Open and closed balls in discrete metric

Question

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.

Answer 1

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 $B(x_0,1) = \{x_0\}.$

Answer 2

$$B(x_0, \varepsilon)=\{y\in X\;|\;d(x_0, y)<\varepsilon\}$$ So if $\varepsilon=1$, then $$B(x_0, \varepsilon)=\{y\in X\;|\;d(x_0, y)<1\}=\{y\in X\;|\;d(x_0, y)=0\}$$ which is the singleton set $\{x_0\}$ by the definition of metric. Open and closed ball in $X$ differ only in the case of $\varepsilon=1$, otherwise they are same.