In Abbott's book (2nd edition) there is an exercise (1.6.3) given in chapter 1 where we are asked to address two rebuttals to cantor's proof that $(0,1)\subseteq R$ is uncountable. I don't understand the solution to the second rebuttal.
background information: the proof uses a contradiction. It assumes (for sake of contradiction) that there is some bijective function $f:N\rightarrow (0,1)$ such that $f(n)=0.a_{n1}a_{n2}a_{n3}...$ It then says that if we let $x=0.b_1b_2b_3...$ where $b_n=\begin{cases} 2\ if\ a_{nn}\neq 2\\ 3\ if\ a_{nn}=2 \end{cases}$ then x will not be in the range of f, and therefore a contradiction is reached and $(0,1)$ is uncountable (since we can't make a bijective function between the natural numbers and the interval).
I understand this proof, but a follow up exercise is given immediately after the proof in which we are tasked with addressing two rebuttals.
rebuttal: "Some numbers have two different decimal representations. Specifically, any decimal expansion that terminates can also be written with repeating 9’s. For instance, 1/2 can be written as .5 or as .4999 . . . . Doesn’t this cause some problems?"
The solution to this rebuttal I obtained from the solutions manual: " By using the digits 2 and 3 in our definition of $b_n$ we eliminate the possibility that the point $x = .b_1b_2b_3 . . .$ has some other possible decimal representation (and thus it cannot exist somewhere in the range of f in a different form.)
I don't understand the significance of choosing 2 and 3 as our digits to switch around. How does this choice address the rebuttal stated above?
Note that the rebuttal above didn't go into detail about how multiple decimal expansions could create problems; part of the exercise is to identify exactly what issue could arise.
There are two key points:
The "antidiagonal" real $x$ needs to be well-defined. Specifically, if $f(n)$ has two different decimal representations $0.a_1a_2...$ and $0.a_1'a_2'...$, it needs to be the case that it doesn't matter which we use when we compute $x$.
The "antidiagonal" real $x$ needs to be not in the range of $f$. We claimed that $x\not=f(17)$ (say) based on a difference in their decimal expansions. But maybe we were looking at the wrong decimal expansions - maybe one of $x$'s decimal expansions is different from one of $f(17)$'s decimal expansions, but $x=f(17)$ nonetheless (e.g. we thought of $f(17)$ as $0.10000...$ and we thought of $x$ as $0.0999999$...).
The point is that by focusing on $2$ and $3$ - and in particular, by staying away from $0$ and $9$ - we've avoided both these problems.