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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.

Arthur
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    The whole line less $0$? – lulu Dec 09 '22 at 18:05
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    @lulu if it contains both $1$ and $-1$ it has to contain $0$. So the whole $\mathbb R$ is an answer to this question, but I think it isn't the one OP is asking for – alvoi Dec 09 '22 at 18:06
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    Regarding the "finite addition", does this also mean that for each $a \in A$, the numbers $a+a$, $a+a+a$, ... $n\cdot a$ for $n \in \mathbb N$ are in $A$? – Torsten Schoeneberg Dec 09 '22 at 18:09
  • Indeed, I want that exactly one of x and -x belongs to A, for every non-zero x. I had not accurately described this – sorry! I now slightly rephrased the question. – Arthur Dec 09 '22 at 18:09
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    @alvoi absolutely, thanks. – lulu Dec 09 '22 at 18:14
  • @Torsten Schoeneberg Yes! I mean that too. Thank you for pointing this out – I had inaccurately described that, too. I've now edited my question. – Arthur Dec 09 '22 at 18:14
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    @Anne Bauval I really want the 2nd condition in my post, if understood that n belongs to ℕ. I've clarified that – thanks for pointing this out. – Arthur Dec 09 '22 at 18:21
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    What kind of choice axiom are you assuming? Do you allow full axiom of choice? – Somos Dec 09 '22 at 19:00
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    @Somos: I am fine with assuming the full axiom of choice. Given the replies by Anne Bauval, PhoemueX and alvoi (Thank you very much – I am studying them now), I expect this to be necessary. – Arthur Dec 09 '22 at 19:10
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    Note that your 2nd condition can be simply written: $A$ is closed under addition (i.e. $A+A\subset A$). – Anne Bauval Dec 09 '22 at 22:33
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    @Anne Bauval: Right, thank you for pointing that out! I've edited the question using your suggestion. – Arthur Dec 09 '22 at 23:28

3 Answers3

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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.

PhoemueX
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  • Oh yeah you ninja'd me, I had a solution similar to yours in mind but I wasn't able to write it down! I think the one I came up with works, too! You clearly need AC to do it – alvoi Dec 09 '22 at 18:48
  • Thank you very much! I do understand why your $A$ is closed under finite additions (2nd condition), and why it contains a strictly positive and strictly negative real number (3rd condition). However, I fail to understand why it satisfies the first condition. Is there no real $y$ that has an infinite (non-zero) decomposition in this Hamel basis? If so, for such $y$, the real $-y$ also has an infinite decomposition, and so neither $y$ nor $-y$ belong to $A$. What do I miss? – Arthur Dec 09 '22 at 19:46
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    @Arthur: Hamel basis means every real number is expressible as a finite $\Bbb{Q}$-linear combination of the basis. – PhoemueX Dec 09 '22 at 19:48
  • Oh, thank you! I'm learning much – really appreciated! – Arthur Dec 09 '22 at 19:48
  • Right, of course! If a vector space has a basis, any vector can be written in a unique way as a finite linear combination of basis vectors. Definition of a basis...

    Thank you for your answer! I am convinced it works.

     – Arthur Dec 09 '22 at 19:58
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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 $A=f(\Bbb R_{\ge0}).$

Anne Bauval
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    Note that not every ordering arises in this way, though (that is, not every ordering is isomorphic to the usual one). For instance, the ordering used in PhoemueX's answer is not isomorphic to the usual one because it is not Archimedean. There are also Archimedean orderings that are not isomorphic to the usual one, since you can take a homomorphism $\mathbb{R}\to\mathbb{R}$ that is injective but not surjective (which gives an Archimedean ordering that is not complete). – Eric Wofsey Dec 09 '22 at 20:08
  • Thank you for these interesting remarks. This is why my solution is simpler. – Anne Bauval Dec 09 '22 at 22:28
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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:

  1. $A_{\alpha}$ contains both a positive and a negative number;
  2. for all $\beta<\alpha$, either $(-a_\beta)∈ A_\alpha$ or $(a_\beta)∈ A_\alpha$;
  3. $A_\alpha$ 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:

  1. $A$ contains both a positive and a negative number;
  2. for all $x∈\mathbb R\setminus\{0\}$, either $(-x)∈ A$ or $(x)∈ A$;
  3. $A$ 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 $A$.

alvoi
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