Open balls in the discrete metric space.

Question

Let $(X,d)$ be a discrete metric space and $x\in X$. Find the following: $B(x,1/2)$, $B(x,3/4)$, $B(x,1)$, $B(x,r)$ with $0<r\leq 1$, $B(x,r)$ with $r>1$.

Here $B(x,r)$ is the open ball centred at $x$ and radius $r$, i.e., $B(x,r) =\{y\in X \mid d(x,y)<r\}$.

My attempt:

$B(x,1/2) = \{x\}$, $B(x,3/4) =\{x\}$, $B(x,1)=\{x\}$.

$B(x,r)$ with $0<r\leq 1$ is $\emptyset$.

$B(x,r)$ with $r>1$ is all of $X$.

Is my answer correct?

You are contradicting yourself by saying that $B(x,r)$ is empty when $0 <r \leq 1$. — geetha290krm, Jun 28 '22 at 11:46
It may help you https://math.stackexchange.com/questions/2497587/open-and-closed-balls-in-discrete-metric — Lion Heart, Jun 28 '22 at 11:57
@geetha290krm Thanks for your answer. My doubts are cleared. — user1234, Jun 28 '22 at 11:57

score 1 · Accepted Answer · answered Jun 28 '22 at 11:47

1

Since $$d(x,y)=\cases{0 & $x=y$\\ 1 & $x\ne y$}$$ we have that$$\mbox{Open ball: }B(x_0, \varepsilon)=\cases{\{x_0\} & $0<\varepsilon \le 1$\\ X & $\varepsilon > 1$}$$

so you are correct except when you say $B(x,r)=\emptyset$ when $0<r\leq 1$. You contradict yourself because you sadi that $B(x_0,1/2)={x_0}$. Be careful

answered Jun 28 '22 at 11:47

Guillermo García Sáez

1

Thanks for your answer. Now my doubts are cleared. But I don't have enough reputations to up vote your answer. – user1234 Jun 28 '22 at 11:59
Don't worry. I'm not here for the upvotes :) – Guillermo García Sáez Jun 28 '22 at 12:01

Open balls in the discrete metric space.

1 Answers1