Question

Show that \(\left|\mathbb{R}^{2}\right|=|\mathbb{R}| .\) Suggestion: Begin by showing \(|(0,1) \times(0,1)|=|(0,1)|\).

Short answer

Yes, \(\left|\mathbb{R}^{2}\right|=|\mathbb{R}|\). This can be shown by establishing a bijection between the intervals (0,1) and (0,1) × (0,1), and extending that to all real numbers via the tangent function.

Step-by-step solution

Understanding the Problem

We are asked to show that the cardinalities of \(\mathbb{R}^{2}\) and \(\mathbb{R}\) are equal. One possible approach, as given in the problem, is to establish a bijection (i.e. one-to-one correspondence) between the intervals (0,1) and (0,1) × (0,1), and to extend that bijection to all real numbers.

Setting up the Bijection

Consider the decimal representations of real numbers in the open interval (0,1). Every number X in (0,1) can be written uniquely as X = 0.a1a2a3a4..., where each ai is a decimal digit (i.e., an integer from 0 to 9) and the dots indicate that the sequence of digits extends indefinitely. We map each X onto a pair of numbers (Y,Z) in the square (0,1) × (0,1) by assigning the odd-numbered digits of X to Y and the even-numbered digits to Z. Specifically, Y = 0.a1a3a5..., Z = 0.a2a4a6... . This map is a bijection between (0,1) and (0,1) × (0,1)

Extending the Bijection

We can extend our bijection to all real numbers using the tangent function. The tangent function is a bijection between the open interval (-π/2, π/2) and all real numbers. Thus, we can say g(x) = tan(x) is a bijection from (-π/2, π/2) onto \(\mathbb{R}\). Similarly, we can establish a bijection f(x) = arc tan(x) from \(\mathbb{R}\) onto (-π/2, π/2). We end up with a bijection between \(\mathbb{R}^{2}\) and \(\mathbb{R}\) by composing these functions. The final bijection can be written as h(x,y) = g [h1 [f(x), f(y)]], where h1 is the bijection from Step 2.