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I think the question sort of speaks for itself. Please no solutions in the answers - I'm mostly looking to see if my logic makes sense:

Let $E$ be a set in a metric space $(X,d)$. Let $E^{\circ}$ bee the interior of $E$, $\partial E$ be the boundary of $E$, $E_{isol}$ be the set of isolated points in $E$, and $E'$ be the derived set of $E$. Characterize the following sets:

$E\cap\partial E=E_{isol}$ -> Boundary points are ones where each $\delta$-ball has a point in $E$ as well as $E^c$. They need not be in $E$, but this intersection forces them to be in $E$. Meanwhile, isolated points are points in $E$ where there exists a $\delta$-ball in which they are the only point of $E$. They are required to be in $E$, and also require any ball to be in $E^c$, by definition. Seems like a fit to me.

$E^{\circ}\cap E'=E^{\circ}$ -> Interior points are ones where some $\delta$-ball about them is contained in $E$. Limit points are ones where each ball about them contains another point in $E$. It seems, by my thinking, that the interior of $E$ is contained in the derived set; if there is some ball about a point entirely contained in $E$, then surely every ball about said point will contain another point of $E$.

$E'\cap E_{isol}=\emptyset$ -> A clash of definitions here. Isolated points must have a ball about them where they are the only element of $E$, while each ball around a limit point must contain something in $E$.

$E^{\circ}\cap E_{isol}=\emptyset$ -> Again, a clash of definitions. No ball with non-zero radius around an isolated point can be contained in $E$.

No formal proofs or anything needed - just developing intuition and working on internalizing the definitions of these sets/points.

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  • $\begingroup$ In general, the intersection of the interior of $A$ and the set of isolated points of $A$ is not empty - consider the set of natural numbers with discrete metric. Every subset of this space is open every point in this space is isolated. $\endgroup$ – Hanul Jeon Sep 15 '14 at 2:15
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Your second and third answers are correct, for the reasons that you give; the other two are not in general correct. I won’t give you the correct answers, but I will give you counterexamples to your two false conjectures.

  1. $E\cap\partial E$: Take $X=\Bbb R$ with the usual metric and $E=[0,1]$; then $\partial E=\{0,1\}$, so $E\cap\partial E=\{0,1\}$, but $E_{isol}=\varnothing$.

  2. $E^\circ\cap E_{isol}$: Let $X=\Bbb Z$ with the usual metric, and let $E$ be any non-empty subset of $X$. Then $E^\circ=E_{isol}=E\ne\varnothing$. (This is essentially tetori’s example in the comments.)

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