Proving that the size of a set containing all unique elements of a series of subsets of a set Ω is equal to this summation?

Question

Let $Ω$ be a finite set. Let $A_1,...,A_n$ be finite subsets of $Ω$. For $S$, a nonempty subset of $\{1,2,...,n\}$ define $A_S = \cap_{i\in S}A_i$.

Let $Y$ be the set of elements of $Ω$ that are in exactly one of the sets $A_1,...,A_n$. Prove that: $|Y| = \sum_{S\subseteq [n]:S \not= \varnothing} (−1)^{|S|−1}|S| · |A_S |$

I have no idea how to do this problem. I attempted to partition the RHS (the sum) along possible values for $|S|$, to get the double sum $\sum_{i=1}^{n}\binom{n}{k}k(-1)^{k-1}\sum_{S\subseteq [n]:|S|=k}|A_S|$, but I think that summation might be slightly off and in any case, I haven't been able to advance past that. Any thoughts?

Answer 1

This is a special case of the Generalised inclusion-exclusion principle: If there are $a_k$ ways to choose $k$ out of $n$ conditions and fulfill them, then there are

$$ \sum_{k=j}^n(-1)^{k-j}\binom kja_k $$

ways to fulfill exactly $j$ of the conditions. In the present case, the conditions are membership in the $A_i$. There are

$$ a_k=\sum_{S\subset[n]\atop |S|=k}|A_S| $$

ways to choose $k$ conditions and fulfil them, so there are

$$ \sum_{k=1}^n(-1)^{k-1}\binom k1a_k=\sum_{S\subset[n]\atop S\ne\emptyset}(-1)^{|S|-1}|S||A_S| $$

ways to fulfil exactly $j=1$ of them.

For more on this generalized inclusion–exclusion principle see the comments and answers under the question linked to above.