Sets And Sequences Accumulation point proof. Q:
Show That the closure of S is the union of S and the set of all accumulation points.
Basically i want to make an argument that the set of all the accumulation points is the equal to the $\partial S$
I do have this Theorem: 
Suppose $S \subset \mathbb {R^{n}}$ and $x \in \mathbb {R^{n}}$ then x belongs to the closure of S iff there is a sequence of points in S that converges to x.
The Theorem isnt in the form i want it to be to make this arguement

It is not the case that the set of accumulation points of $S$ is precisely $\partial S$. Consider for example the real interval $(0,1)$. Its set of accumulation points is $[0,1],$ but its boundary is just $\{0,1\}$. The boundary does, however, contain all accumulation points that aren't in the set.
However, with that Theorem, you're very nearly there. It suffices to show that if $x\notin S$, then $x$ is an accumulation point of $S$ if and only if $x\in\partial S$. Indeed, by definition, since $x\notin S$, the following are equivalent:

$x$ is an accumulation point of $S$.
There is a sequence of points of $S$ converging to $x$.
$x\in\partial S$.

You should be able to justify each equivalence more or less by definition (using the fact that $x\notin S$).
