Can somebody explain to me Cantor's diagonalization argument? [duplicate]

Question

Like..can somebody explain this to me as if I was a 5 year old or something? Every explanation I read repeats the same exact thing that I simply do not understand. This is what my book says:

"The real numbers between 0 and 1 can be listed in some order, say, $r_1, r_2, r_3, ...$Let the decimal representation of these real numbers be

$r_1 = 0.d_{11}$$d_{12}$$d_{13}$$d_{14}$... $r_2 = 0.d_{21}$$d_{22}$$d_{23}$$d_{24}$... $r_3 = 0.d_{31}$$d_{32}$$d_{33}$$d_{34}$... $r_4 = 0.d_{41}$$d_{42}$$d_{43}$$d_{44}$...

Where $d_{ij}$ is an element of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Then for a new real number with decimal expansion r = $d_{1}$$d_{2}$$d_{3}$$d_{4}$... where the decimal digits are determined by the following rule:

$d_{i}$ = {4 if $d_{ii}$ does not equal 4, 5 if $d_{ii}$ = 4}.

And I'm sorry but..what? What in the world is any of this trying to get at? What is the whole r1, r2, r3 thing even mean? Why do we have to create a "new real number"? What is the point? Why? Why are we doing any of this? I don't understand any of the process behind it and I don't understand how it all leads to the conclusion that the real numbers are uncountable. I have absolutely no idea what is going on here.

Answer 1

Every argument is an argument for something. The Cantor diagonal argument is an argument to prove that set of real numbers is uncountable.

What is a countable set? Let's say a set is countable if we can start ordering the elements of a set like the first, the second and so on. Formally we have to find a bijection with natural numbers.

To prove that reals are uncountable we first assume the contrary, namely that set of reals is countable. Then we have to find a contradiction, rendering the assumption false. To do that we find a real number which is not counted. Cantor diagonal argument construct such a real number which is not counted.

So here are the steps:

Goal: Set of real numbers is uncountable.

Step 1: Assume that the set is countable. This means that the set of real numbers can be written as a set with first element, second element and so on, which is $\{r_1,r_2,r_3,\dots\}$.

Step 2: One way to show that the assumption of step 2 is not possible is to find a real number which is not counted there. How? By Cantor diagonal argument.

Suppose that we are going to consider only numbers between 0 and 1. Then the new number is such that it is different from the first number at the first digit, from the second element at the second digit and so on. For instance look at the following: $$ \begin{matrix} 0.& \color{red}9 & 7& 0& 6& \dots\\ 0.& 8 & \color{red}2& 4& 3& \dots\\ 0.& 4 & 5& \color{red}2& 8& \dots\\ 0.& 1 & 2& 5& \color{red}3& \dots\\ \vdots & \vdots & \vdots& \vdots& \vdots& \dots\\ \hline\\ 0.& 8 & 1& 1& 4& \dots\\ \end{matrix} $$ You can see that the number is different from all counted numbers and also you can see that it is constructed by using the diagonal elements of counted numbers written as above, hence the nominalization of Diagonal argument.

Answer 2

The argument is often presented as a proof by contradiction, but it can be presented more directly, which I think makes it a bit clearer:

Theorem. Let $f$ be any function $\mathbb{N} \to \mathbb{R}$. Then there is some real number not in the image of $f$; that is, $f$ is not surjective.

Proof. Given $f : \mathbb{N} \to \mathbb{R}$, construct a real number $x$ via its infinite decimal expansion; take its $n$th digit $x_n$ to be 4 if the $n$th digit of the standard decimal expansion of $f(n)$ is 5, and take $x_n$ to be 5 otherwise. Now for any $n$, we get that $x$ differs from $f(n)$ in the $n$th digit of their decimal expansions; so $x \neq f(n)$. So $x$ cannot be in the image of $f$. $\square$

So there is no surjection $\mathbb{N} \to \mathbb{R}$; so there’s certainly no bijection, so the reals aren’t countable.