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In $\mathbb{R}^2$, the Taxicab metric is defined by $d(x,y) = |x_1 - y_1| + |x_2 - y_2|.$ In this metric, describe/draw the open unit ball, closed unit ball, and the unit sphere which are based at the origin with radius $1$.

Here is my work:

Since it says this problem based at the origin,

Open unit ball is $B(0,1) = \{x \in M : d(x,0) < 1\}$ which means open ball with center origin and radius $1$.

Closed unit ball is $B[0,1] = \{x \in M : d(x,0) <= 1\}$

Unit sphere is $S(0,1) = \{x\in M : d(x,0) = 1\}$

But I'm not sure how to draw them. I draw a dotted circle for the open ball with center $(0,0)\,$ and $r = 1$. But I doubt because I draw within negative quadrant. Like just a circle with $(0,0)\,$ and $r = 1$. I hope you understand my description....

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  • $\begingroup$ I think I learned to call that the Manhattan metric? “Taxicab metric” seems like a bad match in some places, such as Boston and many old European cities. $\endgroup$ – Harald Hanche-Olsen Sep 13 '13 at 20:01
  • $\begingroup$ @Harald: I’ve seen both terms, but taxicab metric is much more familiar. $\endgroup$ – Brian M. Scott Sep 13 '13 at 20:02
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Do you know what the taxicab metric is? Find its definition and try to find lots of points in $\mathbb R^2$ that will have distance precisely $1$ from the origin. That will be the boundary of $B(0,1)$, and it will not look anything like a circle.

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  • $\begingroup$ Then a rectangular $\endgroup$ – therexists Sep 13 '13 at 20:03
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    $\begingroup$ @therexists: In what orientation? $\endgroup$ – Brian M. Scott Sep 13 '13 at 20:03
  • $\begingroup$ | | | ---------------------- $\endgroup$ – therexists Sep 13 '13 at 20:03
  • $\begingroup$ Sorry I tried to draw what I drew. Um, (0,0) and x = 1 , y = 1 $\endgroup$ – therexists Sep 13 '13 at 20:04
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    $\begingroup$ @therexists: But the point $\langle 1,1\rangle$ is $2$ units away from the origin, not $1$. $\endgroup$ – Brian M. Scott Sep 13 '13 at 20:05
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Start with the points on the coordinate axes (either $x=0$ or $y=0$). Then compute the points where $x=\pm y$. If you don't see it yet, choose some values for $x$ between zero and one and compute $y$ through \begin{gather} |y| = 1-|x|. \end{gather} Here, I used $x$ and $y$ as the two components of the vectors in $\mathbb R^2$.

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