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1 Elementary Set Theory 1 Elementary Set Theory Notation: fg enclose a set. f1; 2; 3g = f3; 2; 2; 1; 3g because a set is not defined by order or multiplicity. f0; 2; 4;:::g = fxjx is an even natural numberg because two ways of writing a set are equivalent. ; is the empty set. x 2 A denotes x is an element of A. N = f0; 1; 2;:::g are the natural numbers. Z = f:::; −2; −1; 0; 1; 2;:::g are the integers. m Q = f n jm; n 2 Z and n 6= 0g are the rational numbers. R are the real numbers. Axiom 1.1. Axiom of Extensionality Let A; B be sets. If (8x)x 2 A iff x 2 B then A = B. Definition 1.1 (Subset). Let A; B be sets. Then A is a subset of B, written A ⊆ B iff (8x) if x 2 A then x 2 B. Theorem 1.1. If A ⊆ B and B ⊆ A then A = B. Proof. Let x be arbitrary. Because A ⊆ B if x 2 A then x 2 B Because B ⊆ A if x 2 B then x 2 A Hence, x 2 A iff x 2 B, thus A = B. Definition 1.2 (Union). Let A; B be sets. The Union A [ B of A and B is defined by x 2 A [ B if x 2 A or x 2 B. Theorem 1.2. A [ (B [ C) = (A [ B) [ C Proof. Let x be arbitrary. x 2 A [ (B [ C) iff x 2 A or x 2 B [ C iff x 2 A or (x 2 B or x 2 C) iff x 2 A or x 2 B or x 2 C iff (x 2 A or x 2 B) or x 2 C iff x 2 A [ B or x 2 C iff x 2 (A [ B) [ C Definition 1.3 (Intersection). Let A; B be sets. The intersection A \ B of A and B is defined by a 2 A \ B iff x 2 A and x 2 B Theorem 1.3. A \ (B [ C) = (A \ B) [ (A \ C) Proof. Let x be arbitrary. Then x 2 A \ (B [ C) iff x 2 A and x 2 B \ C iff x 2 A and (x 2 B or x 2 C) iff (x 2 A and x 2 B) or (x 2 A and x 2 C) iff x 2 A \ B or x 2 A \ C iff x 2 (A \ B) [ (A \ C) 1 Definition 1.4 (Set Theoretic Difference). Let A; B be sets. The set theoretic difference A n B is defined by x 2 A n B iff x 2 A and x 62 B. Theorem 1.4. A n (B [ C) = (A n B) \ (A n C) Proof. Let x be arbitrary. Then x 2 A n (B [ C) iff x 2 A and x 62 B [ C iff x 2 A and not (x 2 B or x 2 C) iff x 2 A and (x 62 B and x 6 inC) iff x 2 A and x 62 B and x 6 inC iff x 2 A and x 62 B and x 2 A and x 6 inC iff x 2 (A n B) and x 2 (A n C) iff x 2 (A n B) \ (A n C) Provisional definition of function: Let A; B be sets. Then f is a function from A to B written f : A ! B if f assigns a unique element b 2 B to each a 2 A. But what is the meaning of "assign"? Basic idea is to consider f : R ! R defined by f(x) = x2. We shall identify f with its graph. To generalize this to arbitrary sets A and B we first need the concept of an ordered pair. IE, a mathematical object ha; bi satisfying ha; bi = hc; di iff a = c and b = d. In particular, h1; 2i= 6 f1; 2g. Definition 1.5 (Cartesian Product). If A; B are sets, then their Cartesian Product A × B = fha; bi ja 2 A; b 2 Bg Definition 1.6 (Function). f is a function from A to B iff f ⊆ A × B and for each a 2 A, there exists a unique b 2 B such that ha; bi 2 f. In this case, the unique value b is called the value of f at a, and we write f(a) = b. It only remains to define ha; bi in terms of set theory. Definition 1.7 (Ordered Pair). ha; bi = ffag; fa; bgg Theorem 1.5. ha; bi = hc; di iff a = c and b = d. Proof. Clearly if a = c and b = d then ha; bi = ffag; fa; bgg = ffcg; fc; dgg = hc; di 1. Suppose a = b. Then ffag; fa; bgg = ffag; fa; agg = ffag; fagg = ffagg Since ffagg = ffcg; fc; dgg we must have fag = fcg and fag = fc; dg. So a = c = d, in particular, a = c and b = d. 2. Suppose c = d. Then similarly a = b = c so a = c and b = d. 3. Suppose a 6= b and c 6= d. Since ffag; fa; bgg = ffcg; fc; dgg we must have fag = fcg or fag = fc; dg. The latter is clearly impossible, so a = c. Similarly, fa; bg = fc; dg or fa; bg = fcg. The latter is clearly impossible, and as a = c, then b = d. 2 NB (Note Bene) - It is almost never necessary in a mathematical proof to remember that a function is literally a set of ordered pairs. Definition 1.8 (Injection). The function f : A ! B is an injection iff (8a; a0 2 A) if a 6= a0 then f(a) 6= f(a0) Definition 1.9 (Surjection). The function f : A ! B is a surjection iff (8b 2 B)(9a 2 A) such that f(a) = b Definition 1.10 (Composition). If f : A ! B and g : B ! C are functions, then their composition g ◦ f : A ! C is dined by (g ◦ f)(x) = g(f(x)) Theorem 1.6. If f : A ! B and g : B ! C are surjections, then g ◦f : A ! C is also a surjection. Proof. Let c 2 C be arbitrary. Since g is surjective, 9b 2 B such that g(b) = c. Since f is surjective, 9a 2 A such that g(a) = b. Then, (g ◦ f)(a) = g(f(a)) = c, hence g ◦ f is a surjection. Definition 1.11 (Bijection). A function f : A ! B is a bijection if f is an injection and a surjection. Theorem 1.7. If f : A ! B and g : B ! C are bijections, then g ◦ f : A ! C is a bijection. Proof. Composition of surjections is a surjection, and compositions of injections are injections. Definition 1.12 (Inverse Function). If f : A ! B is a bijection, then its inverse, f −1 : B ! A is defined by f −1(b) = the unique a 2 A such that f(a) = b. Remarks - If f : A ! B is a bijection, it is easily checked that f −1 : B ! A is a bijection. In terms of ordered pairs, f −1 = fhb; ai : ha; bi 2 fg Definition 1.13 (Equinumerous). Two sets, A and B, are equinumerous, writ- ten A ∼ B iff there exists a bijection f : A ! B. Theorem 1.8 (Galileo). Let E = f0; 2; 4;:::g be the even natural numbers. Then, N ∼ E Proof. We can define a bijection f : N ! E by f(n) = 2n Remark 1.1. If is often extremely difficult to explicitly define a bijection f : N ! A. However, suppose that f : N ! A is a bijection. For each n 2 N let an be f(n). Then, a0; a1;::: is a list of the elements of A such that every element occurs exactly once, and conversely, if such a list exists, then we can define a bijection f : N ! A by f(n) = an 3 Theorem 1.9. N ∼ Z Proof. We can list the elements of Z: 0; 1; −1; 2; −2;::: Theorem 1.10. N ∼ Q Proof. We proceed in two steps. First, we prove that N ∼ Q+ = fq 2 Q : q > 0g and consider the following 1 2 3 1 ! 1 1 ! 2 2 3 infinite matrix 1 . 2 % 2 . 1 2 3 # 3 % 3 . 3 Proceed through the matrix along the indicated route adding rational num- bers to your list if they have not already occurred. Second, we declare that N ∼ Q. In the first part, we showed that there exists a bijection f : N ! Q+ hence we can list Q by 0; f(1); −f(1);:::. Definition 1.14 (Power Set). If A is any set, then its power set is P(A) = fB : B ⊆ Ag, so P(f1; 2; : : : ; ng)is of size 2n. Theorem 1.11 (Cantor). N 6∼ P(N) Proof. This method of proof is called the diagonal argument. We must show that there does not exist a bijection f : N ! P(N). Let f : N ! P(N) be any function. So, we shall prove that f is not a surjection. Hence, we must find a set S ⊆ N such that 8n 2 N f(n) 6= S. We do this via a "time and motion study" For each n 2 N we must make the nth decision. That is, if n 2 S. We perform the nth task, that is, ensure f(n) 6= S. We make the nth decision so that it accomplishes the nth task, ie, n 2 S iff n 62 f(n).
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