The Cardinality of infinite series of natural numbers?

Question

Given an infinite sequence $a_1,a_2,a_3,...$,and the map $F(a_1,a_2,a_3...) = {p_1}^{-a_1}{p_2}^{-a_2}{p_3}^{-a_3}...$

Where $p_i$ is the ith prime (chosen by the axiom of choice). Why isn't this map one to one on the the interval $[0,1]$?

Answer 1

As long as you choose $a_i$ big enough such that for some $ε>0$ all $p_i^{-a_i} ≤ 1-ε < 1$ you have $F(a_1, …) = 0$. So $F$ can not be injective.

Answer 2

I assume the elements $a_i$ are natural numbers (including zero).

If there exists such a $N$ that $a_i=0$ for all $i>N$, then the product is obviously a rational number.

If such a $N$ does not exists, the product, if you define $$\prod_{i=1}^\infty p_i^{a_i} = \lim_{N\rightarrow\infty}\prod_{i=1}^Np_i^{a_i},$$

will actually equal zero. Therefore, the mapping only covers some of the rational numbers (not even all of them) and $0$.