Complete Sets of Real Numbers: Find All

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SUMMARY

The discussion centers on identifying complete sets of real numbers, defined as nonempty sets \( A \subseteq \mathbb{R} \) where for all \( a, b \in \mathbb{R} \) such that \( a + b \in A \), it follows that \( ab \in A \). The analysis concludes that the only complete set is \( A = \mathbb{R} \). The reasoning involves demonstrating that \( 0 \) and all negative numbers must be included in \( A \), leading to the inclusion of all positive numbers as well.

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

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