I have learned the definition of the closure of $A$, as the set $\overline A$, of all the points and limit points of $A$. It occurred to me that maybe the closure is also the set of all interior and boundary points of $A$. I did a very long and boring proof which goes as follows :
Let $x\in \overline A$. Suppose $x\in A$. If $x\in int(A)$ then we are done. If $x\notin int(A)$, then since $x\in A$, it must be a boundary point of $A$. Now suppose $x\notin A$. Then $x$ is a limit point of $A$. So each neighborhood of $x$ contains $x$, which is a point not in $A$, and a point in $A$. So it is a boundary point. Conversely, if $y$ is an interior point of $A$, then it is a limit point as well. Suppose $y$ is a boundary point of $A$. Either $y\in A$ or $y\notin A$ and in the latter case, $y$ being a boundary point, must be a limit point of $A$. This completes the proof.
I want to know a simpler, more intuitive reasoning as to why this is true; and whether this is true in other topological spaces. I have used a fact that is true because we are in $\Bbb R$ - interior points are limits points. This leads me to believe the result isn't true in general.
For any subset $A$ of a topological space $X$, there is a three-way partition of the space : $$X = \operatorname{int}(A) \cup \partial A \cup \operatorname{int}(X \setminus A)$$
where all three sets are pairwise disjoint. (If $x \in X$, some neighbourhood of $x$ stays inside $A$, or some neighbourhood of $x$ stays inside the complement of $A$, or finally, all neighbourhoods of $x$ intersect both $A$ and its complement.)
Also, $$X\setminus \overline{A} = \operatorname{int}(X\setminus A)$$
which is clear by the definitions too: $x$ is not in the closure of $A$ iff there is some neighbourhood of $x$ that misses $A$ iff there is some neighbourhood of $x$ that stays inside the complement of $A$.
Combining these two facts indeed:
$$ \overline{A} = \operatorname{int}(A) \cup \partial A$$
and this holds for all spaces $X$ and all subsets $A$ of $X$.