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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 !

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You should say that $G$ is an open subset of $\mathbb{R}$, otherwise there are other possibilities. – Michael Albanese Dec 29 '12 at 12:33
up vote 17 down vote accepted

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 $\mathbb{R} \subseteq G$.

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$G$ 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 $x$.

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I don't think this works. For instance, what is $f(1)$ if $b=2$? And likewise, taking $b=2$ again, we get $f(3) = 3/\lfloor(3/2) \rfloor = \frac{3}{1} = 3 \nleq 2$. – Clive Newstead Jan 2 '13 at 18:06
Indeed. I had to use ceil in stead of floor, then it works. I'll fix it. – barto Jan 4 '13 at 18:30

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