Is it true that $Int(A') = cl(A)' ?$ Is it true that $Int(A') = cl(A)' ?$
The actual question was -
Show that $Int(A') = \bar{A}'$ is true for any subset $A$ of a topological space.
I'm not really interested in proof, but I've this doubt, that interior of a set is always open ( also closed if it is $R$ or empty) and derived set and closure are always closed.
So how can be the above statement be true$?$
May be I'm missing some concept. 
 A: It is true! Let's prove it!
Proposition 1: Let $A$ be a subset of some topological space $X$. Some $x \ \in \ X$ is in $\bar{A}$ if and only if every neighbourhood $U$ of $x$ intersects $A$.
Proof
Let us assume that there exists a neighbourhood of $x$ that is disjoint from $A$, which we will call $V$. Well then $X \ - \ V$ is a closed set of $X$ that contains $A$ but does not contain $x$. Therefore, if $x \ \in \ \bar{A}$, then it follows that every neighbourhood of $x$ intersects $A$.
Conversely, let us assume that $x \notin \bar{A}$. Well, $\bar{A}$ is a closed set itself, and $x \ \in \ X \ - \ \bar{A}$, which is open. This means that $x$ has a neighbourhood that does not intersect $A$. Thus, it follows that if every neighbourhood of $x$ intersects $A$, then $x$ must be in $\bar{A}$.
Proposition 2: For some subset $A$ in some topological space $X$, $\text{Int}(X \ - \ A) \ = \ X \ - \ \bar{A}$
We will show inclusion both ways. Let $x$ be in $X \ - \ \bar{A}$. It follows that $x$ is in some neighbourhood $V$ that does not intersect $A$. Since the interior of $X \ - \ A$ is simply the union of all open sets contained within $X \ - \ A$, and as we just showed, $x \ \in \ V$ with $V$ disjoint from $A$, therefore completely contained in $X \ - \ A$, it follows that $x \ \in \ \text{Int}(X \ - \ A)$.
Now, let $x \ \in \ \text{Int}(X \ - \ A)$. This means $x$ is in an open set disjoint from $U$ (the interior itself), which implies that $x \ \notin \ \bar{A}$, and is therefore in $X \ - \ \bar{A}$.
Therefore, we have shown that the two sets are equal: $\text{Int}(X \ - \ A) \ = \ X \ - \ \bar{A}. \ \blacksquare$
Edit
I'm assuming $A' \ = \ X \ - \ A$. If not, then obviously this proof is misplaced.
