Let $(X,d)$ be the discrete metric space, $x,y\in X$.
I'm reading in one source that the open ball in the discrete metric
$$d(x,y)=\cases{0 & $x=y$\\ 1 & $x\ne y$}$$
are defined as
$$\mbox{Open ball: }B(x_0, \varepsilon)=\cases{\{x_0\} & $0<\varepsilon \le 1$\\
X & $\varepsilon > 1$}$$
-and-
$$\mbox{Closed ball: } B[x_0, \varepsilon]=\cases{\{x_0\} & $0<\varepsilon < 1$\\
X & $\varepsilon \ge 1$}$$
However, I do not understand how, for example, in the open ball it is possible that we have just the singleton when $\varepsilon=1$, and thus what is the difference between open and closed balls in the discrete metric? I think that if $\varepsilon=1$ then the ball should be the entire space $X$.
Would appreciate some clarification.
Best Answer
The open ball $B(x_0,1)$ is defined as all the points whose distance from $x_0$ is less than one. Since any point other than $x_0$ has distance of exactly one from $x_0,$ (and one is not less than one) any point other than $x_0$ is not in $B(x_0,1).$ Thus $B(x_0,1) = \{x_0\}.$