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Math 310 Class Notes 1: Axioms of Set Theory

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Math 310 Class Notes 1: Axioms of Set Theory MATH 310 CLASS NOTES 1: AXIOMS OF SET THEORY Intuitively, we think of a set as an organization and an element be- longing to a set as a member of the organization. Accordingly, it also seems intuitively clear that (1) when we collect all objects of a certain kind the collection forms an organization, i.e., forms a set, and, moreover, (2) it is natural that when we collect members of a certain kind of an organization, the collection is a sub-organization, i.e., a subset. Item (1) above was challenged by Bertrand Russell in 1901 if we accept the natural item (2), i.e., we must be careful when we use the word \all": (Russell Paradox) The collection of all sets is not a set! Proof. Suppose the collection of all sets is a set S. Consider the subset U of S that consists of all sets x with the property that each x does not belong to x itself. Now since U is a set, U belongs to S. Let us ask whether U belongs to U itself. Since U is the collection of all sets each of which does not belong to itself, thus if U belongs to U, then as an element of the set U we must have that U does not belong to U itself, a contradiction. So, it must be that U does not belong to U. But then U belongs to U itself by the very definition of U, another contradiction. The absurdity results because we assume S is a set. Hence S is not a set. The paradox is in spirit parallel to the following statement of Russell: The barber shaves all those who do not shave themselves. The question is: What about the barber himself? If he shaves himself then he does not shave himself. But if he does not shave himself he shaves himself. To overcome Russell's paradox, an axiomatic set theory was proposed in the 1920s, known as the Zermelo-Frankel theory, which introduces nine axioms that form the foundation of mathematics. 1 To begin with, in the system we study, there are two predicates 2, read belongs to, and =, read equals. We also have a symbol ;, read the empty set, reserved for a fixed object. Otherwise, x; y; z; p; q; r; a; b; c, etc., and their uppercase counterparts denote variable objects. The predicate = has the property: x = x for all x; x = y implies y = x for all x; y; x = y and y = z imply x = z for all x; y; z: Definition 1. A is called a set if either there is an a 2 A or else A = ;. If a 2 A, then a is called an element of the set A. We say a set A is a subset of a set B, written A ⊂ B or B ⊃ A, read A is contained in B or B contains A, if a 2 A implies a 2 B for all a 2 A. We will use the notation fa; b; c; · · · g to denote a set when its elements are emphasized. Axiom 1 (Axiom of Extensionality). Given sets A and B, then A = B if and only if they have the same elements. Axiom 2 (Axiom of Schema of Separation). Given a set A and a statement (the term can be defined precisely in Mathematical Logic) Ψ(x) of the elements x of A, there exists a set B such that it consists of all x 2 A that satisfy Ψ(x). We will denote B in Axiom 2 by the symbol B = fx 2 A : Ψ(x)g Question 1: Show that the empty set ; contains no elements at all, as its name suggests. (Hint: Consider B = fx 2 ; : x 6= xg; where the statement Ψ(x) is x 6= x. Axiom 2 says that B is a set. Now apply Definition 1 to assert that B = ;.) Axiom 3 (Union Axiom). Given sets A and B, there is a set C that contains exactly the elements of A and B. This set C is called the \union" of A and B, denoted A [ B. Note: Given sets A and B, Axiom 3 assures the existence of the set A [ B. Then Axiom 2 implies the existence of the set fx 2 A [ B : x 2 A and x 2 Bg; 2 which we call the intersection of A and B, denoted A \ B. Axiom 4 (Pairing Axiom). Given sets A and B, there is a set C whose elements are exactly the two sets. That is, C = fA; Bg: Note: For any set B, Axiom 4 implies the existence of the set fB; Bg, which we denote by fBg for short. Axiom 5 (Sum Axiom). Given a set A, there is a set C that is the collection of elements of the set elements of A. For example, for A = ff1g; f1; 2g; f4; 5g; Mt. Everestg; we have C = f1; 2; 4; 5g. Axiom 6 (Power Set Axiom). Given a set A, there is a set C whose elements are all the subsets of A. For example, for A = f1; 2; 3g, we have C = f;; f1g; f2g; f3g; f1; 2g; f1; 3g; f2; 3g; f1; 2; 3gg: Question 2: Show that the empty set ; is always a subset of any set. Axiom 7 (Axiom of Regularity). Given a set A 6= ;, there is an x 2 A such that when x is a set it satisfies A \ x = ;. For example, for A = ff1g; f1; 2g; 3; 4g; we can choose x = f1g. Axiom of Regularity is devised to assure that A2 = A for any set A, as our intuition expects. Question 3: Show that A2 = A holds for any set A. (Hint: Consider the set fA; fAgg; why is it a set in the first place?) Axiom 8 (Axiom of Infinity). There is a set A such that it contains the empty set ;, and if a set a is in A, then its successor a0 := a [ fag is also in A. This axiom has to do with the notion of natural numbers, as we will see later. Axiom 9 (Axiom of Choice). Given sets I and A, for each i 2 I let Ai be a subset of A (i.e., we are given subsets of A indexed by elements of I). Then there is a set C formed by picking one element from each Ai. 3 Axiom 9 can be made more precise when we define the notion of a function later. In this axiomatic system of sets, one is only allowed to make asser- tions by logical inferences, starting from the axioms. The conclusion of an assertion is called a theorem (in particular, an axiom is a theorem without doing any logical inferences). For example, the statement in Question 1 is a theorem. One can freely quote theorems one has proven thus far to arrive at new theorems, etc. However, the most important issue in an axiomatic system is its consistency. That is, one has to make sure that the axioms do not have intrinsic logical contradiction among themselves. Kurt G¨odel in the 1930s proved that if one assumes that the first eight axioms of set theory above are consistent, i.e., have no contra- diction among them, then the ninth axiom, the Axiom of Choice, is consistent with the previous eight! But then in the 1960s, Paul Co- hen proved that if the first eight axioms of set theory are consistent, then the negation of the Axiom of Choice is also consistent with the previous eight! The conclusion of these two stunning theorems, from which G¨odeland Cohen won the Fields Medal, is therefore that the Axiom of Choice is independent of the first eight axioms, i.e., it cannot be logically derived from the first eight, nor can its negation. It sim- ply has nothing to do with the first eight axioms. Hence, whether we adopt the Axiom of Choice is a matter of taste! Dropping the Axiom of Choice would lead to a different mathematics from ours. This new mathematics would be equally valid as ours, in which we adopt the axiom. Whether the first eight axioms are consistent is unknown. G¨odelalso proved yet another startling theorem, which is named his incompleteness theorem. It states, roughly, that in an axiomatic system (the term can be made precise in Mathematical Logic), if it contains natural numbers and if the system is consistent, then there must be a true statement that cannot be logically deduced from the axioms! Equivalently, if all true statements can be logically deduced from the axioms, then the system is inconsistent! In particular, our mathematical system is incomplete! For a beautiful exposition of this incompleteness theorem, you can read the Pulitzer Prize winner G¨odel,Escher, Bach: an Eternal Golden Braid, 20th anniversary ed. by Douglas R. Hofstadter. 4.
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