Cantor's Diagonalization Method

Cantor published that $R$ is uncountable back in 1874. The proof for the linked theorem is pretty much what he proposed; however, he also proposed this theorem:

(0, 1)

is uncountable

An alternative way to prove this is to show that

f : (0, 1) \to R

where

f (x) = \frac{x - \frac{1}{2}}{x (1 - x)}

is a valid bijection. This would imply

(0, 1) \sim R

so then since

R

is uncountable then

(0, 1)

is as well.

Proof
Assume that there is some function $f : N \to (0, 1)$ that is bijective. For each $m \in N$ then there is a real number $f (m)$ it maps to. We can represent each number with its decimal expansion:

f (m) = 0. a_{m_{1}} a_{m_{2}} a_{m_{3}} \dots

So for each $m, n \in N$ then $a_{m n}$ is the digit from the set ${0, . . ., 9}$ that represents the $n$ -th digit in the decimal expansion of $f (m)$ . The injectivity of $N \to (0, 1)$ via $f$ can be summarized with:

Pasted image 20241006155454.png

The key assumption here is that every real number in $(0, 1)$ is assumed to be somewhere on this list.

Now, define a real number $x \in (0, 1)$ with the decimal expansion $x = 0. b_{1} b_{2} b_{3} \dots$ using the rule:

b_{n} = {\begin{cases} 2 & a_{n n} \neq 2 \\ 3 & a_{n n} = 2 \end{cases}

Notice that clearly $x \neq f (1)$ since $b_{1} \neq a_{11}$ . Similarly $x \neq f (2)$ since $b_{2} \neq a_{22}$ . Repeat this process, so then $x \neq f (n)$ for any $n \in N$ so then $x$ must not be in the list! This is a contradiction since we had assumed that every real number in $(0, 1)$ was in the list but clearly $x$ isn't.
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Rebuttals to the Diagonalization Argument

The following are rebuttals to the above diagonalization argument, and ways to fix that. See Cantor's Problems#1.6.3 for more info on this.