Existence of a series which converge to any arbitrary point depending on sign Informally, my question is as follows:

Given a set of points $\{x_\alpha\}_{\alpha \in \mathcal{I}} \subseteq \mathbb{R}$, do there exists a sequence $(a_n)_{n \in \mathbb{Z}^+}$ such that $\sum_{n=1}^\infty \pm a_n$ converges to any $x_\alpha$ for a suitable choice of $\pm$ for each $a_n$?


This question is motivated by the well-known fact that given a conditionally real-valued convergent series $\sum_{n=1}^\infty a_n$, one can arbitrarily re-arrange the terms so that the series converge to any value in $\mathbb{R}$. Here, I do not allow the re-arranging of terms, but instead allow for the signs of each term to vary.
To formalise the problem, consider an arbitrary set of points $\{x_\alpha\}_{\alpha \in \mathcal{I}} \subseteq \mathbb{R}$. Do there exists a sequence $(a_n)_{n \in \mathbb{Z}^+}$ such that the image of the function:
\begin{align*}
f : \{-1,1\}^{\mathbb{Z}^+} &\to \mathbb{R} \\
(s_1,s_2,s_3,\dots) &\mapsto \sum_{n=1}^\infty s_na_n
\end{align*}
contains $\{x_\alpha\}_{\alpha \in \mathcal{I}}$ as a subset? 
Any insights provided will be appreciated.
 A: Any sequence $(a_n)$ with $\lim_{n\to\infty} a_n = 0$ and $\sum_{n=1}^\infty |a_n| =\infty$ will do.
Given any $x \in \mathbb R$, define $b_n=s_na_n$ and thus $s_n$ recursively by
$$b_n=
\begin{cases}
|a_n|  & \text{, if } \sum_{i=1}^{n-1} b_n \le x \\
-|a_n| & \text{, otherwise.}
\end{cases}
$$
Given an any $\epsilon > 0$, let $N_\epsilon$ be such that $\forall n > N_\epsilon: |a_n| < \epsilon$ (exists because of the limit condition on $(a_n)$).
If $\sum_{i=1}^{N_\epsilon} b_i \le x$, then $b_{N_\epsilon+1}$ will be non-negative, and the next $(b_i)$ will be as well until at some point $N^{(1)}_\epsilon > N_\epsilon$ we have $\sum_{i=1}^{N^{(1)}_\epsilon} b_i > x$ for the first time after $N_\epsilon$. This is due to the absolute series $\sum_{n=1}^\infty |a_n|$ diverging. 
Since $\sum_{i=1}^{N^{(1)}_\epsilon} b_i = b_{N^{(1)}_\epsilon} + \sum_{i=1}^{N^{(1)}_\epsilon-1} b_i$ and $b_{N^{(1)}_\epsilon}=|a_{N^{(1)}_\epsilon}| < \epsilon$ and $\sum_{i=1}^{N^{(1)}_\epsilon-1}b_i \le x$, we get
$$ x-\epsilon < x < \sum_{i=1}^{N^{(1)}_\epsilon} b_i < x+ \epsilon.$$
An analogous argument shows that such an $N^{(1)}_\epsilon$ exists if $\sum_{i=1}^{N_\epsilon} b_i > x$.
It's now easy to see by induction that for all $n \ge N^{(1)}_\epsilon$
$$ x-\epsilon  < \sum_{i=1}^{n} b_i < x+ \epsilon$$
holds. It's true for $n = N^{(1)}_\epsilon$, and if it is true for $n=k$, then if
$$\sum_{i=1}^{k} b_i \le x,$$
we have $b_{k+1}=|a_{k+1}| \ge 0$ and
$$x-\epsilon < \sum_{i=1}^{k} b_i \le  \sum_{i=1}^{k+1} b_i = \sum_{i=1}^{k} b_i + |a_{k+1}| \le x + |a_{k+1}| < x+\epsilon.$$
A similar argument works when $\sum_{i=1}^{k} b_i > x$.
So for the given $\epsilon$, we found an index $N^{(1)}_\epsilon$, such that the partial sums of our series $\sum_{n=1}^{\infty} b_n$ stay within an $\epsilon$-wide corridor of $x$ after $N^{(1)}_\epsilon$. That means
$\sum_{n=1}^{\infty} b_n = x.$
So one can arrange for the whole $\mathbb R$ to be expressable as limit of such a sign-changed series.
