I've been trying to find a way to consistently generate bijections between $\mathbb{N}$ and countable sets of Real numbers in $[0,1]$. Given any such set, by definition there is a bijection to the natural numbers, but I am wondering about a general method that can be used for any such set.
The best idea I had so far is the following: Choose a bijection $f: \mathbb{N}\to\mathbb{Q}\cap[0,1]$ such that $f(1)=0$ and $f(2)=1$. I will attempt to use this bijection to construct, from a countable set $X$ of real numbers in $[0,1]$, a list $x_1,x_2,x_3\dots$ containing every number in $X$. At any step $n\ge2$, the set $\{f(1),f(2)\dots f(n)\}$ partitions the interval $[0,1]$ into a finite number of intervals. At the $n$th step, then, proceed as follows: If $f(n)\in X$, append $f(n)$ to the list. Then, some interval in the partition of $[0,1]$ generated by $\{f(1),f(2)\dots f(n-1)\}$ has been split by $f(n)$ into two intervals $A$ and $B$. If either $A$ or $B$ contains only a finite number of points from $X$, order the points in that set by the standard order on the real numbers, and append the elements of that set to the list. If both $A$ and $B$ contain only a finite number of points, then $A\cup B\cup f(n)$ does as well, so the numbers have already been added and can be ignored.
It seems like this could work. Specifically, it works on such countable subsets as $\{0\}\cup\{1/n:n\in\mathbb{N}\}$ (or on similar subsets of irrational numbers). However, I have no idea how I could show that it works. Can anyone show that this does/does not work? Does anyone know a better method for ordering countable subsets of $[0,1]$?
-Thanks