I just need to clarify that i've sketched these open balls correctly, the metric is given by:
$$d(x,y) = \begin{cases} |x_2-y_2|, & \text{if $x_1 = y_1$;} \\
|x_2| + |y_2| + |x_1-y_1|, & \text{if $x_1 \neq y_1 $} \end{cases}$$
And i have to sketch for $A = \{x \in \mathbb{R}^2 \colon d(x,(0,1)) = 2\}$
also $B = \{x \in \mathbb{R}^2 \colon d(x,(2,1)) \leq 1\}$
Would the first 'ball' be a diamond crossing through $(0,1)$, $(1,0)$, $(0,-1)$, $(-1,0)$ with all points on the boundary of the diamond in the set except the point $(0,1)$? Also $(0,3)$ would be in the set too right?
also for the second 'ball', would we have a diamond, with center point $(2,1)$ and it would include the boundary and all points inside?
Best Answer
I got same as you for the set $A$. But for set $B$ it's only the vertical line segment connecting $(2,0)$ to $(2,2)$ [including endpoints]. Using $(x_1,x_2)$ for the coordinates of $x$ and $(y_1,y_2)=(2,1)$, the condition $x_1 \neq y_1$ becomes $x_1 \neq 2$, and the distance being at most $1$ becomes the inequality $$|x_2|+1+|x_1-2| \le 1,$$ which is only satisfied at $(x_1,x_2)=(2,0).$ This is already in conflict with the fact that we're at this point doing the case $x_1 \neq 2,$ but even throwing it in has no effect as it's already on the segment joining $(2,0)$ to $(2,2).$