Numbers as product of prime numbers raised to exponents $e\in A \subset \mathbb R$

44 Views Asked by At

If $\mathbb A^+\subseteq \mathbb R^+$, then let $\mathbb A^- =\{- a : a\in\mathbb A^+\}$, $\mathbb A^* = \mathbb A^+\cup\mathbb A^-$, $\mathbb A=\mathbb A^*\cup\{0\}$.

Set of positive integers can be written as:

$$\mathbb Z^+=\{x=\prod_{p\text{ prime}} p^{x_i},x_i\in\mathbb N\}=\mathbb N$$

Set of positive rationals can be built on top of that as:

$$\mathbb Q^+=\{x=\prod_{p\text{ prime}} p^{x_i},x_i\in\mathbb Z\}$$

This is useful in, for example, finding explicit bijection between naturals and rationals.

Let $\mathbb A_0^+ := \mathbb Z^+,\mathbb A_1^+ := \mathbb Q^+$, then we can continue building sets with similar properties:

$$\mathbb A_2^+=\{x=\prod_{p\text{ prime}} p^{x_i},x_i\in\mathbb Q\}$$ $$\mathbb A_3^+=\{x=\prod_{p\text{ prime}} p^{x_i},x_i\in\mathbb A_2\}$$ $$\dots$$ $$\mathbb A_n^+=\{x=\prod_{p\text{ prime}} p^{x_i},x_i\in\mathbb A_{n-1}\}$$ $$\dots$$

Notice:

  • $\mathbb Q = \mathbb A_1 \subset \mathbb A_2 \subset \mathbb A_3 \subset \dots \subset \mathbb A_n$ and thus $\mathbb A_n$ is dense in $\mathbb R$, for every $n\ge1$.

  • $\mathbb A_n$ is in bijection with $\mathbb A_{n-1}$, and thus with $\mathbb N$, making it countably infinite, for every $n$.


Questions

  • Can we take the limit $n\to\infty$, and talk about set $\mathbb A=\mathbb A_\infty\cap\mathbb R$, in some sense? Can we explicitly construct elements of $\mathbb A$?

  • Are there any interesting or useful, conclusions or observations, that can be made from this sequence of sets?


Remark

I have no idea how to approach or analyze this, or even if I stated the questions well enough - this is something inspired by the linked question. I'm curious about how can this be analyzed or talked about in mathematical sense - as I don't have knowledge in set theory, but perhaps the most basic one.