Question

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). \(\mathbb{R}\) and (0,1)

Short answer

A bijection between the set of all real numbers (\(\mathbb{R}\)) and the open interval (0,1) can be defined by the function f(x) = \( ( \sin( \frac{π (x - 0.5)}{2}) + 1) / 2 \), which is both injective (one-to-one) and surjective (onto).

Step-by-step solution

Define the Bijection

We need to define a function that is both injective and surjective between the two given sets, i.e., the function must map each real number to a unique number between 0 and 1 and vice versa. A suitable function would be one that sends a number to its sine mapped to the positive domain. One of the simplest functions that meets these requirements is f(x) = \( ( \sin( \frac{π (x - 0.5)}{2}) + 1) / 2 \) .

Prove Injectivity

A function f is injective (or one-to-one) if every y in the codomain is the image of at most one x in the domain. To prove the injectivity of our function, assume that f(x1) = f(x2). Then we have \( ( \sin( \frac{π (x1 - 0.5)}{2}) + 1) / 2 = ( \sin( \frac{π (x2 - 0.5)}{2}) + 1) / 2 \). It must follow then that \( (x1 - 0.5) = (x2 - 0.5) \), and consequently, x1 = x2, which confirms the function's injectivity.

Prove Surjectivity

A function f is surjective (or onto) if every y in the codomain corresponds to at least one x in the domain. Given a number in the range (0,1), we can always find a real number that satisfies the equation \( y = ( \sin( \frac{π (x - 0.5)}{2}) + 1) / 2 \). This confirms the function's surjectivity.