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I am trying to prove the above but am having some difficulty. I already have proved that $int(E)\cup\partial{E}\subset\overline{E}$, but can't get the other direction. That is, I can't figure out how to show that $\overline{E}\subset int(E)\cup\partial{E}$. Below are the definitions that I'm working with.

Closure: $\overline{E}=E\cup cp(E)$; the closure is the set plus its cluster points.

Interior: $int(E) = \{x:\exists r, B_r(x)\subset E\}$ or $int(E)=(\overline{X\setminus E})^{c} $

Boundary: $\partial{E} = \overline{E}\cap\overline{(X\setminus E)}$

Thanks for all your help.

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  • $\begingroup$ Suppose $x \in \overline E$. Then by definition, every ball $B_r(x)$ intersects $E$. Suppose further that $x$ is not an interior point of $E$. Then by definition, every ball $B_r(x)$ intersects $E^c$. What can you conclude? $\endgroup$ – Bungo Oct 7 at 16:39
  • $\begingroup$ @Bungo So, I understand that if $x\in\overline{E}$ must then either be an element of $E$ or an element of $cp(E)$, which I think is what you're going after. If I understand correctly, we are trying to show that if $x\in cp(E)$, and isn't and interior point of E, then it must be in the complement of E. However, I'm not quite sure how to actually show/prove this fact. $\endgroup$ – Tanner Fields Oct 7 at 18:13
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    $\begingroup$ No, if $x$ is a cluster point of $E$ and isn't an interior point of $E$, then it's a boundary point of $E$. It may or may not be in $E$. Also, it's possible that $x \in \overline{E}$ is not an interior point or a cluster point of $E$. It could be an isolated point or $E$. $\endgroup$ – Bungo Oct 7 at 18:17
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Indeed:

\begin{eqnarray} \text{int}(E) \cup \partial E &=& \text{int}(E) \cup (\overline{E} \cap \overline{E^{c}}) \\ &=& (\text{int}(E) \cup \overline{E}) \, \cap \,\ (\text{int}(E) \, \cup \text{int}(E)^{c}) \\ &=& \overline{E} \, \cap X = \overline{E} \end{eqnarray}

where, as you said, $\overline{E^{c}} = \text{int}(E)^{c}$.

I hope I helped you.

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