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My question is

If $x$ is a boundary point of $S$ ($S$ is subset of $R$), does every ball of $x$ contain both interior points and exterior points of $S$?

I think this is false.

Since $R$ is union of $\mathrm{int} \ S$, $\mathrm{ext} \ S$,and boundary of $S$.

Suppose $\mathrm{int} \ S =\emptyset$, then $R$ is union of boundary point and $\mathrm{ext} \ S$.

Then ball of boundary point does no contain any interior points.

Not sure wheter this is valid counter example and $\mathrm{int} \ S$ can be assumed to be empty.

Anyone please correct me if I am wrong and give me any counterexample?

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Yes, this is correct. As for an example of a set with empty interior, consider $\Bbb Z$ the set of integers, as a subset of $\Bbb R$.

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  • $\begingroup$ Hi if Z is a subset of R , I undetstand that Z has no interior points. But, it seems to me that every point in complement of Z isexterior point and no boundary points exist. $\endgroup$ – Kaytlyn Jan 23 '15 at 17:59
  • $\begingroup$ $\text{bd }S=\overline{S}\setminus\text{int }S$, so $\text{bd }\Bbb Z=\Bbb Z$. $\overline S$ is the closure of $S$. $\endgroup$ – Tim Raczkowski Jan 23 '15 at 18:03
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Correct. As a different counter-example, consider that the boundary of $S$ may coincide with the boundary of $R$, as in the case where $R=[0,2]$ and $S=[1,2]$. A ball around $2$ contains no points in the exterior of $S$.

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