Consider the metric space $(\mathbb R^2,\sigma)$, where $\sigma$ is the discrete metric. Let $$E=\lbrace(x,y)\in \mathbb R^2 : x^2+y^2<1 \rbrace$$ i.e. the unit disc. Find the following sets [note that $E'$ denotes the derived set, not the complement]:
i) $E^\circ$
ii) $E_{ext}$
iii) $\partial E$
iv) $E'$
v) $E_{isol}$
Since we are dealing with the discrete metric, the $\delta$-balls about an element $a$ are either the whole metric space [$\mathbb R^2$ in this case], or the singleton set $\lbrace a \rbrace$, depending on the $\delta$.
Thus it seems to me that any set where the logic includes "for all $\delta$" should be empty, while the sets that only mention "some $\delta$" should be as big as possible.
So for i), the points that can be contained in $E$ with a suitable $\delta$ should be the entire set $E$, since we can just pick a $\delta$ such that the ball around it is the singleton set.
Similarly for ii), this should be all points in $E^c$ [the complement], since there certainly exists a suitable $\delta$.
iii) I claim is $\emptyset$; given any point, there exists a $\delta$ where the $\delta$-ball around it is the point itself. So no point can intersect both $E$ and $E^c$ simultaneously for each $\delta$.
iv), I think, should also be $\emptyset$. Given any point we can pick a $\delta$ such that the ball is the point itself, meaning that not every ball will include other points of $E$.
v) is confusing to me, assuming I've got the others right. By what I've deduced earlier on my own, $E_{isol}=E \cap \partial E \implies E_{isol}=\emptyset$. But this is a contradiction, because taking an arbitrary point in $E$, there always exists a $\delta$ where the $\delta$-ball is the point itself. So $E_{isol}$ should be $E$.