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Let $(X,d)$ be a metric space which has exactly $25$ open balls. Then is it necessarily a discrete space? Is $X$ necessarily a finite space? If so why?

Thank you in advance.

Anil Bagchi.
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    I think $X$ will have $24$ elements then. Because any open ball in a discrete space is either a singleton or the whole space. – Anil Bagchi. Jun 26 '20 at 19:38
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    For each $x \in X$, we have $${x}=\bigcap_{r > 0}B(x,r).$$ Since there are only finitely many possible combinations arising from the right-hand side, it follows that $X$ is finite. – Sangchul Lee Jun 26 '20 at 19:43
  • Hence the space is discrete as well. So it is complete and totally bounded and hence compact. – Anil Bagchi. Jun 26 '20 at 19:53
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    @Phibetakappa Compactness is a direct consequence of the presence of only finite balls. – Sahiba Arora Jun 26 '20 at 20:07
  • Yeah I know that @Sahiba because finitely many open balls $\implies$ finitely many open sets. So any open cover is itself a finite collection and hence it can itself be viewed as it's finite subcover and we are through. – Anil Bagchi. Jun 26 '20 at 20:13
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    @Phibetakappa. Not every ball in a discrete space has to be either singleton or the whole space. Take ${0,1,2}$ as a subset of $\mathbb{R}.$ This is a discrete space since every singleton is open. But also $B(0, 1.5) = {0,1}$ and $B(2, 1.5) = {1,2}$ are open balls. – md2perpe Jun 26 '20 at 20:21
  • Yeah @md2perpe you are right. In a discrete space the metric $d$ might not be necessarily discrete. If we have $d,$ a discrete metric over the space $X$ then indeed $X$ will have exactly $24$ elements if there are $25$ open balls in $(X,d),$ what I mentioned in my first comment. But what will happen in your situation? $X$ is a discrete space but the metric endowed with it is not discrete? How many elements can $X$ have in this case? – Anil Bagchi. Jun 27 '20 at 03:01
  • @md2perpe but it is quite clear that $\text {Card} (X) \leq 24.$ I think finding lower bound of $\text {Card} (X)$ is a good exercise. – Anil Bagchi. Jun 27 '20 at 04:22

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