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Theorem:Let $(X,d)$ be a metric space, $x \in X$ and $r >0$. Then the open ball $B(x,r)=\{y \in X :d(x,y) <r\}$ is an open set.

Now if we consider X to be set of all integers under usual metric, then we can find the open balls using the usual metric but those open balls will not be open sets (according to definition: every point is an interior point.}. So how to justify this theorem?

mathematics12
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  • The theorem doesn't make sense unless you specify a topology. What are the open sets to begin with... – Jürgen Sukumaran Aug 12 '20 at 13:01
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    I do not know if this is your doubt but from this definition one can conclude that $\mathbb{Z}$ is opened in $\mathbb{Z}$ but not in $\mathbb{R}$. –  Aug 12 '20 at 13:05
  • If by usual metric on the integers you mean the one induced by the Euclidean metric, then the resulting topology is discrete. Every set is open. – halrankard Aug 12 '20 at 13:20
  • "but those open balls will not be open sets (according to definition: every point is an interior point" But they are open sets and every point is an interior point.... Example $B(5,\frac 14)={y\in \mathbb Z: d(5,y)< \frac 14} = {5}$. for any $\epsilon: 0< \epsilon < 1$ we have $B(5,\epsilon) = {5}\subset {5}$ so $5$ *IS* an interior point of ${5}$. – fleablood Aug 12 '20 at 16:06
  • FWIW IMO this is not a duplicate. The OP is not asking for why the theorem is true, but for why their example is not a counter example..... which isn't to say this question hasn't been asked before (I remember seeing it several times); just that the cited question is a different question than this one. – fleablood Aug 12 '20 at 16:10

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If $X$ is the integers then $B(n, 1) = \{y \in \Bbb Z: |n-y| < 1\}=\{n\}$ as integers are at least $1$ apart. And all singletons are open sets in $(\Bbb Z, d)$ so there is no problem.

Henno Brandsma
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  • Please check for duplicates before duplicating answers to previous questions already asked and answered. – amWhy Aug 12 '20 at 13:29