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. Suppose, for contradiction, that there exists x R such that x > n for every n N. Then x is an upper bound of N, so N has a supremum M = sup N R. Since n M for every n N, we have n 1 M 1 for every n N, which implies that n M 1 for every n N. But then M 1 is an upper bound of N, which contradicts the assumption that M is a least upper bound. By taking reciprocals, we also get from this theorem that for every > 0 there exists n N such that 0 < 1/n < . These results say roughly that there are no infinite or infinitesimal real numbers. This property is consistent with our intuitive picture of a real line R that does not “extend past the natural numbers,” where the natural numbers are obtained by counting upwards from 1. Robinson (1961) introduced extensions of the real numbers, called non-standard real numbers, which form non-Archimedean ordered fields with both infinite and infinitesimal elements, but they do not satisfy Axiom 2.17. The following proof of the uncountability of R is based on its completeness and is Cantor’s original proof (1874). The idea is to show that given any countable set of real numbers, there are additional real numbers in the “gaps” between them. Theorem 2.19. The set of real numbers is uncountable. Proof. Suppose that S = {x1, x2, x3, . . . , xn, . . . } is a countably infinite set of distinct real numbers. We will prove that there is a real number x R that does not belong to S. If x1 is the largest element of S, then no real number greater than x1 belongs to S. Otherwise, we select recursively from S an increasing sequence of real numbers ak and a decreasing sequence bk as follows. Let a1 = x1 and choose b1 = xn1 where n1 is the smallest integer such that xn1 > a1. Then xn / (a1, b1) for all 1 n n1. If xn / (a1, b1) for all n N, then no real number in (a1, b1) belongs to S, and we are done e.g., take x = (a1 + b1)/2. Otherwise, choose a2 = xm2 where 2.7. Properties of the supremum and infimum 31 m2 > n1 is the smallest integer such that a1 < xm2 < b1. Then xn / (a2, b1) for all 1 n m2. If xn / (a2, b1) for all n N, we are done. Otherwise, choose b2 = xn2 where n2 > m2 is the smallest integer such that a2 < xn2 < b1. Continuing in this way, we either stop after finitely many steps and get an interval that is not included in S, or we get subsets {a1, a2, . . . } and {b1, b2, . . . } of {x1, x2, . . . } such that a1 < a2 < · · · < ak < · · · < bk < · · · < b2 < b1. It follows from the construction that for each n N, we have xn / (ak, bk) when k is sufficiently large. Let a = sup kN ak, inf kN bk = b, which exist by the completeness of R. Then a b (see Proposition 2.22 below) and x / S if a x b, which proves the result. This theorem shows that R is uncountable, but it doesn’t show that R has the same cardinality as the power set P(N) of the natural numbers, whose uncountability was proved in Theorem 1.47. In Theorem 5.67, we show that R has the same cardinality as P(N); this provides a second proof that R is uncountable and shows that P(N) has the cardinality of the continuum. 2.7. Properties of the supremum and infimum In this section, we collect some properties of the supremum and infimum for later use. This section can be referred back to as needed. First, we state an equivalent way to characterize the supremum and infimum, which is an immediate consequence of Definition 2.11. Proposition 2.20. If A R, then M = sup A if and only if: (a) M is an upper bound of A; (b) for every M0 < M there exists x A such that x > M0 . Similarly, m = inf A if and only if: (a) m is a lower bound of A; (b) for every m0 > m there exists x A such that x < m0 . We frequently use this proposition as follows: (a) if M is an upper bound of A, then sup A M; (b) if A is nonempty and bounded from above, then for every > 0, there exists x A such that x > sup A . Similarly: (a) if m is a lower bound of A, then m inf A; (b) if A is nonempty and bounded from below, then for every > 0, there exists x A such that x < inf A + . Making a set smaller decreases its supremum and increases its infimum. In the following inequalities, we allow the sup and inf to be extended real numbers. Proposition 2.21. Suppose that A, B are subsets of R such that A B. Then sup A sup B, and inf A inf B. Proof. The result is immediate if B = , when A = , so we may assume that B is nonempty. If B is not bounded from above, then sup B = , so sup A sup B. If B bounded from above, then sup B is an upper bound of B. Since A B, it follows that sup B is an upper bound of A, so sup A sup B. Similarly, either inf B = or inf B is a lower bound of A, so inf A inf B.