Can countability coexist with infinity?

Question

This question concerns the countability of the real numbers. First I will show how I count the numbers between 0 and 1 on the real line. It is done by reversing digits behind the coma, so that e.g. 0,761 maps to 167. Obviously this is 1 to 1 mapping, but there are infinite number of those unique mappings depending on the chosen reasonable radix. In decimal number system I could count numbers like this:

$ 0 , 0.1 , 0.2 , 0.3 ... 0.9 , \\ 0.01 , 0.11 , 0.21 , 0.31 , ... , 0.91 , \\ 0.02 , 0.12 , 0.22 , 0.32 , ... , 0.92 , \\ 0.03 , 0.13 , 0.23 , 0.33 , \dots , 0.93 , \\ \vdots \\ 0.09 , 0.19 , 0.29 , 0.39 , \dots , 0.99 , \\ 0.001 , 0.101 , 0.201 , 0.301 , \dots , 0.901 , \\ \vdots \\ 0.002 , \dots , 0.902, \\ \vdots \\ 0.092, \dots , 0.992 , \\ 0.003 , \dots , 0.903 , \\ \vdots \\ 0.004, \dots , 0,904,\\ \vdots \\ \infty \\ $

Now, given that I can "succeed" to count to infinity, I would also count all irrational numbers. There is no reason to haste. But then all irrational numbers are "somewhere" in the infinity. So either counting to infinity allows me to write irrational numbers backwards, or infinity and countability can not coexist. Which one is true ? How does your solution compare to rational numbers ?

Answer 1

Let's denote the $n^\text{th}$ number in your list by $f(n)$. To use your method to prove that $[0,1)$ is countable, you need to show that $f$ is a surjection, that is, for every real number $x \in [0,1)$ there is a natural number $n$ such that $f(n) = x$. In particular, you have to show that there is a natural number $n$ such that $f(n) = 1/3$. Note that $n$ here is just an ordinary, finite, natural number. So you have to show that $1/3$ appears at some finite stage in your list (e.g. it is the tenth number listed, or the millionth number listed.)

This is not the case, and it is not correct to argue that it must appear sometime because $\mathbb{N}$ is so very large. For no value $n \in \mathbb{N}$ do we have $f(n)=1/3$, any more than we have $f(n) = -1$ or $f(n) = 37$. These numbers simply do not appear in the list.

Answer 2

You have only counted the numbers whose decimal expansion is finite. This covers absolutely no irrational number, and in fact not even all the rational numbers as well.

Furthermore there is absolutely no reason to expect that this sort of process is continuous. That is to say, the set of finite strings of integers is countable, but the set of infinite strings is uncountable. Even if you can approximate one infinite string by its finite initial segments with a countable process, you can't approximate all the infinite strings at the same time with finite approximation with a countable process.

Answer 3

To extend a little Peter's comment, you are assuming that you have the decimal expansion of all real numbers. Even if you find a way to extend your method to reals with infinite decimal places (such as $\pi$), there are still a countable number of them. You are missing actually a number of reals that are uncountable and cannot be written with any algorithm: the uncomputable numbers (that is: most of the real numbers!)