2.2 Some Consequences of the Completeness Axiom

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1 60 CHAPTER 2. IMPORTANT PROPERTIES OF R 2.2 Some Consequences of the Completeness Axiom 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 Archimedean Property 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 2.2. Every non-empty subset S of the integers which is bounded above has a largest element. Proof. Since S is a non-empty subset of Z (hence of R) which is bounded above, by the supremum property (completeness axiom), sup S exists. Let w = sup S. It is enough to that w S. In a problem from the previous section, it was established that if a set contains its supremum, then the supremum is also the largest element. We do a proof by contradiction. Suppose that w / S. Since w < w, by theorem 2..33, there exists m S such that w < m w Since we assumed w / S, we can even write w < m < w We repeat the same argument using the fact that m < w to find n S such that m < n < w. So, we have w < m < n < w The inequality n < w is equivalent to w < n. If we add up the two inequalities w < m and w < n, we obtain < m n which is equivalent to n m <. But since m < n, we have n m > 0 and therefore 0 < n m <. Since both m and n are in S, they are integers. Therefore, n m is also an integer. Since there are no other integers between 0 and, the inequality 0 < n m < is not possible. It means that our assumption w / S cannot happen. Hence, w S. Remark This lemma looks very similar to the least upper bound property. But its conclusion is quite diff erent. The result of this lemma is obviously not true for subsets of R. An example is (0, 5) which is not empty, bounded above but

2 2.2. SOME CONSEQUENCES OF THE COMPLETENESS AXIOM 6 has no largest element. For this example, we would like to say that the largest element is the number "just before 5. The problem is that such a number does not exist. However, in the case of a subset of Z, such a number exists. Every integer has a successor and a predecessor. Lemma Every non-empty subset S of the integers which is bounded below has a smallest element. Proof. Similar to the previous lemma and left as an exercise. Theorem Z is unbounded both above and below. Proof. If Z were bounded above, by lemma 2.2. it would have a largest element, which we know it does not. A similar arguments is used to show that Z is not bounded below. We are now ready to state the Archimedean principle. Theorem (Archimedean Property) For each strictly positive real number x, there exists a positive integer n such that n < x. Proof. Let x be as in the theorem. Then, cannot be an upper bound of Z x since Z is not bounded above. Therefore, we can find n Z and n > 0 such that It follows that n < x. x < n This theorem tells us that by choosing n large enough, we can make n as close to 0 as we want. In other words, inf n : n Z+ = 0 where Z + denotes the set of positive integers. We will prove this in the exercises. The Archimedean property is sometimes stated differently. We give an equivalent statement which proof is left as an exercise. Theorem (Archimedean Property 2) If x and y are real numbers, x > 0, then there exists a positive integer n such that nx > y. Proof. See exercises Denseness of Q in R 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.

3 62 CHAPTER 2. IMPORTANT PROPERTIES OF R Definition (Dense) 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. Proof. Let a and b be two arbitrary real numbers such that a < b. Then, b a > 0. By the Archimedean principle, we can choose n Z such that This is equivalent to saying that 0 < n < b a a < a + n < b (2.) We will finish the proof by showing there exists an integer m such that We define S = a < m n < b {x Z : x n > a } = {x Z : x > na} since n > 0 This set is bounded below by na. It is also non-empty since Z is not bounded above. Hence, by lemma 2.2.3, S has a smallest element. Call it m. We will show m is the integer we are looking for. We have m n > a (2.2) but m a (2.3) n since m is the smallest integer for which m n > a. If we write m n = m + n n then, combining equations 2.2 and 2.4 gives a < m n = m + n n If we use equation 2.3, we obtain (2.4) a < m n a + n Finally, using equation 2. gives a < m n < b which is what we wanted to prove.

4 2.2. SOME CONSEQUENCES OF THE COMPLETENESS AXIOM 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, by the completeness axiom. Example 2.2. [0, ] is complete. If T [0, ] and T then by the completeness axiom, inf T and sup T exist since T R. Furthermore, 0 is a lower bound of T and is an upper bound of T hence 0 inf T sup T hence both inf T and sup T belong to [0, ]. Since T was arbitrary, the result follows. Example (0, ) is not complete since sup ((0, )) = / (0, ). Corollary Q is not complete. Proof. Define { S = x Q : 0 < x < } 2 ( = 0, ) 2 Q Clearly, S is a subset of Q. We will show that Q is not complete by showing that S does not have a supremum in Q. Because 2 is an upper bound of S, we have sup S 2. We want to show that in fact, sup S = 2. We do a proof by contradiction. Suppose that sup S < 2. Since Q is dense in R, we can find a rational number q such that sup S < q < 2. Thus q S. Therefore, we should have q sup S and not sup S < q < 2. So, we must have sup S = 2. We see that S, a subset of Q has a supremum which is not in Q Exercises. Prove lemma Prove that inf n : n Z+ integers. = 0 where Z + denotes the set of positive 3. Find the supremum and infimum of the sets below. (a) A = {, 2, 4, 8 },... = 2 n : n N. 2 + n (b) E = : n N. n 4. Prove theorem using the results of this section. 5. Prove that between any two real numbers there exists an irrational number. This proves that the set of irrational numbers, R \ Q is dense in R.

5 64 CHAPTER 2. IMPORTANT PROPERTIES OF R 6. Prove that every interval which contains more than one element must contain an infinite number of rational as well as irrational numbers (hint: do it by contradiction). 7. Given x R, prove there is a unique (!) n Z such that n x < n. 8. The goal of this problem is to give a "real analysis" proof of the division algorithm which states that if a and b are positive integers, b 0, then there exists integers r and q where 0 r < b such that a = bq + r. We do the proof in several steps. Let a and b be positive integers and define (a) Prove that S. S = {n Z : nb a} (b) Prove that S is bounded above. (c) Explain why S has a largest member that we will call q. (d) Prove that if we define r = a bq then 0 r < b. (Note that this is the same as a = bq + r). 9. Are the integers complete? Justify your answer.

6 2.2. SOME CONSEQUENCES OF THE COMPLETENESS AXIOM Hints for the Exercises For all the problems, remember to first clearly state what you have to prove or do.. Prove lemma Hint: Proof is similar to that of lemma Prove that inf n : n Z+ = 0 where Z + denotes the set of positive integers. Hint: Use the definition of the infimum and the Archimedean property. 3. Find the supremum and infimum of the sets below. (a) A = {, 2, 4, 8 },... = 2 n : n N. Hint: Use the definition of supremum and infimum. 2 + n (b) E = : n N. n Hint: Use the definition of supremum and infimum. For infimum, also use the Archimedean property. 4. Prove theorem using the results of this section. Hint: Consider different cases on y and use Prove that between any two real numbers there exists an irrational number. This proves that the set of irrational numbers, R \ Q is dense in R. Hint: Use the definition of denseness, but instead of applying it to two arbitrary real numbers a and b, apply it to 2a and 2b. 6. Prove that every interval which contains more than one element must contain an infinite number of rational as well as irrational numbers (hint: do it by contradiction). Hint already given. 7. Given x R, prove there is a unique (!) n Z such that n x < n. Hint: You need to prove existence and uniqueness. For existence, define S = {k Z : k > x}. Explain why S is bounded below and not empty. What does this imply? For uniqueness, assume that there are two such numbers, say m and n. Using a proof similar to that of lemma 2.2., prove that m = n by showing that < m n <. 8. The goal of this problem is to give a "real analysis" proof of the division algorithm which states that if a and b are positive integers, b 0, then there exists integers r and q where 0 r < b such that a = bq + r. We do the proof in several steps. Let a and b be positive integers and define S = {n Z : nb a}

7 66 CHAPTER 2. IMPORTANT PROPERTIES OF R No hints since you are already guided through the proof. (a) Prove that S. (b) Prove that S is bounded above. (c) Explain why S has a largest member that we will call q. (d) Prove that if we define r = a bq then 0 r < b. (Note that this is the same as a = bq + r). 9. Are the integers complete? Justify your answer. Hint: Use the definition and the results of this section.