Sequence Covering All Rationals: Existence?

Introduction

The question of whether a sequence $a_n$ can be constructed such that the union of its terms equals the set of rational numbers $Q$ is a fascinating one. In simpler terms, can we create a sequence that, as it goes on infinitely, covers every single rational number? This delves into the properties of rational numbers, sequences, and how we can methodically arrange an infinite set like $Q$ into a sequential order. In this article, we will explore this question and provide a comprehensive explanation.

Understanding the Basics

Before diving deep, let's clarify some key concepts.

• Rational Numbers ($Q$): These are numbers that can be expressed as a fraction $p/q$, where $p$ and $q$ are integers and $q$ is not zero. Examples include $1/2$, $-3/4$, $5$, and $0$.
• Sequence ($a_n$): A sequence is an ordered list of numbers. Each number in the sequence is called a term, and we can refer to the $n$-th term as $a_n$. For example, $a_n = n$ would give the sequence $1, 2, 3, 4, \dots$.
• Union of Sets ($\bigcup$): The union of a collection of sets is a set containing all elements from all the sets. In our context, $\bigcup_{i = 1}^{\infty}{a_i}$ represents the set of all terms in the sequence $a_n$.

So, our main question is: Can we find a sequence $a_n$ such that every rational number appears at least once in the sequence?

The Countability of Rational Numbers

The key to answering this question lies in the countability of rational numbers. A set is countable if its elements can be put into a one-to-one correspondence with the set of natural numbers ($N = {1, 2, 3, \dots}$). In simpler terms, we can list all the elements of the set in a sequence.

Georg Cantor famously proved that the set of rational numbers is countable. This might seem counterintuitive at first because between any two rational numbers, there are infinitely many other rational numbers. However, Cantor's proof provides a method to systematically list all rational numbers.

Cantor's Diagonal Argument

One way to visualize the countability of rational numbers is through Cantor's diagonal argument. Consider a table where the rows represent numerators and the columns represent denominators:

1/1  1/2  1/3  1/4  1/5 ...
2/1  2/2  2/3  2/4  2/5 ...
3/1  3/2  3/3  3/4  3/5 ...
4/1  4/2  4/3  4/4  4/5 ...
5/1  5/2  5/3  5/4  5/5 ...
...

We can traverse this table diagonally, listing the rational numbers. We skip any fractions that are not in their simplest form (i.e., we skip fractions that can be reduced). This gives us a sequence that includes every positive rational number:

$1/1, 2/1, 1/2, 1/3, 3/1, 4/1, 3/2, 2/3, 1/4, 1/5, 5/1, \dots$

To include negative rational numbers and zero, we can modify the sequence as follows:

$0, 1/1, -1/1, 2/1, -2/1, 1/2, -1/2, 1/3, -1/3, 3/1, -3/1, \dots$

This sequence now includes all rational numbers, demonstrating that $Q$ is countable.

Constructing the Sequence $a_n$

Since we know that the set of rational numbers $Q$ is countable, we can explicitly construct a sequence $a_n$ that covers all rational numbers. The process is as follows:

1. List all rational numbers: Use a method like Cantor's diagonal argument to create an ordered list of all rational numbers. Let's denote this list as $q_1, q_2, q_3, \dots$.
2. Define the sequence: Define the sequence $a_n$ such that $a_n = q_n$ for all $n$. In other words, the $n$-th term of the sequence $a_n$ is the $n$-th rational number in our ordered list.

Thus, the sequence $a_n$ is simply an enumeration of all rational numbers. By construction, the union of the terms in the sequence $a_n$ will be equal to the set of rational numbers $Q$:

$\bigcup_{i = 1}^{\infty}{a_i} = Q$

This answers our main question in the affirmative: Yes, there exists a sequence $a_n$ such that $\bigcup_{i = 1}^{\infty}{a_i} = Q$.

Addressing the Additional Information

Now, let's consider the additional information provided:

Assume that $Q = {S_1 \cup S_2 \cup \cdots \cup S_k}$ for sets $S_n$ of the form ${A \cdot \mathbb{N}+B}$ for some positive integers $A$ and $B$. Does there exist a sequence $a_n$ such that $a_n = \dots$

This part of the question introduces a specific structure for representing the set of rational numbers $Q$ as a union of arithmetic progressions. An arithmetic progression is a sequence of numbers such that the difference between consecutive terms is constant. In the form ${A \cdot \mathbb{N}+B}$, $A$ is the common difference, and $B$ is the first term (when $N = 0$, although we typically start $\mathbb{N}$ from 1, the concept remains similar).

The question essentially asks if we can express the set of rational numbers as a finite union of arithmetic progressions. The answer to this is no. Here’s why:

Proof by Contradiction

Suppose that $Q = S_1 \cup S_2 \cup \cdots \cup S_k$, where each $S_n$ is of the form ${A_n \cdot \mathbb{N}+B_n}$. Each $S_n$ is an arithmetic progression. The union of finitely many arithmetic progressions will always have a specific structure, and it cannot cover all rational numbers.

Consider the properties of arithmetic progressions. Each $S_n$ contains numbers that are evenly spaced. When you combine a finite number of these evenly spaced sets, you will still have gaps. The set of rational numbers, however, is dense, meaning between any two rational numbers, there exists another rational number. Arithmetic progressions cannot capture this density in a finite union.

To prove this more rigorously, consider the least common multiple (LCM) of the common differences $A_1, A_2, \dots, A_k$. Let $L = \text{LCM}(A_1, A_2, \dots, A_k)$. Now, consider a rational number of the form $x/L$ for some integer $x$. If we choose a very large $x$, we can find a gap in the union of arithmetic progressions. This contradicts the assumption that the union covers all rational numbers.

Therefore, it is impossible to express the set of rational numbers $Q$ as a finite union of sets in the form ${A \cdot \mathbb{N}+B}$.

Conclusion

In conclusion, while it is indeed possible to construct a sequence $a_n$ such that the union of its terms equals the set of rational numbers $Q$, this relies on the countability of $Q$ and a method to enumerate all rational numbers. However, $Q$ cannot be expressed as a finite union of arithmetic progressions ${A \cdot \mathbb{N}+B}$. The countability allows us to list the rationals in a sequence, but the structure of arithmetic progressions is too restrictive to cover all rational numbers in a finite union.

Understanding these concepts provides insight into the nature of rational numbers, sequences, and the subtle differences between countability and density. This exploration showcases the beautiful intricacies of mathematical structures and their properties.