Sets with Unique Subset Summing to Every Real Do there exists sets of reals such that every real has a unique subset that sums to it. Formally, do there exists sets $S\subset\mathbb{R}$ such that every $r\in\mathbb{R}$ has a unique (up to permutation) representation as $\sum_{i=0}^ms_i$ for some $m\in\mathbb{N}\cup\{\infty\}$ and a sequence ${(s_i)}_{i=0}^m$ such that for all $0\leq i\leq m$ if $m$ is finite or $0\leq i<\infty$ if $m$ is infinite, $s_i\in S$, and for no indices $i\neq j$, $s_i=s_j$? $m=0$ is interpreted as the empty sum with value $0$, thus $0$ is automatically a sum for every set, but one which cannot change other sums, so that the problem is non-trivial. The "up to permutation" comment could just mean the multiset underlying a sequence that gives a sum is unique, not that all sequences with that multiset have to give that sum, but in fact this must be the case by the Riemann Rearrangement Theorem, so the clarification does not affect the answer. I can't show that such an $S$ actually exists, but I can work out some properties. Such sets are weird, $0$ must have a neighbourhood in which it's the only limit point or there are no limit points. In any such neighbourhood $N$, summing over all the negative terms gives a least sum $l$ and the positive terms gives a greatest sum $u$. Then if $d=u-l$, then there must be infinitely many sums from $S\setminus N$ in every interval of length $2d$. So it's a weird set. This question is motivated by the analogue of the non-negative powers of $2$ working for finite sums and the natural numbers, and the powers of $-2$ working for finite sums and the integers.
 A: Yes, such $S$ does exist, choose $S:=\{(-2)^n|n\in \mathbb{Z}\}$.
It represent everything: 
First we'll show that $\{(-2)^0, (-2)^1,...\}$ represent all $\mathbb{Z}$.
lemma we'll prove by induction on $n\in \mathbb{N}$ that:  
if n is odd than: $\{(-2)^0, (-2)^1,...(-2)^n\}$ represent all integers in $[\frac{2-2*4^\frac{n+1}{2}}{3}, \frac{4^\frac{n+1}{2}-1}{3}]$ and if we use $(-2)^n$ we represent all integers in $[\frac{2-2*4^\frac{n+1}{2}}{3}, \frac{-2*4^\frac{n-1}{2}-1}{3}]$
if n is even than: $\{(-2)^0, (-2)^1,...(-2)^n\}$ represent all integers in $[\frac{2-2*4^\frac{n}{2}}{3}, \frac{4^{\frac{n}{2}+1}-1}{3}]$ and if we use $(-2)^n$ we represent all integers in $[\frac{2+4^\frac{n}{2}}{3}, \frac{4^{\frac{n}{2}+1}-1}{3}]$
This proof by induction is immidiate, only the calculation seems scary. the main benefit from this lemma is that we see that $\{(-2)^0, (-2)^1,...(-2)^n\}$ represnt segments of size that always grow and thus $\{(-2)^0, (-2)^1,...\}$ represens $\mathbb{Z}$. In addition, we see that the maximal element (in absolute value) of the representaion uniqly determine in which of the distinct segments the integer which its represnt lie.
Now, we'll prove that $\{(-2)^{-1}, (-2)^{-2},...\}$ represent all the reals in the segment $[-\frac{2}{3},\frac{1}{3}]$, and then we'll finish because every real lies in $\mathbb{Z}+[-\frac{2}{3},\frac{1}{3}]$
lemma 2 $S_n:=\{(-2)^{-1}, (-2)^{-2},...(-2)^{-n}\}$, represnt every number in $T_n:=\mathbb{Z}/{2^n}\cap [-\frac{2}{3},\frac{1}{3}]$
proof for $n=1$ we get that $\{(-2)^{-1}\}$ indeed represnt $\mathbb{Z}/{2}\cap [-\frac{2}{3},\frac{1}{3}]=\{-\frac{1}{2},0\}$
So for $S_{(n+1)}$ we'll use the induction assumption:
if we dont use $(-2)^{-n}$, we get all $T_n$, if we use $(-2)^{-n}$ we get new $2^n$ new elements because $(T_n+(-2)^{-n})\cap T_n=\emptyset$ and we can't get out of $[-\frac{2}{3},\frac{1}{3}]$ since $-\frac{2}{3}$ equal the sum of all negative elements of $\{(-2)^{-1}, (-2)^{-2},...\}$ and $\frac{1}{3}$ equal the sum of all positive elements of $\{(-2)^{-1}, (-2)^{-2},...\}$
and since $|T_{(n+1)}|=2^{(n+1)}$ we get that $S_{(n+1)}$ represent all $T_{(n+1)}$.
Now for general $r\in [-\frac{2}{3},\frac{1}{3}]$, If $r$ is rational with power of 2 denominator we finish (as show above). Else we'll find an infinte series $\{a_n\}\subseteq S$ s.t. $\sum a_n = r$
first we need to find the "right element for $r$ in $T_1$" - namely the unique element in $x_1\in T_1$ s.t. $r\in [x_1-\frac{1}{3}2^{-1},x_1+\frac{2}{3}2^{-1}]$, such $x_1$ exists and is unique since such segements (for differents $x\in T_1$) are distinct and cover exactly all of $[-\frac{2}{3},\frac{1}{3}]$.
After that, choose $x_2\in T_2$ to be the "right element for $r$ in $T_2$" (the only elemnt in $T_2$ s.t. $r\in [x_1-\frac{2}{3}2^{-2},x_1+\frac{1}{3}2^{-2}]$)
and procced like that for all $x_1,x_2,..$
After we have all $\{x_n\}$ we need to define $\{a_n\}\subseteq S$:
for $x_1$ decide if we need to add $(-2)^{-1}$ to our sum to make it eqaul to $x_1$, for $x_2$ decide if we need to add $(-2)^{-2}$ to our sum to make it eqaul to $x_2$ (it's possible since $x_2 \in \{x_1,x_1+(-2)^{-2}\}$)  and so on...
our sum of $\{a_n\}$ convergence to $r$ since the finite sums always eqauls to a number that lie with $r$ in a segement of length that become smaller and smaller.
The represention is unique: uniqness come from the fact that if $\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty b_n$ and $\{a_n\},\{b_n\}\subseteq S$ are sequences that are monotone decreasing (in absolute value) than $a_n=b_n$. Since, as we saw in the first paragraph, the maximal element of the sum choose certein segement from distinct segments that covers $\mathbb{R}$ to be the segment in which the number it represnts lie. Therefore $\sum_{n=2}^\infty a_n = \sum_{n=2}^\infty b_n$ and then $a_2=b_2$, and by induction every $a_n=b_n$.
