TITLE: equivalent definitions of closure
QUESTION [5 upvotes]: This is a topology question.
First of all, I am sorry this is a really dummy question. As a math student, it is a shame I haven't taken any courses in topology.
A closure of a set $A$ is usually defined as $\operatorname{int}(A)\cup \partial A $.
I would like to know how to prove the equivalence of this statement to $A \cup A'$, where $A'$ is the derived set.
Thanks in advance.
========EDIT==========
$\operatorname{int}(A) = \bigcup\{O :\, O\text{ is open and }O\subseteq A \}$
$\partial A$ is defined as for any $x$, an open set containing it intersects both $A$ and $A^{c}$
$A'$ is a set containing all limit points that are defined as every open set containing a limit point also at least contains an element of $A$.
I hope it is clearer now.
And It would be appreciated if a 'standard' definition of closure could be given.
===========UPDATE=========
Thanks Asaf Karagila for editing the text.
I am sorry I only can accept one answer, so I go for the first and most voted one.
But I really appreciate other's contibutions.
Thank you all!
cheers

REPLY [4 votes]: The "standard" definition of closure that I usually encounter is the dual definition of interior. One defines interior as the largest open set that is contained in $A$; this is achieved with the definition you give (the union of all open sets contained in $A$) since the union of open sets is open, and the union of subsets of $A$ is a subset of $A$.
To achieve the dual definition, we define the closure of $A$ to be the smallest closed set that contains $A$, which is then easily seen to be equivalent to:
$$\mathrm{cl}(A) = \bigcap\{ C\mid A\subseteq C\quad\text{and}\quad C\text{ closed}\}.$$
(Or, one can define the closure of $A$ to be the complement of the interior of the complement of $A$,
$$\mathrm{cl}(A) = \left(\mathrm{int}(A^c)\right)^c,$$
since the smallest closed set that contains $A$ is the complement of the largest open set that is contained in $A^c$).

Now, you want to prove that $\mathrm{int}(A)\cup \partial A = A\cup A'$. 
To prove $\mathrm{int}(A)\cup\partial A \subseteq A\cup A'$, note that $\mathrm{int}(A)\subseteq A \subseteq A\cup A'$ by definition, so it suffices to show that $\partial A \subseteq A\cup A'$. Take $x\in\partial A$. If $x\in A$, there is nothing to do. Now assume $x\notin A$; use the definition of $\partial A$ to show that you will necessarily have $x\in A'$.
The converse inclusion is perhaps a bit more delicate, since you don't have any obvious inclusions. I would proceed as follow: first show that $A\subseteq \mathrm{int}(A)\cup \partial A$, by showing that if $a\in A-\mathrm{int}(A)$, then $A\in\partial A$; to do this, show that any open set containing $a$ cannot be completely contained in $A$. Then we need to show that $A'\subseteq \mathrm{int}(A)\cup\partial A$. Show that if $x\in A'-A$, then $x\in \partial A$. Then simply note that if $x\in A'\cap A$, then you're done already, since you already know that $\mathrm{int}(A)\cup \partial A$ contains every point of $A$.

Finally, we probably want to check that the "standard" definition I gave agrees with the ones you were given. 
If $x\notin \mathrm{int}(A)\cup\partial A$, then $x\notin\partial A$, so there is an open set that contains $x$ and is either completely contained in $A$, or completely contained in $A^c$; but if it were completely contained in $A$, then we would have $x\in\mathrm{int}(A)$, a contradiction to the choice of $x$; thus, there is an open set $U$ with $x\in U$ and $U\cap A=\emptyset$. That means that $C=U^c$ is closed and contains $A$, so $\mathrm{cl}(A)\subseteq C$. Since $c\notin C$, then $x\notin \mathrm{cl}(A)$.  We have shown that $\mathrm{cl}(A)\subseteq \mathrm{int}(A)\cup\partial A$. 
Now let $x\notin \mathrm{cl}(A)$. Then there exists a closed set $C$ such that $A\subseteq C$ and $x\notin C$. Let $U=C^c$; then $U$ is open, $U\cap A=\emptyset$, and $x\in U$. In particular, $x\notin A$, and $x\notin A'$, so $x\notin A\cup A'$. Thus, we have shown that $A\cup A'\subseteq \mathrm{cl}(A)$. 
Hence:
$$A\cup A' \subseteq \mathrm{cl}(A) \subseteq \mathrm{int}(A)\cup\partial A = A\cup A',$$
proving the desired equality.