Open Balls in R^n

Measuring Distance from a Center

The previous tutorial introduced norms on $\mathbb{R}^n$, which measure the size of a vector. A norm also gives a distance between two points: the distance from $x$ to $a$ is $\|x-a\|$. An open ball collects the points whose distance from a chosen center is less than a specified positive radius. The norm determines how that distance is measured, so different norms can give balls with different shapes.

Definition: Let $\|\cdot\|$ be a norm on $\mathbb{R}^n$, let $a\in\mathbb{R}^n$, and let $r>0$. The open ball with center $a$ and radius $r$, relative to this norm, is $$ B_r(a;\|\cdot\|)=\{x\in\mathbb{R}^n:\|x-a\|<r\}. $$ When the norm is clear from context, write $B_r(a)$. For the standard one, Euclidean, and infinity norms, the corresponding balls may be written $B_r^{(1)}(a)$, $B_r^{(2)}(a)$, and $B_r^{(\infty)}(a)$.

The radius must be positive, and the inequality is strict. In particular, the center belongs to the ball because $\|a-a\|=0<r$. A point at distance exactly $r$ does not belong. The set of points at exactly that distance is the sphere of radius $r$ about $a$, not part of the open ball.

Three Norms, Three Shapes

The formulas for the three standard norms turn the definition into familiar coordinate conditions. In $\mathbb{R}^2$, a Euclidean ball is a disk, a one-norm ball is a diamond, and an infinity-norm ball is a square with sides parallel to the coordinate axes. The center and radius alone do not determine the ball unless the norm is specified.

Worked Example: A Euclidean Ball in the Plane

Take the Euclidean norm, center $a=(1,-2)$, and radius $3$. A point $(x,y)$ belongs to the ball precisely when

\sqrt{(x-1)^2+(y+2)^2}<3,

or, after squaring the nonnegative sides,

$$ (x-1)^2+(y+2)^2<9. $$

For example, the point $(3,-2)$ has distance $\sqrt{(3-1)^2+(-2+2)^2}=\sqrt{4}=2<3$, so it is inside. The point $(4,-2)$ has distance $\sqrt{(4-1)^2+(-2+2)^2}=\sqrt{9}=3$, so it is on the boundary and is not in the open ball.

Worked Example: An Infinity-Norm Ball Is a Square

Let $a=(1,-1)\in\mathbb{R}^2$ and $r=2$. Since $\|(x,y)-(1,-1)\|_\infty=\max\{|x-1|,|y+1|\}$, membership requires both coordinate differences to be less than $2$:

|x-1|<2 \quad\text{and}\quad |y+1|<2.

The first inequality is equivalent to $-1<x<3$, and the second is equivalent to $-3<y<1$. Thus \[ B_2^{(\infty)}((1,-1)) =\{(x,y):-1<x<3,\ -3<y<1\}. \] This is an open square. For instance, $(2,0)$ is inside because the two coordinate differences have absolute value $1$, while $(3,0)$ is outside because its first coordinate difference has absolute value $2$.

Worked Example: A One-Norm Ball Is a Diamond

At the origin in $\mathbb{R}^2$, the one-norm ball of radius $4$ is described by

B_4^{(1)}((0,0))=\{(x,y):|x|+|y|<4\}.

The point $(1,2)$ belongs to this ball because $|1|+|2|=3<4$. The point $(2,2)$ does not, because $|2|+|2|=4$, so it lies on the boundary. The boundary consists of four line segments joining $(4,0)$, $(0,4)$, $(-4,0)$, and $(0,-4)$; the strict inequality excludes all four segments from the ball.

Basic Geometry of Balls

Translation changes the center without changing the shape or radius. More precisely, adding a fixed vector $v$ to every point of a ball moves its center from $a$ to $a+v$. Scaling changes both the center and the radius. The absolute value of the scalar is essential: a negative scalar reverses directions, but a norm measures length without regard to direction.

Theorem (Translation and Dilation of Norm Balls): Let $\|\cdot\|$ be a norm on $\mathbb{R}^n$, let $a,v\in\mathbb{R}^n$, let $r>0$, and let $\alpha\neq0$. Then $$ B_r(a;\|\cdot\|)+v=B_r(a+v;\|\cdot\|) \quad\text{and}\quad \{\alpha x:x\in B_r(a;\|\cdot\|)\} =B_{|\alpha|r}(\alpha a;\|\cdot\|). $$

For the dilation identity, if $x\in B_r(a;\|\cdot\|)$, absolute homogeneity gives \[ \|\alpha x-\alpha a\|=|\alpha|\,\|x-a\|<|\alpha|r. \] Thus every point $\alpha x$ in the scaled set belongs to $B_{|\alpha|r}(\alpha a;\|\cdot\|)$. Conversely, suppose $y\in B_{|\alpha|r}(\alpha a;\|\cdot\|)$. Since $\alpha\neq0$, set $x=y/\alpha$. Then \[ \|x-a\|=\left\|\frac{y-\alpha a}{\alpha}\right\| =\frac{1}{|\alpha|}\|y-\alpha a\|<r. \] So $x\in B_r(a;\|\cdot\|)$ and $y=\alpha x$, proving the reverse inclusion. $\square$

Worked Example: Scaling by a Negative Number

Use the Euclidean norm and let $a=(1,-2)$. Scaling every point of $B_2^{(2)}(a)$ by $-3$ gives a ball whose center is $-3a=(-3,6)$, not $(-3,-6)$, and whose radius is $3\cdot2=6$:

\{-3x:x\in B_2^{(2)}((1,-2))\} =B_6^{(2)}((-3,6)).

Indeed, if $x\in B_2^{(2)}((1,-2))$, then $\|x-(1,-2)\|_2<2$, and \[ \|-3x-(-3,6)\|_2 =\|-3(x-(1,-2))\|_2 =3\|x-(1,-2)\|_2<6. \] For a direct check, $(1,-2)$ is in the original ball since its distance from the center is $0<2$; it maps to $(-3,6)$, the new center. The point $(2,-2)$ is also in the original ball because its distance is $1<2$; it maps to $(-6,6)$, whose distance from $(-3,6)$ is $3<6$.

The condition $\alpha\neq0$ in the dilation theorem matters. If $\alpha=0$, every point maps to the origin, so the image is the singleton $\{0\}$, not an open ball of positive radius. A negative dilation does not cause this problem: its radius is multiplied by $|\alpha|$, a positive number.

Norm Comparisons Place Balls Inside One Another

The Comparison of the Standard Norms from the previous tutorial gives inequalities between the distances measured by the one, Euclidean, and infinity norms. Those inequalities immediately yield containments between balls with the same center. The radii in these containments may differ; using the same radius for two different norms does not usually produce the same set.

Theorem (Containments Among Standard Norm Balls): For $a\in\mathbb{R}^n$ and $r>0$, $$ B_{r/\sqrt{n}}^{(\infty)}(a) \subseteq B_r^{(2)}(a) \subseteq B_r^{(\infty)}(a) \quad\text{and}\quad B_{r/\sqrt{n}}^{(2)}(a) \subseteq B_r^{(1)}(a) \subseteq B_r^{(2)}(a). $$

Proof. Take $x\in B_{r/\sqrt n}^{(\infty)}(a)$. Then $\|x-a\|_\infty<r/\sqrt n$. The norm comparison $\|z\|_2\leq\sqrt n\,\|z\|_\infty$, applied to $z=x-a$, gives \[ \|x-a\|_2\leq\sqrt n\,\|x-a\|_\infty<r. \] Hence $x\in B_r^{(2)}(a)$. If $x\in B_r^{(2)}(a)$, the comparison $\|z\|_\infty\leq\|z\|_2$ gives \[ \|x-a\|_\infty\leq\|x-a\|_2<r, \] so $x\in B_r^{(\infty)}(a)$.

For the second chain, if $x\in B_{r/\sqrt n}^{(2)}(a)$, then \[ \|x-a\|_1\leq\sqrt n\,\|x-a\|_2<r, \] so $x\in B_r^{(1)}(a)$. Finally, if $x\in B_r^{(1)}(a)$, the comparison $\|z\|_2\leq\|z\|_1$ gives \[ \|x-a\|_2\leq\|x-a\|_1<r, \] and therefore $x\in B_r^{(2)}(a)$. These four implications prove the stated containments. $\square$

The proof uses strict inequalities at the radius threshold. For example, knowing only $\|x-a\|_\infty\leq r/\sqrt n$ would give $\|x-a\|_2\leq r$, which does not ensure membership in the open ball $B_r^{(2)}(a)$. The strict inequality in the smaller ball is what yields the required strict inequality after applying the norm bound.

Why Open Balls Matter

Open balls are the basic neighborhoods used to describe local behavior in a metric space. For a norm $\|\cdot\|$, the associated metric is $d(x,y)=\|x-y\|$, as follows from the norm axioms. A norm ball is open in this metric: every point inside it has a smaller ball around it that still fits inside the original ball.

Theorem (Norm Balls Are Open in Their Induced Metric): Let $\|\cdot\|$ be a norm on $\mathbb{R}^n$, and use the metric $d(x,y)=\|x-y\|$. For every $a\in\mathbb{R}^n$ and $r>0$, the ball $B_r(a;\|\cdot\|)$ is open in this metric.

The norm comparisons also explain why the three standard choices give compatible local notions of closeness in a fixed $\mathbb{R}^n$: a sufficiently small ball for one norm contains a ball for another, as the containments above show. But the balls themselves can have visibly different shapes. When working with a ball, always identify its norm, center, radius, and strict boundary condition before simplifying its description.

Check Your Understanding

Use the definition and results in this tutorial to answer each question.

Write the coordinate condition for the Euclidean ball of radius $5$ centered at $(2,1)$ in $\mathbb{R}^2$.
Describe $B_3^{(\infty)}((0,2))$ in $\mathbb{R}^2$ using inequalities on the coordinates.
What are the center and radius of the image of $B_4(a;\|\cdot\|)$ under multiplication by $-2$?
Explain why a point at distance exactly $r$ from $a$ is not in $B_r(a;\|\cdot\|)$.
Which norm comparison gives $B_{r/\sqrt n}^{(\infty)}(a)\subseteq B_r^{(2)}(a)$?

What You'll Learn