Chapter 4 Basic Set Theory
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Chapter 4 Basic Set Theory There are two good reasons for studying set theory. First, it’s a indispens- able 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 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 CHAPTER 4. BASIC SET THEORY 29 4. 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 = a} b/ a ,anda = a = a = a (each of these is a different6 entity).→ ∈ { } 6 { } 6 {{ }} 6 {{{ }}} Axiom 2 (Empty set): There is a set which has no members. A xx/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 φ1 and φ2 are both empty sets. Because an empty set has no members, it follows that x φ x φ . Given the axiom of ∈ 1 ↔ ∈ 2 extensionality, it follows that φ1 = φ2. 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) ∀ ∀ ∃ ∀ ∈ ↔ ∨ 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 aus- sonderung)): Let U be any set. Then the members of U that satisfy a condi- tion 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 ∪ CHAPTER 4. BASIC SET THEORY 31 2. 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 comprehen- sion 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. Unfortu- nately, 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?Theanswer 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 ∈ ∈ 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.