Supremum Archives - Stumbling Robot

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 $. \qquad \blacksquare$

Stumbling Robot

Tag: Supremum

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