1.3 Natural Numbers 21Theorem 1.3.2 The intersection^5 of any collection of inductive subsets of F
is inductive.Proof. Let C denote a collection of inductive sets. We shall prove that
n C is an inductive set.
(i) For all CE C, 1 EC, since C is an inductive set. Therefore, 1 En C.
(ii) Suppose x E n C. Let C E C. Then x E C. Since C is inductive,
x + 1 E C. Since this is true for all C E C, x + 1 E n C. •We are now ready to define the natural numbers in any ordered field.Definition 1.3.3 The set of natural numbers of an ordered field F is the
intersection of all the inductive subsets of F. In symbols,Np= n S ,
where S denotes the collection of all inductive subsets of F.We call the elements of Np the natural numbers of F.Theorem 1.3.4 The set of natural numbers is the smallest inductive subset
of F, in the sense that Np is an inductive set and every inductive subset of F
contains Np as a subset.Proof. By Theorem 1.3.2, Np is an inductive set. Now, suppose C is an
inductive subset of F. Then C E S , the collection of all inductive subsets of F.
Thus, n S ~ C , since the intersection of a collection of sets is a subset of any
one of the sets in the collection. That is, Np ~ C. •
Theorem 1.3.5 For any ordered field F,
(a) All natural numbers of F are positive.(b) 1 is the smallest natural number of F. That is, Vn E Np, n ?'.: 1.(c) Vn E Np, ifn > 1, then n-1 E Np.
Proof. (a) Recall that the set Pis an inductive subset of F. Therefore, by
Theorem 1.3.4, Np ~ P. Thus, all elements of Np are positive.
(b) Let A = { x E F : x ?'.: 1}. Then
(i) 1 EA, and
(ii) Suppose x E A. Then x ?'.: 1. Hence, x + 1 ?'.: l ;
That is, x + 1 EA. Therefore x EA:::} x + 1 EA.- For a review of sets and intersections, see Appendix B .l.