Prove $\mathrm{ext}\,(A)=\mathrm{int}\,(X\setminus A)$ For a subset $A$ of topological space $X$, the exterior of $A$ is the set $\mathrm{ext}\,(A)=X\setminus\mathrm{cl}(A)$.
What I have so far:
$\text{ext}\,(A)
= X\setminus \text{cl}(A) 
= X\setminus (A \cup A') 
= (X\setminus A) \cap (X\setminus A')   $ (DeMorgan's Laws)
But this is where I am stuck. I am unsure if this proof is a simple transformation, or if i'm approaching it from the wrong angle.
Any advice?
 A: Since$\DeclareMathOperator{\Int}{Int}$ $A \subseteq \overline{A}$, we have $$X\setminus \overline{A} \subseteq X\setminus A$$ Hence, $X\setminus \overline{A}$ is an open set contained in $X\setminus A$, so it is also contained in its interior. Therefore, $X\setminus \overline{A} \subseteq \Int(X\setminus A)$.
Similarly, from $\Int(X\setminus A) \subseteq X\setminus A$ we obtain $$A = X \setminus (X \setminus A) \subseteq X \setminus \Int(X\setminus A)$$
Therefore, $X \setminus \Int(X\setminus A)$ is a closed set which contains $A$, so it also contains its closure:
$$\overline{A} \subseteq X\setminus \Int(X\setminus A)$$
By complementing, we obtain $\Int(X\setminus A)\subseteq X\setminus \overline{A}$.
Therefore:
$$\mathrm{Ext}\,A = X\setminus \overline{A} = \Int(X\setminus A)$$
A: Whenever you’re asked to prove $A=B$, for two sets $A$ and $B$, you assume $x \in A$ and show $x \in B$.  Then you assume $x \in B$ and show $x \in A$ 
If $x \in ext(A)$, then $x$ is in an open set that does not intersect $A$. Hence $x$ is not in the closure of $A$. 
If $x \in X$ and also $x$ is not in the closure of $A$ then it is not a limit point of $A$.  Hence there is an open set containing $x$ that does not intersect $A$.  Hence it’s in the exterior of $A$. 
