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This question is a spin-off of this question. When trying to solve that question, we came up with the idea of construction functions using sets $A\subseteq \mathbb{R}$ having the properties that $2xy\in A$ for all $x,y \in A$ and $xy\in A$ for all $x\in A$ and $y\in\mathbb{R}\backslash A$. How would one characterize the set of such sets $A$? I would literally have no idea where to start, although the problem can be formulated so easily. I know that the sets $A=\varnothing$, $A=\{0\}$, and $A=\mathbb{R}$ qualify, but are these the only ones? Any ideas/suggestions? Thanks!

RMWGNE96
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    If $1\in A$ then also $1\cdot(\Bbb{R}\backslash A)\subset A$ and hence $A=\Bbb{R}$. Similarly, if $x\in A$ and $x^{-1}\notin A$ then $1=x\cdot x^{-1}\in A\cdot(\Bbb{R}\backslash A)\subseteq A$ and hence $A=\Bbb{R}$, a contradiction. So if $x\in A$ is nonzero then $x^{-1}\in A$. So if $A$ contains a nonzero element $x$, then also $2\cdot x\cdot x^{-1}\in 2\cdot A\cdot A\subseteq A$, and hence $2,2^{-1}\in A$. It follows that $2\cdot 4^k\in A$ for all $k\in\Bbb{Z}$. Just thinking 'out loud'... – Servaes May 20 '19 at 19:40
  • Also, if $0\notin A$ but there is a nonzero $x \in A$, then we have $0 \in (\mathbb{R}\backslash A)$ so $x \cdot 0 \in A$, a contradiction. So either $A=\varnothing$ or $0 \in A$. Which makes sense if you consider the original problem where we already determined that $f(0)=0$. – RMWGNE96 May 20 '19 at 19:52
  • Maybe one can argue that $m(A)<\infty$ with $m$ the Lebesgue measure is impossible. – RMWGNE96 May 20 '19 at 19:57

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I think there are no such non-trivial $A$. For simplicity assume $A \subset \mathbb R_+$ (if there is a non-zero number in $A$ then there is positive number in $A$, and if all positive numbers are in $A$ then $A = \mathbb R$).

  1. If $1 \in A$ then $\bar A\subset A \cdot \bar A$ and so $\bar A = \varnothing$. Thus $1 \notin A$.

  2. If $x \in A$ and $\frac{1}{x} \notin A$ then $x \cdot \frac{1}{x} \in A$. So if $x \in A$ then $\frac{1}{x} \in A$. Also if $x \in \bar A$ then $\frac{1}{x} \in \bar A$.

  3. If $x \in A$ and $2x \in A$ then $2 \cdot x \cdot \frac{1}{2x} \in A$. So if $x \in A$ then $2x \notin A$.

  4. If $x \in A$ and $\sqrt{x} \notin A$ then $x \cdot \frac{1}{\sqrt x} \in A$. So if $x \in A$ then $\sqrt{x} \in A$.

Take $x \in A$. From (4), $\sqrt x \in A$. From definition, $2 \cdot \sqrt x \cdot \sqrt x = 2 \cdot x \in A$. At other hand, from (3), $2 \cdot x \notin A$.

mihaild
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