Computing Balls in Euclidean, Taxicab, and Sup Metrics

One Center and Radius, Three Different Balls

A metric ball is defined by distance, but its coordinate shape depends on which metric measures that distance. In the plane, the Euclidean, taxicab, and sup metrics therefore give three different pictures for balls with the same center and radius. The key is to translate each distance inequality into a condition on the coordinates.

We work with \(\mathbb{R}^2\) equipped, in turn, with each of these metrics and its induced topology. For a center \(c\in\mathbb{R}^2\) and radius \(r>0\), the open ball is the set of points at distance strictly less than \(r\) from \(c\). Write \(c=(a,b)\) and \(z=(x,y)\), so the coordinate differences are \(x-a\) and \(y-b\).

Definition: For a metric \(d\), the open ball with center \(c\) and radius \(r>0\) is \(B_d(c,r)=\{z:d(c,z)<r\}\). The strict inequality matters: points at distance exactly \(r\) are not in the open ball.

The three distance formulas on \(\mathbb{R}^2\) are \(d_2(c,z)=\sqrt{(x-a)^2+(y-b)^2}\), \(d_1(c,z)=|x-a|+|y-b|\), and \(d_\infty(c,z)=\max\{|x-a|,|y-b|\}\). They measure, respectively, Euclidean distance, the sum of the coordinate errors, and the largest coordinate error.

Coordinate Descriptions and Their Boundaries

Theorem: For \(r>0\), the three open balls centered at \((a,b)\) have the following coordinate descriptions:

\begin{aligned} B_{d_2}((a,b),r)&=\{(x,y):(x-a)^2+(y-b)^2<r^2\},\\ B_{d_1}((a,b),r)&=\{(x,y):|x-a|+|y-b|<r\},\\ B_{d_\infty}((a,b),r)&=\{(x,y):|x-a|<r\text{ and }|y-b|<r\}. \end{aligned}

Their boundaries are, respectively, the circle \((x-a)^2+(y-b)^2=r^2\), the diamond \(|x-a|+|y-b|=r\), and the square \(\max\{|x-a|,|y-b|\}=r\).

Proof. By definition, \((x,y)\) is in the Euclidean ball exactly when \(\sqrt{(x-a)^2+(y-b)^2}<r\). Both sides are nonnegative, so squaring preserves the inequality and gives \((x-a)^2+(y-b)^2<r^2\). For the taxicab metric, its defining distance formula gives \(|x-a|+|y-b|<r\) directly. For the sup metric, \(\max\{|x-a|,|y-b|\}<r\) holds exactly when both entries in the maximum are less than \(r\), giving the two strict inequalities in the theorem.

To verify the boundary descriptions, use the definition of boundary as the set of points every open neighborhood of which meets both the ball and its complement. In any metric, if \(d(c,z)<r\), then every point \(w\) with \(d(z,w)<r-d(c,z)\) satisfies \(d(c,w)\leq d(c,z)+d(z,w)<r\), so \(z\) is inside the ball rather than on its boundary. If \(d(c,z)>r\), then every \(w\) with \(d(z,w)<d(c,z)-r\) satisfies \(d(c,w)\geq d(c,z)-d(z,w)>r\), so \(z\) is outside the ball and not on its boundary.

It remains to check points with \(d(c,z)=r\). For each of the three metrics, scaling the coordinate difference vector by a nonnegative number \(s\) scales its distance from \(c\) by \(s\). For \(0<t<1\), let \(z_-\) and \(z_+\) have coordinate differences \((1-t)(z-c)\) and \((1+t)(z-c)\), respectively. Then \(d(c,z_-)= (1-t)r<r\) and \(d(c,z_+)=(1+t)r>r\). Also \(d(z,z_-)=tr\) and \(d(z,z_+)=tr\). Taking \(t\) arbitrarily small puts both points in every neighborhood of \(z\), one inside and one outside the ball. Thus every point at distance \(r\) is on the boundary, proving all three descriptions. \(\square\)

The equations also tell us how to sketch. The Euclidean boundary meets the horizontal and vertical lines through its center at distance \(r\), producing a circle. The taxicab boundary has vertices \((a+r,b)\), \((a-r,b)\), \((a,b+r)\), and \((a,b-r)\); its four straight edges form a diamond. The sup boundary is made of the four sides \(|x-a|=r\) or \(|y-b|=r\), forming a square whose sides are parallel to the coordinate axes. In every case, the ball itself consists of the points strictly inside its boundary.

Worked Example: Three Balls Centered at the Origin

Take center \((0,0)\) and radius \(2\). Compute the three balls and describe how to sketch each one.

For the Euclidean metric, substitute \(a=0\), \(b=0\), and \(r=2\) into the coordinate description. The result is \(x^2+y^2<2^2\), or \(x^2+y^2<4\). Its boundary is \(x^2+y^2=4\), a circle with horizontal and vertical intercepts \((2,0)\), \((-2,0)\), \((0,2)\), and \((0,-2)\). For example, the boundary point \((2,0)\) gives \(2^2+0^2=4\).

For the taxicab metric, the condition is \(|x|+|y|<2\), and the boundary is \(|x|+|y|=2\). Setting \(y=0\) gives \(|x|=2\), so the horizontal vertices are \((2,0)\) and \((-2,0)\). Setting \(x=0\) gives \(|y|=2\), so the other vertices are \((0,2)\) and \((0,-2)\). Between these vertices the boundary consists of straight diagonal segments. For instance, \((1,1)\) satisfies \(|1|+|1|=2\), confirming it lies on the boundary, not in the open ball.

For the sup metric, the condition is \(\max\{|x|,|y|\}<2\), equivalently \(-2<x<2\) and \(-2<y<2\). Its boundary consists of the four sides of the square with vertices \((2,2)\), \((2,-2)\), \((-2,2)\), and \((-2,-2)\). At \((2,2)\), the maximum of the two absolute coordinate values is \(\max\{2,2\}=2\), so this vertex is on the boundary and is excluded from the open ball.

Check: the point \((0,0)\) has distance \(0\) from the center in all three metrics, and \(0<2\), so it belongs to each open ball. The point \((1,1)\) has Euclidean distance \(\sqrt{1^2+1^2}=\sqrt{2}<2\), taxicab distance \(|1|+|1|=2\), and sup distance \(\max\{|1|,|1|\}=1<2\). Thus it is inside the Euclidean and sup balls, but not the taxicab ball, as the equations predict.

Worked Example: Translating the Shapes

Find the coordinate descriptions and sketching vertices for radius-\(3\) balls centered at \((1,-1)\).

For the Euclidean ball, the differences from the center are \(x-1\) and \(y-(-1)=y+1\). Hence its condition is \((x-1)^2+(y+1)^2<3^2\), or \((x-1)^2+(y+1)^2<9\). The circle boundary has horizontal intercepts relative to its center at \((1+3,-1)=(4,-1)\) and \((1-3,-1)=(-2,-1)\), and vertical intercepts at \((1,-1+3)=(1,2)\) and \((1,-1-3)=(1,-4)\).

For the taxicab ball, the condition is \(|x-1|+|y+1|<3\). Its diamond vertices are \((1+3,-1)=(4,-1)\), \((1-3,-1)=(-2,-1)\), \((1,-1+3)=(1,2)\), and \((1,-1-3)=(1,-4)\). For example, the vertex \((4,-1)\) gives \(|4-1|+|-1+1|=|3|+|0|=3\), as required for the boundary.

For the sup ball, the condition is \(|x-1|<3\) and \(|y+1|<3\). These inequalities become \(-2<x<4\) and \(-4<y<2\). The square's vertices are therefore \((4,2)\), \((4,-4)\), \((-2,2)\), and \((-2,-4)\). For instance, at \((4,2)\), the coordinate differences are \(4-1=3\) and \(2+1=3\); their maximum absolute value is \(3\), so that vertex is on the boundary.

Check: the center \((1,-1)\) satisfies each ball's inequality: both coordinate differences are \(0\), and each resulting distance is \(0<3\). The point \((2,1)\) has differences \(2-1=1\) and \(1+1=2\). Its Euclidean distance is \(\sqrt{1^2+2^2}=\sqrt{5}<3\), its taxicab distance is \(|1|+|2|=3\), and its sup distance is \(\max\{|1|,|2|\}=2<3\). It belongs to the Euclidean and sup balls but lies on the taxicab boundary.

Worked Example: Testing Membership in All Three Metrics

Let the center be \((2,-1)\), the radius be \(6\), and test \(p=(5,3)\) and \(q=(5,5)\) for membership in each open ball.

For \(p=(5,3)\), the coordinate differences are \(5-2=3\) and \(3-(-1)=4\). Its Euclidean distance is \(\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\), and \(5<6\), so \(p\) is in the Euclidean ball. Its taxicab distance is \(|3|+|4|=3+4=7\), and \(7\) is not less than \(6\), so \(p\) is not in the taxicab ball. Its sup distance is \(\max\{|3|,|4|\}=\max\{3,4\}=4<6\), so \(p\) is in the sup ball.

For \(q=(5,5)\), the coordinate differences are \(5-2=3\) and \(5-(-1)=6\). Its Euclidean distance is \(\sqrt{3^2+6^2}=\sqrt{9+36}=\sqrt{45}\). Since \(45>36=6^2\), this distance is greater than \(6\), so \(q\) is not in the Euclidean ball. Its taxicab distance is \(|3|+|6|=3+6=9>6\), so \(q\) is not in the taxicab ball. Its sup distance is \(\max\{|3|,|6|\}=\max\{3,6\}=6\), so \(q\) is on the sup boundary, not inside the open ball.

Check: the calculated results respect the strict membership test in every metric. In particular, equality occurs for the sup distance of \(q\), and \(6<6\) is false; therefore \(q\) is excluded from that open ball.

A Common Pitfall: Replacing the Metric or the Strict Inequality

A frequent error is to treat every ball in the plane as an ordinary Euclidean disk. For the radius-\(2\) taxicab ball centered at \((0,0)\), the incorrect calculation \(x^2+y^2<4\) would include \((1,1)\), because \(1^2+1^2=2<4\). But the taxicab distance from \((0,0)\) to \((1,1)\) is \(|1|+|1|=2\), exactly the radius. The correct condition is \(|x|+|y|<2\), so \((1,1)\) is on the diamond boundary and is excluded. Also do not replace \(<r\) with \(\leq r\): that would describe the ball together with its boundary, not the open ball.

For a reliable sketch, first write the inequality for the specified metric. Then locate the equality boundary, using intercepts for the circle or diamond and side equations for the square. Finally, shade the region satisfying the strict inequality. The boundaries differ because the metrics aggregate coordinate errors differently: by squares and a square root, by a sum of absolute values, or by a maximum.

Check Your Understanding

Use the strict distance inequality when deciding membership.

Write the three coordinate descriptions for radius-\(1\) balls centered at \((-2,3)\).
What are the four boundary vertices of the radius-\(5\) taxicab ball centered at \((0,0)\)?
For the radius-\(4\) sup ball centered at \((1,2)\), give the four vertices of its boundary square.
For the Euclidean ball centered at \((0,0)\) with radius \(3\), decide whether \((2,2)\) is inside, on the boundary, or outside.
Explain why a point at distance exactly \(r\) from a center is excluded from the open ball of radius \(r\), even though it is on the boundary.