Sets II MATH Sets II. Benjamin V.C. Collins, James A. Swenson MATH 2730

Transcription

1 MATH 2730 Sets II Benjamin V.C. Collins James A. Swenson

2 New sets from old Suppose A and B are the sets of multiples of 2 and multiples of 5: A = {n Z : 2 n} = {..., 8, 6, 4, 2, 0, 2, 4, 6, 8,... } B = {n Z : 5 n} = {..., 15, 10, 5, 0, 5, 10, 15,... } Definition (by example) The union of A and B is A B = {n Z : 2 n or 5 n}. Their intersection is A B = {n Z : 2 n and 5 n}.

3 Venn diagrams A B A B

4 Who thought of Venn diagrams? John Venn ( ) Leonhard Euler ( ) Theorem (Boyer s Law) Mathematical formulas and theorems are usually not named after their discoverers. H.C. Kennedy, Amer. Math. Monthly 79, 66-67

5 More Venn diagrams A \ B B \ A A B

6 More operations Suppose A and B are the sets of multiples of 2 and multiples of 5: A = {n Z : 2 n} = {..., 8, 6, 4, 2, 0, 2, 4, 6, 8,... } B = {n Z : 5 n} = {..., 15, 10, 5, 0, 5, 10, 15,... } Definition (by example) The set difference A \ B = {n Z : 2 n and 5 n}. The set difference B \ A = {n Z : 5 n and 2 n}. The symmetric difference A B = (A \ B) (B \ A). Note: In the video, we mistakenly used the minus sign in the definition of the symmetric difference, rather than the backslash.

7 Sets II Subsets A B A A B B A \ B A B B \ A

8 Complements Thinking outside the disks There are always a lot of objects lying outside A B, but usually, we only care about elements in some universal set U. If we have chosen a particular universal set U, we can define the complement of a set A to be A = U \ A. We can prove identities like A B = A B.

9 Identities for set operations Theorem If A, B, and C are sets, then: A = A A = A B = B A A B = B A (A B) C = A (B C) (A B) C = A (B C) A (B C) = (A B) (A C) A (B C) = (A B) (A C) empty set commutativity associativity distributive laws

10 Reminder: proof templates Proving A B Write Let x A. Leave some space, and write, So x B. Therefore A B. Then fill the gap as usual, making no further assumptions about x. Proving two sets are equal Given sets A and B, it is easy to prove A = B. First, write, and prove A B. Then write, and prove B A.

11 Postponed example: set equality Example Prove: A (B C) = (A B) (A C). Proof. Let A, B, and C be sets. Suppose x A (B C). Thus x A or x B C. We consider two cases. First, suppose x A. In this case, x A B because x A. Likewise, x A C. Otherwise, x B C. Thus x B and x C. So again, x A B, and x A C. Thus x A or x B; also, x A or x C. So, in either case, x A B and x A C. Therefore x (A B) (A C).

12 Postponed example: set equality Example Prove: A (B C) = (A B) (A C). Proof. Now suppose x (A B) (A C). Thus x A B and x A C. We consider two cases. First, suppose x A. Then x A or x B C is true. Otherwise, x A. Since x A B, we have x B. Similarly, since x A C, we have x C. Thus x B C. So, in either case, x A or x B C. Therefore x A (B C).

13 A different kind of operation Definition Given sets A and B, the A B is the set of all ordered pairs with first entry in A and second entry in B: z A B x A, y B, z = (x, y). Example If A = {3, 4, 5} and B = {1, 2}, then A B = {(3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (5, 2)}.

14 Why is called Cartesian? The is named after René Descartes, who invented the idea of using ordered pairs as coordinates for points. René Descartes ( )

15 Why is called a product? As a corollary of the multiplication principle, we get: Proposition If A and B are finite sets, then A B = A B.

16 How many are there? Theorem (Principle of inclusion and exclusion) If A and B are finite sets, then A + B = A B + A B. Plan: We will ask a cleverly-chosen counting question. We will show that the lhs is the answer to the question, then that the rhs is the answer to the same question. Q: Imagine that we put a red sticker on each element of A, and a blue sticker on each element of B. How many stickers will we use? We need A red stickers and B blue stickers, for a total of A + B. On the other hand, each element of A B has a sticker, and each element of A B has an extra sticker, making A B + A B stickers in all.

17 Proof template: combinatorial proof Proving the integer equation lhs=rhs combinatorially Ask a cleverly-chosen counting question. Show that lhs is the answer to the question. Then show that rhs is the answer to the same question.