# How Do Dedekind Cuts Prove Convergence of Rational Sequences?

• jgens
In summary, the conversation discusses proving that every non-decreasing, bounded sequence of rational numbers converges to some real number using Dedekind cuts. The homework equations define a real number as a set of rational numbers with certain properties. The attempt at a solution involves defining a set \alpha and verifying the four conditions for it to be a real number. The conversation ends with discussing how to prove that \alpha is the real number that the sequence converges to.
jgens
Gold Member

## 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!

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

## 1. What are Dedekind cuts and sequences?

Dedekind cuts and sequences are mathematical concepts introduced by German mathematician Richard Dedekind in the 19th century. They are used to construct the real numbers from the rational numbers and provide a rigorous foundation for the real number system.

## 2. How are Dedekind cuts and sequences related?

Dedekind cuts and sequences are closely related as they both involve partitioning the rational numbers into two sets. Dedekind cuts use a partition to define a real number, while sequences use a partition to approximate a real number.

## 3. What is the purpose of Dedekind cuts and sequences?

The purpose of Dedekind cuts and sequences is to provide a rigorous and complete construction of the real numbers. They also allow for the precise definition of concepts such as continuity, limits, and convergence in calculus and analysis.

## 4. Are Dedekind cuts and sequences used in other areas of mathematics?

Yes, Dedekind cuts and sequences have applications in various areas of mathematics, including topology, number theory, and functional analysis. They are also used in the study of irrational and transcendental numbers.

## 5. Are Dedekind cuts and sequences difficult to understand?

The concept of Dedekind cuts and sequences may be challenging for some students to grasp at first, but with practice and a solid understanding of the rational numbers, they can be easily understood. Many introductory calculus and analysis courses cover these concepts in detail.

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