Prove that the rationals satisfy the Archimedean property, but not the least-upper-bound axiom.

by RoRi

July 2, 2015

Recall, the Archimedean property states that if $x > 0$ and $y \in \mathbb{R}$ is arbitrary, then there exists an integer $n > 0$ such that $nx > y$ .
Further, recall that the least upper bound axiom states that every nonempty set $S$ of real numbers which is bounded above has a supremum.
Now, prove that $\mathbb{Q}$ satisfies the Archimedean property, but not the least-upper-bound axiom.

First, we prove that $\mathbb{Q}$ satisfies the Archimedean property.
Proof. This is immediate since if $x,y$ are arbitrary elements in $\mathbb{Q}$ with $x>0$ , then they are also in $\mathbb{R}$ since $\mathbb{Q} \subseteq \mathbb{R}$ . Thus, by the Archimedean property in $\mathbb{R}$ we know there is an $n \in \mathbb{Z}_{>0}$ such that $nx > y$ . Hence, the Archimedean property is satisfied in $\mathbb{Q}. \qquad \blacksquare$

Next, we prove that $\mathbb{Q}$ does not satisfy the least-upper-bound axiom.
Proof. Let

$S = \{ r \in \mathbb{Q} \mid r^2 \leq 2 \}.$

Then $S$ is non-empty since $1 \in S$ . Further, $S$ is bounded above by 4 since $r^2 \leq 2 \ \implies \ r^2 \leq 16 \ \implies \ r \leq 4$ . (Of course, there are better upper bounds available, but we just need any upper bound.)
Now, we must show that $S$ has no supremum in $\mathbb{Q}$ to show that the least-upper-bound property fails in $\mathbb{Q}$ (since this will mean $S$ is a nonempty set which is bounded above, but fails to have a least upper bound in $\mathbb{Q}$ ).
Suppose otherwise, say $b = \sup S$ with $b \in \mathbb{Q}$ . We know $b^2 \neq 2$ (I.3.12, Exercise #11). Thus, by the trichotomy law we must have either $b^2 < 2$ or $b^2 > 2$ .
Case 1: If $b^2 < 2$ , then there exists $r \in \mathbb{Q}$ such that $b < r < \sqrt{2}$ (since the rationals are dense in the reals, see I.3.12, Exercise #6). But then, $r^2 < 2$ (since $r < \sqrt{2} \ \implies \ r \cdot r < \sqrt{2} \cdot \sqrt{2} \ \implies \ r^2 < 2$ ) and $r \in \mathbb{Q}$ implies $r \in S$ with $r > b$ , contradicting that $b$ is an upper bound for $S$ . Hence, we cannot have $b^2 < 2$ .
Case 2: If $b^2 > 2$ , then there exists $r \in \mathbb{Q}$ such that $b > r > \sqrt{2}$ . But then, $r^2 > 2$ , so if $s \in S$ we have $s < r$ ; hence, $r$ is an upper bound for $S$ which is less than $b$ . This contradicts that $b$ is the least upper bound of $S$ . Hence, we also cannot have $b^2 > 2$ .
Thus, there can be no such $b^2 \in \mathbb{Q}$ (since by the trichotomy exactly one of $b^2> 2, \ b^2 < 2, \ b^2 = 2$ must hold, but we have shown these all lead to contradictions).
Hence, $\mathbb{Q}$ does not have the least-upper-bound property $. \qquad \blacksquare$

8 comments

August 18, 2019 at 10:38 am

Hannes says:

For showing that the set of rationals does not possess the least-upper-bound axiom it would be quicker assume it does, and then apply the result of Exercise 6 to the interval of rational numbers between 0 and sqrt 2. Exercise 6 guarantees that we can find another number closer to sqrt 2 than the supposed supremum.

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April 14, 2019 at 11:34 am

Anonymous says:

Aren’t you assuming that the square root of 2 exists in this proof. That is proved only in the next section.

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- August 11, 2019 at 5:14 pm
  
  richardsull says:
  
  Here’s a proof that does not assume that the square root of 2 exists (I hope the LaTeX works okay):
  
  Let $s = \sup S$ with $s \in \mathbb{Q}$ . Choose $n \in \mathbb{Z}^+$ so that $0 < 1/n s^2 - \frac{2s}{n}\\ &> s^2 - 2s\cdot\frac{s^2 - 2}{2s}\\ &= 2 \end{align*} So$ s – 1/n $is also an upper bound of$ S $(since$ (s – 1/n)^2 > 2 $), but$ s – 1/n < s $, contradicting the claim that$ s $is the least upper bound of$ S$.
  
  Reply
  - August 11, 2019 at 5:24 pm
    
    richardsull says:
    
    Okay, that didn’t work. I tried to use an align* environment and it didn’t appear.
    
    Let’s try this again.
    
    Let $s = \sup S$ with $s \in \mathbb{Q}$ . Choose $n \in \mathbb{Z}^+$ so that $0 < 1/n s^2 - (2s)/n > s^2 - (2s)(1/n) = s^2 - (2s)((s^2 - 2)/(2s)) = 2$
    
    So $s - 1/n$ is also an upper bound of $S$ (because $(s - 1/n)^2 > 2$ , but $s - 1/n < s$ , contradicting the claim that $s$ is the least upper bound of $S$ .
  - August 11, 2019 at 5:28 pm
    
    richardsull says:
    
    Okay, I give up with the formatting. Just choose n so that 1/n is less than (s^2 – 2)/(2s). Show that (s – 1/n)^2 is greater than 2. Thus s – 1/n is an upper bound of S, but s – 1/n is less than sup S, a contradiction.
December 13, 2017 at 11:53 pm

Fiestita says:

Hi Rori, i don’t get why you say (Case 2) : b^2 > 2 —> exists r (rational) such that b > r > sqrt(2)
We know that b= -2 satisfies b^2 > 2 but don’t satisfies that b > r > sqrt(2)

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February 11, 2016 at 12:25 pm

martin says:

For the second proof, could we say quickly that the set P (positve integers) is in Q so it’s unbounded above that’s why it doesn’t satisfy axiom 10?

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- February 11, 2016 at 12:37 pm
  
  RoRi says:
  
  No, we can’t do that. The least-upper-bound axiom is not the same as being unbounded above. For instance, $\mathbb{R}$ is unbounded above but possesses the least upper bound axiom.
  
  The critical thing is that if $X$ is the set you want to show satisfies the least upper bound axiom, you need to show that for every nonempty subset $S \subseteq X$ which is bounded above has a least-upper-bound in $X$ . So, the second proof is showing that $\mathbb{Q}$ fails this since we have a non-empty set that is bounded above, but it’s least upper bound is not actually in $\mathbb{Q}$ (since its least upper bound is $\sqrt{2}$ which is not a rational number). Does that make sense?
  
  Reply

Stumbling Robot

Prove that the rationals satisfy the Archimedean property, but not the least-upper-bound axiom.

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