Question

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

Short answer

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

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.

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.

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.

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.