Closure of $A \subset \mathbb{R}$ The closure of $A$ can be equal to $\operatorname{int}(A)\displaystyle\cup\operatorname{bdry}(A)$. Another definition is that the closure is the set of limit points of $A$.
How are these 2 definitions equivalent?
 A: We call $a$ a limit point of $A$ if every neighbourhood of $a$ contains a point of $A$ different from $a$.
We call $a$ a point of closure of $A$ if every neighbourhood of $a$ contains at least one point of $A$.
Let $C$ denote the set of points of closure of $A$. Then you want to show that $\overline{A} = C = \operatorname{int}{(A)} \cup \operatorname{bdry}{(A)}$
$C \subset \operatorname{int}{(A)} \cup \operatorname{bdry}{(A)}$:
Let $a \in C$, i.e. let $a$ be a point of closure of $A$. By the definition of point of closure we know that every neighbourhood of $a$ contains a point $a^\prime \in A$. Assume that $a$ is neither in the boundary nor in the interior of $A$. Then there is an open set $O$ such that $a \in O \subset A^c$. But this would be a contradiction because $O \cap A = \emptyset$ hence there is no $a^\prime \in O$. Hence $a$ has to be either in $\operatorname{int}{(A)}$ or $\operatorname{bdry}{(A)}$.
$\operatorname{int}{(A)} \cup \operatorname{bdry}{(A)} \subset C$:
Let $a \in \operatorname{int}{(A)} \cup \operatorname{bdry}{(A)}$. Then either $a \in \operatorname{int}{(A)}$ or $a \in  \operatorname{bdry}{(A)}$. If $a \in \operatorname{int}{(A)}$ then every neighbourhood of $a \in  \operatorname{int}{(A)} \subset A$ certainly contains $a$ hence $a$ is a point of closure, i.e. $a \in C$. If $a$ is in the boundary of $A$ then by the definition of boundary, every neighbourhood of $a$ has non-empty intersection with both $A$ and $\mathbb{R} \setminus A$ hence again $a$ is in $C$.
As kahen pointed out in the comments, if you use the common definition of limit point you can well have sets that have points that aren't limit points in the usual sense but are in the closure of the set. 
Hope this helps.
A: The closure $\overline{A}$ of $A \subset X$ is the set of so-called adherence points of $A$: all $x$ in $X$ such that every neighborhood of $x$ intersects $A$ (for me, a limit point of $A$ is an $x$ such that every neighborhood of $x$ intersects $A \setminus {x}$). 
If $x$ is such an adherence point, so every neighborhood of $x$ intersects $A$, there are two mutually exclusive options: either some neighborhood of $x$ sits completely inside $A$, and then it is in the interior of $A$ by definition, or every neighborhood of $x$ intersects both $A$ and its complement, and this by definition makes $x$ a point of the boundary of $A$. And clearly both types of points (interior and boundary) are by definition adherence points, so we have the equality that $\overline{A} = bdry(A) \cup int(A)$, as claimed.  
A: Generally speaking if $A$ is a subset of a topological space, and $x\in X$, exactly one of three things can occur:


*

*Every neighborhood $U$ of $x$ meets (intersects novoidly) $A$ and $A^c$  Such points form the boundary of $A$.

*$A$ is a neigbhorhood of $x$.  In this case, we have $x\in A$. Such points form the interior of $A$

*$A^c$ is a neighboorhood of x. Such points form the interior of the complement of $A$.
The point $x$ is a limit point of $A$ if for every neighborhood $U$ of $x$, 
$$U\cap A - \{x\} \not= \emptyset.$$
