equivalent definitions of closure 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
 A: 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. 
A: There are many definitions of 'closure', and one which is especially useful is:
$$
  \overline{A} = \bigcap \{ C \;|\; A \subseteq C, C \;\textrm{closed} \}
$$
Anyway, I'd like to prove $int(A) \cup \partial A = A \cup A'$ directly.  (It's sorta fun, really.)
I.  Show $int(A) \cup \partial A \subseteq A \cup A'$.  Since $int(A) \subseteq A$, all we must show is $\partial A \subseteq A \cup A'$.  Let $x \in \partial A$.  By definition, any open $U$ containing $x$ must intersect both $A$ and $A^c$ nontrivially.  Suppose $x \notin A$ (if $x \in A$, we're already done).  Let $U$ be any open set containing $x$.  Then $U \cap A \neq \emptyset$, which shows there is some point of $A$, not equal to $x$ (since $x \notin A$) living inside $U$, thus proving $x \in A'$.
II.  Show $A \cup A' \subseteq int(A) \cup \partial A$.  Here, we must show both $A \subseteq int(A) \cup \partial A$ and $A' \subseteq int(A) \cup \partial A$.
IIa.  $A \subseteq int(A) \cup \partial A$.  Let $x \in A$ and suppose $x \notin \partial A$.  Then (by def. of $\partial A$), there exists some open set $U$ containing $x$ such that either $U \cap A = \emptyset$ or $U \cap A^c = \emptyset$.  But $x \in U \cap A$ (hence non-empty), so we must have $U \cap A^c = \emptyset$.  This is equivalent to $U \subseteq A$.  Now we have an open set entirely within $A$, containing $x$.  By definition $x \in int(A)$.
IIb.  $A' \subseteq int(A) \cup \partial A$.  Let $x \in A'$ and suppose $x \notin \partial A$.  As in IIa, this implies there exists an open $U$ containing $x$ such that either $U \cap A$ or $U \cap A^c$ is empty.  But by definition of $A'$, every open set containing $x$ must intersect $A$ in an element not equal to $x$.  So $U \cap A \neq \emptyset$, implying that $U \cap A^c = \emptyset$, which (as in IIa) implies $x \in int(A)$.
By the way, most proofs in elementary point-set topology can be churned out in this manner.  Argue from definitions, use set theory, use logic to break down unions into "OR" and intersections into "AND", etc.
Hope this helps!
A: The definition of open set that you gave:
$\operatorname{int}(A) = \bigcup\{O :\, O\text{ is open and }O\subseteq A \}$
Basically describes the interior of $A$ as the largest open set contained in $A$. Can you think of a dual notion for closed sets? Think, though, that counter/dual to the notion of an open set, the closure of a set must contain the set.
