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On $\mathbb{R}^2$ we have a metric defined by $d(x,y)=|x_1- y_1|+ |x_2- y_2|$. Describe and illustrate $B_1(0,0)$, the ball of radius $1$ centered at the origin $(0,0)$.

SOLUTION By definition $B_1(0, 0)=\{(x,y)\in \mathbb{R}^2 : |x-0|+|y-0|<1\}$ Since the ball is at the origin and it has the radius one, then I find that it is the set of all points within the rectangle having the corner points $(0, 1)$, $(1, 0)$, $(0,-1)$, and $(-1,0)$ Is my idea correct? Thanks.

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Yes, it is. $\ \ $ – David Mitra Apr 14 '12 at 15:02
Please don't tag your questions on metric spaces as (functional-analysis), please use (metric-spaces) instead. Functional analysis is a topic taught to advanced undergraduates and graduate students. – t.b. Apr 14 '12 at 15:05
So in the definition of $d$ we have $x = (x_1, x_2)$ and $y = (y_1, y_2)$. Right? – Thomas Apr 14 '12 at 15:17

Your answer is correct.

Though it wasn't asked for in your problem, a justification of your answer would be edifying:

The open unit ball is by definition $$ B(0,1)=\bigl\{\, (x,y)\in\Bbb R^2\,\bigl|\,|x|+|y|<1\,\bigr\}. $$ Towards determining geometrically what $B(0,1)$ is, let's consider cases:

Let's find the portion of $B(0,1)$ in the first quadrant of the Euclidian plane. If $(x,y)$ is in the first quadrant of the Euclidian plane, then $(x,y)\in B(0,1)$ if and only if $x+y<1$. If you sketch the graph of the equation $x+y=1$, you should be able to convince yourself that $Q_1\cap B(0,1)$ is the region depicted below (you want the region in Quadrant 1 with $x+y<1$, this would be the region in Quadrant 1 beneath the graph of the equation $x+y=1$).

enter image description here

If you consider the other cases according to the other three quadrants, you can establish that $B(0,1)$ is as you described and as depicted below (or, you could use a symmetry argument):

enter image description here

Note that since $B(0,1)$ is the open unit ball, points on the the boundary of the diamond above are not in $B(0,1)$.

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Your idea is correct. This metric is commonly known as the "Manhatten metric", because in Manhattan the driving distance from one point to another is roughly the difference in avenues plus the difference in streets (or whatever they call the lanes out there).

In $\mathbb{R}^2$, an open "ball" is "actually" an open rhombus. Now, this is wrong, of course, since under the manhattan metric a rhombus would be yet another form, so strictly speaking I should say that an open ball under the manhattan metric is an open rhombus under the euclidean metric.

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