Math 361 Construction of R 11 Construction of R Theorem 11.1. There does not exist q Q such that q2 = 2. 2 Proof. Suppose that q Q satis¯es q2 = 2. Then we can suppose that q > 0 and express 2 q = a=b, where a; b ! are relatively prime. Since 2 a2 = 2 b2 we have that a2 = 2b2: It follows that 2 a; say a = 2c. Hence j 4c2 = 2b2 2c2 = 2b2 This means that 2 b, which contradicts the assumption that a and b are relatively prime. j Theorem 11.2. There exists r R such that r2 = 2. 2 Proof. Consider the continuous function f : [1; 2] R, de¯ned by f(x) = x2. Then f(1) = 1 and f(2) = 4. By the Intermediate Value !Theorem, there exists r [1; 2] such that r2 = 2. 2 Why is the Intermediate Value Theorem true? Intuitively because R \has no holes"... More precisely... De¯nition 11.3. Suppose that A R. r R is an upper bound of A i® a r for all a A. ² 2 · 2 A is bounded above i® there exists an upper bound for A. ² r R is a least upper bound of A i® the following conditions hold: ² 2 { r is an upper bound of A. { If s is an upper bound of A, then r s. · Axiom 11.4 (Completeness). If A R is nonempty and bounded above, then there exists a least upper bound of A. Deepish Fact Completeness Axiom Intermediate Value Theorem. ) 2006/10/25 1 Math 361 Construction of R Remark 11.5. The analogue of the Completeness Axiom fails for Q. eg consider C = q Q q 0 or q2 < 2 : f 2 j · g Then C is a nonempty subset of Q which is bounded above but has no least upper bound. The least upper bound \should be" p2 RrQ. Our construction of R will be 2 such that q Q q 0 or q2 < 2 = p2: f 2 j · g Target We now want to extend Q to a larger set of \numbers" R so that the Complete- ness Axiom holds. De¯nition 11.6. A Dedekind cut is a subset x Q such that: = x = Q. ² ; 6 6 x is \downward closed"; ie if q x and r < q, then r x. ² 2 2 x has no largest element. ² Example 11.7. For each q QQ, the set ² 2 E(q) = r Q r < q f 2 j g is a Dedakind cut. ² r Q r 0 or r2 < 2 f 2 j · g is a Dedekink cut. De¯nition 11.8. R is the set of Dedekind cuts. De¯nition 11.9. We de¯ne the binary relation <R on R by x <R y i® x ( y: Theorem 11.10. <R is a linear order on R. Proof. Suppose that x <R y and y <R z. Then x ( y and y ( z. It follows that x ( z and so x <R z. Thus <R is transitive. Now we shall prove that <R satis¯es trichotomy. Let x; y R. Clearly at most one of the following holds: 2 x ( y; x = y; y ( x: Suppose that the ¯rst two fail; ie x y. We must prove that y ( x. 6 Since x y, there exists r xry. Consider any q y. If r q, then r y since y is downward6closed, which is a2contradiction. Thus q <2r. Since·x is downw2ard closed, q x. Thus y ( x. 2 2006/10/25 2 Math 361 Construction of R Theorem 11.11. If A R is nonempty and bounded above, then there exists a least upper bound for A. Proof. We shall prove that A is the least upper bound of A. Claim. A R. 2 S Proof. LetS x A. Then x A and so A = . Now let z R be an upper bound 2 6 ; 2 of A. Then x z for all x A and so A z. Since z = Q, it follows that A = Q. Now suppose that q 2ASand r < q. SThen there exists6 x A such that q x6 and clearly r x. Thus r big2 cupA. So SA is downward closed. 2 S2 Finally2 suppose that2 qS A is the largest element. Then there exists x A such that q x and clearly q is2the largestS element of xm which is a contradiction.2 Hence A has2 no largest element. S S Since x A for all x A, it follows that A is an upper bound for A. By the 2 above argument, if z R is any upper bound of A, then A z. Hence A is the least upper boundS of A2. S S S We next want to de¯ne an additive operation on R. Note that E(1) = r Q r < 1 f 2 j g E(2) = s Q s < 2 f 2 j g and hence E(3) = t Q t < 3 = r + s r E(1) and s E(2) : f 2 j g f j 2 2 g This suggests we make the following de¯nition. De¯nition 11.12. We de¯ne the binary operation +R on R by x + y = q + r q x and r y R f j 2 2 g . Lemma 11.13. If x; y R, then x + y R. 2 R 2 Proof. Since x; y = , it follows that x + y = . To see that x + y = Q, choose some 6 ; R 6 ; R 6 q0 Qrx and r0 Qry. Then we have that: 2 2 q < q0 for all q x ² 2 r < r0 for all r y ² 2 2006/10/25 3 Math 361 Construction of R It follows that if q x and r y, then 2 2 q + r < q0 + r0 and so q0 + r0 = x + y. Thus x + y = Q. 2 R R 6 To see that x +R y is downward closed, suppose that t x +R y and p < t. Then there exist q x and r y such that t = q + r. Since 2 2 2 p < q + r it follows that p q < r and so p q y. Hence ¡ ¡ 2 p = q + (p q) x + y: ¡ 2 R Finally we check that x +R y has no largest element. Let t x +R y be any element. Then there exists q x and r y such that t = q + r. Since x2 has no largest element, 2 2 there exists q < q0 x and so 2 t = q + r < q0 + r x + y: 2 R Theorem 11.14. For all x; y; z R, we have that 2 x +R y = y +R x (x +R y) +R z = x +R (y +R z): Proof. We just check the ¯rst equality. x + y = q + r q x; r y R f j 2 2 g = r + q r y; q x f j 2 2 g = y +R x De¯nition 11.15. 0 = r Q r < 0 R f 2 j g Theorem 11.16. (a) 0 R. R 2 (b) For all x R, we have x + 0 = x. 2 R R Proof. (a) is clear! To prove (b) we must prove that r + s r x and s < 0 = x: f j 2 g 2006/10/25 4 Math 361 Construction of R Claim. r + s r x and s < 0 x. f j 2 g Proof. Suppose that r x and s < 0. Then r + s < r. Since x is downward closed, r + s x. 2 2 Claim. x r + s r x and s < 0 . f j 2 g Proof. Let p x. Since x has no largest element, there exists p < r x. Thus p r < 0 and 2 2 ¡ p = r + (p r): ¡ Problem 11.17. How should we de¯ne p2? Recall that ¡ p2 = q Q q < 0 or q2 < 2 : f 2 j g We cannot de¯ne p2 = q Q q p2 ¡ f 2 j ¡ 2 g because this is not downward closed. How about p2 = q Q q = p2 ¡ f 2 j ¡ 2 g First Candidate: x = r Q r = x ¡ f 2 j ¡ 2 g Fatal aw! Consider E(1) = q Q q < 1 . Then f 2 j g E(1) = r Q r < 1 ¡ f 2 j ¡ 6 g = r Q r 1 = R f 2 j · ¡ g 2 since it has a greatest element. De¯nition 11.18. For any x R, we de¯ne 2 x = r Q ( s > r) s = x : ¡ f 2 j 9 ¡ 2 g Example 11.19. E(1) = r Q r < 1 R: ¡ f 2 j ¡ g 2 Theorem 11.20. For every x R, we have: 2 (a) x R ¡ 2 (b) x + ( x) = 0 R ¡ R 2006/10/25 5 Math 361 Construction of R Proof. To see that x = , choose some t = x and let r = t 1. Then ( t) = x and r < t. Hence r ¡x.6 ; 2 ¡ ¡ ¡ ¡ 2 ¡ 2 ¡ To see that x = Q, choose any p x. 6 2 Claim. p = x ¡ 2 ¡ Proof. If s > p, then s < p and so s x. ¡ ¡ ¡ 2 To see that x is downward closed, suppose that r x and p < r. Then there exists s > r such¡that s = x. But clearly s > p and so p2 ¡ x. To see that x has ¡no largest2 element, suppose that r 2 ¡x. Then there exists s > r ¡ 2 ¡ such that s = x. Since Q is a dense linear order, there exists p Q such that s > p > r and clearly¡p2 x.