Alternative injection from $(0,1)$ to $p(\mathbb{N} )$

Question

I know the standard injection is to consider the binary expansion of a number in the interval, but I was wondering if it is possible to create an injection using a decimal expansion.

To that effect first denote the decimal expansion of a $r\in (0,1)$ as $r=0.d_1d_2d_3d_4\dots$. Choose the decimal expansion of all $x\in (0,1)$ as above except that $\forall n\in \mathbb{N} \exists m>n$ such that $d_m\in \{0,1,2,3,4,5,6,7,8\}$. (This does not exclude any element in the interval).

Now consider the function $f:(0,1)\to p(\mathbb{N})$ where $f(x)=\{2^{d_1},3^{d_2},5^{d_3},\dots,p_i^{d_i},\dots\}$. I believe that Euclid's theorem and the fundamental theorem of arithmetic ensure that $f$ is an injection, but I would like some confirmation.

Answer 1

This almost works, but if you have more than one $0$ digit, you’ll get multiple copies of $1$ in the description of $f(x)$. To avoid this problem, define

$$f(x)=\left\{p_i^{d_i+1}:i\in\Bbb Z^+\right\}\;.$$

Added: As Henning Makholm notes below in the comments, the original idea of setting

$$f(x)=\left\{p_i^{d_i}:i\in\Bbb Z^+\right\}$$

actually does work; it just isn’t quite as elegant. If no multiple of some prime $p_i$ appears in $f(x)$, then $d_i=0$, so the $i$-th digit is still reconstructible. In that case the $1\in f(x)$ doesn’t actually tell us anything that we can’t already learn from $f(x)\setminus\{1\}$.