Math 361 Axiom of Choice 17 Axiom of Choice De¯nition 17.1. Let be a nonempty set of nonempty sets. Then a choice function for is a function f sucFh that f(S) S for all S . F 2 2 F Example 17.2. Let = (N)r . Then we can de¯ne a choice function f by F P f;g f(S) = the least element of S: Example 17.3. Let = (Z)r . Then we can de¯ne a choice function f by F P f;g f(S) = ²n where n = min z z S and, if n = 0, ² = min z= z z = n; z S . fj j j 2 g 6 f j j j j j 2 g Example 17.4. Let = (Q)r . Then we can de¯ne a choice function f as follows. F P f;g Let g : Q N be an injection. Then ! f(S) = q where g(q) = min g(r) r S . f j 2 g Example 17.5. Let = (R)r . Then it is impossible to explicitly de¯ne a choice function for . F P f;g F Axiom 17.6 (Axiom of Choice (AC)). For every set of nonempty sets, there exists a function f such that f(S) S for all S . F 2 2 F We say that f is a choice function for . F Theorem 17.7 (AC). If A; B are non-empty sets, then the following are equivalent: (a) A B ¹ (b) There exists a surjection g : B A. ! Proof. (a) (b) Suppose that A B. Then there exists a injection f : A B. Fix some a A). Then we can de¯ne a¹surjection g : B A by ! 0 2 ! g(b) = the unique a A such that f(a) = b, if such an a exists 2 = a0; if no such a exists. (b) (a) Suppose that g : B A is a surjection. Then for each a A, ) ! 2 S = b B g(b) = a = a f 2 j g 6 ; Let = Sa a A and let h be a choice function for , ie h(Sa) Sa for all a A. ThenF we fcanj de¯ne2 ga function k : A B by k(a) = hF(S ). We claim2 that k is2 an ! a injection. To see this, suppose that a = a0 A. Let k(a) = b and k(a0) = b0. Then 6 2 b S , b0 S and so g(b) = a, g(b0) = a0. Hence b = b0. 2 a 2 a0 6 2006/12/06 1 Math 361 Axiom of Choice Theorem 17.8. If A is any in¯nite set, then ! A. ¹ Proof. Let f be a choice function for = (A)r . Then we can de¯ne a function F P f;g g : ! A ! by recursion via g(0) = f(A) and g(n + 1) = f(Ar g(0); : : : ; g(n) ): f g Clearly g is an injection and so ! A. ¹ Corollary 17.9 (AC). If · is any in¯nite cardinal, then ·. @0 · Proof. Let card A = ·. Then A is an in¯nite set and so ! A. Hence ·. ¹ @0 · Corollary 17.10 (AC). A set A is in¯nite i® there exists a proper subset B A such that B A. ½ ¼ Proof. ( ) If there exists a proper subset B A such that B A, then A is clearly not ¯nite.( Hence A is in¯nite. ½ ¼ ( ) Suppose that A is in¯nite. Then ! A and so there exists an injection f : ! A. De¯ne) a function g : A A by ¹ ! ! g(f(n)) = f(n + 1) for all n ! 2 g(x) = x for all x = ran f. 2 Then g is a bijection between A and B = Ar f(0) . f g Corollary 17.11 (AC (Remark: doesn't really need AC)). If A is a nonempty set, then A is countable i® there exists a surjection f : ! A. ! Proof. This follows since A is countable i® A !. ¹ Theorem 17.12 (AC). A countable union of countable sets is countable; ie if is countable and each A is countable, then is also countable. A 2 A A Proof. If = , then = is countable.SHence we can suppose that = . We can also suppA ose; that =A , since; would contribute nothing to . A 6 ; ;S2 A ; A Claim. There exists a surjection f : ! ! . £ ! A S S 2006/12/06 2 Math 361 Axiom of Choice Proof. Since is countable, there exists a surjection g : ! . For each n !, let A ! A 2 An = g(n). Then = A ; A ; : : : ; A ; : : : : A f 0 1 n g (Of course, the sets An are not necessarily distinct!) Since each An is countable, we can choose a surjection h : ! A . Then we can de¯ne a surjection n ! n f : ! ! £ ! A by [ f(n; m) = hn(m): Finally, let k : ! ! ! be a surjection. Then ' = f k : ! is a surjection. Hence is countable.! £ ± A A S QuestionS 17.13. Where did we use (AC) in the above proof? Answer. Since each An is countable, we have that S = h h: ! A is a surjection = : n f j ! n g 6 ; We have applied (AC) to obtain a choice function for = S n ! . F f n j 2 g De¯nition 17.14. If A is a set, then a ¯nite sequence in A is a function f : n A, where n !. ! 2 Remark 17.15. 1. If n = 0 = , then we obtain the empty sequence, f = . ; ; 2. Suppose n > 0 and f : n A. For each l n, let al = f(l). Then we often write f as ! 2 a0; a1; : : : ; an 1 : h ¡ i De¯nition 17.16. If A is a set, then Sq(A) is the set of ¯nite sequences in A. Theorem 17.17. card(Sq(!)) = . @0 Proof. We must show that Sq(!) !. First we can de¯ne an injection g : ! Sq(!) by g(n) = n . Thus ! Sq(!).¼Next we can de¯ne an injection h: Sq(!)! ! as h i ¹ ! follows. Let p0; p1; : : : ; pn; : : : be the increasing enumeration of the primes. Then h( ) = 1 ; a0+1 a1+1 an 1+1 h( a0; : : : ; an 1 ) = p0 p1 : : : pn ¡1 : h ¡ i ¡ Thus Sq(!) !. By SchrÄoder-Bernstein, Sq(!) !. ¹ ¼ 2006/12/06 3 Math 361 Transcendental Numbers Corollary 17.18. If A is a countable set, then card(Sq(A)) = . @0 Proof. Once again, we must show that Sq(A) !. Fix some a A. Then we de¯ne an injection f : ! Sq(A) by ¼ 2 ! n times f(n) = a; : : : ; a ; n 1 h i ¸ = ; n = 0 ;z }| { Thus ! Sq(A). Next we can de¯ne an injection h: Sq(A) Sq(!) as follows. Let k : A ¹! be an injection. Then ! ! h( ) = ; ; h( a0; a1; : : : ; an 1 ) = k(a0); k(a1); : : : ; k(an 1) h ¡ i h ¡ i Thus Sq(A) Sq(!). Since Sq(!) !, it follows that Sq(A) !. By SchrÄoder- Bernstein, Sq(¹A) !. ¼ ¼ ¼ 18 Transcendental Numbers De¯nition 18.1. Let r R be a real number. 2 (a) r is algebraic i® there exists a polynomial p(x) = a + a x + : : : + a xn; a = 0 0 1 n n 6 with integer coe±cients such that p(r) = 0. (b) Otherwise, r is transcendental. Example 18.2. (a) By considering p(x) = 2x 1, we see that 1=2 is algebraic. More generally, each ¡ q Q is algebraic. 2 (b) By considering p(x) = x2 2, we see that p2 is algebraic. ¡ (c) It is known (but hard to prove) that e and ¼ are transcendental. Theorem 18.3. There exist uncountably many transcendental numbers. This is an easy corollary of the following result. Theorem 18.4. There exist only countably many algebraic numbers. 2006/12/06 4 Math 361 Well-orderings Proof. Let P be the set of polynomials with integer coe±cients. Then we can de¯ne an injection f : P Sq(Z) ! by f(a + a x + : : : + a xn) = a ; a ; : : : ; a : 0 1 n h 0 1 ni Thus P Sq(Z) and so P is countable. Note that each p(x) P has ¯nitely many roots. Th¹us the set of algebraic numbers is the union of countably2 many ¯nite sets and hence is countable. Puzzle. Determine whether there exists: (a) an uncountable set of circular discs in R2, no two of which intersect. A (b) an uncountable set of circles in R2, no two of which intersect. B (c) an uncountable set of ¯gure eights in R2, no two of which intersect. C 19 Well-orderings De¯nition 19.1. The set W is said to be well-ordered by i®: Á (a) is a linear ordering of W ; and Á (b) every nonempty subset of W has a {least element. Á Example 19.2. (a) The usual ordering < on N is a well-ordering. (b) The usual ordering < on Z is not a well-ordering. Theorem 19.3. Let be a linear order on the set A. Then the following are equivalent: Á (a) is a well-ordering. Á (b) There do not exist elements a A for n ! such that n 2 2 a a a : : : a a : : : 0  1  2   n  n+1  Proof. (a) = (b) We shall prove that not(b) = not(a). So suppose that there exists elements)a A for n ! such that ) n 2 2 a a : : : a a : : : 0  1   n  n+1  2006/12/06 5 Math 361 Well-orderings Let = B = an n ! A. Then clearly B has no {least element. Thus is not a well-ordering.; 6 f j 2 g Á Á (b) = (a) We shall prove that not(a) = not(b).