Show that $Int(A)=X\setminus\overline{X\setminus A}$. 
Let $A$ be subset in topological space of $(X,\tau)$, show that $\operatorname{Int}(A)=X\setminus\overline{X\setminus A}$.

The definition of interior provided is "largest open set in A", which I don't know how to utilize. Please give me some idea.
 A: You want to show that the set
$$B=X\setminus\overline{X\setminus A}$$
is the interior of $A$.
If your definition of interior is the largest open subset of $A$, then you should show:


*

*a) $B$ is open;

*b) $B$ is a subset of $A$;

*c) If $U\subseteq A$ is open, then $U\subseteq B$.


a) Since $\overline{X\setminus A}$ is closed, the complement $B=X\setminus\overline{X\setminus A}$ is open.
b) Clearly, $X\setminus A\subseteq \overline{X\setminus A}$ and by de Morgan's laws we get 
$$B=X\setminus\overline{X\setminus A} \subseteq X\setminus(X\setminus A) = A.$$
c) If $U$ is an open subset of $A$, then $X\setminus U$ is closed and $X\setminus U\supseteq X\setminus A$. This implies 
$$X\setminus U\supseteq \overline{X\setminus A}$$
(directly from the definition of closure as the smallest closed superset). 
From this we get (again by de Morgan's laws)
$$U = X\setminus(X\setminus U) \subseteq X\setminus\overline{X\setminus A} = B.$$
A: You want to show that the interior of $A$ is the complement of the closure of the complement.  Closure is the smallest closed set containing the set.  Interior is the largest open set contained in the set.  Open and closed sets are complements...
Suppose to the contrary that there was an element, $x$, in the interior of $A$ which was not in the complement of the closure of the complement.  So $x$ is in the closure of the complement.  So $x$ is in every closed set that contains the complement.  In particular $x$ is in the complement of the interior of $A$, because the complement of the interior of $A$ is closed and contains the complement of $A$.  This is a contradiction, because $x$ cannot be in $A$ and its complement.
Suppose to the contrary that there was an element, $x$, in the complement of the closure of the complement of $A$, which was not in the interior of $A$.
A: $$
x \in \operatorname{Int}(A) \subset A \quad \Rightarrow \quad x \notin X \setminus A \subset \overline{X \setminus A} \quad \Rightarrow \quad  x \notin \overline{X \setminus A}
$$
$$
x \in X\setminus \overline{X \setminus A} \quad \Rightarrow \quad \operatorname{Int}(A) \subset X \setminus \overline{X \setminus A}
$$
Think of the other implication. 
