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!
On
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 $X\setminus A$.
On
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 $X\setminus A$.
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:
$$\bigcap_{U \in \mathcal{A}} (X \setminus U) = \bigcap \{ U^c : U \in \mathcal{A} \} = \bigcap \{ F \subseteq X : F \supseteq X \setminus A \ \& \ F \text{ is closed } \}.$$