Lecture 5: The Archimedian Property, Density of the Rationals, and Absolute Value

18.100A: Complete Lecture Notes Lecture 5: The Archimedian Property, Density of the Rationals, and Absolute Value For all x, y ∈ R and x < y, there exists an r ∈ R such that x < r < y (take r = x+y 2 ). Question 1. Can we find r ∈ Q such that x < r < y? Theorem 2 The answer is yes! i) (Archimedian Property) If x, y ∈ R and x > 0, then ∃n ∈ N such that nx > y. ii) (Density of Q) If x, y ∈ R and x < y then ∃r ∈ Q such that x < r < y. Proof : i) Suppose that x, y ∈ R and x > 0. Then we wish to show that ∃n ∈ N such that n > xy . Suppose this is not the case. Then, ∀n ∈ N, n ≤ xy . In other words, N is bounded above by xy . Hence, ∃a = sup N ∈ R. Since a is the least upper bound for N, a − 1 cannot be an upper bound for N. Hence, ∃m ∈ N such that a − 1 < m =⇒ a < m + 1 ∈ N. However, this is a contradiction, because then a is not an upper bound for N. Therefore, ∃n ∈ N such that n ≥ xy . ii) Suppose x, y ∈ R and x < y. Then, there are three cases: • 0 ≤ x < y, • x < 0 < y, and • x < y ≤ 0. For the second case, take r = 0 ∈ Q. So, assume that 0 ≤ x < y. Then, by the Archimedian Property, ∃n ∈ N such that n(y − x) > 1. Again by the Archimedean property, ∃l ∈ N such that l > nx. Thus, consider the set S = {k ∈ N | k > nx}. By the well-ordering property of N, S has a least element, m ∈ S =⇒ nx < m =⇒ x < Since m − 1 ∈ / S, m − 1 ≤ nx =⇒ m ≤ nx + 1 < ny. Hence, x< m n < y. Therefore, m < y. n If instead we have x < y ≤ 0, then 0 ≤ −y < −x =⇒ ∃r̃ ∈ Q such that −y < r̃ < x =⇒ x < −r̃ < y by the previous case. 1 m n ∈ Q. Theorem 3  1 = sup 1 − 1 n |n∈N . Proof : If n ∈ N, then 1 − n1 < 1 =⇒ 1 is an upper bound of this set. Suppose that x is an upper bound for the set {1 − 1/n | n ∈ N}. We now prove that x ≥ 1. For the sake of contradiction, assume that x < 1. By the Archimedean property, there exists an n ∈ N such that 1 < n(1 − x). Therefore, ∃n ∈ N such that x < 1 − 1/n. Hence, x is not an upper bound for the set {1 − 1/n | n ∈ N} if x < 1. Thus, if x is an upper bound, x ≥ 1. Therefore,   1 sup 1 − | n ∈ N = 1. n We now begin proving some theorems about supremums and infinimums which will make them easier to use. Theorem 4 Suppose that S ⊂ R is nonempty and bounded above. Then, x = sup S if and only if 1. x is an upper bound for S. 2. for all  > 0, ∃y ∈ S such that x −  < y ≤ x. Proof : This is left as an exercise in Assignment 3. Notation 5 For x ∈ R and A ⊂ R, define x + A := {x + a | a ∈ A} xA := {xa | a ∈ A}. Theorem 6 Using this new notation, we have the following theorems: 1. If x ∈ R and A is bounded above, then x + A is bounded above and sup(x + A) = x + sup A. 2. If x > 0 and A is bounded above then xA is bounded above and sup(xA) = x sup A. Proof : 1. Suppose that x ∈ R and A is bounded above. Therefore, sup A ∈ R by the least upper bound property of R. Then, ∀a ∈ A, a ≤ sup A. Hence, ∀a ∈ A, x + a ≤ x + sup A. Hence, x + sup A is an upper bound for x + A. Let  > 0. Then, ∃y ∈ A such that sup A −  < y ≤ sup A =⇒ (x + sup A) −  < y + x ≤ x + sup A. 2 Therefore, by our previous theorem, x + sup A = sup(x + A). 2. Suppose that x > 0 and A is bounded above. Thus, sup A ∈ R. Then, ∀a ∈ A, a ≤ sup A and thus xa ≤ x sup A. Hence, x sup A is an upper bound of xA. Let  > 0. Then ∃y ∈ A such that sup A −  < y ≤ sup A =⇒ x sup A −  < xy ≤ x sup A. x Therefore, by the previous theorem, sup(xA) = x sup A. Theorem 7 Let A, B ⊂ R such that ∀x ∈ A, ∀y ∈ B, x ≤ y. Then, sup A ≤ inf B. Proof : The proof of this is left to the reader. Absolute Value Definition 8 If x ∈ R we define  x, |x| := −x, x≥0 . x≤0 Theorem 9 We can prove a bunch of theorems about the absolute value function that we usually take for granted: 1) |x| ≥ 0 and |x| = 0 ⇐⇒ x = 0. 2) ∀x ∈ R, | − x| = |x|. 3) ∀x, y ∈ R, |xy| = |x||y|. 4) |x2 | = x2 = |x|2 . 5) If x, y ∈ R, then |x| ≤ y ⇐⇒ −y ≤ x ≤ y. 6) ∀x ∈ R, x ≤ |x|. Proof : 1) If x ≥ 0 then |x| = x ≥ 0. If x ≤ 0, then −x ≥ 0 =⇒ |x| = −x ≥ 0. Thus, |x| ≥ 0. Now suppose x = 0. Then, |x| = x = 0. For the other direction, suppose |x| = 0. Then, if x ≥ 0 =⇒ x = |x| = 0. If x ≤ 0, then −x = |x| = 0. Therefore, x = 0 ⇐⇒ |x| = 0. 2) If x ≥ 0 then −x ≤ 0. Thus, |x| = x = −(−x) = | − x|. If x ≤ 0 then −x ≥ 0 and thus | − x| = | − (−x)| = |x|. 3) If x ≥ 0 and y ≥ 0, then xy ≥ 0 and |xy| = xy = |x||y|. If x ≤ 0 and y ≤ 0, then xy ≤ 0 =⇒ |xy| = −xy = (−x)y = |x||y|. 4) Take x = y in 3). Then, |x2 | = |x|2 . Since x2 ≥ 0, it follows that |x2 | = x2 . 5) Suppose |x| ≤ y. If x ≥ 0, then −y ≤ 0 ≤ x = |x| ≤ y. Therefore, −y ≤ x ≤ y. If x ≤ 0, then −x ≥ 0 and | − x| ≤ y. Hence, −y ≤ −x ≤ y =⇒ −y ≤ x ≤ y. 3 6) Take y = |x| in 5). On its own, these properties of the absolute values may not seem all that useful, but in the next lecture we will prove the extremely important Triangle Inequality. 4