TITLE: Can somebody explain to me Cantor's diagonalization argument?
QUESTION [5 upvotes]: 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.

REPLY [8 votes]: 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.