Problem 23
Question
Show that the set \(Z\) of all integers is countable.
Step-by-Step Solution
Verified Answer
The set of all integers \(Z\) is countable as there exists a bijection from \(Z\) to \(N\), the set of natural numbers. The bijection can be defined as \(f(n) = 2n\) for \(n ≥ 0, f(n) = -2n + 1\) for \(n < 0\). This shows that each natural number is associated with one and only one integer, and each integer is associated with one and only one natural number.
1Step 1: Define what is needed
A set is called countable if there exists a bijection from the set to the set of natural numbers \(N\). Natural numbers \(N\) = {1, 2, 3, 4, …}. So, what has to be done here is to find a bijection from the set of integers \(Z\) to the set of natural numbers \(N\).
2Step 2: Find a bijection from \(Z\) to \(N\)
Integers \(Z\) = {…, -3, -2, -1, 0, 1, 2, 3,…}. Now, arrange all members of \(Z\) in a sequence that each integer is listed once, and no integer is skipped, such as 0, 1, -1, 2, -2, 3, -3... This arrangement not only lists out every integer, but also pairs each integer with a unique natural number (the order in which they appear in this sequence). We can then define the bijection \(f\) from \(Z\) to \(N\) as follows: \(f(n) = 2n\) for \(n ≥ 0, f(n) = -2n + 1\) for \(n < 0\).
3Step 3: Verify the bijection
It can be observed that every natural number is associated with one and only one integer through the function \(f\), and conversely, each integer is associated with one and only one natural number. This shows that \(f\) is indeed a bijection from \(Z\) to \(N\). Therefore, \(Z\) is countable according to the definition.
4Step 4: Conclusion
The set of all integers \(Z\) is countable according to the definition of countability because there exists a bijective function from \(Z\) to the set of Natural numbers \(N\). This shows that even though \(Z\) seems to have more elements than \(N\) because it includes negative numbers and zero, the two sets are 'the same size' in terms of cardinality.
Key Concepts
BijectionIntegersNatural NumbersCardinality
Bijection
Understanding the concept of a bijection is crucial in proving that a set is countable. A bijection is a type of function where every element in one set is paired with a unique element in another set, and vice versa. This means each element from the first set matches exactly one element in the second set and each element in the second set matches exactly one element in the first set.
In the context of the given exercise, a bijection was constructed between the set of integers, denoted as \( Z \), and the set of natural numbers, \( N \). The specific function defined was as follows:
In the context of the given exercise, a bijection was constructed between the set of integers, denoted as \( Z \), and the set of natural numbers, \( N \). The specific function defined was as follows:
- \( f(n) = 2n \) for non-negative integers \( n \)
- \( f(n) = -2n + 1 \) for negative integers \( n \)
Integers
The integers, often represented as \( Z \), include all whole numbers ranging from negative infinity to positive infinity. This set includes negative numbers, zero, and positive numbers: \( \{ ..., -3, -2, -1, 0, 1, 2, 3, ... \} \).
Integers are an extension of the natural numbers that also include negative values.
For example:
Integers are an extension of the natural numbers that also include negative values.
For example:
- Negative integers: \( -1, -2, -3, ... \)
- Zero: \( 0 \)
- Positive integers: \( 1, 2, 3, ... \)
Natural Numbers
The natural numbers, denoted by \( N \), are the set of positive whole numbers beginning from 1 and extending indefinitely: \( \{ 1, 2, 3, 4, ... \} \).
Natural numbers are typically the first numbers we learn to count and form the basis of elementary arithmetic.
Some key features:
Natural numbers are typically the first numbers we learn to count and form the basis of elementary arithmetic.
Some key features:
- They have no negative numbers.
- They start from 1.
- They continue without any end, referred to as 'infinite'.
Cardinality
Cardinality refers to the "size" or "number of elements" in a set. In a non-mathematical sense, you can think of it as the way to understand the count or size of a collection.
Two sets have the same cardinality if there exists a bijection between them. This concept allows for comparing the size of infinite sets, which might appear intuitive for finite sets but can reveal surprising results for infinite sets.
In this exercise where a bijection was found between the integers \( Z \) and the natural numbers \( N \), the cardinality of both sets was shown to be equal, proving that the set of integers is countable. This demonstrates that despite the inclusion of negative numbers in \( Z \), both sets, \( Z \) and \( N \), are considered to have the same size when dealing with their infinity extension.
Two sets have the same cardinality if there exists a bijection between them. This concept allows for comparing the size of infinite sets, which might appear intuitive for finite sets but can reveal surprising results for infinite sets.
In this exercise where a bijection was found between the integers \( Z \) and the natural numbers \( N \), the cardinality of both sets was shown to be equal, proving that the set of integers is countable. This demonstrates that despite the inclusion of negative numbers in \( Z \), both sets, \( Z \) and \( N \), are considered to have the same size when dealing with their infinity extension.
Other exercises in this chapter
Problem 21
Suppose that \(a, b, A, B\) are all \(>0\). Is it always true that $$ \frac{a+b}{A+B} \leq \frac{a}{A}+\frac{b}{B} $$
View solution Problem 21
Let \(a_{1}=0, a_{2}=1\), and \(a_{n+2}=\frac{n a_{n+1}+a_{n}}{n+1}\) (a) Calculate the value of \(a_{6}\) and \(a_{7}\). (b) Prove that \(\left\\{a_{n}\right\\
View solution Problem 23
Let \(\left\\{x_{n}\right\\}\) be a bounded real sequence and set \(\beta=\lim \sup _{n \rightarrow x} x_{n} .\) Show that for any \(\varepsilon>0\), \(x_{n} \l
View solution Problem 24
If \(b \leq x_{n} \leq c\) for all but a finite number of \(n\), show that \(b \leq \lim \inf _{n \rightarrow x} x_{n}\) and \(\lim \sup _{n \rightarrow x} x_{n
View solution