Sets identities on topological space I am trying to show the following identities
Suppose $X$ is a topological space and let $A \subset X$, then:
$(a) int(X \setminus A)=X \setminus \overline{A}$
$(b) \overline{X \setminus A}=X \setminus intA$
For the first one I couldn't do anything In (b), if I call the topology $\tau$, I can express $$X \setminus intA=X \setminus (\bigcup_{U \in \tau, U \subset A}U)$$. But $X \setminus (\bigcup_{U \in \tau, U \subset A}U)=\bigcap_{U \in \tau, U \in A} (X \setminus U)$.
Since $U \subset A$, then $X\setminus A \subset X \setminus U$, so $X \setminus A \subset \bigcap_{U \in \tau, U \in A} (X \setminus U)$. Since this subset is closed, by definition of closure we have $\overline{X \setminus A} \subset \bigcap_{U \in \tau, U \in A} (X \setminus U)$.
In order to complete th proof in (b) I have to show the other inclusion, I couldn't do that part. Any help with this and with (a) would be appreciated.
 A: To finish (b) note that
$$X \setminus int A = X \setminus(\bigcup_ {U \in \tau, U \subset A}U)=\bigcap_{U \in \tau, U \subset A} (X \setminus U) \, =\bigcap_{V \text{closed}, X \setminus A \subset V} V = \overline{X \setminus A}$$
The equality of intersections can be argued as follows:
Let $\mathcal{U}= \{U \in \tau: U \subset A\}$ and $\mathcal{V}= \{V: X \setminus A \subset V, \,\,V \text{closed}\}$. Then $U \in \mathcal{U} \iff X \setminus U \in \mathcal{V}.$
Alternatively, it is clear that 
$$\bigcap_{V \text{closed}, X \setminus A \subset V} V \subset\bigcap_{U \in \tau, U \subset A} (X \setminus U) .$$
Suppose $x \in  \bigcap_{U \in \tau, U \subset A} (X \setminus U) $ and $x \notin  \bigcap_{V \text{closed}, X \setminus A \subset V} V .$ Then $x \notin U$ for every open set $U \subset A,$ but there exists a closed set $V$ with $A \subset V$ such that $x \notin V$. Hence, $x \in X \setminus V \subset A,$. Since $X \setminus V $ is an open set, there is a contradiction.
Therefore,
$$\bigcap_{U \in \tau, U \subset A} (X \setminus U) \subset \bigcap_{V \text{closed}, X \setminus A \subset V} V .$$
To start (a) note that 
$$X \setminus \overline{A}=X \setminus\bigcap_{V \text{closed}, A \subset V} V =\bigcup_{V \text{closed}, X \setminus V \subset X \setminus A} (X \setminus V) \, \subset\bigcup_ {U \in \tau, U \subset X \setminus A}U = int (X \setminus A)$$
