Chapter 4. Basic Set Theory. 4.1 The Language of Set Theory

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1 Chapter 4 Basic Set Theory There are two good reasons for studying set theory. First, it s a indispensable tool for both logic and mathematics, and even for other fields including computer science, linguistics, and so on. But there is also a second reason it s philosophically interesting. For many, the philosophy of mathematics is really just the philosophy of set theory. This is probably a mistaken view, but it is not wholly implausible, since there seems to be counterparts in set theory for all mathematical objects (i.e., numbers are sets, functions are sets, integrals and derivatives are sets, and so on). In any case, familiarity with set theory is essential. 4.1 The Language of Set Theory There are two basic notions, set and member. Strictly speaking these must remain undefined, since otherwise we would end up giving definitions that are either circular or lead to an infinite regress. Nevertheless, we can step outside of the theory and give what Frege called an explication of the concepts. Things such as a flock of birds or a pack of wolves are sets. The individual birds and wolves are members of the sets. Examples such as these help us to get a grip on set and member. But beware. A pack of wolves has a location, whereas a set does not. Sets are abstract entities that exist outside of spacetime. They have no mass, or charge, or shape, or any other physical property. Sometimes it s useful to distinguish a set of bricks that make up a wall from the physical collection of those bricks that make up the wall. The former is located no where and won t harm you, while the latter can give you 27

2 CHAPTER 4. BASIC SET THEORY 28 a nasty bump if you run into it. We will use the symbol for the membership relation, and use A, B, C,..., {},a,b,c,... for both sets and their members. Curley brackets are a common way to specify sets. Thus, {a, b} is the set with the two members a and b. We will define many more important set-theoretical concepts as we move along. Though it is convenient to use upper case letters for sets and lower case for members, sets can be members of other sets; so A B,B a, and a b are perfectly legitimate. We will also use the standard symbols from logic, such as,,,, and so on. We adopt the same formation rules as before, with the following additions. When A is a set and a is a member of it, then the following are wffs: a A, {a} = A, and so on. Exercise 11 Which of the following are wffs? (Don t worry about which are true.) 1. a b 2. A a 3. a {b, c} 4. {b, c} a 5. x X(x X) 6. a/ A (this means (a A)) 7. A 8. a B b A 9. {a, b} = x(x {b, a}) HINTS & ANSWERS 1. yes 2. yes (don t be misled by the letters, A could be a member and a aset) 3. yes

3 CHAPTER 4. BASIC SET THEORY yes 5. yes (we ll let quantifiers range over set variables such as X) 6. yes 7. no (here A is a set, not a sentence) 8. yes 9. no (the left side is a set, the right side is a sentence) 4.2 Axioms Axiom 1 (Extensionality): Two sets are identical, if and only if they have the same members. Symbolically: A B x((x A x B) A = B)) This axiom implies the following: a {a}, {a, b} = {b, a} (order is irrelevant), {a} = {a, a, a, a, a} (there is just one yhing is in the set; repeating it is merely redundant), b 6= a b/ {a}, anda 6= {a} 6= {{a}} 6= {{{a}}} (each of these is a different entity). Axiom 2 (Empty set): There is a set which has no members. A x x/ A (It is usually called the empty set, or the null set, and is denoted φ.) Each of the following is equal to the empty set: φ = the set of humans over 10 feet tall = the set of unicorns = the set of even prime numbers that are greater than 2. You shouldn t think of these as different sets that happen to be empty. There is really only one empty set (and these are different ways of describing it). Theorem 1 The empty set is unique. Proof. Suppose that φ 1 and φ 2 are both empty sets. Because an empty set has no members, it follows that x φ 1 x φ 2. Given the axiom of extensionality, it follows that φ 1 = φ 2. Axiom 3 (Pairing): For any a and b, there is a set which has both of them and nothing else as members. a b A x(x A x = a x = b)

4 CHAPTER 4. BASIC SET THEORY 30 Given that Bob exists and the proposition P exists, then there exists a set containing both of them, {Bob, P}. We can use the axiom again. given that Bob exists and the set {Bob, P} exists, then there exists a set {Bob, {Bob, P}}. Obviously, we can build up all sorts of new sets in this way. Axiom 4 (Union): For any sets A and B, there is a set C whose members are exactly the members of A and B. A B C x (x C x A x B) For example, if A = {a, b, c} and B = {1, 2}, thenthereisasetc, such that C = {a, b, c, 1, 2}. Notation 1 {x : P (x)} means the set of all x such that the condition P(x) holds. (For example, {x : x is red} is the set of all red things; {y : y Z y>27} is the set of all integers that are greater than 27.) Axiom 5 (Comprehension (also know as specification, separation, and aussonderung)): Let U be any set. Then the members of U that satisfy a condition P determine a set. A(A = {x : x U P (x)} If we start with the class members as the set U, andletf mean is a woman and P mean is a philosophy student, then {x : x U Fx Px} is the set of women philosophy students in the class. Exercise 12 Let A = {1, 2, 3,...},B = {x : x is a prime number},c = {a, b, c,..., z},d = x : x is an even number},e = x : x is an odd number}. 1. What is the result of applying the Union Axion to E and D? 2. What is the result of applying the Pairing Axiom to E and D? 3. What is {x : x A x<8}? 4. What is {x : x B x C}? 5. What is {x : x D x E}? ANSWERS: 1. E D = A

5 CHAPTER 4. BASIC SET THEORY {E,D} 3. {1, 2, 3, 4, 5, 6, 7} 4. {a, b, c,..., z, 2, 3, 5, 7, 11, 13,...} 5. A 4.3 Russell s Paradox Let s take a moment to look at Russell s paradox. The axiom of comprehension has the slightly odd form that it does have in order to get around this problem. The current axiom is due to Zermelo, who proposed it to block a number of paradoxes that had arisen in set theory in its early days (in the late 19th and early 20th century). A very natural and intuitive principle of reasoning that was widely used is this: Every condition determines a set. If the condition (or property) is being red then there is a set of red things, and if the condition is being a prime number, then there is a set of prime numbers, and so on. The principle seems completely self-evident. Unfortunately, it leads to a contradiction. (Set theory with this principle is often called naive set theory. ) Notice that some sets seem to be members of themselves and others not. The set of apples, for instance, is not an apple, so it is not a member of itself. On the other hand, the set of abstract entities is an abstract entity, so it would be a member of itself. So far, so good. There is nothing problematic about the conditions being an apple or being an abstract entity. But now consider the condition of being not a member of itself. The set of apples seems to satisfy this condition while the set of abstract entities does not. Let s form the set which corresponds to the condition (that is, the set of all things that are not members of themselves), and call it R. R = {x : x/ x} It looks like a legitimate set. The set of apples will be in R, butthesetof abstract entities will not. Now let s ask the question: Is it true or false that R R? The answer must be either Yes or No. Let s assume Yes: But if R R, thenr {x : x/ x}, and so must satisfy the condition x/ x, so R/ R. Now let s assume No. But if R/ R, thenr does satisfy the condition x/ x, so it must be a

6 CHAPTER 4. BASIC SET THEORY 32 member after all; hence R R. Either assumption leads to its opposite, so we have an outright contraction, R R R/ R. There have been a number of reactions to Russell s paradox. Zermelo s modification of the axiom of comprehension is the most popular. The key idea is that the defining condition is limited to the background set; it is not allowed to apply to everything. After almost 100 years, no one has been able to derive a contradiction. Exercise 13 Try to derive Russell s paradox using the axioms given so far as a way to convincing yourself that Zermelo s modification is plausible. Prior to Zermelo s axiomatization (in 1908), there was thought to be a universal set, that is, a set that contained everything. (Naturally, it would have to contain itself, as well.) This is no longer believed to be the case. The reasoning behind Russell s paradox is used to prove there is no universal set. Theorem 2 There is no universe. That is, U x x U. Proof. Suppose there is a universal set U that contains all sets. Define asetr as follows: R = {x : x U x/ x}. Since U contains everything, R U. By the same reasoning as that involved in Russell s paradox, we have R R R/ R. Since this is a contradiction, we blame the premiss that lead to the absurdity, namely, the assumption that U exists. Thus, there is no such set U. 4.4 Some Key Concepts Using the concepts, axioms, and notation developed so far, we can now define a number of very central concepts. Definition 9 (Union): A B = {x : x A x B} For example, {a, b} {1, 2, 3} = {a, b, 1, 2, 3} Definition 10 (Intersection): A B = {x : x A x B} For example, {a, b} {b, c, 2} = {b} Definition 11 (Subset): A B ( x x A x B) For example, {a, b} {a, b, c}; notice two special cases, A A, φ A

7 CHAPTER 4. BASIC SET THEORY 33 Definition 12 (Proper subset)a B (A B A 6= B) Definition 13 (Difference): A B = {x : x A x/ B} For example, {a, b} {b, c} = {a} Definition 14 (Complement): A 0 = {x : x U x/ A} For example, if U is the set of natural numbers and A is the set of even numbers, then A 0 is the set of odd numbers. Exercise 14 Let U = {1, 2, 3, 4, 5} be the background set and let A = {1, 2}, B = {2, 3}, and C = {3, 4, 5}. What are each of the following? 1. A 0 2. B 0 3. A B 4. A 0 B 0 5. C 0 A 6. U A 7. A U 8. A φ 9. B φ 10. U φ A (A B) B φ Answers: (1) A 0 = {3, 4, 5} = C, (4)A 0 B 0 = {1, 3, 4, 5}, (11)φ 0 = U, (14) B φ = B

8 CHAPTER 4. BASIC SET THEORY More Axioms Axiom 6 (Power Set): For any set A, there is a set whose members are exactly the subsets of A. (The power set of A is usually denoted A.). A B x (x B x A) (Here B is the powerset A) For example, if A = {a, b} then A = {A, φ, {a}, {b}}; and if B = {1, 2, 3}, then B = {B,φ, {1, 2}, {1, 3}, {2, 3}, {1}, {2}, {3}}. Note the special case, φ = {φ}. Notice the pattern. If a set has n members, then its power set has 2 n members. This is why it s called the power set. Exercise 15 (1) What is the powerst of {Bob,Alice}? (2) How many members in the powerset of the alphabet {a, b, c,..., z}? Axiom 7 (Infinity): There is a set with infinitely many members. Sincewehaven tyet defined infinite, the axiom does not really make sense yet. Later we will define it, but for now an intuitive understanding will do. So far I have stated nine axioms. There are more, but we won t be needing them. So I ll only mention some of them: There is the axiom of choice, the axiom of replacement, the axiom of regularity. These are pretty much standard and are in common use by logicians and mathematicians. There are still others, for example, so-called large cardinal axioms, that are quite controversial. These are not part of regular mathematics (at least not yet), but rather are the subject of ongoing research. Now we shall develop some of the key concepts. 4.6 The Algebra of Sets We will now develop the notions of intersection, subset, and so on. This basic part of set theory is known as the algebra of sets. Theorem 3 (Commutative law): A B = B A Proof. x A B x A x B Def x B x A tautology x B A Def

9 CHAPTER 4. BASIC SET THEORY 35 A B = B A Ax. of Extensionality This is a typical proof in the algebra of sets. Notice how it works. We need to show that two sets are identical, which we do by showing that they have exactly the same members. The crucial step in the proof is to use a fact from elementary logic, namely that P Q is equivalent to Q P. A proof does not have to be laid out exactly as above. You may be a bit more or less formal. The crucial point is to include the relevant information in a way that is as clear to the reader as possible. You might find something like this better: Proof. Let us assume that an arbitrary x is a member of the set A B. By the definition of union, this means that x is a member of A or is a member of B. Clearly, this is the same as saying that x is a member of B or x is a member of A. Again, by the definition of union, this means that x is a member of B A. The equality of these two sets follows from the Axiom of Extensionality. Pick the style of writing up proof that you like best. Proofs must be correct. After that make it as easy on your reader as you can. Theorem 4 (Associative laws): Theorem 5 (Distributive laws) Theorem 6 (DeMorgan s laws) A (B C) = (A B) C A (B C) = (A B) C A (B C) = (A B) (A C) A (B C) = (A B) (C B) (A B) 0 = A 0 B 0 (A B) 0 = A 0 B 0

10 CHAPTER 4. BASIC SET THEORY 36 Proof. (of the first) x (A B) 0 x/ A B (x A B) Def / (x A x B) Def x/ A x/ B tautology, Def / x A 0 x B 0 Def 0 x A 0 B 0 Def (A B) 0 = A 0 B 0 Axiom of Extensionality Noticeonceagainthatthekeystepisatransformationinlogic. Inthis proof it is the step from (P Q) to P Q. It is no accident that this equivalence is called De Morgan s in logic as well as in set theory. Theorem 7 A φ = A and A φ = φ Proof. (of the first) x A φ x A x φ Def x A (since x/ φ) A φ = A Axiom of Extensionality Theorem 8 A 00 = A Hint: P P Theorem 9 A B A C B C Proof. A B (x A x B) Def (x A x C x B x C) tautology (x A C x B C) Def A C B C Def Theorem 10 A B A C B C

11 CHAPTER 4. BASIC SET THEORY 37 Theorem 11 A B B 0 A 0 Theorem 12 φ A Hint: P Q is true if P is false. Theorem 13 A A = φ and A φ = A Proof. (of the first) x A A x A x/ A Def x φ (equivalent to a contradiction) A A = φ Axiom of Extensionality Exercise 16 Prove each of the theorems above that were not proven. 4.7 Counter-Examples Theorems, of course, are true for any set whatsoever. A sentence which is not a theorem will have counter-examples, that is, example sets which make the sentence false. Of course, there might be other sets which make it true, in which case the sentence is satisfiable. When faced with a sentence that might be a theorem, try to prove it. If you can t, then try to find a counter-example. For example, we might wonder if A B = A B. To show that it is not atheoremweleta = {a},b = {b}. Then A B = {a} {b} = {a, b}. But A B = {a} {b} = φ 6= A B. So, A and B defined this way provide a counter-example to the alleged theorem. Exercise 17 Prove or give a counter-example to each of the following. 1. A B = B A 2. A 0 B 0 = B 0 A 0 3. A B = B 0 A 4. A B 6= φ A 6= φ 5. A B 6= φ A 6= φ

12 CHAPTER 4. BASIC SET THEORY A C B C A B C 7. (B A) A = A 8. (A B) A = φ 9. (A (B C) =(A B) (A B) 10. (A B B A) A = B Hints and Answers: (4) Counter-example: Let B = {a} and A = φ, then A B = {a} 6= φ, (5) Proof. Suppose A B 6= φ, then x x A B; thus, x A x B. Therefore, A is not empty, ie, A 6= φ. 4.8 Further Reading Enderton, Elements of Set Theory Potter, Set Theory and Its Philosophy Halmos, Naive Set Theory (Thisisagoodshortbook; butitis not naive in the technical sense.)