The Real Numbers. Here we show one way to explicitly construct the real numbers R. First we need a definition.

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1 The Real Numbers Here we show one way to explicitly construct the real numbers R. First we need a definition. Definitions/Notation: A sequence of rational numbers is a funtion f : N Q. Rather than write f and f(n) for the sequence and its values, we typically write {a n } and a n. A sequence {a n } is called a Cauchy sequence if, for all ɛ > 0 (which we may take to be a rational epsilon for now), there is an N N such that a n a m < ɛ for all n, m N. For any rational number c, we let c (with the quotes) denote the sequence {a n } defined by a n = c for all n N. In other words, c is the sequence consisting entirely of the number c. This is clearly a Cauchy sequence. We say a sequence {a n } converges to the (rational) number c, and write {a n } c, if for all ɛ > 0, there is an N N such that a n c < ɛ for all n N. Lemma: If {a n } and {b n } are Cauchy sequences, then so are {a n + b n } and {a n b n }. All Cauchy sequences are bounded. Proof: Let ɛ > 0. By definition, there are N 1, N 2 N such that a n a m < ɛ/2 for all n, m N 1 and b n b m < ɛ/2 for all n, m N 2. Let N = max{n 1, N 2 }. Then for all n, m N we have (a n + b n ) (a m + b m ) = (a n a m ) + (b n b m ) a n a m + b n b m < ɛ/2 + ɛ/2 = ɛ. This shows that {a n + b n } is a Cauchy sequence. Next, get N 3, N 4 N such that a n a m < 1 for all n, m N 3 and b n b m < 1 for all n, m N. Set B 1 = max n N3 { a n + 1} and B 2 = max n N4 { b n + 1}. One readily verifies that a n < B 1 and b n < B 2 for all n N (this shows that all Cauchy sequences are bounded). Note that B 1, B 2 > 0 by construction. Now there are N 5, N 6 N such that a n a m < ɛ/(2b 2 ) for all n, m N 5 and b n b m < ɛ/(2b 1 ) for all n, m N 6. Let N = max{n 5, N 6 }. Then for all 1

2 n, m N we have a n b n a m b m = a n (b n b m ) + b m (a n a m ) a n (b n b m ) + b m (a n a m ) = a n b n b m + b m a n a m < B 1 ɛ/(2b 1 ) + B 2 ɛ/(2b 2 ) = ɛ. This shows that {a n b n } is a Cauchy sequence. Definition: We say two Cauchy sequences {a n } and {b n } of rational numbers are equivalent, and write {a n } {b n }, if the sequence {a n b n } 0. Lemma: This is an equivalence relation. Proof: For any sequence {a n } of rational numbers, {a n a n } = 0 0, so that {a n } {a n } for any Cauchy sequence {a n }. Suppose {a n } {b n } and let ɛ > 0. Then for some N N, a n b n < ɛ for all n N. Thus b n a n < ɛ for n N and {b n a n } 0. In other words, {b n } {a n }. Suppose {a n } {b n } and {b n } {c n } and let ɛ > 0. Then for some N N, a n b n < ɛ/2 for all n N and for some M N, b n c n < ɛ/2 for all n M. This implies that a n c n = a n b n + b n c n a n b n + b n c n < ɛ 2 + ɛ 2 = ɛ for all n max{n, M}. Thus {a n c n } 0 and {a n } {c n }. Definition: The real numbers R is the set of equivalence classes of Cauchy sequences of rational numbers. We ll write [{a n }] to denote the equivalence class which contains the sequence {a n }. We view Q as a subset of R by identifying the rational number c with [ c ]. Definition: Define addition and multiplication of real numbers by [{a n }] + [{b n }] = [{a n + b n }] and [{a n }] [{b n }] = [{a n b n }]. Lemma: These operations are well-defined, i.e., depend only on the equivalence classes and not the particular elements of the equivalence classes used. 2

3 Proof: Suppose {a n } {a n} and {b n } {b n} are all Cauchy sequences. By the lemma above, both {a n + b n } and {a n b n } are Cauchy sequences. Let ɛ > 0. Then for some N 1, M 1 N, a n a n < ɛ/2 for all n N 1 and b n b n < ɛ/2 for all n M 1. This implies that (a n + b n ) (a n + b n) = (a n a n) + (b n b n) a n a n + b n b n < ɛ 2 + ɛ 2 = ɛ for all n max{n 1, M 1 }. Thus, {a n + b n } {a n + b n} and addition is well-defined. By a previous lemma, all four sequences {a n }, {a n}, {b n } and {b n} are bounded. Thus, there is a B > 0 such that a n, a n, b n, b n B for all n N. For some N 2, M 2 N we have a n a n < ɛ 2B for all n N 2 and b n b n < ɛ 2B for all n M 2. This implies that 2 a n b n a nb n = (a n a n)(b n + b n) + (a n + a n)(b n b n) (a n a n)(b n + b n) + (a n + a n)(b n b n) = a n a n b n + b n + a n + a n b n b n a n a n ( b n + b n ) + ( a n + a n ) b n b n < ɛ 2B (2B) + ɛ 2B (2B) = 2ɛ for all n max{n 2, M 2 }. Thus, a n b n a nb n < ɛ for all n max{n 2, M 2 } and {a n b n } {a nb n}. This shows that multiplication is well-defined. Theorem 1: The real numbers are a field. Proof: Since the rational numbers are a field, ([{a n }] + [{b n }]) + [{c n }] = [{(a n + b n ) + c n }] = [{a n + (b n + c n )}] = [{a n }] + ([{b n }] + [{c n }]) for any Cauchy sequences {a n }, {b n } and {c n }, and similarly for multiplication. Also, [{a n }] + [{b n }] = [{a n + b n }] = [{b n + a n }] = [{b n }] + [{a n }], and similarly for multiplication. Next, ([{a n }]+[{b n }]) [{c n }] = [{(a n +b n ) c n }] = [{(a n c n )+(b n c n )}] = ([{a n }] [{c n }])+([{b n }] [{c n }]). 3

4 Since 0 is the additive identity element of Q, [ 0 ] + [{a n }] = [{0 + a n }] = [{a n }] for any [{a n }] R, so that [ 0 ] is an additive identity for R. Clearly [{a n }] + [{ a n }] = [{a n + ( a n )}] = [ 0 ] for any [{a n }] R, so that [{ a n }] is an additive inverse for [{a n }]. Since 1 is the multiplicative identity element of Q [{a n }] [ 1 ] = [{a n 1}] = [{a n }] for all [{a n }] R. Clearly 0 1. Suppose that [{a n }] [ 0 ], i.e., {a n } 0. This means that for some ɛ > 0, there are infinitely many n N such that a n ɛ. Since {a n } is a Cauchy sequence, there is an N N such that a n a m < ɛ/2 for all n, m N. Since there must be an n 0 N such that a n0 ɛ, then a m a n0 a n0 a m > ɛ ɛ 2 = ɛ 2 for all m N. In particular, a m 0 for m N. Further, it is not difficult to see that the sequence {b n } defined by { an if n N, b n = if n N a n is a Cauchy sequence equivalent to {a n }. Thus, we may assume without loss of generality that a n 0 for all n N. Now since the rational numbers are a field, so that [{a n }] has a multiplicative inverse. {a n }. [{a n }] [{a 1 n }] = [{a n a 1 n }] = [ 1 ], Definition: For [{a n }] R, [{an }] = [{ an }]. Lemma: This is well-defined, i.e., [{an }] R and depends only on the equivalence class of Proof: Let ɛ > 0. Then for some N N we have a n a m < ɛ for all n, m N. Since an a m an a m, this implies that an a m < ɛ for all n, m N and { an } is a Cauchy sequence. 4

5 Suppose {a n} {a n } and let ɛ > 0. Then for some N N, a n a n < ɛ for all n N. As above, this implies that a n a n < ɛ for all n N, so that { a n } { a n }. Definition: We say a real number [{a n }] is greater than zero (or positive), and write [{a n }] > 0, if there is an N N and an ɛ > 0 such that a n ɛ for all n N. Lemma: This is well-defined. Proof: Suppose {a n} and {a n } are equivalent Cauchy sequences. Suppose further that there is some N N and an ɛ > 0 such that a n ɛ for all n N. There is an M N such that a n a n < ɛ/2 for all n M. This implies that a n a n a n a n > ɛ ɛ 2 = ɛ 2 for all n max{n, M}. Definition: We say a real number [{a n }] is greater than a real number [{b n }], and write [{a n }] > [{b n }], if [{a n }] [{b n }] > 0. Theorem 2: For any [{a n }], [{b n }] R, the following three properties hold: a) [{a n }] [ 0 ], with equality if and only if [{a n }] = [ 0 ]; b) [{an }] [{b n }] = [{an }] [{b n }] ; c) [{a n }] + [{b n }] [{a n }] + [{b n }]. In other words, is an absolute value on R. Proof: Starting with the first property, we have a n 0 for all n. Suppose that {a n } 0. Then there must be an ɛ > 0 such that a n ɛ for infinitely many n N. Let N N be such that a n a m < ɛ/2 for all n, m N. Since there is an n 0 N with a n0 ɛ, we have a m a n0 a n0 a m > ɛ ɛ/2 = ɛ/2 for all m N. Thus, [{a n }] = [{ a n }] > [ 0 ]. For the second property, using a n b n = a n b n for all n, we have [{an }] [{b n }] = [{an b n }] = [{ a n b n }] = [{ a n }] [{ b n }] = [{a n }] [{b n }]. Now for the third property - the triangle inequality. Suppose first that there is some ɛ > 0 and some N N such that a n + b n + ɛ a n + b n for all n N. Then by the definitions, [{a n }] + [{b n }] = [{ a n + b n }] < [{ a n + b n }] = [{a n }] + [{b n }]. So suppose this is not the case and let ɛ > 0. Then there are infinitely many n N such that a n + b n + ɛ/2 > a n + b n. Also, there is an N N such that an a m, bn b m, an + b n a m + b m < ɛ/6 5

6 for all n, m N. Choose an n N such that a n + b n + ɛ/2 > a n + b n. Then for all m N we have a m + b m a m + b m < a n + ɛ 6 + b n + ɛ 6 a n + b n + ɛ 6 = a n + b n a n + b n + ɛ 2 < ɛ. By the triangle inequality for rational numbers, a m + b m a m + b m. Thus, a m + b m a m + b m < ɛ for all m N and { a n + b n } { a n + b n }. By the definitions, this means that [{an }] + [{b n }] = [{an }] + [{bn }]. Lemma: Suppose [{a n }] and [{b n }] are two unequal real numbers. Then there is a rational number c such that [{a n }] [ c ] < [{a n }] [{b n }]. Proof: Since {a n } {b n }, {a n b n } 0. By Theorem 2, [{a n }] [{b n }] = [{a n b n }] > 0 so that there is an ɛ > 0 and an N N such that a n b n ɛ for all n N. Also, for some M N we have a n a m < ɛ/2 for all n, m M. Let c = a N+M. Then for all n N + M, a n c < ɛ/2 and a n b n ɛ. In particular, a n b n a n c ɛ/2 for all n N + M. By the definitions, this means that [{a n }] [ c ] < [{a n }] [{b n }]. Theorem 3: Every Cauchy sequence of real numbers converges, i.e., the real numbers are a topologically complete field. [Technically speaking, the ɛ in the definition of Cauchy sequence of real numbers is allowed to be real. However, it suffices to restrict to the case where ɛ is a positive rational number by the lemma above.] Proof: Let {r n } be a Cauchy sequence of real numbers. If there is an r R and an N N such that r n = r for all n N we are done, since in this case {r n } r. So suppose this is not the case. For each n N let n N be least such that n > n and r n r n. For each n N choose (via the lemma above) an a n Q such that r n [ a n ] < r n r n. Let ɛ > 0. Then there is an N N such that r n r m < ɛ/3 for all n, m N. By the triangle 6

7 inequality, we have a n a m = [ a n ] [ a m ] = [ a n ] r n + r n r m + r m [ a m ] [ a n ] r n + r n r m + r m [ a m ] < r n r n + r n r m + r m r m < ɛ 3 + ɛ 3 + ɛ 3 = ɛ for all n, m N. This shows that {a n } is a Cauchy sequence (of rational numbers), i.e., [{a n }] R. Let ɛ > 0 again. Then there are is an N N such that r n r m, a n a m < ɛ/3 for all n, m N. In particular, r N [ a N ] < ɛ/3 and also [ a N ] [{a m }] < ɛ/3. Using the triangle inequality once more, r n [{a m }] = r n r N + r N [ a N ] + [ a N ] [{a m }] r n r N + r N [ a N ] + [ a N ] [{a m }] < ɛ 3 + ɛ 3 + ɛ 3 = ɛ for all n N. Thus, {r n } [{a m }] R. Lemma: The sum and product of two positive real numbers is positive. Proof: Suppose [{a n }] and [{b n }] are positive real numbers. Then by definition there are N 1, N 2 N and ɛ 1, ɛ 2 > 0 such that a n ɛ 1 for all n N 1 and b n ɛ 2 for all n N 2. This implies that a n + b n ɛ 1 + ɛ 2 > 0 and a n b n ɛ 1 ɛ 2 > 0 for all n max{n 1, N 2 }. In other words, [{a n }] + [{b n }] = [{a n + b n }] > 0 and [{a n }] [{b n }] = [{a n b n }] > 0. Theorem 4: The real numbers are totally ordered by <. Proof: Suppose x < y. By definition, this means y x > 0. Since (y + z) (x + z) = y x, we have y + z > x + z. Suppose x < y and y < z. Then z x = (z y) + (y x) is positive, being the sum of two positive real numbers. Suppose x < y and z > 0. Then z(y x) is positive, so that zy > zx. Suppose x y and y x, and write x = [{a n }], y = [{b n }]. Let ɛ > 0. There are infinitely many n N such that a n b n < ɛ/3 and infinitely many n N such that b n a n < ɛ/3. Also, there are N, M N such that a n a m < ɛ/6 for all n, m N and b n b m < ɛ/6 for all 7

8 n, m M. Choose an n 0, m 0 max{n, M} such that a n0 b n0 < ɛ/3 and b m0 a m0 < ɛ/3. Then b n0 a n0 < b m0 + ɛ/6 a m0 + ɛ/6 < 2ɛ/3. For any n max{n, M} we have a n b n = a n a n0 + a n0 b n0 + b n0 b n a n a n0 + a n0 b n0 + b n0 b n < ɛ 6 + 2ɛ 3 + ɛ 6 = ɛ. Thus {a n b n } 0 and [{a n }] = [{b n }]. This shows that either x < y or y < x or x = y. Finally, suppose x > y and y < x. Then [ 0 ] = (x y) + (y x) is positive since it is the sum of two positive numbers. Similarly, if x = y and either x > y or y > x, then [ 0 ] is positive, which is clearly not the case, so that at most one of x > y, y > x and x = y holds. Theorem 5: Suppose S R, S, and there is a B R with x B for all x S. (We say B is an upper bound for S.) Then there is an upper bound b R for S such that for all c < b there is an x S with x > c. (We say b is a least upper bound for S.) Proof: Chose an x 0 S and let m 0 N be largest such that B m 0 is an upper bound for S. Then m 0 exists since B 0 is an upper bound for S and B n is not an upper bound for S whenever n > B x 0. We claim that there is a sequence B 1, B 2,... of upper bounds for S such that B n 2 n is not an upper bound for S and 0 B n 1 B n 2 n for all n N. We prove this claim by induction. Let m 1 N be largest such that B 0 m is an upper bound for S. Then m 1 is either 0 or 1 since B 0 0 is an upper bound for S and B 0 1 = B is not. Let B 1 = B 0 m Then B 1 is an upper bound for S and 0 m = B 0 B Also, B = B 0 (m 1 + 1)2 1 in not an upper bound for S by the definition of m 1. Now suppose B 1,..., B n have been chosen which satisfy the requirements above. Let m n+1 be the largest integer such that B n m n+1 2 n 1 is an upper bound for S. By the induction hypothesis, B n is an upper bound for S and B n 2 n is not, so that m n+1 is either 0 or 1. Let B n+1 = B n m n+1 2 n 1. Then B n+1 is an upper bound for S and 0 m n+1 2 n 1 = B n B n+1 2 n 1. Finally, B n+1 2 n 1 = B n (m n+1 +1)2 n 1 is not an upper bound for S by the definition of m n+1. We claim that this is a Cauchy sequence. Indeed, let ɛ > 0 and suppose n, m N with n > m. 8

9 Then B n B m = (B n B n 1 ) + + (B m+1 B m ) 2 n m 1 = 2 m 1 ( m+1 n ) m 1 1 2m n = = 2 m (1 2 m n ) < 2 m. Thus, if n, m N where 2 N ɛ, then B n B m < ɛ. Let b R be the limit of this Cauchy sequence. Let x S. Suppose x > b and let ɛ = x b. For some n N, B n b < ɛ. But this implies that x b > B n b, so that x > B n. This contradicts the fact that all B n s are upper bounds for S. Thus, b is an upper bound for S. Finally, suppose c < b is an upper bound for S. Then b c 2 n 0 for some n 0 N, which implies that b 2 n 0 is an upper bound for S. Choose N N such that B n b < 2 n 0 1 for all n N. We then have B n 2 n 0 1 > b 2 n 0 for all n N. But B n 2 n is not an upper bound for any n, so that B n 2 n 0 1 B n 2 n is not an upper bound for any n > n 0. This contradition shows that c b, so that b is a least upper bound for S. 9

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