= Ball
{disambiguate=mathematics}
{title2=$B_d(a,r)$}
{wiki}
= Ball
{synonym}
In a <metric space> $(X,d)$, an <open ball> of radius $r>0$ about $a$ is $B_d(a,r)=\{x\in X:d(a,x)<r\}$. A <closed ball> instead uses $d(a,x)\leq r$. The strict inequality matters: a point at distance exactly $r$ belongs to the <closed ball> and is excluded from the <open ball>. In particular, $(1,0)$ is excluded from all three <open unit balls> above, and included in all three <closed unit balls>.
The three <closed unit balls> centred at zero are $|x_1|+|x_2|\leq1$, $x_1^2+x_2^2\leq1$, and $\max\{|x_1|,|x_2|\}\leq1$. Their shapes are a diamond (a rotated <square>), a <disk> with a <circle> as its <boundary>, and an axis-aligned <square>. A <ball> need not look round! The corresponding <open balls> are the <interiors> of these shapes. At $(0.8,0.8)$, <Manhattan distance> from zero is $1.6$ and <Euclidean distance> is $\sqrt{1.28}\approx1.131$, whereas <Chebyshev distance> is $0.8$: only the <square>'s <open unit ball> contains this point.
\Image[/metric-unit-ball-comparison.png]
{title=Three closed unit balls on the same coordinate scale}
These shapes illustrate $B_1(0,1)\subseteq B_2(0,1)\subseteq B_\infty(0,1)$, with either the <open ball> or <closed ball> convention used consistently. Their outlines are included in the displayed <closed balls>. In an arbitrary <metric space>, however, points at distance exactly $r$ need not form the topological <boundary>; the <discrete metric> below shows why this distinction matters.
Back to article page
