Prove Uncountability of Reals in an infinite base number system (without diagonalization) So I was trying to study some infinite dimensional spaces when I came across the idea:
What if I didn't choose a base, or rather, what if my base was infinite?
I did some thinking and decided to let each number be the most simplified positive fraction $\left[\frac{p}{q}\right]$. This makes talking about rational numbers easy, since all rational numbers are now "whole" numbers in this system. But then I started thinking about how to make decimals and representing any real number.
I used the following example as motivation on how to consistently write/represent real numbers:
$$\pi = [3].[3.1][3.14][3.141]...$$
So, basically I'm representing irrational numbers as their unique rational decimal expansion sequence. Clearly all operations compute in the same fashion as decimal.
But this makes me wonder: When you express real numbers in this fashion, it makes it to where the diagonalization argument doesn't work because, for example: $[3].[3.1][\mathbf{4.14}][3.141]...$ isn't a number because it's not a a rational decimal expansion of any number.
Obviously the diagonalization argument is constructive and fails to work here because I'm not representing things in exactly the same way as you do in typical numerical systems with finite bases.
I guess I'm wondering if this means that real numbers are countable/listable in this numerical system, and if not, why not? Since every real number can be written as a limit of rational numbers, and you can't use a diagonalization argument to show that you can construct a number not listed in the "set of real numbers" like you do in the normal case.
 A: I believe what you are trying to do is to construct a Hamel basis.
If $V$ is a vector space over a field $K$, then a Hamel basis of $V$ over $K$ is a subset $\mathcal{B} \subset V$, such that every $v\in V$ can be written as
$$ v = \sum_{w\in B} \alpha_w(v) w $$
where $\alpha_v(w) \in K$ for all $w\in\mathcal{B}$, $|\{\alpha_w(v) : \alpha_w(v) \neq 0\}|<\infty$ (then the sum is well defined) and if
$$ v = \sum_{w\in B} \alpha_w(v) w = \sum_{w\in B} \beta_w(v) w,  $$
then $\alpha_w(v) = \beta_w(v)$ for all $w\in B$ (the representation is unique).
The existence of such basis are equivalent to the axiom of choice.
In your case let $K = \mathbb{Q}$ and $V = \mathbb{R}$. In fact, what happens is that since the real numbers are uncountable, the basis $\mathcal{B}$ must be uncountable. Otherwise we would have a bijection between each real $v$ and a finite subset of $\mathcal{B}$, if $\mathcal{B}$ is countable then the set of all finite subsets of $\mathcal{B}$ is countable, therefore we would have that $\mathbb{R}$ is countable (see this).
If you want to know more about Hamel basis, see this.
If you want to see an amazing consequence of the existence of Hamel basis, see this. Spoiler: every polynomial of degree $n$ is the sum of $n+1$ periodic functions.
A: The argument that Cantor gave for the uncountability of the reals was not based on any "base representation" (decimal or binary or otherwise) of real numbers. Some popularisations of it have adopted that. His original argument was a topological one, essentially, using completeness of the reals. Later he showed the irrationals in $(0,1)$ are uncountable because they have a unique continued fraction expansion (and because we use all integers it comes closest to your "infinite base" idea IMO).
So use continued fractions for the irrationals only and the argument goes through.
