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Would it be possible to present a set in a discrete metric space as a ball? More concretely, suppose our metric space is $X = (1,2] \cap \Bbb{Q}$ equipped with the discrete metric, where the discrete metric is defined as:

$$d(x,y)=\begin{cases} 0, &\text{if}\;x=y;\\\\ 1,&\text{otherwise}.\end{cases}$$

Would it be possible to find a ball $B$ in $X$ such that $B = X$, for $X = (1,2] \cap \Bbb{Q}$ as above?

Other examples or general statements would also be welcome.

Aqqqq
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    As @Math_QED says in their answer, the ball $B_r(x)$ is the singleton ${ x }$ when $r < 1$, and it equals $X$ when $r \geq 1$. So, you can write any discrete space as a single ball. More generally, any bounded metric space is an open ball. –  Sep 29 '18 at 07:03
  • @Brahadeesh Am I correct that "bounded metric space" is just the metric space in which the distances between two elements have an upper limit? – Aqqqq Sep 29 '18 at 07:10
  • That's right. The definition that I have seen of a bounded metric space is that there is a sufficiently large (but finite) $r > 0$ such that $X = B_r(x)$ for some $x \in X$. This is equivalent to your statement that a bounded metric space is one for which ${ d(x,y) : x,y \in X }$ is bounded above. –  Sep 29 '18 at 07:18

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If the set is a singleton or the entire space, yes.

These are the only options for a ball in the discrete space, so in all other options the answer is no.

Henno Brandsma
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  • Would the set given in my question be a singleton or the entire space?(If yes, would you mind if you also write down the ball?) Would you mind if you give an example for the ball for a set of entire space? – Aqqqq Sep 29 '18 at 06:55
  • Are you working in the metric space $(1,2] \cap \mathbb{Q}$? Then this set is equal to $B(2,\epsilon)$ where $\epsilon$ is your favorite number strictly greater than $1$. –  Sep 29 '18 at 07:32
  • Thank you. What about the case in which the whole set is $(0,10] \cap \Bbb{Q}$ (definition of the metric remains the same)? – Aqqqq Sep 30 '18 at 12:45
  • Look at my answer: is that set a singelton? Then yes. Is it the entire space? Then yes. Is it something else? Then no. –  Sep 30 '18 at 12:46
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In general, if $(X,d)$ is a bounded metric space, then it can be expressed as an open ball. A bounded metric space is one for which there exists an $r > 0$ such that $X = B_r(x)$ for some $x \in X$.

If $d$ is the discrete metric on $X$, then $(X,d)$ is a bounded metric space, because for any $r \geq 1$, $X = B_r(x)$ for any $x \in X$.