91Ó°ÊÓ

When dealing with countably infinite sets, the Cartesian product is a fundamental concept. Consider two sets: \(A = \{a_1, a_2, a_3, \dots \}\) and \(B = \{b_1, b_2, b_3, \dots \}\). The Cartesian product, denoted by \(A \times B\), results in a new set that contains all possible ordered pairs \((a, b)\), where \(a\) belongs to set \(A\) and \(b\) belongs to set \(B\).
But how does this relate to countably infinite sets? Both \(A\) and \(B\) are countably infinite, meaning they can be mapped one-to-one to the natural numbers \(\mathbb{N}\). The challenge arises in proving that \(A \times B\) is also countably infinite. This can be achieved by finding a bijective function from \(\mathbb{N}\) to \(A \times B\). A function is bijective if it pairs every element of one set uniquely to an element of another set, capturing both injectivity and surjectivity.
Visualize the Cartesian product like a grid, where each point \((a_i, b_j)\) on the grid corresponds to the selection of one element from \(A\) and one from \(B\). This visualization helps understand how the vast number of combinations in \(A \times B\) can still be brought into correspondence with the countably infinite natural numbers.

A bijection function is pivotal when proving that a set is countably infinite. In our context, we need to establish a bijection between the natural numbers \(\mathbb{N}\) and the Cartesian product \(A \times B\). A function \(f\) is a bijection if it is both injective (no two inputs give the same output) and surjective (every possible output is matched by some input).
In our problem, we defined the bijection function as \(f(a_n, b_m) = 2^{n-1}(2m-1)\). Here's why this choice works:

• It takes the index \(n\) from set \(A\) and \(m\) from set \(B\) and merges them into a unique natural number.
• This ensures that every natural number directly corresponds to one pair \((a_i, b_j)\), and vice versa, ensuring no overlaps occur.

As a result, the function is both injective and surjective. Thus, it acts as a perfect bridge from \(\mathbb{N}\) to \(A \times B\), allowing us to map infinity in a controlled way.
Understanding bijection helps us grasp how infinite combinations can be organized and matched meticulously, affirming that \(A \times B\) remains countably infinite.

For a function to be bijective, it must satisfy two conditions: injectivity and surjectivity.**Injectivity**
A function is injective (or "one-to-one") if each element of the function's domain maps to a unique element of its range. In mathematical terms, if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\).

In our example, injectivity requires that if \(f(a_n, b_m) = f(a_p, b_q)\), then indeed \(n = p\) and \(m = q\). The function \(f(a_n, b_m) = 2^{n-1}(2m-1)\) achieves injectivity since different pairs \((a_n, b_m)\) produce different outputs.**Surjectivity**
A function is surjective (or "onto") if every element of the function's range has a corresponding element in the domain. This means that for every element in the range, there is at least one element in the domain that maps to it.

Our problem required showing that for any pair \((a_t, b_s)\) in \(A \times B\), there exists a natural number \(k\) such that \(f(k) = (a_t, b_s)\). Given the defined function, any combination can be achieved, demonstrating that every pair \((a_t, b_s)\) is represented.Checking both injectivity and surjectivity certifies the bijection is valid, thereby proving that the Cartesian product \(A \times B\) is countably infinite.