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I know that $\mathbb{R}^2$ with the post office metric is not separable. And the post office metric is defined by $m(x,y) = d(x,0) + d(y,0)$ for distinct points $x$ and $y$, and $m(x,x) = 0$ where $d$ is metric on $\mathbb{R}^2$.

My idea for the proof: For every $x \in \mathbb R^2$ we can find an $r_x$ such that $x$ is seperated from other elements of $\mathbb R^2$ (for instance if $r_x = d(x,0)$). And for a $D$ which is dense in $\mathbb R^2$ we can find a $y \in D$ st $y \in B(x,r_x)$. Then, we can match each $x \in \mathbb R^2$ with a $y \in D$. And since $\mathbb R^2$ is uncountable, $D$ is also uncountable. Thus, $\mathbb R^2$ is not separable since any dense set in it is uncountable.

Is my proof correct? I am not sure since $r_x$ depends on $x$.

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There's nothing wrong with $r_x$ depending on $x$. However, I think you're slightly obscuring the idea by talking about $y\in D\cap B(x,r_x)$. The point is that this $y$ must be $x$ itself (and you should say so explicitly); otherwise it might be that the same $y$ worked for many $x$'s and ruined the conclusion. Also, $\mathbb R^2\setminus\{0\}$ is actually dense, so you'll have to explicitly exclude $x=0$ from you argument. –  Henning Makholm Nov 17 '11 at 14:21
    
My answer has the same content of the comment. Unfortunately I was writing it while Henning posted. –  Giovanni De Gaetano Nov 17 '11 at 14:25
    
Thank you both for your help! –  marvinthemartian Nov 17 '11 at 14:42
    
sory but i didnt understand a partıcular point. every point in R^2 isolated wrt post office metric the only dense point is origin. So we have countable dense point and this means r^2 is seperable wrt post office metric. What is wrong with my idea –  user19663 Nov 17 '11 at 22:55
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It is almost correct. A small detail is that your machinery doesn´t work for the point $(0,0)$, exclude it! Choosing $r_x=\frac{1}{2} d(x,0)$ the unique point in $B(x,r_X)$ is x, so y=x. This is a bijection between $D$ and $\mathbb{R}^2 - {(0,0)}$, so the thesis.

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