(Inverse) mapping from $\mathbb{R}$ to subsets of $\mathbb{N}$ There's the usual mapping $2^\mathbb{N}\mapsto\mathbb{R}_{[0,1)}$ (where $2^\mathbb{N}$ denotes the powerset of $\mathbb{N}$) given by $x\in2^\mathbb{N}\mapsto\sum_{i\in x}\frac{1}{2^i}\in\mathbb{R}$. I'm interested in the inverse $\mathbb{R}_{[0,1)}\mapsto2^\mathbb{N}$, i.e., given $r\in\mathbb{R}$, what's the corresponding $x_r\in2^\mathbb{N}$ that generates it? (Note: I'm just guessing appropriate tags)
Is that $x_r\in2^\mathbb N$ explicitly constructible? Or its characteristic function, i.e., for $r\in\mathbb{R}_{[0,1)}$ and $x_r\mapsto r$, $\chi_r(i)=\left\{{1\ if\ i\in x_r\atop0\ otherwise}\right.$?
Also, poset ordering question as follows. While it's clear that $x_1\subseteq x_2\Longrightarrow r_{x_1}\le r_{x_2}$, what about the inverse? That is, $r_{x_1}\le r_{x_2}\overset?\Longrightarrow x_1\subseteq x_2$. And if not true, got an explicit counterexample?
__Edit__
Firstly, thanks very much Lubin, Manmaid, Ross, Daniel for your elaborate and thoughtful answers/comments. I haven't had time to fully digest everything, but one apparent take-away is the pesky appearance of infinite subsets. And that's got to be an unnecessary artifact: there's no reason why, say, $\frac13=.333\ldots=\sum_{i\in x}\frac1{2^i}$, should require some stupid-large $x\subseteq\mathbb N$ for its representation, whereas the information content of $\frac13$ is ultimately pretty minimal.
And it was Ross' mention of https://en.wikipedia.org/wiki/Dyadic_rational that suggested the culprit. It's the explicit "base 2" representation/expansion which I naively introduced in the above question that I'm now thinking is the problem. Daniel's answer introduces a slightly different expansion, but the same underlying problem remains.
So here's a tentative conjecture for a more general representation, which hopefully maps "simple $r\in\mathbb R$'s" to "simple $x_r\in2^\mathbb N$'s". But the representation algorithm's somewhat more complicated than Daniel's or mine above. And the meaning of the subset-order $x_1\subseteq x_2$ isn't at all obvious.
Consider the usual pair enumeration https://en.wikipedia.org/wiki/Pairing_function whereby $\mathbb{N\times N}\rightarrow\mathbb{N}$. And we can then iterate that for $\mathbb{N^3\rightarrow N}$ (or any finite power, but we just need $3$). So consider some $x\in2^\mathbb N$ and interpret every $i\in x$ as three-component $i=(j,k,m)\mapsto j/k^m$. That is, we decompose $r\in\mathbb R$ into rational components of the form $j/k^m\in\mathbb Q$, with any mix of base k's that's convenient.
Now, as Lubin and Ross previously pointed out, it's still possible to represent, e.g., $\frac12=\frac1{2^2}+\frac1{2^3}+\cdots$. So we need some formal way to distinguish a "canonical representation" $\frac12=\frac1{2^1}$. So I haven't figured that out, nor how to interpret the poset order $x_1\subseteq x_2$, etc.
But I believe the premise, that these pesky infinite subsets should go away (except for transcendental reals) is necessary for any sensible $\mathbb R_{[0,1)}\Leftrightarrow2^\mathbb N$. As per https://en.wikipedia.org/wiki/Scott_domain#Explanation, the poset order is an information order, so you shouldn't have teensy-weensy elements on one side of the $\Leftrightarrow$ corresponding to biggy-wiggy elements on the other side. Somehow, that's got to go. But is the preceding conjecture really the simplest or most straightforward way?
__Edit#2__
Reply (too lengthy for comment) to Ross's comment below, "With your edit I understand even less what you are asking..."
Yeah, sorry about that. In a preceding comment replying to you, I mentioned

"Actually, the constructible bijection's for an explicit correspondence between the universal domains $\mathcal P\omega\Leftrightarrow\mathcal U$ ( where the rhs denotes the universal interval domain, as per https://books.google.com/books?id=epb39s2wgt0C&pg=PA162 )

And I'd thought I'd solved that to my own satisfaction, modulo the kind of bijection I asked for here. That bijection's ultimate purpose is thus a correspondence between sets of integers on the $\mathcal P\omega\equiv2^\mathcal N$ side and half-open real intervals with rational endpoints on the $\mathcal U$ side.
By the way, note that in that $\mathcal U$ domain, any real is the lub of intervals including it. Put another way, the real line corresponds to the collection of total (aka maximal) elements of $\mathcal U$ wrt reverse-inclusion ordering. That is, the smaller the interval, the more information you have about the real it refers to.
The upshot is that for the purposes of this question, I was perfectly happy discussing rational approximations to reals. As per a comment to Manmaid,

...Actually, I simplified the question by neglecting to discuss intervals. If you represent $r_{x_n}\in\mathbb R$ by an $x_n=\lbrace i_1,i_2,\ldots,i_n\rbrace$ containing the first $n$ smallest $i$'s in order, with $i_n$ the largest among them, then you're really representing an interval $[r_{x_n},r_{x_n}+1/2^{i_n})$.

So every time I mentioned "real", my underlying thinking was really "interval" as an element of that $\mathcal U$ domain. But I didn't think it necessary to mention that in the context of the original question. I thought I just needed the requested bijection wrt reals, and then everything would just fall into place wrt $\mathcal U$.
So where'd that go wrong? The bijection's subject to a pesky additional constraint wrt that reverse-inclusion information ordering, which it must respect. Suppose the bijection's $b:\mathcal U\rightarrow2^{\mathcal N}$,
and write $u_1\sqsubseteq u_2$ for the $u_1,u_2\in\mathcal U$ poset order, and $x_1\subseteq x_2$ for the $x_1,x_2\in2^\mathcal N$ order. Then $b$ must satisfy $u_1\sqsubseteq u_2\Longrightarrow b(u_1)\subseteq b(u_2)$. (and given that $b$'s one-to-one, the arrow has to go both ways)
And that didn't just automatically happen like I'd expected (more accurately, I didn't even think about it until I noticed it wasn't happening). And that unexpected problem got me asking additional stuff seemingly unrelated to the original question, which I guess accounts for your "understand even less what you are asking" remark. (and I hope this helps clear that up a little)
 A: First, for any $x\in \mathbb{R}_{[0,1)}$ consider these steps. Let's call it Structure:


*

*Calculate $2x$, if $2x\geq 1$, then  fix first number $1$, and if $2x<1$, fix first number $0$. Now calculate $\{2x\}$, the fraction part of $2x$.


*Calculate $2\{2x\}$, if $2\{2x\}\geq 1$, then  fix second number $1$, and if $2\{2x\}<1$, fix second number $0$. Now calculate $\{2\{2x\}\}$, the fraction part of $2\{2x\}$.
Continue this process. Finally you will get an element of $2^{\mathbb{N}}$.

The construction of a bijective function:
Let $f:\mathbb{R}_{[0,1)}\rightarrow 2^{\mathbb{N}}$
Suppose $x\in \mathbb{R}_{[0,1)}$ is not of the form $\frac{n}{2^m}$, where $n<2^m$ is odd natural number and $m\in \mathbb{N}$, then $f(x)=$ the final sequence you get from the Structure for $x$.
Now if $x\in \mathbb{R}_{[0,1)}$ is of the form $\frac{n}{2^m}$, where $n<2^m$ is odd natural number and $m\in \mathbb{N}$, then values looks like this:
$$\frac{1}{2}\\\frac{1}{4}\space\space\frac{3}{4}\\\frac{1}{8}\space\space\frac{3}{8}\space\space\frac{5}{8}\space\space\frac{7}{8}\\\frac{1}{16}\space\space\frac{3}{16}\space\space\frac{5}{16}\space\space\frac{7}{16}\space\space\frac{9}{16}\space\space\frac{11}{16}\space\space\frac{13}{16}\space\space\frac{15}{16}\\\dots\\\dots\\\dots$$
First fix $f(\frac{1}{2})=(1,1,1,\dots)$. Now calculate the sequence you get from Structure for taking $x=1/2$. Surely you get $(1,0,0,0,\dots)$. Now fix $f(\frac{1}{4})=(1,0,0,0,\dots)$ and $f(\frac{3}{4})=(0,1,1,1,\dots)$. Again calculate the sequence you get from Structure for taking $x=1/4$ and $x=3/4$. Sequence you get are $(0,1,0,0,0,\dots)$ and $(1,1,0,0,0,\dots)$ respectively. You immediately fix $f(\frac{1}{8})=(0,1,0,0,0,\dots)$, $f(\frac{3}{8})=(0,0,1,1,1,\dots)$, $f(\frac{5}{8})=(1,1,0,0,0,\dots)$ and $f(\frac{7}{8})=(1,0,1,1,1,\dots)$.Continue in this way and finally you have $f$ is a bijective mapping.
A: It looks as if you mean for $\Bbb N$ to be $\{1,2,3,\cdots\}$.
Naively, a function has an inverse only if it’s one-to-one. But your map $2^{\Bbb N}\to[0,1\rangle$, is not one-to-one, since $\frac12$ is the image both of $(1,0,0,\cdots)$ (i.e. the function that’s zero everywhere but at $n=1$) and $(0,1,1,1,1,\cdots)$ (i.e. the function that’s $1$ everywhere but at $n=1$. In more familiar terms, $\frac12=\frac14+\frac18+\frac1{16}+\cdots$
This is just the familiar fact that every real with a terminating decimal (or terminating binary expansion) has a different, nonterminating expansion.
The moral? You can not form an inverse to your function.
A: Consider the Cantor set $C=\{\sum_{n=1}^{\infty}2f(n)3^{-n}: f\in F\}$ where $F$ is the set of all functions $f:\mathbb N\to \{0,1\}.$ For $x\in C$ there is a unique $f_x\in F$ such that $x=\sum_{n=1}^{\infty}2f_x(n)3^{-n},$ and $f_x=f_y\implies x=y.$
So for $x\in C$ let $G(x)=\{n\in \mathbb N: f_x(n)=1\}.$ For $x\in [0,1]$ \ $C$ let $G(x)$ be any member of $2^{\mathbb N}$ that you want, as $G:C\to 2^{\mathbb N}$ is already surjective.
Remark: $G$ is not injective, but neither is the surjection $H(x)=\sum_{i\in x}2^{-i}$ from $2^{\mathbb N}$ to $[0,1]$ in your Q. For example $H(\mathbb N$ \ $\{1\})=H(\{1\})=1/2.$
A: We can construct an explicit bijection between $[0,1)$ and $2^{\Bbb N}$ as follows.  As long as $x$ is not a rational of the form $\frac a{2^n}$ with $1 \le a \lt 2^n$ it has a unique binary expansion so we can match them up.  One problem is that numbers of the form $\frac a{2^n}$ have two binary representations, one ending in all $0$s and one ending in all $1$s.  The one with all $0$s starts with the finite expansion of $a$ and the one with all $1$s starts with the finite expansion of $a-1$, both padded on the left with zeros to make $n$ bits.  The other problem is that we have excluded $1$ from the real interval so we have nothing to match with $1,1,1,1,\ldots $ 
Given a rational of the form $\frac a{2^n}$ with $a$ odd so the fraction is in lowest terms, let $m$ be the minimum value so that $2^m \gt a$.  If $n-m$ is even we will make a correspondence to a sequence with an infinite run of $0$s while if $n-m$ is odd we will make a correspondence to a sequence with an infinite run of $1$s.  We let $k=\lfloor \frac {n-m}2 \rfloor$, then the real $\frac a{2^n}$ corresponds to the binary expansion of $\frac a{2^{n-k}}$ using the terminating version if $k$ is even and the infinite $1$s if $k$ is odd.  Finally if $a=1$ we push all the expansions down by $1$ to make $1,1,1,1,\ldots $ correspond to $\frac 12$.
