If $0 \in G$ is an open subset of $\mathbb{R}$ and if $ x + y \in G$ for all $x, y\in G $ then $G=\mathbb{R}$. How, I can  prove !
If $0 \in G$ is an open subset of $\mathbb{R}$ and if $ x + y \in G$ for all $x, y\in G $ then  $G =\mathbb{R}$.
Plz help !
 A: Since $0 \in G$ and $G$ is open there exists $\varepsilon>0$ such that $B(0,\varepsilon) \subseteq G$ (where $B(0,\varepsilon) := \{x \in \mathbb{R}; |x| <\varepsilon\}$ denotes the open ball of radius $\varepsilon > 0$ centered at $0$).
Now let $x \in \mathbb{R}$ arbritary. There exists $k \in \mathbb{N}$ such that $\left| \frac{x}{k} \right| < \varepsilon$, hence $\frac{x}{k} \in  B(0,\varepsilon) \subseteq G$. Now we can write
$$x = \underbrace{\frac{x}{k} + \ldots+ \frac{x}{k}}_{k \, \text{times}}$$
and conclude $x \in G$ since each of the summands is in $G$ (and $G$ is closed under addition by assumption). Hence $\mathbb{R} \subseteq G$.
A: $G$ has a negative number and a positive number, because $G$ is open. It is possible to find a closed interval $[a,b]\subset G$ with $a<0$ and $b>0$. Now because for any strict positive number $x$, $0< \frac{x}{\lceil\frac{x}{b}\rceil}\leq b$, so $f(x)=\frac{x}{\lceil\frac{x}{b}\rceil}\in G$.
For every positive integer $z$ we have that $zy\in G$ if $y\in G$, so letting $y=f(x)$ and $z=\lceil\frac{x}{b}\rceil$ makes $x\in G$.
The same reasoning holds for negative numbers $x$.
