Question

Discuss order-completeness of the set of rational numbers

Answer 1

The set of rational numbers is not order complete. Let us consider, for example, a set :

$A:=\{x\in\mathbb{Q} : x^2<2\}$

We can clearly see that $A$ is bounded in $\mathbb{Q}$ , for example we have $\forall x\in A, x<2$ , as we know that for all $y\geq2, y^2\geq2^2=4>2$. But the set $A$ does not have a precise upper bound (supremum), as if we had $x_0 = \sup A$ , then (we could prove that) $x_0^2=2$. But we know that there is no such rational number for arithmetical reasons. Therefore $\mathbb{Q}$ is not order complete.

In fact the set of real numbers $\mathbb{R}$ is obtained from $\mathbb{Q}$ by order-completing it.