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Mathematics LibreTexts

2.5: Some Consequences of the Completeness Axiom

  • Page ID
    19032
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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 \((a, b) .\)