14

Can $\mathbb{R}^{+}$ be divided into two disjoint nonempty sets so that each set is closed under both addition and multiplication?

I know if we only require both sets to be closed under addition then this can be done. For example, this post gives an answer.

Thank you very much!

Community
  • 1
Elf
  • 169
  • 1
    Well, at least up to the last point, the same proof seems to go through. – Berci May 17 '17 at 21:40
  • a clarification: Closed under addition means that if $a,b\in A$ then $a+b\in A$, do we allow the possibility that $a=b$? – Mirko May 17 '17 at 23:14
  • @Mirko Yes, I think so. – Elf May 17 '17 at 23:17
  • @Mirko Yes of course. The set ${1}$ is not closed under addition since $1 + 1 = 2$ isn't in it. The addition operator works on all pairs of elements of the given set. – user4894 May 18 '17 at 00:04
  • Say $A,B$ was such a partition with $1\in A$. It follows that $\frac1n\in A$ for all $n$, so $\mathbb Q\subset A\supseteq\mathbb Q A^j$, and $B\supseteq\mathbb Q B^j$ for all $j\ge1$. (Here to be precise, $\mathbb Q$ must denote only the positive rationals.) – Mirko May 18 '17 at 00:23
  • I think that if we look at $\mathbb R$ as a vector space over $\mathbb Q$ we have that both sets are $\mathbb Q$-convex. If we could prove that the only sets such that both the set and the complement are $\mathbb Q$ convex are the semihyperplanes we would have that $A$ and $B$ are the preimages of the non-negative and the positive reals under a suitable linear transform (such that the kernel is hyperplane $H$ that divides $\mathbb R$ in two halves. Picking a vector $v$ which extends the basis from a basis of $H$ we would have that $A$ and $B$ are of the form described in my solution. – Asinomás May 18 '17 at 00:31
  • Oh wait, that is false, at least maybe we can prove that they are the semihyperplanes such that the face has some missing dots. – Asinomás May 18 '17 at 00:33
  • 4
    Congratulations and welcome to MSE! It a truly marvelous achievement to become a member 11 months ago, not post anything since then till three hours ago and remember your password all that time, and then immediately get 10+ upvotes :) – Mirko May 18 '17 at 00:54

2 Answers2

5

This is an adaptation of the construction in the paper by Daniel Kane (linked in Jorge's answer).

Claim: There exists a non-trivial $\mathbb{Q}$-linear derivation on $\mathbb{R}$.

Before giving the proof, a non-trivial derivation $D$ lets us define a partition of $\mathbb{R}^+$ by $$ A=\big\{x\in \mathbb{R}^+:D(x) \geq 0\big\},\,\,\,\, B=\big\{x\in \mathbb{R}^+:D(x) < 0\big\}. $$ The sets $A$ and $B$ are closed under addition because $D$ is $\mathbb{Q}$-linear, and are closed under multiplication by the Leibniz rule. The partition is non-trivial because $D$ is non-trivial: if $x\in\mathbb{R}^+$ satisfies $D(x)\neq 0$, then $D(x)$ and $D(x^{-1})=-x^{-2}D(x)$ have opposite sign.

Proof of the claim: We construct a derivation with $D(\pi)=1$. Consider the set of pairs $(A,D)$, where $A$ is a subring of $\mathbb{R}$ containing $\pi$ and $D:A\to\mathbb{R}$ is a derivation satisfying $D(\pi)=1$. This set is non-empty because it contains $(\mathbb{Q}[\pi],\frac{d}{d\pi})$, and is partially ordered by extension. The set satisfies the hypotheses of Zorn's Lemma, so contains a maximal element $(A,D)$. It must be the case that $A=\mathbb{R}$, because if $x\in\mathbb{R}\backslash A$ we can extend $D$ to $A[x]$: if $x$ is transcendental over $A$ we set $D(x)=0$, and if $x$ is algebraic over $A$ with minimal polynomial $f$ we set $D(x)=-D(f)(x)/f'(x)$, where $D(f)$ is the polynomial obtained by applying $D$ to the coefficients of $f$.

Julian Rosen
  • 16,142
  • Thank you for this answer! I think this is clearer and easier to understand than the one in Kane's paper :) – Elf May 18 '17 at 17:22
4

This was answered three years ago at MO here. It is indeed possible.

The relevant links are this one, and this one, apparently a paper which characterizes the solutions is in preparation and will be published here, although its been in preparation for a long time now.

I'll leave my lame failed attempt:

Current attempt:

Think of $\mathbb R$ as a vector space over $\mathbb Q$, let $A$ and $B$ be the desired partition.

Clearly $span(A)\cup span(B)=\mathbb R$.

It follows wlog that $span(A)=\mathbb R$.

Pick a basis $a_i$ with index family $I$ of $\mathbb R$ consisting of elements of $A$.

It follows that every vector consisting of non-negative coefficients is in $A$.

Now suppose that a vector $a\in A$ contains a non-negative coefficient $\lambda_i$, then the vector $-a_i$ is also in $A$.

We can thus characterize $A$ as follows: Select a subset $X\subseteq I$, then $A$ is the set of vectors in $\mathbb R^+$ in which the coefficient belonging to each $a_i$ is non-negative for all $x\in X$.

We conclude that $B$ is the set of vectors in $\mathbb R^+$ such that at least one coefficient $a_x$ is negative. Of course if $|I|\neq 1$ we arrive at a contradiction, since we would be able to find two elements of $B$ whose sum is in $A$.

Conclusion: the sets $A$ and $B$ are formed in the following way:

Pick a basis of $\mathbb R$ with a distinguished vector $v$.

let $A$ be the subset of $\mathbb R^+$ containing non-negative $v$ coefficient.

let $B$ be the subset of $\mathbb R^+$ containing non-negative $v$ coefficient.

Asinomás
  • 105,651
  • If $span(A)=\Bbb R$ then we can add elements of $A$ to get an element of $B$. – Ross Millikan May 17 '17 at 23:40
  • If $a\in A$ has a non-negative coefficient $\lambda_i$, why must $-a_i$ be in $A$? – Julian Rosen May 17 '17 at 23:44
  • Thank you for the links and the attempt :) – Elf May 18 '17 at 17:26
  • sure no problem, I really didn't think it would be possible. I thought that the basis for $\mathbb R$ over $\mathbb Q$ would get really messed up by the transform $\mathbb R^+\rightarrow \mathbb R$ given by $x\rightarrow \log(x)$. (this is the isomorphism between addition and multiplication) – Asinomás May 18 '17 at 17:29