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 $B = sup S$.


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 $\mathbb{Q}$.

  • $\begingroup$ 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
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    $\begingroup$ @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

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

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    $\begingroup$ 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
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    $\begingroup$ 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

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.


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