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In $\mathbb{R}^2$ sketch $B((1,2),3)$, the open ball of radius $3$ at the point $(1,2)$, with the following metrics:

a.) the post-office metric given by $$d(x,y) = \left\{ \begin{array}{l l} \sqrt{x_1^2+x_2^2}+\sqrt{y_1^2+y_2^2}, & \quad \text{if $x\neq y$}\\ 0, & \quad \text{if $x=y$} \end{array} \right.$$ for $x=(x_1,x_2), y=(y_1,y_2)\in\mathbb{R}^2$.

b.) the metric $$d(x,y)=\frac{5\| x-y\|_2}{1+\| x-y\|_2}.$$

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Should it be $\|x-y\|^2$? – Sigur Feb 22 '13 at 2:09
@Sigur I think there is no problem with assuming it is the Euclidean distance. This does define an equivalent distance. – 1015 Feb 22 '13 at 2:11
According to the question it's $\parallel x-y\parallel_2$. – Luis_G Feb 22 '13 at 2:12

(a) circle of radius $\sqrt{0.58}$ centered at $(0,0)$

(b) circle of radius $\frac12$ centered at $(1,2)$
See here for the calculation in (b).

enter image description here

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