How Do Dedekind Cuts Prove Convergence of Rational Sequences?

[jgens]

Homework Statement

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

Homework 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

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!

 

[HallsofIvy]

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