elementary set theory - Explicit bijection between $Bbb R$ and permutations of $Bbb N$

The set of permutations of the natural numbers has the cardinality of the continuum. I've got injections in both directions, no problem. The Schröder–Bernstein theorem tells us that this implies the existence of a bijection. I'm wondering if it's possible to construct one explicitly.

(In what follows, I'm using the convention that $0\not\in\Bbb N$. This obviously doesn't change anything, cardinality-wise.)

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For $\Bbb R\to S_\Bbb N$, we note that every real number is the limit of some rearrangement of the alternating harmonic series. If $\alpha\in\Bbb R$, we start with positive terms: $1+\frac13+\frac15+\cdots$, until we obtain a partial sum greater than $\alpha$. We then add negative terms until our partial sum is less than $\alpha$, then we switch back to the positive terms, starting with the first unused one, etc.

In this way, we construct a series, and if we take the absolute values of the reciprocals of the terms of it, we have a permutation of $\Bbb N$. This mapping is not onto, because many permutations of the series converge to $\alpha$ - all we have to do is "overshoot" at some point, and then continue converging to $\alpha$ as usual. (More trivially, we only get permutations with $\sigma(1)=1$ using this particular construction.)

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In the other direction, if we have a permutation $\sigma\in S_\Bbb N$, we can write the continued fraction $[\sigma(1);\sigma(2),\sigma(3),\ldots]$. This actually injects $S_\Bbb N$ into $\Bbb R\setminus\Bbb Q$, because non-terminating continued fractions represent irrational numbers. Thus, this mapping is not onto; it is not even onto the irrationals.For example, no permutation of $\Bbb N$ maps in this way to any quadratic irrational, or to $e$, or to any other irrational whose c.f. expansion has repeated terms.

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So, those injections are fun and all, but finding an explicit bijection seems hard. Clearly, a bijection between $S_\Bbb N$ and any interval would suffice, because there are standard, elementary ways to construct bijections between $\Bbb R$ and any interval.

I have Googled in vain for a solution, and I don't believe this question is a duplicate. I will be happy for a hint or a full solution, or an explanation of why such an explicit construction is impossible.

Full disclosure: someone online claimed that this is "not hard", but refused to explain how, other than mentioning the Cauchy sequence construction of the reals. I don't see how that's useful, and I think he's mistaken or bluffing. I'm not too proud to admit I'm stumped. :/

Answer

It's not exactly pretty, but we can do something like:

1. Get rid of the rationals by mapping $q+n\sqrt2$ to $q+(n+1)\sqrt2$ for every $q\in\mathbb Q$ and $n\in\mathbb N_0$


2. Map the irrationals to the irrationals in $(0,1)$ by standard techniques, e.g. by $$ x\mapsto\begin{cases} 1/(2+x) & \text{for }x>0 \\ 1-1/(2-x) & \text{for }x<0 \end{cases} $$


3. Write each irrational in $(0,1)$ as a continued fraction. Ignoring the leading $0$ in each continued fraction, they become all the infinite sequences of positive integers.


4. Now convert each infinite sequence of positive integers to a bijection between the odd naturals and the even naturals by the "back-and-forth" technique. This construction proceeds in steps, by alternating between
   
   
   
   
   • Take a number $n$ from the sequence and map the first odd natural that has not yet been paired, with the $n$th even natural that has not yet been paired.
   
   
   • Take a number $m$ from the sequence and map the $m$th odd natural that has not yet been paired, with the first even natural that has not yet been paired.
   
   


5. Finally convert each bijection $f$ from odds and evens (which was only about odds and evens in order to make the previous step easier to describe) to a permutation of $\mathbb N$, namely $n\mapsto f(2n+1)/2$.