Calculate the cardinality of the union of all two-intersections

Question

Suppose we have the classic inclusion-exclusion problem setup: we have a superset $S$, and $k$ sets $A_1,A_2,\dots,A_k$. We know $|S|, \forall_i (|A_i|=c_1), \forall_{i,j}(|A_i\cap A_j|=c_2), \dots, |\bigcap_{i=1}^k A_{i}|=c_k$. Now I want to calculate $$\vert\bigcup_{i<j}(A_i\cap A_j)\rvert$$ When $k=2$ it is trivial and the answer is $c_2$, when $k=3$, it seems the answer is $3c_2-2c_3$, but for larger $k$, I can't get a correct answer. I have also tried to work backward from the union of all $k-1$ intersections, etc. Any help is appreciated.

Answer 1

You can do it by induction on $k$. When the $(A_i)$ verify the properties you described, let us say that they check the property $P_k(c_1,\ldots,c_k)$. We want to prove that in this case, $$ \left|\bigcup_{1 \leqslant i < j \leqslant k} A_i \cap A_j\right| = f_k(c_1,\ldots,c_k), $$ for some function $f_k$. As you wrote it, we have $f_2(c_1,c_2) = c_2$.

Assume that such an $f_{k - 1}$ exists at rank $k - 1$. We have, $$ \bigcup_{1 \leqslant i < j \leqslant k} A_i \cap A_j = \bigcup_{1 \leqslant i < j \leqslant k - 1} A_i \cap A_j \cup \bigcup_{1 \leqslant i \leqslant k - 1} A_i \cap A_k. $$ Let $$ B = \bigcup_{1 \leqslant i < j \leqslant k} A_i \cap A_j, \qquad C = \bigcup_{1 \leqslant i < j \leqslant k - 1} A_i \cap A_j, \qquad D = \bigcup_{1 \leqslant i \leqslant k - 1} A_i \cap A_k. $$ Therefore, $B = C \cup D$. Notice that $(A_1,\ldots,A_{k - 1})$ verify $P_{k - 1}(c_1,\ldots,c_{k - 1})$, thus by induction, $|C| = f_{k - 1}(c_1,\ldots,c_{k - 1})$. To compute $|D|$, use the Poincaré formula, $$ |D| = \sum_{j = 1}^{k - 1} (-1)^{j - 1}\sum_{1 \leqslant i_1 < \cdots < i_j \leqslant k - 1} |A_{i_1} \cap \cdots \cap A_{i_j} \cap A_k| = \sum_{j = 1}^{k - 1} (-1)^{j - 1}\binom{k - 1}{j}c_{j + 1} $$ And for the cardinality of $C \cap D$, notice that $x \in C \cap D$ is equivalent to $x \in A_i$ and $x \in A_j$ for two distinct $1 \leqslant i < j \leqslant k - 1$ and $x \in A_i \cap A_k$ for some $i$. $x \in A_i$ is automatically fullfilled by the first condition, thus $x \in C \cap D$ is equivalent to : $x$ belongs to at least two $A_i$ ($i \leqslant k - 1$) and $x \in A_k$. Finally, $$ C \cap D = \bigcup_{1 \leqslant i < j \leqslant k - 1} A_i \cap A_j \cap A_k = \bigcup_{1 \leqslant i < j \leqslant k - 1} (A_i \cap A_k) \cap (A_j \cap A_k) $$ Now, notice that $(A_1 \cap A_k,\ldots,A_{k - 1} \cap A_k)$ verify $P_{k - 1}(c_2,\ldots,c_k)$ so by induction, $|C \cap D| = f_{k - 1}(c_2,\ldots,c_k)$. Finally, $f_k$ exists and, $$ f_k(c_1,\ldots,c_k) = f_{k - 1}(c_1,\ldots,c_{k - 1}) - f_{k - 1}(c_2,\ldots,c_k) + \sum_{j = 1}^{k - 1} (-1)^{j - 1}\binom{k - 1}{j}c_{j + 1}. $$ For example, we find the result you gave with $k = 3$, $$ f_3(c_1,c_2,c_3) = f_2(c_1,c_2) - f_2(c_2,c_3) + 2c_2 - c_3 = c_2 - c_3 + 2c_2 - c_3 = 3c_2 - 2c_3 $$ Moreover, a trivial induction show that $f_k$ is a linear form in $(c_2,\ldots,c_k)$ with coefficients in $\mathbb{Z}$ (it doesn't depend on $c_1$ and the hypothesis that all $A_i$ have the same cardinality is actually useless). However, the recursion formula is quite complicated and I don't think we can explicite $f_k$ for all $k$, but we can do it at least for small values of $k$ thanks to the induction formula.