# Homework Help: Dedekind Cuts and Sequences

1. Dec 25, 2009

### jgens

1. The problem statement, all variables and given/known data

Prove that every non-decreasing, bounded sequence of rational numbers converges to some real number using Dedekind cuts.

2. Relevant equations

A real number is a set $\alpha$, of rational numbers, with the following four properties:
• If $x \in \alpha$ and $y$ is a rational number with $y < x$, then $y \in \alpha$.
• $\alpha \neq \emptyset$
• $\alpha \neq \mathbb{Q}$
• There is no greatest element in $\alpha$; in other words, if $x \in \alpha$, then there is some $y \in \alpha$ with $y > x$

3. The attempt at a solution

This seems to be a straight forward proof, but I can't seem to complete it satisfactorily; any advice and/or suggestions are appreciated. Well, here goes nothing . . .

Let $\{a_n\}$ be a non-decreasing, bounded sequence of rational numbers and define the set $\alpha \subset \mathbb{Q}$ such that $\alpha = \{x \in \mathbb{Q}: x < a_n \;\mathrm{where}\; n \in \mathbb{N}\}$. Now, to prove that $\alpha$ is a real number, we need only verify that our four conditions hold:
• Suppose $x\in\alpha$. Then $x < a_n$ for some $n \in \mathbb{N}$. Now, let $y < x$, then $y < a_n$ and consequently $y\in\alpha$
• Clearly $\alpha \neq \emptyset$
• Since $\{a_n\}$ is bounded above, this means that there must be some rational number $x > a_n$ for all $n\in\mathbb{N}$. Thus $x \notin \alpha$ and consequently $\alpha \neq \mathbb{Q}$
• Suppose $x\in\alpha$. Then, for some $n\in\mathbb{N}$ we have that $x < a_n$. Since $x,a_n\in\mathbb{Q}$ we know that $\varepsilon = \frac{a_n - x}{2} \in \mathbb{Q}$. Consequently, $x < x + \varepsilon < a_n$. Therefore, $\alpha$ contains no greatest element.
Now, I just need to prove that $\alpha$ is the real number that this sequence converges to. Intuitively, it seems like it should be, but how might I articulate an appropriate argument? Thanks!

Last edited: Dec 26, 2009
2. Dec 26, 2009

### HallsofIvy

First show that every n, $\alpha_n< \alpha$. Then show that for any $epsilon$, there exist n such that $\alpha- \alpha_n< \epsilon$.