Axiom of completeness counterexample for $A\subseteq \mathbb{Q}$

Question

In Abbot's "Understanding Analysis," the Axiom of Completeness is stated as "every nonempty set of real numbers that is bounded above has a least upper bound." He then gives $S = \{r \in \mathbb{Q}| r^2 < 2\}$ as an example of a set of rational numbers such that the Axiom of Completeness isn't valid. But since there exists a rational number in $S$ arbitrarily close to root 2, isn't root 2 a supremum of $S$? Since a supremum of a set doesn't have to be part of the set, why does the supremum of $S$ have to be a rational number for the Axiom to be valid?

Answer 1

The Axiom of Completeness is an axiom that is valid for the real numbers. This is distinct from the definition of completeness, which may or may not apply to a set:

An ordered set $S$ is complete if every non-empty subset of $S$ that is bounded above has a least upper bound in $S$.

The Axiom of Completeness then just says that $\Bbb R$ is complete. And using the above definition, it is clear that $\Bbb Q$ is not complete.

Answer 2

We need to keep in mind where we are taking a supremum. In this case we have a subset $S\subseteq \mathbb{Q}$ and we want to find a supremum in $\mathbb{Q}$. Because the "Axiom of Completeness" for $\mathbb{Q}$ would say:

Every nonempty set of the rational numbers that is bounded above has a least upper bound in $\mathbb{Q}$.