Closure of the complement of a subset in a topological space I am trying to get a quick introduction on topology in order to study measure theory. Nevertheless, I am at a halt since I cannot really validate this proposition: 

Shouldn't the complement of an open set contained in A be every set that is either closed or not contained in A, so not necessarily those which are both closed and containing the contrary of A?
Help would be much appreciated!
 A: You're mistaking complement at two different levels.
Consider a family 
$$\mathcal{A} = \{ U \subseteq X : U \subseteq A \ \& \ U \text{ is open } \}$$


*

*The complement of this family (in $\mathcal{P}(X)$) is 
$$\mathcal{A}^c = \{ U \subseteq X : U \not \subseteq A \text{ or } U \text{ is not open }\}.$$
It is the set of those $U$ which do not belong to $\mathcal{A}$.


*The set of complements of sets in this family is 
$$\{ U^c : U \in \mathcal{A} \} = \{ F \subseteq X : F \supseteq X \setminus A \ \& \ F \text{ is closed } \}.$$
It is the set of those $F$ whose complement $F^c$ belong to $\mathcal{A}$.

Those are two different notions. You're thinking of the first while the proof goes with the second:
$$\bigcap_{U \in \mathcal{A}} (X \setminus U) = \bigcap \{ U^c : U \in \mathcal{A} \} = \bigcap \{ F \subseteq X : F \supseteq X \setminus A \ \& \ F \text{ is closed } \}.$$
A: The complement of a set $A$ which is a subset of a set $X$ is$$X\setminus A=\{x\in X\,|\,x\notin A\}.$$So, the complement of an open set $O$ contained in $A$ is a closed set (since it is the complement of an open one) containing $X\setminus A$.
A: You could also reason like this. 
$\mathring A$ is the largest open subset of $A$ in the sense that it contains every open subset of $A$.
Then automatically $X\setminus\mathring A$ is the smallest closed superset of $X\setminus A$ in the sense that it is contained in every closed superset of $X\setminus A$.
This states exactly that $X\setminus\mathring A$ is the closure of $X\setminus A$.
