Interior or border point of a set I am working on a classical real analysis problem as follow:

Find $int (S)$ and $bd (S)$ if $S = \{ \frac{1}{n} | n \in \mathbb N \}= \{ \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ... \}$.

The answers are respectively $ int (S) = \emptyset$ and $bd (S) = \{0\} \cup \{ \frac{1}{n} | n \in \mathbb N \}$. And here are my textbook's definition of interior point and border point:

Let $S \subseteq \mathbb R$. A point $x$ in $\mathbb R$ is an interior point of S if there exists a neighborhood $N$ of $x$ such that $N \subseteq S$. If for every neighborhood of $N$ of $x$, $N \cap S \neq \emptyset$ and $N \cap S^c \neq \emptyset$, then $x$ is a border point of $S$.

And then there is also this:

Every point $x \in S$ is either interior or border point of $S$.

To my inexperienced mind, the answer to $int (S) = \emptyset $ is understandable because while neighborhood $N (x | \epsilon >0)$ is an open set (per textbook), but $S = \{ \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ... \}$ is incomplete, meaning that there is gap between any two consecutive elements of $S$. Therefore by technicality $N (x | \epsilon )$ does not exist inside $S$. (Am I correct here?)
The answer to $bd (S) = \{0\} \cup \{ \frac{1}{n} | n \in \mathbb N \}$ is understandably the direct consequence of $int (S) = \emptyset $, i.e., if none of the elements of $S$ is interior point, then all of them must be border points, plus the zero.
But what troubles me is this: What happens if I instead answer $int(S)$ first and then $bd (S)$ second? Am I going to get the same answers? 
(i) For $\{0\}$, I was tempted to conclude that zero is indeed a border point because $N (0 | \epsilon) \cap S \neq \emptyset$ and $N (0 | \epsilon) \cap S^c \neq \emptyset$ for all $\epsilon >0$. But had I just concluded that $N (x | \epsilon)$ does not exist in $S$? 
(ii) How about for the rest of the points, i.e., $\frac{1}{n}$? How do I conclude $N (\frac{1}{n} | \epsilon ) \cap S \neq \emptyset$ and $N (\frac{1}{n} | \epsilon ) \cap S^c \neq \emptyset$ for all possible $\epsilon >0$?
I think I must have missed something here. Any help would therefore be very much appreciated. Thank you for your time.
 A: Personally, I like the following definitions:


*

*A point $p$ is an interior point of $X$ if $\exists r>0$ such that $B_r(p) \subset X$. The interior of $X$ is the set of all interior points.

*A point $q$ is a boundary point of $X$ if $\forall r>0$ $B_r(q) \cap X \neq \emptyset$ and $B_r(q) \cap X^C \neq \emptyset$.
(With this, open sets have none of their boundary points, closed sets have all of their boundary points, giving at least some dichotomy to open-closed. Can we call sets that have some of their boundary points "ajar"? :) )
Your sequence has no interior points because any small neighborhood around $\frac{1}{n}$ will contain points that are not in the sequence. Therefore, they are all boundary points. However, you can also show that every neighborhood of $0$ has infinitely many points $\frac{1}{n}$ as well as negative numbers, so it also satisfies the definition of boundary point. All other points in $\mathbb{R}$ are interior points of the complement, so not boundary points of the sequence.
