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Consider $\mathbb{R}$ with its usual topology. For an arbitrary set of $A$ of $\mathbb{R}$, let $A'$ be the complement of $A$ and $A^-$ be the closure of $A$. Finally, let $\partial A$ be the boundary of the set $A$.

Let $E=\lbrace 1/n : n\in\mathbb{N} \rbrace$. What is $\partial E$?

At first, I thought that the boundary would be the empty set. But then, using the definition of the boundary $\partial E=E^-\cap E^{'-}$ I have that, since all the points in $E$ are isolated, $E^-=E$. Further, some limit points in the complement $E^-$ can be the elements of $E$, i.e. $E\subset E^{'-}$. Therefore $\partial E = E$. Does this make sense?

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1 Answer

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$\bar E=E\cup\{0\}$. And $\overline {E'}=\Bbb R$.

Thus $\partial E=E\cup\{0\}$.

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