# Ngô Quốc Anh

## October 15, 2021

### Least upper bound axiom, completeness axiom, and Archimedean property and Cantor’s intersection theorem are equivalent

Filed under: Giải Tích 1 — Ngô Quốc Anh @ 17:03

This note concerns the equivalence between the three properties usually taken as an axiom in synthetic constructions of the real numbers. We start with the least upper bound property, call L, which is usually appeared in construction of the real numbers.

(L, least upper bound axiom): If $A$ is a non-empty subset of $\mathbf R$, and if $A$ has an upper bound, then $A$ has a least upper bound $u$, such that for every upper bound $v$ of $A$, there holds $u \leq v$.

The second property, call C, is the completeness of reals.

(C, completeness axiom): If $X$ and $Y$ are non-empty subsets of $\mathbf R$ with the property $x \leq y$ for any $x \in X$ and $y \in Y$, then there is some $c \in \mathbf R$ such that $x \leq c \leq y$ for any $x \in X$ and $y \in Y$.

The third, also last, property, call AC, is the set of two results: the Archimedean property and Cantor’s intersection theorem. These two results often appear as consequences of the construction of reals.

Archimedean property: For any real numbers $x$ and $y$ with $x>0$, there exists some natural number $n$ such that $nx > y$.

Cantor’s intersection theorem: A decreasing nested sequence of non-empty, closed intervals in $\mathbf R$ has a non-empty intersection.

Our aim is to prove that in fact the above three properties (L), (C), and (AC) are equivalent. Our strategy is to show the following direction:

(L) ⟶ (C) ⟶ (AC) ⟶ (L).

In the last part of this note, we present a simple application.

PROOF OF (L) ⟶ (C)

Let $X$ and $Y$ be non-empty subsets of $\mathbf R$ with the property $x \leq y$ for any $x \in X$ and $y \in Y$. Clearly, any element of $Y$ is an upper bound for $X$. This tells us that $\sup X$ exists in $\mathbf R$. We denote

$c=\sup X$

and prove that $c$ is the desired real number. The inequality $x \leq c$ for all $x \in X$ is obvious by means of sup. The only possibility for which the inequality $c \leq y$ for all $y\in Y$ fails to hold is that $c$ is not a lower bound for $Y$, namely there is some $y \in Y$ such that $y< c$. But this violates the definition of $c$ because there always exists some $x\in X$ in such a way that $y.

PROOF OF (C) ⟶ (AC)

First, we prove Cantor’s Intersection Theorem. Let $[a_i,b_i]$ for $i \geq 1$ be a decreasing nested sequence of non-empty, closed intervals in $\mathbf R$, namely $b_i > a_i$ and

$[a_{i+1}, b_{i+1}] \subset [a_i, b_i]$

for all $i \geq 1$. To make use of (C) we construct

$X = \{a_1, a_2, a_3, ...\}, \quad Y = \{b_1, b_2, b_3,..\}.$

Clearly, the above sets $X$ and $Y$ fulfill the hypotheses of (C). Hence there is some $c \in \mathbf R$ such that

$x \leq c \leq y$

for all $x \in X$ and $y\in Y$. In particular, $a_i \leq c \leq b_i$ for all $i \geq 1$. Thus

$\displaystyle c \in \bigcap_{i \geq 1} [a_i, b_i].$

This proves Cantor’s Intersection Theorem.

Second, we prove the Archimedean property. By way of contradiction, there exist $x>0$ and $y \in \mathbf R$ such that

$n x < y$

for all $n \in \mathbf N$. To make use of (C) we construct

$X = \{x, 2x, 3x, ...\}, \quad Y = \text{ the set of upper bounds for } X.$

Then there is some $c \in \mathbf R$ such that

$nx \leq c \leq y$

for any $n \in \mathbf N$ and any $y \in Y$. But then this implies that $c$ is also an upper bound for $X$. Hence, $c$ belongs to $Y$. This extra information tells us that $c$ is a lower bound for $Y$, hence $c$ is the least upper bound for $X$. As $x>0$, there must exist some element in $X$, say $mx$ for some $m\in \mathbf N$, in such a way that

$c-x

which is equivalent to

$c<(m+1)x.$

Thus, we have just shown that

$mx \leq c < (m+1)x.$

PROOF OF (AC) ⟶ (L)

Given the (AC) property, we now prove the L property. Let $X$ be any non-empty, bounded from above subset of $\mathbf R$. We show that $\sup X$ exists. (If $X$ is bounded from below, then $\inf X$ exists, and the proof follows similarly.) Our strategy is to construct a sequence of nested closed intervals whose intersection is $\sup X$. Initially, we choose some $a_1 \in X$ and some $b_1 \notin X$ with $a_1. Such points $a_1$ and $b_1$ exist because $X$ is non-empty and bounded from above. From the closed interval $[a_1, b_1]$ we construct $[a_2, b_2]$ via the rule

$\displaystyle [a_2, b_2] = \begin{cases}\big[c, b_1\big] & \text{ for some } X \ni c \geq \frac{a_1+b_1}2 \text { if } \frac{a_1+b_1}2 \text{ is not an upper bound for } X,\\ \Big[a_1, \frac{a_1+b_1}2 \Big] & \text{ if } \frac{a_1+b_1}2 \text{ is an upper bound for } X.\end{cases}$

Obviously, $[a_2, b_2] \subset [a_1, b_1]$ and

$\displaystyle | [a_2, b_2] | \leq \dfrac 12 |[a_2, b_2] |.$

Repeating the above process we can construct a sequence of nested closed intervals $[a_i, b_i]$ with $i \geq 1$ with the following properties

$\displaystyle | [a_{i+1}, b_{i+1}] | \leq \dfrac 12 |[a_i, b_i] |$

and $a_i \in X$ and $b_i$ is an upper bound of $X$. By Cantor’s Intersection Theorem, there exists

$\displaystyle c \in \bigcap_{i \geq 1} [a_i, b_i].$

We show $c=\sup X$. First, $c$ is an upper bound for $X$. If this were not true, then we would have some $x\in X$ such that $c. By the Archimedean property, there is some $n \in \mathbf N$ such that

$|a_n - b_n| \leq 2^{1-n} |a_1 - b_1| < |c-x|.$

As $c \in [a_n, b_n]$ we deduce that $b_n \in (c,x)$ which immediately implies that $b_n \in X$.

$\rule{2cm}{0.5pt} \, a_n \, \rule{1.5cm}{0.5pt} \, c \, \underbrace{\, \rule{1cm}{0.5pt} \, b_n \, \rule{3cm}{0.5pt}}_{\text{length } = \, |c-x| > |a_n-b_n|} \, x \, \rule{2cm}{0.5pt}$

This contradict the choice of $b_n$. Now we show that $c$ is the least upper bound for $X$. Again if this were not true, then we would have some $y\in\mathbf R$ such that

$y

Again by the Archimedean property, there is some $n \in \mathbf N$ such that

$|a_n - b_n| = 2^{1-n} |a_1 - b_1| < |c-y|$

which implies $a_n \notin X$, otherwise $a_n \leq y.

$\rule{2cm}{0.5pt} \, y \, \underbrace{\, \rule{3cm}{0.5pt} \, a_n \, \rule{1cm}{0.5pt} }_{\text{length } = \, |y-c| > |a_n-b_n|}\, c\, \rule{1.5cm}{0.5pt} \, b_n\, \rule{2cm}{0.5pt}$

This contradict the choice of $a_n$. Hence, we complete the proof.

APPLICATION

As application, we show that the equation $x^2=2$ has solution in $\mathbf R$, denoted by $\sqrt 2$, given one of the above three equivalent properties.

Assuming (L). We let

$X = \{ x \in \mathbf R : x \geq 0, \, x^2 < 2\}.$

Clearly the set $X$ is non-empty because $1 \in X$ and is bounded from above by $2$. Then the desired solution to $x^2=2$ is nothing but $\sup X$.

Assuming (C). We construct

$X = \{ x \in \mathbf R : x \geq 0, \, x^2 < 2\}, \quad Y = \{ y \in \mathbf R : y \geq 0, \, y^2 > 2\}.$

Then the desired solution to $x^2=2$ is nothing but the real number $c$, namely the one satisfying

$x \leq c \leq y$

for all $x\in X$ and all $y \in Y$.

Assuming (AC). We construct the sequence of nested closed intervals $[a_{i+1}, b_{i+1}] \subset [a_i, b_i]$ as in the above proof staring from $[a_1, b_1] = [1,2]$. Then the desired solution is $\bigcap_{i \geq 1} [a_i, b_i]$.