# Proof since s is bounded below by the completeness

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Proof. Since S is bounded below, by the Completeness Axiom there exists w = inf S . If w 2 S then we are done. So assume w 62 S , in order to derive a contradiction. We apply the two properties of the inf: first, w n for all n 2 S . Since w 62 S , in fact we have the slightly stronger w < n for all n 2 S , and w + 1 is not a lower bound for S , so there exists k 2 S Z for which k < w + 1 . Since S has no minimum element, k is not the smallest element of S , so there must be j 2 S with j < k . Putting these all together, w < j < k < w + 1 . In particular, 0 < k - j < ( w + 1) - j < ( w + 1) - w = 1, but k - j 2 Z and the distance between any two integers is at least one! So this is impossible, and we conclude that S has a minimum element. Applied to subsets of the natural numbers this property is called the “well-ordering principle”: Corollary 1.13 (Well-Ordering Principle) . Any nonempty subset of N has a minimum ele- ment. The Corollary follows from the previous Theorem, since N is bounded below (by 1), and so any nonempty subset of N is bounded below. We’re now ready to prove the Density Theorem. 10
Proof of the Density Theorem. First we choose the denominator of r = m/n . By the Archimedean Property, there exists n 2 N with n > 1 ( b - a ) , that is 1 n < ( b - a ) . (1.2) To find the numberator, consider the set S = { k 2 Z : k > na } . Since S Z and S is bounded below by na , by Theorem 1.12 S contains a minimal element m , and so m 2 S and m - 1 62 S . This implies: m > na and m - 1 na. (1.3) Putting (1.2) and (1.3) together we get: a < m n = m - 1 n + 1 n a + 1 n < a + ( b - a ) = b. Then the conclusion holds with r = m/n . 11
2 Sequences A sequence of real numbers is a function f : N ! R ; for each counting number n 2 N we associate to it a real number f ( n ) = x n . There are many di erent ways of denoting the sequence. Here are a few which you may see: ( x 1 , x 2 , x 3 , . . . ) = ( x n ) = ( x n ) n 2 N = ( x n : n 2 N ) . The text also writes X = ( x n ), using a single capital letter for the sequence as a whole. Example 2.1. (a) The function f can be explicitly given. For instance, x n = 2 n 2 +1 n 2 , 2 n 2 + 1 n 2 n 2 N = 3 , 9 4 , 19 9 , 33 16 , . . . , . (b) ( x n ) = ( ( - 1) n ) n 2 N = ( - 1 , 1 , - 1 , 1 , - 1 , . . . ) . (c) We may define sequences by iteration. Let g : R ! R be a given real-valued function. Choose an initial value x 1 2 R and then define the sequence iteratively, x n +1 = g ( x n ) , n = 1 , 2 , 3 , . . . This is a natural way to define sequences, for example to approximate solutions to equa- tions. For a more specific example, take g ( x ) = 1 2 ( x + 2 x ) , and generate the sequence ( x n ) by iteration: x 1 = 2 and x n +1 = 1 2 x n + 2 x n , n = 1 , 2 , 3 , . . . . The first few values are: ( x n ) = (2 , 1 . 5 , 1 . 41 ¯ 6 , 1 . 414215686 . . . , 1 . 414213562 . . . , . . . ) Later, we will prove that this sequence converges to p 2 . Note that a sequence is not the same thing as a set. A sequence is an infinite ordered list of numbers. In a sequence the same number may appear several times, and changing the order of the elements of a sequence creates an entirely di erent sequence. A set has no order, and there is no point in repeating the same value several times. Taking Example (b) above, { x n : n 2 N } = { - 1 , +1 } is a set with two elements. The sequence ( x n ) is not the same thing. If we let { y n } n 2 N = ( - 1 , - 1 , 1 , 1 , - 1 , - 1 , 1 , 1 , . . . ), this is a di erent sequence than { x n } n 2 N , yet it takes values in the same set { y n : n