# Let $a\in\mathbb R^n$ be a fixed point. How to prove $B(a, 1/2)\cap B(g+a, 1/2)=\emptyset$ where $g\in \mathbb Z^n-\{0\}$?

Let $a\in\mathbb R^n$ be a fixed point. How to prove $B(a, 1/2)\cap B(g+a, 1/2)=\emptyset$ for some $g\in \mathbb Z^n-\{0\}$?

It sounds a silly question, and obvious one, but it is a fact I need for proving a result envolving group action..

I know I should proceed by contradiction but I wasn't able to show it. I have a feeling the contradiction will be $g=0$. Can anyone help me..

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Draw a picture. –  1015 Mar 24 '13 at 20:49
No, I need a formal proof despite the result being obvious.. –  PtF Mar 24 '13 at 20:50
A good picture is often the first step towards a formal proof. –  1015 Mar 24 '13 at 20:52
I did several pictures and nothing ahaha –  PtF Mar 24 '13 at 20:56

Take $x$ in the intersection. That is $\|x-a\|<\frac{1}{2}$ and $\|x-a-g\|<\frac{1}{2}$. Then $$\|g\|=\|(x-a)-(x-a-g)\|\leq\quad ?$$

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$\|g\|\leq \|x-a\|+\|x-(a+g)\|<1$ then? –  PtF Mar 24 '13 at 20:54
@PtF There you go. Triangular inequality. And since $g\neq 0$... –  1015 Mar 24 '13 at 20:54
but what contradiction that gives me? –  PtF Mar 24 '13 at 20:55
@PtF Since $g\neq 0$ has integer coordinates, $\|g \|^2=k_1^2+\ldots+k_n^2\geq 1$ because at least one $k_j$ is a nonzero integer. –  1015 Mar 24 '13 at 20:56
Now I see ashashas, thanks, how didn't I think about it? Thanks for the help, that saved my life hahaha –  PtF Mar 24 '13 at 20:58

Hint:

Assume $x\in B(a,1/2) \cap B(g+a,1/2)$. Then $\left|x - a\right| < 1/2$ and $\left|x - (g + a)\right| < 1/2$. Use the triangle inequality to conclude something about $\left| g\right|$.

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I did that but I got nowhere... –  PtF Mar 24 '13 at 20:51
I've added more details. –  azimut Mar 24 '13 at 20:54
I dit that too and I get only $|g|<1$, but I can't see why this is a contradiction.. –  PtF Mar 24 '13 at 20:57
You're almost done then. There is some precondition on $g$... –  azimut Mar 24 '13 at 20:57
I can't believe I didn't see what Julien showed me: since $g\neq 0$ is an integer $g=(g_1, \ldots, g_n)$ with some $g_i\neq 0$ therefore $\|g\|^2=g_1^2+\ldots+g_n^2\geq 1$, a contradiction.. Thanks anyway =D –  PtF Mar 24 '13 at 21:01