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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a "non trivial" solution of the Cauchy functional equation, i.e. $f$ is not of the form $f(x)=cx$ for any $c\in\mathbb{R}$ and satisfies the relation $$f(x+y)=f(x)+f(y)\quad\forall\,x,y\in\mathbb{R}$$ Define $$A=\{x\in\mathbb{R}:f(x)\geq 0\},\quad B=\{x\in\mathbb{R}:f(x)< 0\}$$ Now $\{A,B\}$ is a partition of $\mathbb{R}$ into two non empty, closed by sum subsets, and it's easy to see that it's different from the trivial ones $\left(\{\mathbb{R}_{\geq 0},\mathbb{R}^-\},\{\mathbb{R}_{\leq 0},\mathbb{R}^+\}\right)$. In particular $$\mathbb{R}^+=(A\cap\mathbb{R}^+)\cup(B\cap\mathbb{R}^+)$$ is a partition of $\mathbb{R}^+$ into two non empty, closed by sum subsets.

My question: It's possible to partition $\mathbb{R}^+$ into two non empty set, both closed by sum and product?

The analogous question for $\mathbb{R}$ has negative answer, and the argument is pretty easy: we can assume $0\in A$, and suppose that there exists $x\in B$. Now $-x\in A$ (otherwise $-x+x=0\in B$), and then $x^2=(-x)^2\in A$, which is a contradiction.

I guess the answer to my question is negative (it seems related to the absence of nontrivial field automorphism of $\mathbb{R}$), but i don't find any efficient argument.

Thank you for all suggestions.

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Yes, it is possible to construct such a partition. See this note by Daniel M. Kane. Any proof would have to involve the axiom of choice, otherwise you can't partition the positive reals into subsets closed under addition, let alone both addition and multiplication.

Added later: There is a recent paper, "Decomposing the real line into Borel sets closed under addition", where the authors show that the only partitions of $\mathbb R$ into countably many (Borel) sets closed under addition are of the form $\mathbb R_+\cup \{0\}\cup \mathbb R _-$, etc. And they mention that the structure of all partitions of $\mathbb R_+$ into two sets closed under addition and multiplication was determined in a paper by G. Kiss, G. Somlai, T. Terpai, which is still in preparation. (Here is G. Kiss's website.)

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