Open ball of the Discrete Metric

Question

Let $(X,d)$ be the discrete metric space, such that for $x,y\in X$,

$$d(x,y)=\cases{0 & $x=y$\\ 1 & $x\ne y$}$$

The open ball is defined as

$$B_r(x)=\cases{\{x\} & $0<\varepsilon \le 1$\\
X & $\varepsilon > 1$}$$

I am having trouble on understanding how we come to this definition of the open ball of the discrete metric. Any explanation would be highly appreciated.

Answer 1

No, the open ball is not defined that way. The open ball $B_r(x)$ is the set $\{y\in X\mid d(x,y)<r\}$, for any metric space. But, if $d$ is the discrete metric, then:

• if $r\leqslant1$ then $d(x,y)<r\iff d(x,y)=0\iff y=x$, and therefore $B_r(x)=\{x\}$;
• if $r>1$, then, for each $y\in X$, $d(x,y)\leqslant1<r$, and therefore $B_r(x)=X$.