What's the boundary of a sequence of integers? 
What's the boundary of a sequence of integers?

I'm trying to learn about topology, particularly boundaries and open and closed sets.  I think I correctly grasp the idea that some point in the real line can be a limit point of a sequence and therefore such a sequence may not contain its boundary.
But I'm struggling to get to grips with the boundary of a sequence of integers in $\mathbb{N}$.  It would seem $\mathbb{N}$ has the discrete topology.  If we take the sequence $3,4,5,6$ then is its boundary in $\mathbb{N}$,  $\{2,7\}$ or $\{3,6\}$?  It would seem we can define this set as an open or closed set with two different boundaries, but the set itself is unchanged so it makes no sense to say the set itself is open or closed as it's either, and we can choose at will what the boundary of the same set is.  Am I being daft or overlooking something obvious?
 A: A point $a$ of a topological space $X$ is called a boundary point of $A\subseteq X$ if for each nbhd $U$ of $a$ we have $U\cap A,U\cap (X\setminus A)\neq \emptyset$. So as already mentioned in comments you need to know what the underlying topology is. You can give $\mathbb{N}$ a discrete topology and consider it as a subset of itself, and in that case you can see that $\partial\mathbb{N}=\emptyset$. If $\mathbb{N}$ is considered as a subset of $\mathbb{R}$ with the usual topology then $\partial\mathbb{N=N}$. For take any point $n\in\mathbb{N}$ and a nbhd $U$ of $n$. Then $U\cap \mathbb{N}\neq\emptyset$ and $U\cap (\mathbb{R\setminus N})\neq\emptyset.$ So the point is that the boundary of a set depends on the open sets or rather the underlying topology. Does this help?
A: For a subset $S$ of a space $X,$ the boundary $\partial S$  (also written $Fr(S),$ as  $Fr$ stands for Frontier)  is $$\partial S=\overline S \cap \overline {X \setminus S}.$$  Sometimes we write $\partial_X S$ because  it depends on the space $X.$
If we take $X=\mathbb N$ where the topology on $\mathbb N$ is its subspace topology as a subspace of $\mathbb R$ (with the usual topology on $\mathbb R),$ then $X$ is a discrete space: Every subset of $X$ is closed. So if $S\subset \mathbb N$ then $\partial_{\mathbb N}S= \overline S \cap \overline {X \setminus S}=S \cap (X$ \ $S)=\phi.$
If we take $X=\mathbb R$ with the usual topology, then $\mathbb N$ is a closed discrete subspace of $X$ so every $S\subset \mathbb N$ is closed in $X.$.... And if $S\subset \mathbb N$ then $\overline {X \setminus S}\supset$ $ \overline {X \setminus \mathbb N}=\overline {\mathbb R \setminus \mathbb N}=\mathbb R=X,$ so $\overline {X \setminus S}=X.$ Hence for any $S\subset \mathbb N$ we have $\partial_{\mathbb R}S=\overline S \cap \overline {X \setminus S}=S \cap  X=S.$ 
A: If  $\mathbb N $ has the discrete topology, the convergent sequences are eventually constant. ..  
Your example is of a finite list of natural numbers .  As a set, $\{3,4,5,6\} $ would have boundary $\emptyset $.  
Sequences have limits, or don't converge, as the case may be. ..
The boundary of a set is the set of limit points, that is, points such that there is a sequence of points in the set converging to them, intersected with the closure of its complement. Alternatively,  it is the closure minus the interior.  
