2.5: Some Consequences of the Completeness Axiom
The ancient Greek geometer and scientist Archimedes was first to observe that even a large distance \(y\) can be measured by a small yardstick \(x ;\) one only has to mark \(x\) off sufficiently many times. Mathematically, this means that, given any \(x>0\) and any \(y,\) there is an \(n \in N\) such that \(n x>y .\) This fact, known as the Archimedean property , holds not only in \(E^{1}\) but also in many other ordered fields. Such fields are called Archimedean . In particular, we have the following theorem.
Any complete field F \(\left(e . g ., E^{1}\right)\) is Archimedean.
That is, given any \(x, y \in F(x>0)\) in such a field, there is a natural \(n \in F\) such that \(n x>y .\)
- Proof
-
(by contradiction) Suppose this fails. Thus, given \(y, x \in F(x>0)\), assume that there is no\(n \in N\) with \(n x>y\).
Then
\[(\forall n \in N) \quad n x \leq y\]
i.e., \(y\) is an upper bound of the set of all products \(n x(n \in N) .\) Let
\[A=\{n x | n \in N\}\]
Clearly, \(A\) is bounded above \((\) by \(y)\) and \(A \neq \emptyset ;\) so, by the assumed completeness of \(F, A\) has a supremum, say, \(q=\sup A\).
As \(q\) is an upper bound, we have (by the definition of \(A )\) that \(n x \leq q\) for all \(n \in N,\) hence also \((n+1) x \leq q ;\) i.e.,
\[n x \leq q-x\]
for all \(n \in N(\) since \(n \in N \Rightarrow n+1 \in N)\).
Thus \(q-x\) (which is less than \(q\) for \(x>0\)) is another upper bound of all \(nx\) i.e., of the set \(A .\)
This is impossible, however, since \(q=\sup A\) is the least upper bound of \(A .\)
This contradiction completes the proof. \(\square\)
In any Archimedean
(
hence also in any complete
)
field
\(F,\)
the set \(N\) of all natural elements has no upper bounds, and the set \(J\) of all integers
has neither upper nor lower bounds. Thus
\[(\forall y \in F)(\exists m, n \in N) \quad-m<y<n\]
- Proof
-
Given any \(y \in F,\) one can use Archimedean property (with \(x=1 )\) to find an \(n \in N\) such that
\[n \cdot 1>y, \text{ i.e., } n>y.\]
Similarly, there is an \(m \in N\) such that
\[m>-y, \text{ i.e., } -m<y.\]
This proves our last assertion and shows that \(n o y \in F\) can be a right bound of \(N(\) for \(y<n \in N),\) or a left bound of \(J(\) for \(y>-m \in J). \square\)
In any Archimedean ( hence also in any complete ) field \(F,\) each left ( right ) bounded set \(A\) of integers \((\emptyset \neq A \subset J)\) has a minimum ( maximum , respectively \)).
- Proof
-
Suppose \(\emptyset \neq A \subseteq J,\) and \(A\) has a lower bound \(y\).
Then Corollary 1 (last part) yields a natural \(m,\) with \(-m<y,\) so that
\[(\forall x \in A) \quad-m<x,\]
and so \(x+m>0\).
Thus, by adding \(m\) to each \(x \in A,\) we obtain a set \((\) call it \(A+m)\) of naturals .
Now, by Theorem 2 of \(§§5-6, A+m\) has a minimum; call it \(p .\) As \(p\) is the least of all sums \(x+m, p-m\) is the least of all \(x \in A ;\) so \(p-m=\min A\) exists, as claimed.
Next, let \(A\) have a right bound \(z .\) Then look at the set of all additive inverses \(-x\) of points \(x \in A ;\) call it \(B .\)
Clearly, \(B\) is left bounded \((\mathrm{by}-z),\) so it has a minimum, say, \(u=\min B\). Then \(-u=\max A .\) (Verify!) \(\square\)
In particular, given any \(x \in F(F\) Archimedean), let \([x]\) denote the greatest integer \(\leq x\) (called the integral part of \(x ) .\) We thus obtain the following corollary.
Any element \(x\) of an Archimedean field \(F\) has an integral part \([x] .\) It is the unique integer \(n\) such that
\[n \leq x<n+1\]
( It exists, by Theorem 2.)
Any ordered field has the so-called density property:
If \(a<b\) in \(F,\) there is \(x \in F\) such that \(a<x<b ;\) e.g. take
\[x=\frac{a+b}{2}.\]
We shall now show that, in Archimedean fields, \(x\) can be chosen rational , even if \(a\) and \(b\) are not. We refer to this as the density of rationals in an Archimedean field
(density of rationals) Between any elements \(a\) and \(b\) \((a<b)\) of an Archimedean field \(F\) (such as \(E^{1}\)), there is a rational \(r \in F\) with
\[a<r<b.\]
Let \(p=[a]\) (the integral part of \(a ) .\) The idea of the proof is to start with \(p\) and to mark off a small "yardstick"
\[\frac{1}{n}<b-a\]
several \((m)\) times, until
\[p+\frac{m}{n} \text{ lands inside } (a, b)\]
then \(r=p+\frac{m}{n}\) is the desired rational.
We now make it precise. As \(F\) is Archimedean, there are \(m, n \in N\) such that
\[n(b-a)>1 \text{ and } m\left(\frac{1}{n}\right)>a-p\]
We fix the least such \(m\) (it exists, by Theorem 2 in \(§§ 5-6 ) .\) Then
\[a-p<\frac{m}{n}, \text{ but } \frac{m-1}{n} \leq a-p\]
(by the minimality of \(m ) .\) Hence
\[a<p+\frac{m}{n} \leq a+\frac{1}{n}<a+(b-a),\]
since \(\frac{1}{n}<b-a .\) Setting
\[r=p+\frac{m}{n},\]
we find
\[a<r<a+b-a=b. \square\]
Note. Having found one rational \(r_{1}\),
\[a<r_{1}<b,\]
we can apply Theorem 3 to find another \(r_{2} \in R\),
\[r_{1}<r_{2}<b,\]
then a third \(r_{3} \in R\),
\[r_{2}<r_{3}<b,\]
and so on. Continuing this process indefinitely, we obtain infinitely many rationals in \((a, b) .\)