I recently learnt Cantor's argument that proves $(0, 1)$ is uncountable. Basically, the argument goes: first assume the contrary, therefore there is a bijection $$f: \mathbb{N} \rightarrow (0, 1)$$ that $$f(1) = 0.a_{11}a_{12}a_{23}…$$ $$f(2) = 0.a_{21}a_{22}…$$ and in general $$f(n) = 0.a_{n1}a_{n2}…$$ where $a_{nm} \in \mathbb{N}, 0 \le a_{nm} \le 9$. Then we construct a number $$s = 0.b_{11}b_{12}…$$ where each $b_{1n} \ne a_{nn}$. The argument then goes that since the new number $s$ differs with any $f(n)$ with at least one digit, it cannot equal to any $f(n)$, therefore this function $f$ is not surjective, leads to a contradiction.
However, there are cases that even if two numbers $a$ and $b$ in $(0, 1)$, when written in decimal form, differs in some digits, they can still be equal to each other. For example: $$0.1000000… = 0.0999999…$$ How, then, can this argument still hold then?
Best Answer
To be a useful bijection, $f$ needs to be $f: \mathbb{N} \rightarrow (0,1)$. Note that, since $f$ is assumed to be bijective, each value of $f(n)$ is unique, regardless of how it is represented.
Assuming we are using base $10$, the $b_{nn}$ can be chosen so they are not equal to $0$ or $9$. Then, no matter how the list is represented, it will not include the new number.