clarification on accumulation points 
Let $A=\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4},\ldots\}$. Then the point $0$ is an accumulation point of $A$. The limit point $0$ does not belong to $A$. Also, $A$ does not contain any other limit points.  

Why doesn't $0$ belong to $A$, is it because there are an infinite amount of open sets $G$ with $0 \in G$? (An accumulation point will never belong to its associated set?)  
What is the difference between open set and neighborhood, isn't every neighborhood an open set?  
How can you tell that there are no more accumulation points in $A$?  
Many thanks.
 A: Note that the element of $A$ are the numbers that have the form $\frac{1}{n}$ so saying $0\in A$ means that theres's some $n\in\mathbb N$ such that 
$$\frac{1}{n}=0$$
which's obviously impossible.
An accumulation point belong to the closure of the set and a set with all it's accumulation points is exactly the closure of the set.
A: An accumulation point of a set can belong to the set. Change your example to the set $B$, where $B$ is $A$ together with $0$. Then $0$ is an accumulation point of $B$ and is in $B$. 
But an accumulation point of a set need not belong to the set. Your $A$ gives one such example.
That there are no other accumulation points of $A$ can be shown as follows. Suppose that $a$ is an accumulation point of $A$. Then for any $\epsilon$, there are infinitely many elements of $A$ that are within $\epsilon$ of $a$. 
Suppose that first that $a\ne 0$. We show that $a$ cannot be an accumulation point of $A$. Let $\epsilon=|a|/2$. Then there are no elements of $A$ within $\epsilon$ of $a$. So $a$ cannot be an accumulation point of $A$.
If $a$ is positive, pick $\epsilon=a/2$. Then every number which is within $\epsilon$ of $a$ is $\gt a/2$. There are only finitely many $n$ such that $\frac{1}{n}\gt a/2$, so $a$ cannot be an accumulation point of $A$.
