Cantor's Proof: Uncountability of Real Numbers

Let $w + iv$ be a regular function of $x + iy$.

The set of real numbers, denoted by $\mathbb{R}$, is the set of all numbers that can be represented on a number line.

Proof Objective

To prove that the set $\mathbb{R}$ of real numbers is uncountable.

We will use Cantor's diagonalization argument to prove that the set of real numbers between 0 and 1 (denoted as $(0, 1)$) is uncountable. Since $(0, 1)$ is a subset of $\mathbb{R}$, if $(0, 1)$ is uncountable, then $\mathbb{R}$ must also be uncountable.

Step 1: Assume $(0, 1)$ is Countable

Assume, for the sake of contradiction, that the set $(0, 1)$ is countable. This means we can list all the real numbers in $(0, 1)$ in a sequence, say $x_1, x_2, x_3, \dots$.

Step 2: Decimal Expansion Representation

Each real number $x_i$ in the sequence can be represented as a decimal expansion:

• $x_1 = 0.a_{11}a_{12}a_{13}\dots$
• $x_2 = 0.a_{21}a_{22}a_{23}\dots$
• $x_3 = 0.a_{31}a_{32}a_{33}\dots$

where each $a_{ij}$ is a digit from 0 to 9.

Step 3: Construct a New Real Number

We will construct a real number $y = 0.b_1b_2b_3\dots$ such that $y$ is in $(0, 1)$ but is not in the list $x_1, x_2, x_3, \dots$. We define the digits $b_i$ as follows:

$$b_i = \begin{cases} 1, & \text{if } a_{ii} \neq 1 \\ 2, & \text{if } a_{ii} = 1 \end{cases}$$

This construction ensures that $b_i \neq a_{ii}$ for all $i$.

Step 4: Show Difference from Every List Member

The number $y$ differs from $x_1$ in the first decimal place (since $b_1 \neq a_{11}$), from $x_2$ in the second decimal place (since $b_2 \neq a_{22}$), and in general, from $x_i$ in the $i$-th decimal place (since $b_i \neq a_{ii}$). Therefore, $y$ is different from every number in the list $x_1, x_2, x_3, \dots$.

Step 5: Contradiction Found

Since $y$ is a real number between 0 and 1, it should be in the list if our assumption that $(0, 1)$ is countable is correct. However, we constructed $y$ such that it is not in the list. This constitutes a contradiction.

Step 6: Final Conclusion

Therefore, our initial assumption that the set of real numbers between 0 and 1 is countable must be false. Hence, the set $(0, 1)$ is uncountable. Since $(0, 1)$ is a subset of $\mathbb{R}$, the set of real numbers $\mathbb{R}$ is also uncountable.

The set of real numbers $\mathbb{R}$ is uncountable, as proven by Cantor's diagonalization argument.