3
$\begingroup$

This exercise is from Tom Apostol's Calculus Vol.1:

I 3.12 Exercises (page 29)

12) The Archimedian property of the real-number system was deduced as a consequence of the least-upper-bound axiom. Prove that the set of rational numbers satisfies the Archimedian property but not the least-upper-bound property. This shows that the Archimedian property does not imply the least-upper-bound axiom.


I don't get what I have to do, and what it means -- to satisfy the least-upper-bound-property. I know it is an axiom, I didn't know it is a property. My closest guess is that if I have two rational numbers $x < y$, I can find an integer $n$ such that $nx > y$. And I don't need to refer to that axiom, because I can find this $n$ by playing with the integers inside the rational numbers.


Previously from the book:

The Archimedian property (page 26): If $x > 0$ and if $y$ is an arbitrary real number there exists a positive integer $n$ such that $nx > y$.

The least-upper-bound axiom (page 25): Every nonempty set $S$ of real numbers which is bounded above has a supremum; that is, there is a real number $B$ such that $B = sup S$.


$\endgroup$
5
$\begingroup$

One way to construct the field of real numbers is axiomatically. In this approach, you have a collection of axioms you want to be true, one of them is that the real numbers satisfy the least upper bound property.

To show that $\mathbb{Q}$ satisfies the Archimedian axiom, you need to show that if $x \in \mathbb{Q}$, $x > 0$, then for any $y \in \mathbb{Q}$, there is $n \in \mathbb{N}$ such that $nx > y$.

To show that $\mathbb{Q}$ does not satisfy the least upper bound property, you need to find a subset of $\mathbb{Q}$ which is bounded above, but has no least upper bound in $\mathbb{Q}$.

$\endgroup$
  • $\begingroup$ Your last paragraph is new to me. Because so far in the book I have not read what it means for a set to satisfy some property. $\endgroup$ – Graduate Dec 29 '13 at 15:36
  • $\begingroup$ It seems that's because you've only seen the least upper bound property expressed for $\mathbb{R}$, but given any partially ordered set, you can ask whether or not it has the least upper bound property. In particular, you can ask the question of $\mathbb{Q}$. $\endgroup$ – Michael Albanese Dec 29 '13 at 15:39
  • $\begingroup$ @Graduate This is actually a reasonably fair point, since it isn't a clear definition. Essentially you are expected to generalize the Archimedean and l.u.b. properties to other sets $F$ by replacing all instances of "real number" to "element of $F$" and all instances of $\mathbb R$ with $F$. $\endgroup$ – Erick Wong Dec 29 '13 at 15:41
  • $\begingroup$ For me it's right to express the supremum for any set by any real number. For example a subset of Q can be bounded by an irrational number. There is nothing wrong with it since I can compare them. That's my understanding. $\endgroup$ – Graduate Dec 29 '13 at 15:42
  • 1
    $\begingroup$ @Graduate Yes, but notice that in the substituted version it would require $B$ to be rational (an element of $\mathbb Q$) rather than real. $\endgroup$ – Erick Wong Dec 29 '13 at 15:44
4
$\begingroup$

For example, the set $\{ x\in\mathbb{Q} : x^2 <2 \} $ does not have a least upper bound in $\mathbb{Q} .$

$\endgroup$
  • 2
    $\begingroup$ My understanding is that this set has the least upper bound, but it is an irrational number (a square root of $2$). An irrational number is also a real number. $\endgroup$ – Graduate Dec 29 '13 at 15:23
  • 2
    $\begingroup$ But You have to prove that the set of rational numbers satisfies the Archimedian property but not the least-upper-bound property. $\endgroup$ – user110661 Dec 29 '13 at 15:31
2
$\begingroup$

Here I prove the claim made in a previous answer, that the set $$ \mathcal{S} = \left\{x\in\mathbb{Q} : x^2<2 \right\} $$ does not have a least upper bound in $\mathbb{Q}$.

Since $\mathbb{Q}\subseteq\mathbb{R}$, $\mathcal{S}$ is a subset of $\mathbb{R}$. Furthermore, $\mathcal{S}$ is non-empty (e.g. $1^2 < 2$) and is bounded above by definition, so by the completeness axiom it has a least upper bound in $\mathbb{R}$ - say $\sup\mathcal{S}=s$. We now show that $s = \sqrt{2}$.

There are only 3 possibilities: $s>\sqrt{2}$, $s<\sqrt{2}$, or $s=\sqrt{2}$.

The first possibility can be eliminated from the definition of $\mathcal{S}$. Clearly $\sqrt{2}$ is an upper bound of $\mathcal{S}$, since $x^2 < 2$ implies $x < \sqrt{2}$. Therefore any number larger than $\sqrt{2}$ cannot be the least upper bound.

To eliminate the second possibility, assume for contradiction that $s<\sqrt{2}$. Since the rationals are dense in $\mathbb{R}$, there is a rational $q$ such that $s<q<\sqrt{2}$. But this implies $q^2 < 2$ and so $q\in\mathcal{S}$, which means that $s<q$ cannot be an upper bound of $\mathcal{S}$.

We are left with $s=\sqrt{2}$. Since $\sqrt{2}\notin\mathbb{Q}$, $\mathcal{S}$ does not have a least upper bound in $\mathbb{Q}$ and so we have found a counterexample which shows that $\mathbb{Q}$ is not complete.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.