Inclusion exclusion principle
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This document contains the proof of two problems using the Inclusion/Exclusion Principle (IEP). For the first problem, it is proven that if the size of the union of two finite sets A and B is less than the sum of their individual sizes, then their intersection is empty. For the second problem, a formula is derived for the size of the union of a finite set A and the complement of another finite set B within a universal set U. The formula obtained is the size of U minus the size of B plus the size of the intersection of A and B.
Inclusion exclusion principle
- 1. Inclusion/Exclusion Principle Tyler Murphy April 30, 2014 This follows from the in-class work on 30 April 2014 Problem 1 Let A and B be finite sets. Prove that if | A ∪ B |<| A | + | B |, then A ∩ B = ∅ Proof. Directly. By The inclusion/Exclusion Principle, | A ∪ B |=| A | + | B | − | A ∩ B |. Suppose that | A ∪ B |<| A | + | B |. Then, by I.E.P., we have | A | + | B | − | A ∩ B |<| A | + | B |. Subtracting | A | + | B | from both sides, we get - | A ∩ B |< 0. So, | A ∩ B |> 0. So A ∩ B = ∅. Problem 2 Let A and B be finite sets in a universal set U. Find a formula for | A ∪ Bc | . By I.E.P., | A ∪ Bc |=| A | + | Bc | − | A ∩ Bc | . Now, we need to ask ourselves what each of these pieces involving the compliment is. First consider what Bc is. That is everything in U that is not in B. So | Bc |=| U − B |=| U | − | B |. So now we have | A ∪ Bc |=| A | + | U | − | B | − | A ∩ Bc | . Now we need to think about what | A ∩ Bc | is. First, consider what this represents. In | A ∪ B |=| A | + | B | − | A ∩ B |, the last piece is the amount of stuff we have overcounted by. 1
- 2. So, ask yourself what we have overcounted by in this problem. If we have | A | + | U | − | B |, we counted all of A once when we counted itself and again when we counted U. But when we took out B, we also took out A ∩ B. So in order to fix this, we need to take out one | A | and add in one | A ∩ B | So we have | A ∪ Bc |=| A | + | U | − | B | − | A | + | A ∩ B | . This simplifies to | U | − | B | + | A ∩ B |. So, | A ∪ Bc |=| U | − | B | + | A ∩ B |. So think about what we have said here. We have said that | A ∩ Bc |=| A − A ∩ B |. Let’s try to justify that. Consider what A∩Bc is. It is everything in A AND everything NOT in B. So this is all the things in A, but not in the intersection of A and B because those things are in B. So we have everything in A minus the things in both. So A∩Bc = A−A∩B. So it is ok to make our claim. The idea with all of this is to think about what the I.E.P. means. It simply means take all the things in the first set, add it to all the things in the second set, and take out everything you overcounted. So if you ever get stuck with trying to figure out what the last piece of the I.E.P. looks like, just step back and think about what you have overcounted. 2