Proving simple set properties...

Transcription

1 Proving simple set properties... Part 1: Some examples of proofs over sets Fall 2013 Proving simple set properties... Fall / 17

2 Introduction Overview: Learning outcomes In this session we will... Outline proof schemes commonly used to prove primitive properties of abstract sets. Provide examples of some common, but important proofs using sets. Proving simple set properties... Fall / 17

3 Introduction Relationship between sets and formal logic Note that the definitions of constructions in sets are expressed using logical operators, such as,, and, in conjunction with the operator. For example, the intersection of two sets, A and B, is defined as: A B = {e e A e B} Proving simple set properties... Fall / 17

4 Introduction Relationship between sets and formal logic Note that the definitions of constructions in sets are expressed using logical operators, such as,, and, in conjunction with the operator. For example, the intersection of two sets, A and B, is defined as: A B = {e e A e B} Such definitions enable constructive (direct) proofs of many properties. Proving simple set properties... Fall / 17

5 De Morgan s Laws De Morgan s laws... Recall from earlier in the semester De Morgan s laws were used to transform conjunctive and disjunctive statements by distributing negation: (a b) a b and (a b) a b. Proving simple set properties... Fall / 17

6 De Morgan s Laws De Morgan s laws... Recall from earlier in the semester De Morgan s laws were used to transform conjunctive and disjunctive statements by distributing negation: (a b) a b and (a b) a b. It so happens that the same law appears in abstract sets: A B Ā B and A B Ā B. Proving simple set properties... Fall / 17

7 De Morgan s Laws De Morgan s laws... Recall from earlier in the semester De Morgan s laws were used to transform conjunctive and disjunctive statements by distributing negation: (a b) a b and (a b) a b. It so happens that the same law appears in abstract sets: How would we prove it? A B Ā B and A B Ā B. Proving simple set properties... Fall / 17

8 De Morgan s Laws Ways of seeing De Morgan s laws Proof. The proof is directly obtained by definition: We prove the first part and leave the second part as an exercise. To show that A B = Ā B: A B = {x (x (A B))} = {x (x A x B)} = {x (x A) (x B)} = {x (x A) (x B)} = {x (x Ā) (x B)} = Ā B. Proving simple set properties... Fall / 17

9 De Morgan s Laws Revisiting that last proof... Notice that early in the sequence of reductions, we used the logical equivalent of De Morgan s laws to replace the disjunction with a conjunction. The step is justified by truth-values, but is somewhat troubling as it appears to assume what it s trying to prove. = {x (x A x B)} and then... = {x (x A) (x B)} Proving simple set properties... Fall / 17

10 De Morgan s Laws Revisiting that last proof... Notice that early in the sequence of reductions, we used the logical equivalent of De Morgan s laws to replace the disjunction with a conjunction. The step is justified by truth-values, but is somewhat troubling as it appears to assume what it s trying to prove. = {x (x A x B)} and then... = {x (x A) (x B)} An alternative proof is obtained by showing that A B = Ā B by showing that each set (occurring on the left and the right of the equals sign) is a subset of the other hence are equal. Proving simple set properties... Fall / 17

11 De Morgan s Laws Arguably, a more set-theoretic proof... Show that A B = Ā B by showing that each set is a subset of the other thus, we can say that they are equal. Proof. Let x A B, then x A B, meaning that x A or x B. Hence, x Ā or x B, which is to say that x Ā B. Thus, A B Ā B. To prove the other side: Suppose that x Ā B, then x Ā or x B. But then, x A or x B, which means that x A B. Therefore, x A B. Because A B Ā B and Ā B A B, they are equal. Proving simple set properties... Fall / 17

12 Preliminary observations Definition The power set of the set S, written P(S) is the set of all subsets contained within S. Thus, the power set is a set of sets. Later, when we explore functions, we will re-visit powersets in fuller generality, but for the present, we are asked to prove that the number of subsets contained within the powerset of a set is 2 S, where S is the cardinality of the set S. Proving simple set properties... Fall / 17

13 Proving by induction The proof is by induction on the cardinality of S. Basis: Observe that when S =, then S = 0, thus P(S) = { } = 1 = 2 0. Proving simple set properties... Fall / 17

14 Proving by induction The proof is by induction on the cardinality of S. Basis: Observe that when S =, then S = 0, thus P(S) = { } = 1 = 2 0. Hypothesis: Assume that P(k) holds for some k 1; show that P(k + 1) follows. Written in terms of the problem, and we need to show S = k P(S) = 2 k, S = k + 1 P(S) = 2 k+1. Proving simple set properties... Fall / 17

15 Power sets, inductively... Let S = k + 1. Proving simple set properties... Fall / 17

16 Power sets, inductively... Let S = k + 1. Let x S and consider the set S which is S {x}. Note that S = k. Proving simple set properties... Fall / 17

17 Power sets, inductively... Let S = k + 1. Let x S and consider the set S which is S {x}. Note that S = k. Rejoin x to each subset of S. Proving simple set properties... Fall / 17

18 Power sets, inductively... Let S = k + 1. Let x S and consider the set S which is S {x}. Note that S = k. Rejoin x to each subset of S. By the induction hypothesis, S contains 2 k elements. But, adding x to each of these give 2 k + 2 k elements, which is 2 k+1 elements in S. Proving simple set properties... Fall / 17

19 Power sets, inductively... Let S = k + 1. Let x S and consider the set S which is S {x}. Note that S = k. Rejoin x to each subset of S. By the induction hypothesis, S contains 2 k elements. But, adding x to each of these give 2 k + 2 k elements, which is 2 k+1 elements in S. This last expression is what I needed to show and we re done. Proving simple set properties... Fall / 17

20 Power sets, inductively... Let S = k + 1. Let x S and consider the set S which is S {x}. Note that S = k. Rejoin x to each subset of S. By the induction hypothesis, S contains 2 k elements. But, adding x to each of these give 2 k + 2 k elements, which is 2 k+1 elements in S. This last expression is what I needed to show and we re done. Thus, for any set S, P(S) = 2 S. Proving simple set properties... Fall / 17

21 Proofs with constructions A great number of proofs about sets utilize set constructions, union, intersection, set complement, etc. At least three methods are commonly employed when constructing proofs about sets: Arbitrary Particular Choose an arbitrary particular and use the definition of the constructs to prove the assertion. Derivation Employ the definitions and logical connectives to display the derivation. Diagrams For three or fewer sets, draw Venn diagrams. Proving simple set properties... Fall / 17

22 Examples of proof schemas... Proposition Suppose that A, B and C are sets. Show that A (B C) (A B) C. Proof. Choose x A (B C). Then x A and x B C. Proving simple set properties... Fall / 17

23 Examples of proof schemas... Proposition Suppose that A, B and C are sets. Show that A (B C) (A B) C. Proof. Choose x A (B C). Then x A and x B C. But, since x B C, it must be that either x B or x C: Proving simple set properties... Fall / 17

24 Examples of proof schemas... Proposition Suppose that A, B and C are sets. Show that A (B C) (A B) C. Proof. Choose x A (B C). Then x A and x B C. But, since x B C, it must be that either x B or x C: Case: x B. Because x A B then x (A B) C. Proving simple set properties... Fall / 17

25 Examples of proof schemas... Proposition Suppose that A, B and C are sets. Show that A (B C) (A B) C. Proof. Choose x A (B C). Then x A and x B C. But, since x B C, it must be that either x B or x C: Case: x B. Because x A B then x (A B) C. Case: x C. Then x (A B) C. Proving simple set properties... Fall / 17

26 Examples of proof schemas... Proposition Suppose that A, B and C are sets. Show that A (B C) (A B) C. Proof. Choose x A (B C). Then x A and x B C. But, since x B C, it must be that either x B or x C: Case: x B. Because x A B then x (A B) C. Case: x C. Then x (A B) C. Because x was arbitrary chosen in A (B C), we may conclude that A (B C) (A B) C. Proving simple set properties... Fall / 17

27 Sometimes, pictures help Example Construct two Venn diagrams: one for the left hand side, the other for the right hand side of the subset relation. Compare Proving simple set properties... Fall / 17

28 Proving equivalence If we choose an arbitrary particular, as we did in the previous slide, then we need to show both sides, i.e., show that the proof works both ways because equality is symmetric. Consider: Proposition Suppose A, B and C are sets. Then A (B C) = (A B) C. Instead of choosing an arbitrary x A (B C) and showing that it is found in (A B) C, and then vice versa, we will start with one side and derive the other, using definitions, axiom, and the properties of boolean logic. Proving simple set properties... Fall / 17

29 Deriving a proof of equality... Proof. Let x be arbitrary, then x A (B C) x A x B C x A x B x C x (A B) x C x (A B) C. Since for all x A (B C) x (A B) C, we may conclude that A (B C) = (A B) C. Proving simple set properties... Fall / 17

30 Convince yourself... Reconstruct the proof by choosing an arbitrary element in the left hand expression, showing that it appears in the right hand side. Then, show the proof the other way. Which requires less writing? Which is clearer? (By what criteria?) Which do you find more convincing? Proving simple set properties... Fall / 17

31 [Optional] Other constructions Power sets and Cartesian Products are often subjects of non-trivial propositions. Time permitting, let s try some of these: Proposition (Intersections over Power Sets) For any sets A and B: P(A B) = P(A) P(B). Proposition (Intersections over Products) For any sets A, B and C: A (B C) = (A B) (A C). Proposition (Products are Nilpotent) For any set A: A = = A. Proving simple set properties... Fall / 17