Subset of the reals that contains $x$ or $-x$ for every non-zero $x$, and is closed under finite addition Is there a subset $A$ of the real numbers such that

*

*$(\forall \ x\in\mathbb{R}\setminus\{0\})$ exactly one of $x$ and $-x$ belongs to $A$;

*$A$ is closed under finite addition: $A+A\subseteq A$, or, in other words, for all $x_1, \ldots , x_n$ in $A$ with $n\in\mathbb{N}$, their sum $\sum_{k=1}^nx_n$ belongs to $A$; and

*$A$ contains a strictly positive and a strictly negative number.

I'd be very grateful for any hints or answers!
edit: Clarifications and improvements thanks to Torsten Schoeneberg and alvoi.
 A: Let $(x_i)_i$ be a Hamel basis of $\Bbb{R}$ where at least one $x_i$ is positive and at least one $x_i$ is negative, choose a well ordering $\prec$ on the index set $I$ and let
$$
A
= \bigg\{ \sum_{i \in I} c_i x_i : \text{ all but finitely many $c_i$ are zero and  the nonzero $c_i$ with the smallest $i$ is positive}\bigg\}.
$$
I think this should work, but don't have the time right now to check the details.
A: Your question is clearly equivalent to: does there exist another total ordering of the group $(\Bbb R,+)$ than the usual one or its opposite?
Thus, to give an example a subset $A$ satisfying your three properties, it suffices to take (using A.C.) a non-"trivial" automorphism $f$ of that group (i.e. $f$ not of the form $x\mapsto ax$) and put $A=f(\Bbb R_{\ge0}).$
A: I think the answer is yes. Let's well-order $\mathbb R\setminus\{0\}$ as $\langle a_\alpha.\ \alpha<2^{\aleph_0}\rangle$. We want to construct sets $A_\alpha$ for $\alpha<2^{\aleph_0}$ with the following conditions:

*

*$A_{\alpha}$ contains both a positive and a negative number;

*for all $\beta<\alpha$, either $(-a_\beta)∈ A_\alpha$ or $(a_\beta)∈ A_\alpha$;

*$A_\alpha$ is closed under finite additions.

Obviously if we can do this, then $A:=\bigcup_{\alpha<2^{\aleph_0}} A_\alpha$ is a set such that:

*

*$A$ contains both a positive and a negative number;

*for all $x∈\mathbb R\setminus\{0\}$, either $(-x)∈ A$ or $(x)∈ A$;

*$A$ is closed under finite additions.

We define the $A_\alpha$'s by transfinite induction. To start, we have to build $A_0$. We take
$$A_0:=\{n\sqrt2-m\sqrt3.\ n,m\ \text{non negative integers not both zero}\}$$
We basically forced conditions 1. and 3. to be true. Condition 2. is vacuously true.
Now, if we have defined $A_{\alpha}$ for $\alpha<2^{\aleph_0}$, to define $A_{\alpha+1}$ we consider $a_\alpha$. There are three possibilities:
a. if $a_\alpha∈ A_\alpha$, we take $A_{\alpha+1}:=A_\alpha$;
b. if $(-a_\alpha)∈ A_\alpha$, we take $A_{\alpha+1}:=A_\alpha$;
c. if $a_\alpha\not∈ A_{\alpha}$ and $(-a_\alpha)\not∈ A_\alpha$, we define
$$A_{\alpha+1}:=\{na_\alpha+mz. n,m\ \text{non negative integers not both zero},\ z∈ A_{\alpha}\}$$
In cases a. and b. the conditions are still true for inductive hypotesis, in case c. we forced $A_{\alpha+1}$ to contain $a_\alpha$ (so 2. is true) and to be closed under finite additions (so 3. is true). 1. is always true since $A_\alpha\subseteq A_{\alpha+1}$.
If instead $\beta<2^{\aleph_0}$ is a limit ordinal, just define $A_\beta:=\bigcup_{\alpha<\beta} A_\alpha$. Then 1. is trivially true, 2. is true by induction and 3. is true since $\langle A_\alpha.\ \alpha<\beta\rangle$ is an ascending chain. So we finished
If you don't suppose AC I don't think you can build this set $A$.
