Showing Rationals are Countable - Confusing Notation

Question

Section 1.2 of Ash's Real Variables With Metric Space Topology deals with countability and finite and infinite sets.

In the chapter, there is an example: Show that the real numbers between $0$ and $1$ are not countable.

The proof of this result supposes that this set is countable, and proceeds to list the numbers comprising it in decimal form, as follows:

$\begin{align}r_{1} = .a_{11}a_{12}a_{13}\cdots \\ r_{2} = .a_{21}a_{22}a_{23}\cdots \\ r_{3} = .a_{31}a_{32}a_{33}\cdots\\ \end{align}$

$\vdots$

It then arrives at a contradiction by exhibiting the real number $r = .b_{1}b_{2}b_{3}\cdots$, where $b_{1} \neq a_{11}$, $b_{2} \neq a_{22}$, $b_{3}\neq a_{33}$, $\cdots$ as an example of a real number between $0$ and $1$, but one that cannot appear on this list.

This is not the part that's confusing me. What is confusing me is an exercise at the end of the section that references this example.

The exercise states as follows:

Suppose that the rational numbers between $0$ and $1$ are listed as in $(4)$.

[Side note: (4) is how the textbook refers to the sequence $\begin{align}r_{1} = .a_{11}a_{12}a_{13}\cdots \\ r_{2} = .a_{21}a_{22}a_{23}\cdots \\ r_{3} = .a_{31}a_{32}a_{33}\cdots\\ \vdots \end{align}$

]

We then pick a rational $r = .r_{1}r_{2}r_{3}\cdots$ with $r_{n}\neq a_{nn}$, $n = 1,2,\cdots$. Why doesn't this show that the rationals are uncountable?

This is confusing me because aren't $r_{1}$, $r_{2}$, etc. themselves decimals between $0$ and $1$? So how can they be digits in the decimal form of $r$?

The back of the book does have some hints about answers, and for this one it said "$r$ might not be rational", which didn't explain any more to me about why decimal expansions are doing the job of digits in $r$. Also, how could $r$ fail to be rational in this situation?

Could somebody please explain this problem to me? The very idea that $r$ could consist of other decimals is very foreign to me and I am very confused to say the least.

Thank you ahead of time.

Answer 1

It's really bad writing. The exercise should have said

"We then pick a rational $s = .s_1 s_2 \cdots$ with $s_n \ne a_{nn}$...."

Now it's clear that the $s_i$ are meant to be digits, and not the previously named $r_i$ which were real (or rational) numbers. The author has simply re-used the name $r_i$ by mistake.

Answer 2

It could fail to be rational because it could be infinite and non repeating.

Basically the number is not on the list but the list of rationals are not the list of all possible decimals.

All this argument does it provide a a decimal that that is not a rational. That is not a contradiction as have no reason to believe any expressible decimal is rational. (Cheating: Consider $\pi - 3 = .1415926.....$. This is cheating because it is jumping the gun but we know that decimal expression is not rational.)

If fact, we can prove that such a decimal can not be rational but... why bother.