Problems with Cantor's Diagonal Argument

Question

Using Cantor's Diagonal Argument to compare the cardinality of the natural numbers with the cardinality of the real numbers we end up with a function $f: \mathbb{N} \to (0, 1)$ and a point $a \in (0, 1)$ such that $a \notin f((0, 1))$; that is, $f$ is not bijective.

My question is: can't we find a function $g: \mathbb{N} \to (0, 1)$ such that $g(1) = a$ and $g(x) = f(x-1)$ for $x > 1$? This function would be bijective, so the cardinality of the two sets would be the same. Actually, if we found a countably infinite set of points that weren't in $f((0,1))$, using Hilbert's Hotel argument we could find a bijective function.

Answer 1

Okay, thanks to all of you for your responses. I think I understood why you can never find such a bijective function.

With the aim of helping others with the same problem, I will synthetize my reasoning here.

Let $f: \mathbb{N} \to (0, 1)$ be an injective function. Using Cantor's Diagonal Argument, we can show that there exists some $a \in (0, 1)$ such that $f(n) = a$ for some $n \in \mathbb{N}$. Then, we create the funcion $g: \mathbb{N} \to (0, 1)$ as follows \begin{equation} g(x) = \left\{ \begin{matrix} a & \textrm{if}\ x = 1\\ f(x-1) & \textrm{if}\ x > 1 \end{matrix} \right. \textrm{.} \end{equation}

However, it isn't bijective, because using Cantor's Diagonal Argument again we can still find some number $b \in (0, 1)$ such that $g(n) = b$ for some $n \in \mathbb{N}$.

We could keep going on finding more functions. Let $h: \mathbb{N} \to (0, 1)$ be a function applying the previous process infinitely many times. Again, using Cantor's Diagonal Argument we see that $h$ is not bijective. Therefore, we can't find the bijective function $b: \mathbb{N} \to (0, 1)$ that would make $\textrm{card}\ \mathbb{N} = \textrm{card}\ (0, 1)$.

$\square$

Please, correct me if I'm wrong.

EDIT: I've just realised that this isn't needed for the proof of Cantor's Diagonal Argument to be completed. When we consider $f: \mathbb{N} \to (0, 1)$, it is an arbitrary function. Therefore, $\nexists\ g: \mathbb{N} \to (0, 1)$ such that $g$ is bijective, because $f$ is any function, even $g$.

Answer 2

This was described by another contributor here as follows: You are $(0,1)$ and you are accused of being countable. The prosecutor presents a witness $f:\Bbb N \to (0,1),$ an alleged bijection. Your lawyer presents $x \in (0,1)\setminus f(\Bbb N).$ The prosecutor then presents $f^*:\Bbb N \to (0,1)$ with $f^*(\Bbb N)\supset \{x\}\cup f(\Bbb N).$ Your lawyer presents $x^*\in (0,1)\setminus f^*(\Bbb N)$.... This may go on for a while until the judge asks the prosecutor " Can you present a witness who can overcome the Cantor Diagonal Defense?" Prosecutor: "No." Judge: "Case dismissed".

Any $f(\Bbb N)\subset (0,1)$ will omit "almost all" of $(0,1).$ There will be uncountably many other $x \in (0,1).$ But we do not need to directly prove it, but can infer it after proving that $(0,1$) is uncountable.... The existence of at least one $x$ in a set is logically sufficient to show it's not empty. That is, $\forall f:\Bbb N\to (0,1)\; (\, (0,1)\setminus f(\Bbb N)\ne \emptyset\,).$