The Sphere of Discrete Metric Space Let $(X,d)$ denotes the usual discrete metric space. Specify the spheres for $\epsilon = 1, \frac12, 2$.
I strictly used the definition of the sphere such that whenever the metric is not equal to epsilon then we get empty set (all red colors). Can anyone check my result to approve or disprove my results?
We find all the spheres, $\partial{B}_\epsilon(x)$, as follows
\begin{align*}
d(x,y) & = \left.
\begin{cases}
{\color{red}1} {\color{red}\not= {\color{red}2,}} & {\color{red}x \ \not={\color{red}y}} \\ {\color{red}0 \not= {\color{red}2},} &\  {\color{red}x = {\color{red}y}}
\end{cases}
\right\} = \emptyset,  \text{ for } \epsilon = 2, \ \partial B_\epsilon(x) = \emptyset.
 \\
d(x,y) &=  \left.
\begin{cases}
1 = 1, & x \ \not=y \\ {\color{red}{0 \not= 1}}, & {\color{red}{x = y}}
\end{cases}
\right\} =  \{x \not= y\}, \text{ for } \epsilon = 1, \ \partial B_\epsilon(x) = \{x \not= y\} = \{x,y\}.
 \\
d(x,y) = \epsilon & =  \left.
\begin{cases}
{\color{red}1} {\color{red}\not= {\color{red}{1/2},}} & {\color{red}x \ \not={\color{red}y}} \\ {\color{red}0 \not= {\color{red}{1/2}},} &\  {\color{red}x = {\color{red}y}}\end{cases}
\right\} =  \emptyset, \text{ for } \epsilon = 1/2, \ \partial B_\epsilon(x) = \emptyset.
\end{align*}
 A: You should not be limiting yourself to just two points. For each $x\in X$ the sphere of radius $1$ about $x$ is $X\setminus\{x\}$: every point of $X$ except $x$ itself is at distance $1$ from $x$. The spheres of radius $2$ and radius $\frac12$ are therefore empty, since there is no point at distance $2$ or $\frac12$ from $x$.
The closed balls of radii $\frac12$, $1$, and $2$ around $x$, however, are $\{x\}$, $X$, and $X$, respectively, and the corresponding open balls are $\{x\}$, $\{x\}$, and $X$.
A: In the discrete metric, the spheres of radii $1$ and $2$ and center $\{x\}$ are equal to $X\setminus\{x\}$ and $\emptyset$ respectively.  The sphere of radius $1/2$ centered at $\{x\}$ is $\emptyset$.
A: The quantity $d(x,y)$ is a number, not a set. It does not make sense to write $d(x,y)=\cdots =\emptyset$. And the expression $\{x\ne y\}$ does not make any sense.

Recall the definition of sphere  of radius $r$ centered at $x$:
$$
\partial B(x,r)=\{y\in X: d(x,y)=r\}\tag{1}
$$
On the other hand, given any $x,y\in X$, $d(x,y)$ is either $0$ or $1$. So the set in (1) must be empty whenever $r\ne 1$ and $r\ne 0$.
For $r=1$, observe that $d(x,y)=1$ for every $y\ne x$ and thus
$$
\partial B(x,1)=\{y\in X: d(x,y)=1\}=\{y\in X: y\ne x\}=X\setminus\{x\}.
$$
A: Ok, although others have already pointed out, I will try to be direct and precise.
Sphere is defined as follows, for a metric space $(X,d)$:
$$s(x_o,r) = \{x \in X : d(x,x_o)=r\}$$
Radius $r >0$ and centered at $x_o$.
Discrete metric is the usual $$d(x,y) =1, x \neq y
\\  =0, x=y.$$
So, all the $x$ which have a distance $1$ from $x_o$ are precisely all the points in $S = X - \{x_o\}$. Now, any $r \neq 1$ means that the $x  \in X$ at a distance $r$ from $x_o$ does not exist because of the ‘discreteness’ of the metric. Hence, for any $r \neq 1$ the sphere will always be $\phi$.
If $r=0$ then sphere would be a singleton. Namely, the centre of the sphere i.e. $\{x_o\}$.
