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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.

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    Oh... I see. "aren't r1, r2, etc. themselves decimals between 0 and 1?" No. They are not. Such an expression $0.r_1r_2r_3....$ where $0 < r_i < 1$ simply would not make any sense. The book is, due to lazy proof-reading, simply starting over again and letter $r_i$ now be digits $0,1,....,8$ or $9$ and the book has completely forgotten that it had previously used those variables to mean something else, and it never occurred to the book that it would be confusing. Just replace those $r_i$ with $t_i$ or any other variable. Does the argument make sense now? – 2017-02-10

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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.

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    in this case, how are we not guaranteed that $s$ will be a rational? – 2017-02-10
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    Well..it's possible that $s = .1415926535 ... etc., i.e., $\pi - 3$ which is not rational. There's no guarantee that $s$ is eventually all zeroes. (If it were, then it'd definitely be rational). But if it's not...then bad things like my example could potentially happen, depending on the order the rationals came in. – 2017-02-10
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    To the OP: Does the book cover the difficulty that a number in $(0,1)$ can have more than representation? Eg. in base two, $1/2=0.1\overline 0=0.0\overline 1$ So if $r$ differs from $r_n$ in the $n$th place, that does not, by itself, imply $r\ne r_n.$ – 2017-02-10
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    @JohnHughes fix your LaTeX. I think you forgot some dollar signs somewhere. – 2017-02-10
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    @user254665 yes, it does mention that, but I had not thought of that possibility myself. – 2017-02-10
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    I did...and I can't edit it any more because I made the comment more than 5 min ago. Sigh. – 2017-02-10
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    Actually, the entire point is that $s = .s_1s_2....$ can *not* be rational as it is not on the list that contains all rationals. Which is fine. It's a *decimal* that is not rational that's not on the list. No-one ever said that all *decimals* are on the list. THis does however prove that eithere 1) the rationals are uncountable (false) OR there are decimals that are not rationals (true). – 2017-02-10
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    Agree, fleablood. I was just answering OP's question, which was "How could $r$ fail to be rational in this situation?" – 2017-02-10
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    @JohnHughes completely unrelated. But every time you answer one of my questions, I think of this guy: https://en.wikipedia.org/wiki/John_Hughes_(filmmaker) – 2017-02-10
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    Fortunately (for both me and you), I'm not dead yet. :) – 2017-02-10
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    @JohnHughes true! And, at least as far as I know, you never made any films with Molly Ringwald. – 2017-02-10
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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.