TITLE: Show that $Int(A)=X\setminus\overline{X\setminus A}$.
QUESTION [2 upvotes]: 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.

REPLY [8 votes]: 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)
$$U = X\setminus(X\setminus U) \subseteq X\setminus\overline{X\setminus A} = B.$$
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