Completeness and the real numbers

Section 3.4 Completeness and the real numbers

Definition 3.4.1. Completeness.

A totally ordered set $(S, <)$ is complete if and only if, for every subset $T \subseteq S$ with an upper bound in $S,$ there is a supremum of $T$ in $S .$

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Equivalently,

S

is complete if and only if every subset of

S

that has a lower bound has an infimum. We have seen above that

Q

is not complete. Nevertheless, we can complete

Q .

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Theorem 3.4.2.

There exists a complete ordered field. Furthermore, it is unique in the following sense: if $F$ and $F^{'}$ are complete ordered fields, then there exists a unique bijection $φ : F \to F^{'}$ such that

$φ (0) = 0 .$
$φ (1) = 1 .$
For every $x, y \in F,$ one has $φ (x + y) = φ (x) + φ (y) .$
For every $x, y \in F,$ one has $φ (x \times y) = φ (x) \times φ (y) .$
For every $x, y \in F,$ if $x < y,$ then $φ (x) < φ (y) .$

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We won't prove this theorem, as the proof is quite long and tedious, and there's not much to be learned from it. So there is a complete ordered field, which, up to a simple renaming of the elements, is unique. We call it

R

or the real line, and we call its elements real numbers.

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This

R

contains a copy of the natural numbers:

0

and

1

are in there because of the field axioms. Note that

0 < 1 .

(Why?) for every natural number

n,

if we can regard

n

as a real number, then we can regard the successor of

n

as the real number

n + 1 .

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Since

R

is a field, we can take the additive inverses of the natural numbers to get a copy of

Z

lying in

R .

We can also take products of integers and multiplicative inverses of integers to get a copy of

Q

lying in

R .

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We are happy to pretend that we have literal inclusions

N \subset N_{0} \subset Z \subset Q \subset R .

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The following theorem implies that there are enough natural numbers. This will be used to show that in fact every real number can be arbitrarily well approximated by rational numbers.

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Theorem 3.4.3. No infinite real numbers.

Let $x \in R .$ Then there exists a natural number $n$ such that $n > x .$

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Proof.

Suppose not. Then $x$ is an upper bound for $\NN_0 \subseteq\RR\text{.}$ By completeness, it follows that there is a real number $N = \sup\NN_0\text{.}$ Now consider $N-1\text{;}$ for any natural number $n\text{,}$ the real number $n+1$ is also a natural number, and therefore $n+1\leq N\text{,}$ so $n\leq N-1\text{.}$ Thus $N-1$ is also an upper bound for $\NN_0\text{.}$ But this contradicts the fact that $N$ is a least upper bound.

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Theorem 3.4.4. Archimedean property.

If $x, y \in R$ are positive real numbers, then there exists a natural number $n \in N_{0}$ such that $n y > x .$

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Proof.

By the previous theorem, choose a natural number $n\gt x/y\text{.}$

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Theorem 3.4.5. Square roots.

If $x \in R$ and $x \geq 0,$ then there exists a unique $y \in R$ such that $y \geq 0$ and $y^{2} = x .$

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Proof.

Let's prove the existence first. Consider the set $S = \{z\in \RR : (z \geq 0)\wedge (z^2\geq x)\}\text{.}$ Observe that $S$ has a lower bound, namely $0\text{.}$ Now by completeness, we may define $y = \inf S\text{.}$ We claim that $y^2=x\text{.}$ Suppose not; then either $y^2 \gt x$ or $y^2 \lt x\text{.}$

If $y^2\gt x\text{,}$ then use the Archimedean property to find a natural number $n$ such that

\begin{equation*} n(y^2-x) \gt 2y \end{equation*}

Thus $y^2-(2/n)y\gt x\text{.}$ Now let $u=y-1/n\text{.}$ We have

\begin{equation*} u^2 = (y-1/n)^2 = y^2-(2/n)y+1/n^2 \gt y^2 - (2/n)y \gt x\text{.} \end{equation*}

Thus $u\in S\text{,}$ and this contradicts our assumption that $y$ is a lower bound.

If on the other hand $y^2\lt x\text{,}$ then use the Archimedean property to find a natural number $n$ such that

\begin{equation*} n(x-y^2) \gt 2y+1 \end{equation*}

Thus $y^2+(2/n)y+1/n^2 \lt y^2+(2/n)y+1/n \lt x\text{.}$ Now let $u=y+1/n\text{.}$ We have

\begin{equation*} u^2 = (y+1/n)^2 = y^2+(2/n)y+1/n^2 \lt x\text{.} \end{equation*}

Now if $z\in S\text{,}$ then $z^2>u^2\text{.}$ This implies that $u\lt z\text{.}$ (Indeed if $u \geq z\text{,}$ then $u^2\geq z^2\text{.}$) Thus $u $ is a lower bound for $S\text{,}$ which contradicts our assumption that $y$ is the greatest lower bound.

This shows that $y^2=x\text{.}$ Now, to prove uniqueness, suppose $v\geq0$ is a positive real number; if $v\lt y\text{,}$ then $v^2 \lt y^2 = x\text{,}$ and if $y\lt v\text{,}$ then $y^2 \lt v^2\text{.}$ Hence $v^2=y^2$ if and only if $v=y\text{.}$