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Chapter 14: Problem 9

Prove or disprove: The set \(\\{0,1\\} \times \mathbb{N}\) is countably infinite.

Short Answer

Expert verified
The set \(\{0,1\} \times \mathbb{N}\) is countably infinite, and this has been proved by establishing a one-to-one correspondence (bijection) with the set of natural numbers through a bijective function.

Step by step solution

01

Defining the set

Start by considering the set \(\{0,1\} \times \mathbb{N}\). This is a cartesian product of the set \(\{0, 1\}\) and the set of natural numbers \(\mathbb{N}\). This means that this set consists of ordered pairs where the first element is either 0 or 1 and the second element is a natural number.
02

Determining a bijection

The aim is to establish a one-to-one correspondence (bijection) between the above set and the set of natural numbers. This can be done by arranging the ordered pairs in the following way: (0,1), (1,1), (0,2), (1,2), (0,3), (1,3), (0,4), (1,4), etc. Now, this sequence can be mapped to the series of natural numbers: 1, 2, 3, 4, 5, 6, 7, 8, etc.
03

Building the function

We need a bijective function \(f: \mathbb{N} \rightarrow \{0,1\} \times \mathbb{N} \). We can consider a function like this: f(n) = (0, n) for all odd numbers and f(n) = (1, n/2) for all even numbers.
04

Proving the function as a bijection

Given any ordered pair, there is exactly one natural number that maps onto it through the function f, and given any natural number, there is exactly one unique ordered pair it maps onto. Therefore, the function f serves as a bijection, and thus, the set \(\{0,1\} \times \mathbb{N}\) is countably infinite.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cartesian Product
The cartesian product is a fundamental concept in set theory. It is the set of all possible ordered pairs formed by taking one element from each of two sets. If you have two sets, say \(A\) and \(B\), the cartesian product \(A \times B\) is created by pairing every element in \(A\) with every element in \(B\).

In the exercise, \(\{0,1\}\) and \(\mathbb{N}\) are combined to create \(\{0,1\} \times \mathbb{N}\). This results in pairs like (0,1), (1,1), (0,2), and so on. Here:
  • \(\{0,1\}\) provides the first element of each pair which can be either 0 or 1.
  • \(\mathbb{N}\), the set of natural numbers, provides the second element.
This strategy allows us to systematically pair elements from \(\{0,1\}\) and \(\mathbb{N}\), forming a new set of ordered pairs.
Countable Sets
Countable sets are those that can be matched one-to-one with the natural numbers \(\mathbb{N}\). This means that the elements of the set can be arranged in a sequence where each element appears exactly once, just like counting numbers.

In the context of \(\{0,1\} \times \mathbb{N}\), finding a way to pair each ordered pair with a unique natural number helps in proving that this set is countably infinite. This involves listing the pairs in a sequence:
  • (0,1) corresponds to 1
  • (1,1) corresponds to 2
  • (0,2) corresponds to 3
  • (1,2) corresponds to 4
This kind of systematic pairing demonstrates that each ordered pair has a unique corresponding natural number, confirming its countability.
Bijection
A bijection is a special type of function that ensures each element in one set pairs with exactly one element in another set, with nothing left out in either.

For \(\{0,1\} \times \mathbb{N}\), the goal was to define a function that maps exactly with natural numbers. The function \(f\) works as follows:
  • For any odd \(n\), \(f(n) = (0, \frac{n+1}{2})\) pairs an odd number with a pair starting with 0.
  • For any even \(n\), \(f(n) = (1, \frac{n}{2})\) maps it to a pair starting with 1.
This mapping accurately reflects a one-to-one correspondence, where each natural number matches uniquely to a pair. Such a function ensures that \(\{0,1\} \times \mathbb{N}\) is not only a proper set but also countably infinite, as every single element is multiplied by a unique natural number.

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Most popular questions from this chapter

Prove that if \(A\) and \(B\) are finite sets with \(|A|=|B|\), then any surjection \(f: A \rightarrow B\) is also an injection. Show this is not necessarily true if \(A\) and \(B\) are not finite.

Prove or disprove: The set \(\mathbb{Q}^{100}\) is countably infinite.

Prove or disprove: The set \(\left\\{\left(a_{1}, a_{2}, a_{3}, \ldots\right): a_{i} \in \mathbb{Z}\right\\}\) of infinite sequences of integers is countably infinite.

Show that the two given sets have equal cardinality by describing a bijection from one to the other. Describe your bijection with a formula (not as a table). The set of even integers and the set of odd integers

Prove or disprove: If there is an injection \(f: A \rightarrow B\) and a surjection \(g: A \rightarrow B\), then there is a bijection \(h: A \rightarrow B\).
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