
# 2.5: Some Consequences of the Completeness Axiom

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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.

Theorem $$\PageIndex{1}$$

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. $$\square$$

corollary $$\PageIndex{1}$$

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 $$y>-m \in J). \square$$

Theorem $$\PageIndex{2}$$

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.

corollary $$\PageIndex{2}$$

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

Theorem $$\PageIndex{3}$$

(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 $$(a, b) .$$