Set Operations. Combining sets into new sets

Transcription

1 Set Operations Combining sets into new sets

2 Union of Sets The union of two sets is the set of all elements that are in one or the other set: A B = {x x A x B} The union is the set theoretic equivalent of the inclusive or. To the right is a Venn diagram illustration of the union. Example: If A = 1,2,3 and B = 3,4,5, then A B = {1,2,3,4,5}.

3 Intersection of Sets The intersection of two sets is the set of all elements that are in both sets: A B = {x x A x B} The intersection is the set theoretic equivalent of the and. To the right is a Venn diagram illustration of the intersection. Example: If A = 1,2,3 and B = 3,4,5, then A B = {3}. If the intersection of A and B is empty, A and B are called disjoint. This situation is represented by a Venn diagram containing two circles that do not intersect. A collection of sets is called pairwise disjoint if any two of them are disjoint.

4 Do not confuse the logical operators with the corresponding set operators While there is a correspondence between logical and set operators, they cannot be used interchangeably. If A and B are sets, then union and intersection must not be written as A B and A B. These notations are meaningless because and act on statements, not on sets. Likewise, if p and q are statements, do not write meaningless expressions like p q or p q.

5 Do not confuse sets with statements that define the sets Joe is a vegetarian does not mean the same as vegetarians. Likewise, x A x B does not have the same meaning as A B. x A x B is the statement that some object x is a member of the set A, or a member of the set B. A B is the set the collection of objects that are in A, or are in B. Statements and sets are different categories of objects. A statement cannot be equal to a set.

6 Symmetric Difference of Sets The symmetric difference of two sets is the set of all elements that are contained in exactly one of the sets: A B = A B = {x (x A) (x B)} The symmetric difference is the set theoretic equivalent of the exclusive or. To the right is a Venn diagram illustration of the symmetric difference. Example: If A = 1,2,3 and B = 3,4,5, then A B = {1,2,4,5}.

7 Set Operator Example Proof 1 Prove that 1,2 1,3 = 1,3. To prove equality of two sets, we must show that either is a subset of the other. We therefore give our proof into two parts: 1,2 1,3 1,3 : Suppose x 1,2 1,3. By definition of union, that means that x 1,2 or x 1,3. By definition of open intervals, the first case implies 1 < x < 2, and the second case implies 1 < x < 3. If x < 2, then x < 3 as well. Hence, the first case implies the same condition as the second case. Therefore, either way, 1 < x < 3. By definition of open interval, this means x 1,3. 1,3 1,2 1,3 : Suppose x 1,3. Since we already know this to be true, (x 1,3 or x 1,2 ) is true as well. By definition of union, this means x 1,2 1,3.

8 Set Operator Example Proof 2 Prove that 1,2 2,3 is the empty set. Proof: Suppose, to get a contradiction, that there exists x 1,2 2,3. By definition of intersection, that means x 1,2 and x 2,3. By definition of open interval, this means 1 < x < 2 and 2 < x < 3. The first condition implies x < 2. The second condition implies x > 2. That is a contradiction.

9 Difference of Sets The difference of one set minus another set is the set of all elements that are in the first but not the second set: A B = A\B = {x x A x B} To the right is a Venn diagram illustration of A B. Unlike the previous set operations, the difference is not symmetric. A B and B A are not equal. There is no single logical operator that we named that corresponds to the difference operation. Example: if A = 1,2,3 and B = 3,4,5, then A B = 1,2 and B A = 4,5. Observe that we can use the difference to give an equivalent definition of the symmetric difference: A B = (A B) (B A).

10 Complement By the complement of a set S, we mean all objects that are not contained in S. We write Sҧ for the complement of S. Some authors also use the notations S c or S. The complement is the set theoretic negation. [Be careful with spelling: compliment is something else entirely.] Unlike the previously defined set operators, which all act on the sets they are explicitly said to be acting on, the complement operation requires an additional input that is not acknowledged in notation. Sҧ is meaningless unless we specify the surrounding set. We need to say what all is. We will illustrate the problem with an example first. Suppose S is the set of all men. Then what the set of all objects that are not in S is depends on the type of objects under consideration. If the set of all objects is all people, then Sҧ is all women. But if the set of all objects is all life on Earth, then assuming that S is all human men, Sҧ includes not only human women, but also plants and animals.

11 ҧ The Universal Set The set of all objects that the complement depends on is called the universal set, and usually denoted U. With that, we can give a formal definition of complement in terms of a set difference: ҧ A = U A We can also give an equivalent direct definition of complement that does not rely on the concept of difference: x U(x A x A ). On the right above, there is a Venn diagram visualization of the complement. Example: if U is the set of real numbers, then 0,1 = (, 0) (1, ). If we had defined the operation of complement first, we could have used it to define difference: A B = A തB. When we are subsets of real numbers, the default universal set is always R.

12 Avoiding a Common Mistake Let us take A = 2,3. If you are stuck in an integer mindset, you might think that the complement of A is, 1 [4, ). We are dealing with real numbers though, and there are many real numbers between 1 and 3, and 3 and 4.The correct complement is ҧ A =, 2 3,. We can prove this as well. x Aҧ means (x A), which is equivalent to 2 x 3 or 2 x x 3. By applying de Morgan, we simplify this to x < 2 x > 3. That is the meaning of x, 2 3,.

13 Set Identities Since the set operators are all defined in terms of logical operators between membership conditions, there is a corresponding set equivalence for each logical equivalence. For example, by applying the commutative law for conjunction, p q = q p, to p = (x A) and q = (x B), we conclude that there is a corresponding commutative law for intersection: A B = x x A x B = x x B x A = B A In set identities, the empty set plays the role of the constant False, the universal set plays the role of the constant True. The next slide shows a list of common set identities and the corresponding logical equivalences they are based on.

14 ҧ ҧ A list of set identities Set Identity Corresponding Logical Equivalence Name of the Identity A B = A തB A B = A തB 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) A = A A U = A A U = U A = A A = A A A = A p q p q (p q) p q p q q p p q q p p (q r) (p q) r p (q r) (p q) r p (q r) (p q) (p r) p q r (p q) (p r) p F p p T p p T T p F F p p p p p p De Morgan s Laws Commutative laws Associate laws Distributive Laws Identity Laws Domination Laws Idempotent Laws ҧ A = A ( p) p Double Negation Law A A B = A A (A B) = A p p q p p (p q) p Absorption Laws

15 Set Identity Proofs (Example 1) Let us give a careful proof one of the absorption laws for sets: A A B = A Observe that this is not technically a statement since A, B are free variables. What is missing in this abridged version of the law is the quantification. The correct mathematical statement of the absorption law is: Theorem: For all sets A, B, the set A A B is equal to A. Proof: suppose A and B are arbitrary sets. To show that A A B = A, let us take x A A B. By definition of union, this is logically equivalent to x A x A B. By definition of intersection, that is logically equivalent to x A (x A x B). By the absorption law of logic, that statement is logically equivalent to x A. We have shown that x x A A B x A and thereby verified the definition of equality of sets. Since A and B were arbitrary, we have proved the theorem.

16 ҧ ҧ Set Identity Proofs (Example 2) Let us prove one of De Morgan s laws: For all sets A, B, the set A B is equal to ҧ A തB. Proof: suppose A and B are arbitrary sets. To show that A B = A B. ҧ A തB, let us take x By definition of complement, x A B is logically equivalent to (x A B). (x A B) is logically equivalent to (x A x B) by definition of union. Now we apply the law of De Morgan for logic p q p q. (Observe we are not assuming the conclusion here. De Morgan s laws for logics and De Morgan s laws for sets are distinct laws, and we are using the former to prove the latter). By De Morgan, (x A x B) is logically equivalent to (x A) (x B), which is logically equivalent to (x A) (x തB) by definition of complement. Finally, by definition of intersection, that statement is equivalent to x A തB. Therefore, we have shown that x x A B x Aҧ തB and thereby verified the definition of equality of the two sets. Since A and B were arbitrary, we have proved the theorem.

17 Unions and Intersections of more than Two Sets Given an arbitrary collection of sets, the union of those sets is the set of all elements that are contained in at least one of the sets. The intersection of those sets is the set of all elements that are contained in all of the sets. Notations: if the collection of sets is A i, where i goes from 1 to n, then the union of all these sets is written as and the intersection is written as n ራ i=1 ሩ i=1 Just like in the sigma notation for sums, it is understood in these notations that the index variable goes through consecutive integers. Unions and intersections can be taken over infinitely many sets as well. n A i A i

18 Example: Unions and Intersections of ራ[0, 1 n ] = 0,1 n=1 Infinitely Many Sets Let us consider the intervals [0, 1 ] for various values of n the positive integer n. As n gets larger, the intervals [0, 1 ] n shrink: each is a proper subset of the next, and they are all proper subsets of the first one, which is [0,1]. Therefore, the union of all the sets is just the first of the intervals, [0,1]. ሩ[0, 1 n ] = {0} n=1 Calculus teaches us that lim 1 n n = 0. This means that the right endpoint of those intervals will eventually move to the left of any positive real number. This means that no positive real number can be in the intersection of all those sets. On the other hand, the number zero is in all of the intervals [0, 1 n ]. Therefore, it is in the intersection. None of the intervals contain negative numbers, hence neither does the intersection. It follows that zero is the only element in the intersection.