Consequences of the Completeness Property Philippe B. Laval KSU Today Philippe B. Laval (KSU) Consequences of the Completeness Property Today 1 / 10
Introduction In this section, we use the fact that R is complete to establish some important results. First, we will prove that Z is unbounded and establish the Archimedean principle. Second, we will prove that the rational numbers are dense in R. Finally, we will prove that Q is not complete. Philippe B. Laval (KSU) Consequences of the Completeness Property Today 2 / 10
Z Unbounded We first establish that Z is unbounded. While this result seems obvious, it turns out that it is not as easy to prove. It depends upon the completeness property of R. We establish this result by proving several lemmas and then use these lemmas to establish the main result. Lemma Every non-empty subset S of the integers which is bounded above has a largest element. How might we prove this? Philippe B. Laval (KSU) Consequences of the Completeness Property Today 3 / 10
Z Unbounded We first establish that Z is unbounded. While this result seems obvious, it turns out that it is not as easy to prove. It depends upon the completeness property of R. We establish this result by proving several lemmas and then use these lemmas to establish the main result. Lemma Every non-empty subset S of the integers which is bounded above has a largest element. How might we prove this? Let S be the subset in question, what can we say about S (remembering what we did before this)? Philippe B. Laval (KSU) Consequences of the Completeness Property Today 3 / 10
Z Unbounded We first establish that Z is unbounded. While this result seems obvious, it turns out that it is not as easy to prove. It depends upon the completeness property of R. We establish this result by proving several lemmas and then use these lemmas to establish the main result. Lemma Every non-empty subset S of the integers which is bounded above has a largest element. How might we prove this? Let S be the subset in question, what can we say about S (remembering what we did before this)? Let w = sup S. How can we prove max S exists? Philippe B. Laval (KSU) Consequences of the Completeness Property Today 3 / 10
Z Unbounded We first establish that Z is unbounded. While this result seems obvious, it turns out that it is not as easy to prove. It depends upon the completeness property of R. We establish this result by proving several lemmas and then use these lemmas to establish the main result. Lemma Every non-empty subset S of the integers which is bounded above has a largest element. How might we prove this? Let S be the subset in question, what can we say about S (remembering what we did before this)? Let w = sup S. How can we prove max S exists? How can we relate w with elements of Z at this point? Philippe B. Laval (KSU) Consequences of the Completeness Property Today 3 / 10
Z Unbounded Remark The above lemma looks very similar to the least upper bound property. But its conclusion is quite different. Lemma Every non-empty subset S of the integers which is bounded below has a smallest element. Theorem Z is unbounded both above and below. We are now ready to state the Archimedean principle. We state two versions of it. Philippe B. Laval (KSU) Consequences of the Completeness Property Today 4 / 10
Archimedean Principle Theorem For each strictly positive real number x, there exists a positive integer n such that 1 n < x. Hint for the proof: Since Z is unbounded, what can we say about 1 x? Philippe B. Laval (KSU) Consequences of the Completeness Property Today 5 / 10
Archimedean Principle Theorem For each strictly positive real number x, there exists a positive integer n such that 1 n < x. Hint for the proof: Since Z is unbounded, what can we say about 1 x? What can t 1 x be to Z if Z is unbounded? Philippe B. Laval (KSU) Consequences of the Completeness Property Today 5 / 10
Archimedean Principle Theorem For each strictly positive real number x, there exists a positive integer n such that 1 n < x. Hint for the proof: Since Z is unbounded, what can we say about 1 x? What can t 1 be to Z if Z is unbounded? x What does this imply? Philippe B. Laval (KSU) Consequences of the Completeness Property Today 5 / 10
Archimedean Principle Remark This theorem tells us that by choosing n large enough, we can make 1 as close to 0 as we want. In other words, { n } 1 inf n : n Z+ = 0 where Z + denotes the set of positive integers. We will prove this in the exercises. Theorem If x and y are real numbers, x > 0, then there exists a positive integer n such that nx > y. Philippe B. Laval (KSU) Consequences of the Completeness Property Today 6 / 10
Denseness We have already mentioned the fact that if we represented the rational numbers on the real line, there would be many holes. These holes would correspond to the irrational numbers. If we think of the rational numbers as dots on the real line and the irrational numbers as holes, one might ask how the holes are distributed with respect to the dots. The next concept gives us a partial answer. Definition A subset S of R is said to be dense in R if between any two real numbers there exists an element of S. Another way to think of this is that S is dense in R if for any real numbers a and b such that a < b, we have S (a, b). Theorem Q is dense in R. That is, between any two real numbers, there exists a rational number. Philippe B. Laval (KSU) Consequences of the Completeness Property Today 7 / 10
Denseness We outline the proof. Write what we have to prove. Philippe B. Laval (KSU) Consequences of the Completeness Property Today 8 / 10
Denseness We outline the proof. Write what we have to prove. a, b R, a < b, m, n Z, n 0 : a < m n Archimedean principle to b a. < b. Apply the Philippe B. Laval (KSU) Consequences of the Completeness Property Today 8 / 10
Denseness We outline the proof. Write what we have to prove. a, b R, a < b, m, n Z, n 0 : a < m < b. Apply the n Archimedean principle to b a. Let S = {x Z : x } n > a. Prove that S is bounded below and not empty. What does it imply? Philippe B. Laval (KSU) Consequences of the Completeness Property Today 8 / 10
Denseness We outline the proof. Write what we have to prove. a, b R, a < b, m, n Z, n 0 : a < m < b. Apply the n Archimedean principle to b a. Let S = {x Z : x } n > a. Prove that S is bounded below and not empty. What does it imply? Put everything together to finish the proof. Philippe B. Laval (KSU) Consequences of the Completeness Property Today 8 / 10
Completeness Definition A set S is said to be complete if every non-empty bounded subset of S has both a supremum and an infimum in S. Example R is complete. Example [0, 1] is complete while (0, 1) is not. Example Q is not complete. Hint, Let S = ( 0, 2 ) Q. Show 2 = sup S. Philippe B. Laval (KSU) Consequences of the Completeness Property Today 9 / 10
Exercises See the problems at the end of my notes on some consequences of the completeness property of R. Philippe B. Laval (KSU) Consequences of the Completeness Property Today 10 / 10