91Ó°ÊÓ

A bijection, or bijective function, is a function that is both injective (one-to-one) and surjective (onto). In other words, every element in the function's codomain is mapped by exactly one element in its domain. This means that a bijection has no repeated mappings and covers every element in the codomain.
To show that the function \( f(x) = \frac{2x - 1}{2x(1 - x)} \) is bijective, we need to demonstrate that it is both injective and surjective.

• Injective: we assume \( f(x_1) = f(x_2) \) and show that it must follow that \( x_1 = x_2 \).
• Surjective: we show that for every \( y \in \mathbf{R} \) (real numbers), there exists a corresponding \( x \in (0, 1) \).

Proving these properties confirms that the function is bijective, illustrating the equivalence of the sets \((0,1)\) and \( \mathbf{R} \).

The Schröder-Bernstein theorem is a theorem in set theory that provides a way to prove two sets have the same cardinality. The theorem states: If there exist injective functions (one-to-one) from set A to set B and from set B to set A, then there exists a bijective function between A and B, showing that A and B have the same cardinality.
To apply the Schröder-Bernstein theorem to our problem:

• We already have the injection \( f: (0, 1) \rightarrow \mathbb{R} \) given by \( f(x) = \frac{2x - 1}{2x(1 - x)} \).
• Consider the injection \( g: \mathbb{R} \rightarrow (0, 1) \) given by \( g(y) = \frac{1}{2} + \frac{1}{\pi} \arctan y \).

Because both injections exist, the Schröder-Bernstein theorem tells us that there is a bijection between \( (0, 1) \) and \( \mathbb{R} \), proving that these sets have the same cardinality.

An injective function, also known as a one-to-one function, is a function where each element in the domain maps to a unique element in the codomain. No two distinct elements in the domain can map to the same element in the codomain.
To prove that \( f(x) = \frac{2x - 1}{2x(1 - x)} \) is injective, assume \( f(x_1) = f(x_2) \). After equating and simplifying, we should be able to show that \( x_1 = x_2 \). This shows that different inputs lead to different outputs, ensuring the function is one-to-one.
In our specific problem, proving the injectivity of \( f(x) \) is necessary for establishing it as part of a bijection, thereby supporting the conclusion that \( (0, 1) \) and \( \mathbf{R} \) have the same cardinality.

A surjective function, also known as an onto function, is one where every element in the codomain is mapped by at least one element in the domain. In other words, the function's image is its entire codomain.
To show \( f \) is surjective, we need to show that for every \( y \in \mathbb{R} \), there exists an \( x \in (0, 1) \) such that \( f(x) = y \).
Starting with the equation \( y = \frac{2x - 1}{2x(1 - x)} \), solve for \( x \). This involves manipulating the equation into a quadratic form and solving for \( x \), ensuring the solution falls within the interval \((0,1)\).
Successfully proving surjectivity means every real number is the image of some number in \((0,1)\), completing the requirements for showing our function \( f(x) \) is bijective.