Real numbers and order completeness

The real numbers correspond to all possible measurements of lengths of a line. They differ from the rationals in that the order is complete. The completeness of the order on the reals is used in proving such results in calculus as the Intermediate Value Theorem, the Theorem on the Existence of Maxima and Minima, Convergence of Monotone Bounded Sequences.

There are at least two major constructions of the real numbers from the rationals. We will look briefly at one of them, the one identifying real numbers with Dedekind cuts:

Definition 59   A Dedekind cut on the rationals consists of two subsets $L$ and $U$ of such that

1. both $L$ and $U$ are nonempty
2. $q\in U$ if and only if there is a rational $r<q$ with $r\in U$ ($U$ is an open interval going to $+\infty$)
3. $q \in L$ if and only if there is a rational $r>q$ with $r \in L$ ($L$ is an open interval going to $-\infty$)
4. If $p\in L$ and $q\in U$ then $p<q$. (In particular note that $L\cap U =\emptyset$.)
5. For any $n\in \ensuremath{\mathbb{N}}$ there are rationals $p\in L$ and $q\in U$ with $q-p<\frac1n$. ($L$ and $U$ are arbitrarily close together.)

Example:

Any rational number gives a Dedekind cut: given $q\in \ensuremath{\mathbb{Q}}$ we let $L=\{r\vert r<q\}$ and $U=\{r\vert r>q\}$. Such a rational cut will have $L\cup U = \ensuremath{\mathbb{Q}}\setminus \{q\}$. An irrational cut will have $L\cup U = \ensuremath{\mathbb{Q}}$

An example of an irrational cut is $L=\{q\vert \mbox{ there is a rational $r$ such that $q<r$ and }r^2<2\}$ and $U=\ensuremath{\mathbb{Q}}\setminus L=\{q\vert<q\mbox{ and } q^2>2\}$.$\diamondsuit$

It is reasonably easy to define addition of Dedekind cuts and an order relation:

Definition 60   The sum of the Dedekind cuts $(L_1,U_1)+(L_2,U_2)$ is the cut with lower set $L=\{q\vert\mbox{ there are }q_1\in L_1 \mbox{ and }q_2 \in L_2 \mbox{ with }q=q_1+q_2\}$ and upper set given by $U=\{q\vert\mbox{ there are }q_1\in U_1 \mbox{ and }q_2 \in U_2 \mbox{ with }q=q_1+q_2\}$

Proposition 129   The sum of Dedekind cuts is again a Dedekind cut.

(3 Points )

The properties of addition of real numbers are fairly easy to prove because this definition uses the addition of so transparently. Multiplication does not behave so nicely with respect to order and as a result is much harder to define and prove things about for reals as Dedekind cuts. Usually one uses topological notions of the density of the rationals in the reals.

Definition 61   The order relation $(L_1,U_1)<((L_2,U_2)$ holds whenever $U_1 \cap L_2$ is not empty.

Notice that this makes the order relation on the cuts obtained from rational numbers the same as the order on the rationals themselves.

Proposition 130   The order relation on Dedekind cuts is transitive.

(2 Points )

Proposition 131   Trichotomy holds for order on Dedekind cuts.

(5 Points )

What sets the real numbers aside is the completness of the order, often stated as the least upper bound axiom: If $S\subseteq \ensuremath{\mathbb{R}}$ is non empty and has an upper bound, then it has a least upper bound.

Proposition 132   If $(L_1,U_1)<(L_2,U_2)$ then $U_2 \subseteq U_1$.

(2 Points )

This suggests the partial ordering $(L_1,U_1)\leq (L_2,U_2)$ whenever $U_2 \subseteq U_1$. This is the order we use in exploration of order completeness of the real numbers. Dedekind cuts have this property:

Theorem 133   Is $S=\{(L_\lambda, U_\lambda)\vert\lambda\in \Lambda\}$ is a nonempty set of Dedekind cuts and $(L_b,U_b)$ has $(L_\lambda, U\lambda)\leq (L_b,U_b)$ then

$$(\bigcup_{\lambda \in \Lambda} L_\lambda, \bigcap_{\lambda\in \Lambda}U_\lambda)$$

is a Dedekind cut which is the least upper bound of the set $S$.

(5 Points )

The least upper bound axiom for is extremely important for analysis. Knowing that may help you prove the following:

Proposition 134   The least upper bound of the set $\{q\vert q\in \ensuremath{\mathbb{Q}}\mbox{ and } q^2<2\}$ is $\sqrt{2}$.

(4 Points )

Theorem 135   Every bounded monotone increasing sequence converges to its least upper bound.

(5 Points )

Theorem 136   If $f:[a,b]\to \ensuremath{\mathbb{R}}$ is continuous and $f(a)<0$ and $f(b)>0$ then there is a $c\in (a,b)$ with $f(c)=0$.

(5 Points )

Hint: Let $c$ be the least upper bound of the set $\{x\vert a\leq x \leq b, f(x)<0\}$. $\spadesuit$

Total for section: 176.