Closure of set $\{1/n:n\in\mathbb{N}\}$ Let $E:=\{1/n:n\in\mathbb{N}\}\subset \mathbb{R}$. I argued that $E^\circ$ does not exist because there does not exist an open set $V\subset E$ since $E$ consists entirely of separated points (singletons), which are all closed subsets of $E$.
However, I'm in doubt: could we say that the interior of $E$ is the empty set?
Also, is it correct to say that, since we can find a $B_\varepsilon(1/n)$ for some $\varepsilon > 0$ about any $n\in \mathbb{N}$, and $B_\varepsilon(1/n)\cap E\ne \emptyset$ and $B_\varepsilon(1/n)\cap E^c\ne \emptyset$, $∂E=E$?
And, finally, if the interior of $E$ does not exist, then the closure of $E$ does not exist either?
 A: Firstly, the definition of the interior of a set $S$ is the set of all points for which you can find an open ball around them fully contained in $S$. You are right to point out here that no such points exist, but just as the interior of the empty set $\emptyset$ and of $\mathbb{Q}$ is the empty set, so it is for your $E$.
In terms of your second question, a common definition of $\partial E$ is $\overline{E} \setminus E^o$, or the closure minus the interior. Since the interior is the empty set here, $\partial E = \overline{E}$. However, $\overline E \not= E$ since $0$ is a limit point for $E$ which is not contained in the set. Thus $0 \in \overline{E}$ but $0 \not\in E$.
$0 \in \overline{E}$ because we can define the closure of $E$ as the union of $E$ and of all its limit points. A limit point $p$ of $E$ is defined as a point for which $\forall \epsilon >0$, $\exists s \in E$ for which $s \in B_\epsilon(p)$ and $s \not= p$. This intuitively means that for something to be a limit point, it has to be arbitrarily close to elements of the set $E$.
A: The interior of $E$ = {interior points of $E$} = $\emptyset$.  It exists.  It's just empty.
In your third paragraph:  You have a mistake in your definition of $\partial E$.
For a point $x$ if for every $\epsilon > 0$, $(B_{\epsilon}(x)-\{x\}) \cap E \ne \emptyset$ and $(B_{\epsilon}(x)-\{x\}) \cap E^c$ then $x$ is a point of $\partial E$.  But NOTE you have to exclude the point $x$ itself.  $x$ will always be in $E$ or it will be in $E^c$ so one of those two conditions will always be true.  So we are only considering points other than x.
So for every $1/n$ we can find $0 < \epsilon < 1/n - 1/(n+1)$ where $(B_{\epsilon}(x)-\{x\}) \cap E = \emptyset$. so $1/n \not \in \partial E$.
So what points outside E are in $\partial E$?  Well, either $0$ jumps out at you or it doesn't.  For all $\epsilon > 0$ we can find an $1/n$ such that $0< 1/n < \epsilon$ so $0$ is a limit point of E.  It's easy to show this is the point in $\partial E$.
So $\partial E = \{0\}$ and $\overline E = \partial E + E = E \cup \{0\}$.  
