PIE: proof by counting

Recall the setup: we have a universal set U and a collection of subsets A_1, A_2, A_3, and so on, up to A_n. PIE claims that we can compute the number of elements of U that are in none of the A_i (that is, |U| - |A_1 \cup A_2 \cup \dots \cup A_n|) if we take the sizes of every possible intersection of some of the sets (including none of them, which yields U itself), and alternately add (if we intersected an even number of sets) or subtract (if odd). That is, using formal notation,

\displaystyle \left| |U| - \bigcup_{i = 1}^n A_i \right| = \sum_{S \subseteq \{1, \dots, n\}} (-1)^{|S|} \left|\bigcap_{i \in S} A_i\right|

Last time, we proved this using some algebra. This time, we’re going to prove it by counting. Consider a particular (arbitrary) element x \in U. We want to see how many times x gets counted: we count 1 for each intersection containing x that gets added, and -1 for each intersection containing x that gets subtracted. In the end, x should end up being counted exactly once if it’s not in any of the A_i, and exactly zero times otherwise.

So, suppose x is in exactly k of the sets A_i. Let’s consider intersections of different numbers of sets and see how many times x gets counted in each.

  • There is only one way to intersect 0 of the sets, which gives U itself. x is definitely in there, so including |U| means x gets counted once.

  • “Intersecting one set” sounds nonsensical, but really it just means we are considering the sets A_1, A_2, etc. by themselves; these all get subtracted. By assumption, x is in exactly k of them, so it gets counted -k times.

  • There are a bunch of different ways to intersect two sets, A_i \cap A_j. How many of them contain x? Of course x is in A_i \cap A_j exactly when it is in both A_i and A_j, so the number of two-set intersections containing x is the number of different ways to pick two out of the k sets containing x, that is, \binom{k}{2}. These all get added, so x is counted \binom{k}{2} times.

  • Likewise with intersections of three sets: the only way for x to end up in an intersection of three sets is if x is contained in all of them, and there are \binom{k}{3} ways to choose three out of the k sets containing x. These get subtracted.

In general, there are \binom{k}{i} different intersections of exactly i sets which contain x, up through i = k. So, all in all, x gets counted a grand total of

\displaystyle\binom{k}{0} - \binom{k}{1} + \binom{k}{2} - \binom{k}{3} + \dots \pm \binom{k}{k}

times. Now, when k = 0, this is just \binom{0}{0} = 1, which makes sense: if x is in none of the A_i, then it gets counted exactly once (as a part of U itself), just as it should be. Otherwise, we have an alternating sum of successive binomial coefficients—that is, we add up a row of Pascal’s triangle with alternating signs. Such an alternating sum of binomial coefficients is always zero, which I proved in another blog post. And that’s also as it should be: if x is in at least one of the A_i then we ultimately want it to be counted zero times.

About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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1 Response to PIE: proof by counting

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