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Artin1Sets.pdf - Appendix Background Material Historically...

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Unformatted text preview: Appendix Background Material Historically speaking, it is of course quite untrue that mathematics is free from contradiction; non-contradiction appears as a goal to be achieved, not as a God-given quality that has been granted us once for all. Nicolas Bourbaki 1. SET MORY This section reviews some conventions about set theory which are used in this book, as well as some facts which will be referred to occasionally. First, a remark about definitions: Any definition of a word or a phrase will have roughly the form (1.1) xxx if @#&$% , where xxx is the word which is being defined and @#&$% is its defining property. For example, the sentence “An integer n is positive if n > 0” defines the notion of a positive integer. In a definition, the word if means if and only if. So in the definition of the positive integers, all integers which don’t satisfy the requirement n > 0 are ruled out. The notation (1.2) {s E S I @#&$%} stands for the subset of S consisting of all elements s such that @#&$% is true. Thus if Z denotes the set of all integers, then N = {n E Zln > 0} describes N as the set of positive integers or natural numbers. Elements a1,..., an of a set are said to be distinct if no two of them are equal. A map (p from a set S to a set T is any function whose domain of definition is S and whose range is T. The words function and map are used synonymously. We re- quire that a function be single-valued. This means that every element s E S must have a uniquely determined image <p(s) E T. The range T of (p is not required to be 586 Background Material Appendix the set of values of the function. By definition of a function, every image element <p(s) is contained in T, but we allow the possibility that some elements t E T are not taken on by the function at all. We also take the domain and range of a function as part of its definition. If we restrict the domain to a subset, or if we extend the range, then the function obtained is considered to be different. The domain and range of a map may also be described by the use of an arrow. Thus the notation go: S —> T tells us that go is a map from S to T. The statement that t = go (s) may be described by a Wiggly arrow: s-> t means that the elements E S is sent to t E T by the map under consideration. For example, the map go: Z —-> Z such that go(n) = 2n + 1 is described by n->2n + 1. The image of the map go is the subset of T of elements which have the form (p(s) for some s E S. It will often be denoted by im go, or by (p(S): (1.3) im go = {t E TIt = (p(s) for some s E S}. In case im go is the whole range T, the map is said to be surjective. Thus (p is surjec- tive if every t E T has the form cp(s) for some s E S. The map go is called injective if distinct elements s1, s2 of S have distinct im- ages, that is, if s1 9E s2 implies that (p(s1) at cp(s2). A map which is both injective and surjective is called a bijective map. A permutation of a set S is a bijective map from S to itself. Let go: S —-> T and 1/1: T —> S be two maps. Then (II is called an inverse function of cp if both of the composed maps go 0 ill: T -—> T and lb 0 go: S —> S are the identity maps, that is, if go(¢(t)) = t for all t E T and l/l((p(S)) = s for all s E S. The in- verse function is often denoted by go“. (1.4) Proposition. A map go: S —> T has an inverse function if and only if it is bi- jective. Proof. Assume that go has an inverse function :11, and let us show that (p is both surjective and injective. Let t be any element of T, and let s, = i/x(t). Then (p(s) = gp(i/1(t)) = t. So t is in the image of go. This shows that (p is surjective. Next, let s1, s; be distinct elements of S, and let t.- = go (Si). Then [#00 = si. So t1, t2 have dis- tinct images in S, which shows that they are distinct. Therefore go is injective. Con- versely, assume that (p is bijective. Then since (p is surjective, every element t E T has the form t = go(s) for some s E S. Since go is injective, there can be only one such element 5. So we define III by the following rule: Mt) is the unique element s E S such that ilr(s) = t. This map is the required inverse function. I: Let go: S —> T be a map, and let U be a subset of T. The inverse image of U is defined to be the set (15) (WW) = {s E S | MS) E U} - This set is defined whether or not (p has an inverse function. The notation go“, as used here, is symbolic. A set is called finite if it contains finitely many elements. If so, the number of its elements, sometimes called its cardinality, will be denoted by IS I. We will also Section 1 Set Theory 587 call this number the order of S. If S is infinite, we write IS I = 00. The following theorem is quite elementary, but it is a very important principle. (1.6) Theorem. Let (p: S —> T be a map between finite sets. (a), If go is injective, then ISI S ITI. (b) If 90 is surjective, then ISI 2 ITI. (c) If IS I = IT I, then (p is bijective if and only if it is either injective or surjec- tive. El The contrapositive of part (a) is often called the pigeonhole principle: If I SI > IT I, then (p is not injective. For example, if there are 87 socks in 79 drawers, then some drawer contains at least two socks. An infinite set S is called countable if there is a bijective map (p: N -> S from the set of natural numbers to S. If there is no such map, then S is said to be uncount- able. (1.7) Proposition. The set IR of real numbers is uncountable. Proof. This proof is often referred to as Cantor’s diagonal argument. Let (p: N —> IR be any map. We list the elements of the image of go in the order (p(l), (p (2), (p(3),. .., and we write each of these real numbers in decimal notation. For example, the list might begin as follows: ¢(1)=82.§5470984534... ¢(2)= .12390345700... ¢(3)= 5.90§40598675... ¢(4)=12.874§5264444... (p(5)= .00145100349... We will now determine a real number which is not on the list. Consider the real number u whose decimal expansion consists of the underlined digits: u = .3 2 8 3 4 . We form a new real number by changing each of these digits, say v=.45142.... Notice that o 9% (p(l), because the first digit, 4, of v is not equal to the correspond- ing digit, 3, of go(1). Also, 1) #= 49(2), because the second digit, 5, of v is not equal to the corresponding digit of <p(2). Similarly, v ah 4901) for all n. This shows that (p is not surjective, which completes the proof, except for one point. Some real numbers have two decimal expansions: .99999... is equal to 1.00000... , for example. This creates a problem with our argument. We have to choose 0 so that infinitely many of its digits are different from 9 and 0. The easiest way is to avoid these digits altogether. n ...
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