Ball (mathematics, $B_d(a,r)$)

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 $d(a,x)\le r$. 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 $|x_1|+|x_2|\le 1$, $x_1^2+x_2^2\le 1$, and $\max \{|x_1|,|x_2|\}\le 1$. 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 $\sqrt{1.28}\approx 1.131$, whereas Chebyshev distance is $0.8$: only the square's open unit ball contains this point.

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.