Complete Sets of Real Numbers: Find All

[Evgeny.Makarov]

Call a nonempty (finite or infinite) set $A\subseteq\Bbb R$ complete if for all $a,b\in\Bbb R$ such that $a+b\in A$ it is also the case that $ab\in A$. Find all complete sets.

 

[Opalg]

Call a nonempty (finite or infinite) set $A\subseteq\Bbb R$ complete if for all $a,b\in\Bbb R$ such that $a+b\in A$ it is also the case that $ab\in A$. Find all complete sets.

[sp]Since $A$ is nonempty it contains some real number $a$. Then $0+a = a\in a$, so the completeness condition implies that $0 = 0a\in A.$

Next, $x + (-x) = 0 \in A$, for every real number $x$. Therefore $-x^2 = x(-x) \in A$. But every negative number is of this form. Therefore $(-\infty,0] \subseteq A.$

Finally, $ \frac12y + \frac12y = y$. So if $y\in A$ then $\frac14y^2 =\bigl(\frac12y\bigr)^2 \in A$. But every positive number is of the form $\frac14y^2$ for some negative number $y$, and therefore belongs to $A$.

Conclusion: $A = \Bbb R.$
[/sp]

 

[Evgeny.Makarov]

Correct. This is a problem from the regional round of the 2016 Russian mathematical olympiad for junior high school students.