\le(\kappa-1)N$, then we're done before we have to start the $\kappa$-th row. Thus each element is in at most $\kappa-1$ subsets, so the intersection is empty, proving that the minimum in this case is $0$.

On the other hand, if $\sum_iN(A_i)\gt(\kappa-1)N$, the $\kappa$-th row contains $\sum_iN(A_i)-(\kappa-1)N$ elements, and in each of these columns there are elements from all $\kappa$ different subsets, so this is the size of the intersection. We cannot have any fewer elements in the intersection than this, since each subset can contribute at most $N-N(A_i)$ to the complement of the intersection, so the intersection has at least $N-\sum_i(N-N(A_i))=\sum_iN(A_i)-(\kappa-1)N$ elements.