Generalization: $x=\text{sup}\{q\in \mathbb{Q}:q<x\}$

Question

How do I prove that $x=\text{sup}\{q\in \mathbb{Q}:q<x\}$? Provided that $x\in\mathbb{R}$...

Answer 1

You need to show that

1. $\text{sup}\{q\in \mathbb{Q}:q<x\} \le x$

and

1. For any $y < x$ there is a $r \in \{q\in \mathbb{Q}:q<x\}$ such that $r > y$.

Answer 2

HINT: Let $\alpha=\sup\{q\in\Bbb Q:q<x\}$; you want to show that $\alpha=x$. One way to do this is to rule out the possibility that $\alpha<x$ and the possibility that $\alpha>x$.

• Suppose that $\alpha<x$; then there a rational number in the interval $(\alpha,x)$. Why is that impossible?

• Suppose that $\alpha>x$; can it then actually be true that $\alpha$ is the least upper bound of $\{q\in\Bbb Q:q<x\}$?