MATH 162, SHEET 7: THE REAL NUMBERS. 7A. Construction of the real numbers as Dedekind cuts

Transcription

1 MATH 162, SHEET 7: THE REAL NUMBERS This sheet is concerned with proving that the continuum R is an ordered field. Addition and multiplication on R are defined in terms of addition and multiplication on Q, so we will use and for addition and multiplication of real numbers, to make sure that there is no confusion with + and on Q. 7A. Construction of the real numbers as Dedekind cuts Definition 7.1. A subset A of Q is said to be a cut (or Dedekind cut) if it satisfies the following: (a) A Ø and A Q (b) If r A and s Q satisfies s < r, then s A (c) If r A then there is some s Q with s > r, s A. We denote the collection of all cuts by R. Lemma 7.2. Let A be a Dedekind cut. Then x A if, and only if, x is an upper bound for A. Definition 7.3. We define on R as follows. Let A, B R be Dedekind cuts. Define A B = {a + b a A and b B} 0 = {x Q x < 0} 1 = {x Q x < 1}. Exercise 7.4. (a) Prove that A B, 0, and 1 are all Dedekind cuts. (b) Prove that {x Q x 0} is not a Dedekind cut. (c) Prove that {x Q x < 0} {x Q x 2 < 2} is a Dedekind cut. Hint: if x Q, x 0, and x 2 < 2, then there is some δ Q, δ > 0 such that (x+δ) 2 < 2. Remember that you should prove any hint that you use. Exercise 7.5. (Homework) Prove that if A R then A = 0 A. Lemma 7.6. Let A R. Then define Then A = {r Q r A but r is not the least element of Q \ A}. 1

2 (a) A R (b) ( A) = A Exercise 7.7. (Homework) Show that if p Q and A = {x Q x < p}, then A = {x Q x < p}. Lemma 7.8. a) If A R, r Q, r > 0, then there exists s A such that s+r A. Hint: Argue by contradiction. b) If A R, r Q, r > 0, then there exists s A such that (s + r) ( A). Hint: First apply a) with r 2. Theorem 7.9. If A R then A ( A) = 0. Definition If A, B R, we say that A < B if A is a proper subset of B. (Recall from Script 1 that A is a proper subset of B if A B but A B. ) Exercise Prove that if A < B then there is some u, u 0 B such that a < u 0 < u, for all a A. Exercise Show that < satisfies (a) If A, B R then exactly one of the following holds: (trichotomy) A < B, A = B, B < A. (b) If A, B, C R with A < B and B < C then A < C. (transitivity) (c) If A, B, C R with A < B then A C < B C. (additivity) Hint for c): Argue by contradiction: suppose that A C = B C. Use Exercise 7.11 and Lemma 7.8 with an appropriate choice of r. Exercise (Homework) Show that if A R and 0 < A, then 0 A. Corollary (Homework) A < 0 0 < A. 2

3 7B. The real numbers form an ordered field Definition For A, B R, 0 < A, 0 < B, we define A B = {ab a A, b B, a > 0, b > 0} {x Q x 0}. If A = 0 or B = 0 we define A B = 0. If A < 0 but 0 < B we replace A with A and use the definition of multiplication of positive elements. Hence, in this case, Similarly, if 0 < A but B < 0, then A B = [( A) B]. A B = [A ( B)] and if A < 0, B < 0 then A B = [( A) ( B)] Exercise (a) Show that if A, B R, then A B R. (b) Show that if A, B R, 0 < A, 0 < B = 0 < A B. (c) Show that if A R, then 1 A = A. Definition If A R, 0 < A, then define A 1 = {r Q r 0} {r Q r > 0 and 1 r A, but 1 r is not the smallest element of Q\A}. For A < 0 we have 0 < A and so ( A) 1 can be defined as above. Then, for A < 0, we define A 1 = [( A) 1 ]. Lemma Let A R. Then, (a) If 0 < A, A 1 = {r Q r 0} {r Q r > 0 and ɛ > 0, ɛ Q such that (b) If A 0, A R, then A 1 R. 1 r + ɛ A}. Exercise Show that if p Q with p 0 and A = {x Q x < p}, then A 1 = {x Q x < 1 p }. Lemma If A R, A > 0, r Q, r > 1, then there exist s A such that rs A. Theorem If A R, A 0, then A A 1 = 1. Theorem R is an ordered field. 3

4 7C. The Archimedean property of the real numbers For every rational number q Q, define the corresponding real number as the Dedekind cut i(q) = {x Q x < q}. For example, 0 = i(0). This gives a well-defined injective function i : Q R. We identify Q with its image i(q) R so that the rational numbers Q are a subset of the real numbers R. (Hence N and Z are also subsets of R.) Note that in this case, this is the same map i that you identified in Theorem 6.27, so that in fact, Q is a subfield of R. Exercise Show that R satisfies Axioms 1-3. Lemma A nonempty subset of R that is bounded above has a supremum. Exercise Show that R satisfies Axiom 4. Lemma N is an unbounded subset of R. Theorem (The Archimedean Property) Let A R be a positive real number. Then there exist nonzero natural numbers m, n N such that 1 n < A < m. Corollary If A R is a real number, then there is an integer n such that n 1 A < n. Definition Consider a subset X C of the continuum. We say that X is dense in C if every p C is a limit point of X. Lemma A subset X C is dense in C if, and only if, X = C. Lemma A subset X C is dense in C if, and only if, for every non-empty open set U C, we have U X Ø. Lemma Given A, B R with A < B, there exists p Q such that A < p < B. Theorem Q is dense in R. 7D. The continuum is isomorphic to the real numbers Axiom 5. The continuum contains a countable dense subset. Exercise Let K C be a countable dense subset of C. Construct an order-preserving bijection f : Q K. (A map f : Q K is order-preserving if r < s in Q = f(r) < C f(s) in K. ) Exercise Let f : Q K be an order-preserving bijection, as found in the previous exercise. Let A R. Then A Q and so f(a) K C. Define F : R C by Show F (A) = sup f(a). 4

5 1. sup f(a) exists, so F is well-defined. 2. F is injective and order-preserving. Theorem Suppose that C is a continuum satisfying Axioms 1-5. Then C is isomorphic to the real numbers R. i.e. If we let < be the order on R and < C be the order on C, then there is a bijection F : R C such that, for all A, B R, A < B = F (A) < C F (B). Axioms 1-5 completely characterize the continuum in the sense that there is only one model for them, namely the real numbers. The main consequence for us is that we will no longer refer to C as anything but R. In addition, now that we have constructed R and proved the fundamental facts about it, we will forget about Dedekind cuts and think of elements of R as numbers. From now on, we will agree to use lower-case letters like x for real numbers and to write + and for and like people usually do. 5