Show that if $A$ is any subset of a topological space $(X,T)$, then $Int(A) = \complement ( \overline {\complement (A)})$
My reasoning went as follows:
$\overline {\complement (A)} = \complement (A) \cup [\complement (A)]^{'}$
Taking the complement of the above and applying De Morgan's, we arrive at
$A\cap C$ where $C = \{y : y$ is not a limit point of $\complement (A)\}$
Let $x \in A\cap C$, then $x\in A$ and $x$ is not a limit point of $\complement (A)$ so $x \in A$ and $\exists$ a $U$ that is open and $U \cap \complement (A) = \{x\}$ or $\emptyset$ but since $x\in A$ then $U \cap \complement (A) = \emptyset$, so $U \subseteq A$, But these are precisely the open sets in the union that makes $Int(A) \blacksquare$
So,
1. Is this correct?
2. Can it be written up more elegantly?
3. Is this the kind of question you would expect to find in an advanced undergraduate level or grad level course?
Thanks.
Best Answer
Here is a slightly different method: I denote closure by "Cl" and interior by "Int", and complement by "$c$".
Since $A^c \subset $ Cl$(A^c)$, it follows that Cl$(A^c)^c \subset A$. Then Cl$(A^c)^c$ is an open set contained in $A$, so Cl$(A^c)^c \subset $ Int$(A)$.
On the other hand, we know that Int$(A) \subset A$, so $A^c \subset $ Int$(A)^c$. Thus Int$(A)^c$ is a closed set containing $A^c$, so Cl$(A^c) \subset $ Int$(A)^c$. Hence Int$(A) \subset $ Cl$(A^c)^c$.
We showed that Int$(A) \subset $ Cl$(A^c)^c$ and Cl$(A^c)^c \subset $ Int$(A)$, so it follows that Cl$(A^c)^c = $ Int$(A)$, which is what we wanted to show.
This is easier than what you'd find in a grad course.