Two definitions of Closure Define the closure of a set as the intersection of all the closed sets that contain it. i.e;
$$\mathrm{cl}(A) = \bigcap\{ C\mid A\subseteq C\quad\text{and}\quad C\text{ closed}\}.$$
Prove that the closure of a set A is equal to  $\operatorname{int}(A)\cup \partial A $
Where:


*

*$\operatorname{int}(S)$ is the set of all interior points of $S$. An interior point $x$ of $S$ is such that $x$ has a neighborhood that is contained in $S$.

*$\partial(S)$ is the set of all boundary points of $S$. A boundary point $x$ of $S$ is such that every neighborhood of $x$ is both in $S$ and its complement.


Using previous results I have managed to show this is true for all closed sets. However, i don´t know how to prove that these two definitions are equivalent for any set. How would you prove it for any set?
 A: First let me state that your definition of $\partial S$ is not correct, or at least ambiguous. It should state that a boundary point $x$ of $S$ is such that each neighborhood $N$ of $x$ is not in $S$ and is not in $S^c$. This in the sense that the statements $N\subseteq S$ and $N\subseteq S^c$ are both not true. Another way to say this is that the sets $N\cap S$ and $N\cap S^c$ are both not empty. That comes close to what your saying, but is not the same. You are demanding that $N\subseteq S$ and $N\subseteq S^c$ ("in $S$ and in its complement") which is not possible. In your definition "is in" must be changed into "meets" ($=$ has a non-empty intersection).
It is clear that $\operatorname{int}\left(A\right)\subseteq A\subseteq\operatorname{cl}\left(A\right)$
so the statement $\operatorname{cl}\left(A\right)=\operatorname{int}\left(A\right)\cup\partial A$
is a direct consequence of the (more informative) statement that $\partial A=\operatorname{cl}\left(A\right)\backslash\operatorname{int}\left(A\right)$. The last statement will be proved in this answer.
If $x\in\partial A$ then it has no neighborhood contained by $A$
so that $x\notin\operatorname{int}\left(A\right)$. Secondly, if $C$ is a
closed set with $A\subseteq C$ and $y\notin C$ then $C^{c}$ is
a neighborhood of $y$ that has an empy intersection with $A$. This
tells us that $y$ cannot belong to $\partial A$. We conclude that
$\partial A\subseteq C$. This is true for each closed set with $A\subseteq C$
so the stronger conclusion $\partial A\subseteq\operatorname{cl}\left(A\right)$
is justified. Proved is now that $\partial A\subseteq\operatorname{cl}\left(A\right)\backslash\operatorname{int}\left(A\right)$.
Let $x\in\operatorname{cl}\left(A\right)\backslash\operatorname{int}\left(A\right)$
and let $U$ be an open set with $x\in U$. From $x\notin\operatorname{int}\left(A\right)$
it follows that $U\subseteq A$ is not true. If $U\subseteq A^{c}$
then $A\subseteq U^{c}$ and consequently $\operatorname{cl}\left(A\right)\subseteq U^{c}$
since $U^{c}$ is a closed set. This however contradicts $x\in\operatorname{cl}\left(A\right)\cap U$.
Our conclusion is that $U\subseteq A^{c}$ is not true either and
proved is now that $x\in\partial A$.
A: I think we can first prove $\mathrm{cl}(A)$ is exactly the point $x \in X$ such that every neighborhood of $x$ intersects A by showing $(\mathrm{cl}(A))^c=\mathrm{int}(A^c)$.
Then using this definition, $\operatorname{int}(A)\cup \partial A \subset \mathrm{cl}(A)$ is clear. For a point $x\in\mathrm{cl}(A)$, we can discuss whether it has a neighborhood contained by $A$ to determine whether $x$ is in $\operatorname{int}(A)$ or $\partial A$ 
