Does $A$ open, non-empty, $\frac {A+A} 2=A+1$ imply $A=\mathbb R$? This is elementary and it may have an easy proof. Does $A$ open,non-empty in $\mathbb R$, $\frac {A+A} 2=A+1$ imply $A=\mathbb R$? There are many sets satisfying this equation: $\mathbb Q$, dyadic rationals etc. I can show that if $A$ satisfies this equation and $A$ is closed then $A=\mathbb R$. I believe that the same is true if $A$ is open. I can show that $\sup A =\infty$ and $\inf A=-\infty$, but I have not been able to show that $A=\mathbb R$. Notations: $\frac {A+A} 2 =\{\frac {a+b} 2:a,b \in A\}$, $A+1=\{a+1:a \in A\}$. Thanks in advance. 
 A: The statement is true, and the assumption that $A$ is open can be relaxed to $A$ having non-empty interior.
Let $I\subseteq A$ be an open interval. Since $A\subseteq \frac{A+A}{2}$, we get $I\subseteq A+1$, or in other words, $I-1\subseteq A$. Similarily, $I-n\subseteq A$ for any $n\in \Bbb N$ (in other words, starting at $0$, any integer translation of $I$ towards $-\infty$ is contained in $A$).
Now we get $\frac{(I-n) + (I-m)}{2} = I-\frac{n+m}2\subseteq A+1$, which gives that $I-\frac{n}2\subseteq A$ for any $n\geq 2$ (in other words, starting at $1$, any half-integer translation of $I$ towards $-\infty$ is contained in $A$).
Continuing this way, we find denser and denser translations of $I$ contained in $A$ (starting at $2$, any quarter-integer translation of $I$ towards $-\infty$ is contained in $A$, and so on). Eventually they are so closely packed that they overlap, which means that there is some $r\in \Bbb R$ such that $(-\infty, r)\subseteq A$
If the complement of $A$ is empty, we are done. If not, take an $x\notin A$. Then $x+1\notin A+1$. But then we can find $a, b\in A$ such that $\frac{a+b}2 = x+1$, which is a contradiciton.
Finding $a, b$ can be done as follows: first find $b\in A$ such that $\frac{r+b}{2}>x+1$, which can be done since $A$ is unbounded above ($\sup A = \infty$). Now set $a = r-2\left(\frac{r+b}2-(x+1)\right)$. We get $a<r$, so $a\in A$, and $\frac{a+b}2 = x+1$.
