2. DEDEKIND CUTS
1. The Natural Numbers, the Integers, and the Rational Numbers in the Real Number Field should all be recoverable once the latter has
will however be based not on the system of axioms proposed by HILBERT (in
§13 of [13], where it is called "the theory of ratios," following the tradition set by EUCLID in his Elements), but on the axioms (Rl)—(R3) of §2.
1. The Natural Numbers, the Integers, and the Rational Numbers
satisfied when elements of K instead of rational numbers are taken) then a contains a maximum element.
(c) Every monotonically decreasing sequence, bounded Below, converges in K.
(c') Every monotonically increasing sequence, bounded above, converges in K.
(d) Thefield K has an Archimedean ordering and every fundamental se- quence (Cauchy sequence) of elements of K converges in K.
(e) The field K has an Archsmedean ordering and for every sequence of nested intervals 1o D Jz
j ...
in K, for which the lengths of I,, converge to zero with increasing n, there exists one and only ones lying in all the intervals
(a) and (a') are obviously equivalent: if and only if M is bounded below is —M = {—z:z M} bounded above, and —infM = sup(—M). Similarly (c) and (c') are equivalent. The complete equivalence of all the assertions will follow from the implications (a) (b) —. (c) (d) —+ (e) —'(a) which we shall prove in turn.
(a) —+ (b): The set /3 is bounded below, every a a is a lower bound.
By (a) /3 has an infimum. Since /3 has no minimum, inf/3 E a. Since a < B
holdsforallaEoandb€13,wehavea<inf/3forallaEa,thatisinffl
is the maximum of a.
(b) —. (c): Let (ba) be a monotonically decreasing sequence, bounded below. We can define a cut (a,13) by a = {x:z <b, for all n} and /3 = {y:
there is an n such that <y). By (b) the set a has a maximum s. We can now show that (ba) converges to s. To prove this suppose e > 0 be given, then there is an index k such that bk <a+ c because ifs + c were < b,, for all k, we should have s + c E a, in contradiction to a = maxa. As is monotonically decreasing, bm b,, for all m k, and since a <bm for Sil m, we therefore have s < bm <bk <8+c
for all m k.
(c) (d):The Archimedean property of the ordering of K can be proved as follows. Let o,b be > 0, and suppose that no B for all n N. Then (no) would be a monotonically increasing sequence bounded above, which by (c) would converge to some a. There would therefore also be an index k such that s — a < no < s for all n k. Between a — a and s there is however room for only one term no of the sequence (no).
To prove that every fundamental sequence converges, we need two lem- mas.
(1) Every sequence (as) has a monotonic subsequence.
(2) Every fundamental sequence is bounded.
We shall postpone the proof of (1) and (2) for a moment and first show that every fundamental sequence (an) converges. Let (an) be a monotonic subsequence. It is bounded and hence s = exists. We assert
that s =
for given any > 0, one can choose the index k sothat 1Gm— < for all m, n k, and there is then a j such that rz 2 k and —SI < It now follows that — —an,l+ Ian, — si
<c
for all n k.
Proof of Lemma (1). We shall say that the sequence (an) has a peak ak for the index k, if for all n k. If there is an infinity of peaks then they form a monotonic non-increasing sequence. If there are no peaks or only a finite number of peaks, there is a last index k beyond which there are no peaks. We begin our subsequence with = k + 1. Since
a is an n1 > n0, for which a peak,
there is an > n1, such that > and so on. We have thus found by recursion a monotonically increasing subsequence (as,).
Proof of Lemma (2). Let (an) be a fundamental sequence. There is an
index k such that lam — < 1 for all m, n k. In particular therefore all subsequent terms for n k lie within the bounded interval (ak — 1,ak + 1). The finitely many initial terms a0,...
,ak_I of the sequence obviously also form a bounded set, and consequently the set of all terms with n E is also bounded.(d) —+ (e): Let ([an,bnJ) be a sequence of nested intervaLs. Then (an) is a fundamental sequence, because, for every k and all m, n k, Gm,Gn lie in [ak,bk], and hence lam — < — ak. Since — = 0 we can therefore ensure that lam — < e by choosing k large enough. By (d),
8 =
exists. Since (an) increases monotonely, $ for all n. As for all k and n, we also have s < 6,, for all n, and thus s [a,,, b,]for every n. Since b,, — a,, becomes arbitrarily small as n increases, a is defined unambiguously.
(e) (a):
Let M be a non-empty subset of K, bounded below. We
can construct a sequence of nested intervals ([a,,, 6,,]), in which all the a,, are lower bounds of M, while none of the b,, are lower bounds of M. We begin with any lower bound a0 and a which is not. We then proceed recursively: having already defined [a,,, 6,,] we form the arithmetic mean d,, = + b,,) and defineb J —
f
[dn,b,,J, jf d,, is a lower boundn+1
— j [a,,,d,,], if d, is not a lower bound.
Then b,,+i — a,,+j = — so that b,, = 2"(b0 — ao). As the ordering is Archimedean, lim(b,, —a,,) = 0. By (e) there is just one s which lies in all the intervals [a,,,b,,j. Now c is a lower bound of M, for otherwise
there would be an x E M with x < c, and since every < z we should
have 6,, — a,, c —a,, c —x which would contradict lim(6,, —a,,)
=
0.This c is the greatest of the lower bounds, because if 6> c were a lower bound, we should have to have 6,, > 6 and b,, —a,, > 6 —a,, > b — c in contradiction to lim(b,, —a,,) = 0.
The list (a)—(e) of equivalent statements by no means exhausts all the possible formulations. One could for example also mention the HEINE—
BOREL covering property or the fact that every bounded infinite subset contains a limit point. The student learns about these and other results, as consequences of the property of completeness, in every introductory course on analysis.
There are totally ordered fields in which every fundamental sequence converges, but in which the ordering is not Archimedean. An example of this will be given in Chapter 12 where the real numbers will be extended to the field *R of non-standard numbers. In this extended field there are infinitely small and infinitely large numbers, and for this reason is not Archimedean, while every fundamental sequence is constant and therefore convergent. Just how much the Archimedean axiom restricts the possibili- ties is shown clearly by the following result due to HoLDER [13a], see also CARTAN [6]: An ordered group is Archimedean if and only if it is isomor- phic to a subgroup of the additive group of real numbers. One does not even have to assume that the group is commutative; it follows from the other hypotheses.