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

Question

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!

Answer 1

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 $f$.

Answer 2

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.

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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.