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By Derek G. Ball and C. Plumpton (Auth.)
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Example text
Thus L* c L and so (L*, £*) =^ (L, Λ). Thus (L*, R*) is the least upper bound for the set A. If Ais a non-empty set of real numbers which is bounded below, then A has a greatest lower bound. Proof. Let the set A be {w, x, y9 . . } . Consider the set B = {—w, —x, —y9 . . } consisting of all those real numbers which are negatives of members of A. Then B is bounded above and so has a least upper bound, say M. Then — M will be the greatest lower bound for A. D COROLLARY. DEFINITION. If a set of numbers is such that any of its bounded subsets has a least upper bound, the set is called complete.
Thus it is defined on a set A satisfying conditions (a) and (b) of (iv). ] DEFINITION. We define m-nby induction on m. Firstly we define l-n = n. Now, assuming that m*n is defined, we define s(m)*n = m>n+n. DEFINITION. Exercises 1. e. that (m+ri)+p = m+(n+p) by induction on p. ] 2. Prove by induction on m that s(n)-\-m = s(n + m) and hence prove that « + m = m+n. 3. Using the definition of multiplication and the result of question 2, prove that (m+l)(p+q) = (m+\)p+(m+l)q. Hence prove that m(p-{-q) = mp+mq by induction on m.
This contradicts the definition of L and L*. Thus L' does not have a greatest member. £') is a Dedekind section. D If the rationals are to be regarded as a subset of the reals, it is necessary not only that to every rationale there should be a corresponding real rational a*, but also that the addition and multiplication of real rationals should mirror the addition and multiplication of rationals. We prove the relevant theorem for addition. The corresponding result for multiplication is left as an exercise.