$\newcommand{\Rr}{\mathbb{R}}$ $\newcommand{\Zz}{\mathbb{Z}}$ $\newcommand{\Nn}{\mathbb{N}}$ $\newcommand{\Qq}{\mathbb{Q}}$ $\newcommand{\ve}{\varepsilon}$
Consequences of the Least Upper Bound Property¶
Read Section 1.2 and think about the following questions:
- How do sup and inf behave with respect to various field operations?
- What is the archimedean property?
- Can you find an example of an ordered field that does not possess the archimedean property?
Proof. Suppose not, then $u:=\sup \Nn$ exists. Since $u-1 <u$, so it cannot be an upper bound of $\Nn$. Thus, there exists $n \in \Nn$ with $u-1 < n$. But then $u < n+1 \in \Nn$, contradicting $u$ is an upper bound of $\Nn$.
C2. (The Archimedean Property of $\Rr$)
For any $x \in \Rr$ and $\ve > 0$, $N\ve > x$ for some $N \in \Nn$.
Proof. By C1, $x/\ve < N$ for some $N \in \Nn$. Equivalently, since $\ve > 0$, $x < N\ve$.
Take $x=1$ in this corollary and we get
Proof. By C1 there exists $N \in \Nn$ so that $N > |x|$, i.e. $-N < x < N$. By going through the integers in
$$S:=\{-N, -N+1, \ldots, 0, \ldots, N\},$$
from the smallest to the largest, eventually one finds the largest $n$ in $S$ so that $n \le x$.
Because $N$ is bigger than $x$ so $n$ must be strictly less than $N$. Therefore, $n+1$ is still in $S$.
But that means, because of our choice of $n$, $n+1$ must exceed $x$. Thus, $ n \le x < n+1.$
By C3, there exists $n \in \Zz$ such that $n \le x < n+1$. If $y \le n+1$, then
$$ n \le x < y \le n+1$$
and we would have $y-x < (n+1)-n = 1$ contradicting the assumption. Thus, $m:=n+1$ is an integer such that $x < m < y$.
Proof. Pick any two real numbers. Call the bigger one $y$ and the smaller one $x$.
By the archimedean property, there exists $N \in \Nn$ such that $N(y-x) > 1$.
Hence by C4, there exists and integer $m$ such that $Nx < m < Ny$ and so $m/N$ is a rational number between $x$ and $y$.
Existence of Square Root (optional)¶
Proposition. Every positive number has a unique positive square root.
Idea. Since $f(x) =x^2$ is strictly increasing on the set of positive reals, a positive real number can have at most one positive square root. Suppose $a > 0$, argue that the set $S = \{x \in \Rr \colon x >0, x^2 \le a\}$ is nonempty and bounded above, so $\sup S$ exists. Finally, argue that $(\sup S)^2 =a$. (Sketch of graph of $f(x) =x^2$).
Proof. Note that for positive $x,y \in \Rr$ if $x < y$ then $x^2 < y^2$. Thus, any real number can have at most one positive square root.
It remains to show that every positive real $a$ has a root square root. For this purpose consider the set $$ S = \{x \in \Rr \colon x > 0, x^2 \le a\}. $$
Note that if $a \gt 1$, then $a \in S$ and if $0 \lt a \lt 1$, then $a \in S$. That means $\min\{1,a\} \in S$. So, $S \neq \emptyset$.
$S$ is also bounded above because if $x > M:=\max\{a,1\}$ then $x^2 = x\cdot x > a \cdot 1 = a$.
Therefore, $M$ is an upper bound of $S$.
So by the least upper bound property, $u:=\sup S \in \Rr$ exists. We will show that $u > 0$ and $u^2 = a$.
First, $u$ is an upper bound of a non-empty set of positive reals, so $u$ must be positive.
Second, for any $0 < \ve < u$, since $u-\ve \lt u$, so $u-\ve$ is not an upper bound of $S$. Therefore, there exists $x \in S$ such that $u-\ve < x \le u$. So, $(u-\ve)^2 \lt x^2 \le a$. Note also that $(u+\ve)^2 > a$, otherwise $u+\ve$ would be in $S$ contradicting $u=\sup S$.
Putting these together we have $$ (u-\ve)^2 < a, u^2 < (u+\ve)^2. $$
Thus, $$ 0 \le |u^2 -a | < (u+\ve)^2 - (u-\ve)^2 = 4u\ve. \tag{*} $$
Since $0< \ve$ can be arbitarily small and $4u$ is a fix number, $4u\ve > 0$ can be arbitarily small.
It now follows from ($*$) that $|u^2-a|$ must be zero, i.e. $u^2 = a$.
Proposition. The ordered field of rational numbers $\Qq$ does not have the least upper bound property.
Ans. The set $A :=\{x \in \Qq \colon x > 0, x^2 \le 2\}$ is non-empty and bounded above in $\Qq$ because $1 \in A$ and $2 \in \Qq$ is an upper bounded of $A$.
Suppose the set of rational upper bounds of $A$ has a least element, say $u$.
Since $\Qq$ is dense in $\Rr$, it can be shown that among all the real upper bounds of $A$, $\sqrt{2}$ is the least. (See HW 03, Q.6). Thus, $u \ge \sqrt{2}$. If $u$ were strictly greater than $\sqrt{2}$, then there is some $r \in \Qq$ such that $\sqrt{2} < r < u$. Hence, $r$ cannot be a rational upper bound of $A$. That means there exists some $s \in A$, such that $r < s$ but that $2 < r^2 < s^2$, contradiction.
Hence $A$ is a non-empty subset of $\Qq$ that is bounded above, but the set of rational upper bounds of $A$ has no least element. Therefore, we have shown that the ordered field of rational numbers does not have the least upper bound property.
Theorem. Every ordered field with the least-upper-bound property is isomorphic to $\Rr$.
Here is a proof.