I'm trying to extend this two formulas : $$M \backslash \overline{A} = \text{int } (M \backslash A)$$ and $$M \backslash \text{int } (A) = \overline{M \backslash A} $$ Let's say we just take a look at the first one, because the second follow the first. They can be applied in a topological space on a set $M$ and with $A \subset M$.
What I'm searching for, is a way to write this formula in the case where the topological space is not $M$ but a bigger space, let's call it $\Omega$.
I know two demonstrations for the formula :
- In a metric space $(M,d)$ and $A \subset M$
$\begin{split} z \not \in \overline A & \iff ( \exists r>0, B(z,r) > \cap A = \emptyset) \\ & \iff ( \exists r>0, B(z,r) \subset M > \backslash A) \\ & \iff z \in \text{int } (M\backslash A) \\ \end{split}$
I think the problem is in the second line. Here it is implicit that we are using the complement of $A$ in $M$ and this cannot be used in a bigger space.
- In a topological space $(X, \tau)$ and $A \subset X$
$\begin{split} X \backslash \overline{A} & = X \backslash \cap \{ F \text{ with } A \subset F \text{ closed set } \} \\ &= \cup X \backslash \{ F \text{ with } A \subset F \text{ closed set } \} \\ &= \cup \{ X \backslash F \text{ with } X \backslash A \subset X \backslash F \text{ open set } \} \\ &= \cup \{ U \text{ with } X \backslash A \subset U \text{ open set } \} \\ &= \text {int } ( X \backslash A ) \end{split}$
I'm suspecting the third line to be false in a bigger space, but I'm not sure about this one.
So the question is, is it possible to use those formula in $\Omega$ and if yes, under what form, how do you prove this and how in all of this, the induced topology is appearing ? Because I was thinking that it should be easy to use the induced topology of $M$ inside of $\Omega$ in order to prove this, but maybe I'm wrong ? since I can't find anything.