Topology - Prove that $[A] = \displaystyle\overline{(X - A)^{\circ}}$ Prove that $[A] = \displaystyle\overline{(X - A)^{\circ}}$ where $[A]$ is the closure of $A$ and $\circ$ is denoted by the interior. Also there is supposed to be a bar over all of $(X - A)$ but it did not do that for some reason. The bar in this case represents the complement.
I have seen this done by using the definition of closure, (the intersection of all closed sets containing $A$) and then using that to get to the result. However I was wandering if proving it would work by choosing an arbitrary point $x \in [A]$ and then using set theory logic to get the result. I am curious how it goes if anyone knows.
 A: Another characterization of $[B]$, the closure of $B$, is that $x \in [B]$ iff every open set $U$ containing $x$ contains some point in $B$. A characterization of $B^{\circ}$, the interior of $B$, is that $x \in B^{\circ}$ iff there exists an open set $U$ containing $x$ such that $U \subseteq B$.
Using these characterizations, we can produce a proof that $[A] = \overline{(X-A)^{\circ}}$ with your desired set-theoretic flavor.
First we show $[A] \subseteq \overline{(X-A)^{\circ}}$. If $x \in [A]$, we know that each open set $U \ni x$ contains some point $y \in A$ (possibly $x$, but that doesn't matter). Equivalently, we can say that, for arbitrary $x \in [A]$, there is no open set $U \ni x$ such that $U \cap A = \emptyset$. This means that $x$ cannot be in the interior of $X - A$, and therefore we must have that $x \in \overline{(X-A)^{\circ}}$. Thus $x \in [A]$ implies $x \in \overline{(X-A)^{\circ}}$, and we have $[A] \subseteq \overline{(X-A)^{\circ}}$.
Conversely, we take arbitrary $x \in \overline{(X-A)^{\circ}}$, so $x \not\in (X - A)^{\circ}$. By our characterization of the interior of $X - A$, we know that this means that any open set $U$ containing $x$ does not satisfy $U \subseteq X - A$. Therefore, for each open set $U$ containing $x$, $U$ contains some point $y \in A$, so $x \in [A]$. Therefore we have $ \overline{(X-A)^{\circ}} \subseteq [A]$.
From the previous two paragraphs, we conclude that $[A] = \overline{(X-A)^{\circ}}$.
A: *

*$[A] \subset \overline{(X-A)^\circ}$

If $x \in [A]$ then $\forall B$ closed such that $A \subset B$ we have $x \in B$. If $x \in B$, a closed set then $x \not\in \overline{B}$ an open set so that for every $B$ we have that $x \in X - \overline{B}$ that is to say $x \in X - (X - B)$ so $x \in \bigcap \left( X - (X - B) \right)$ which by DeMorgans laws we get that $x \in \bigcup (X - B)$ so that $x \in (X - A)^\circ$ (since $A \subset B \implies X - B \subset X - A$).


*$[A] \supset \overline{(X-A)^\circ}$

If $x \in \overline{(X-A)^\circ}$ then $x \not\in \bigcup \{C ; C \text{ is open and } C \subset (X - A) \}$ so again DeMorgans laws tell us that $x \in \bigcap \{\overline{C} ; C \text{ is open and } C \subset X-A\}$ and now since $C \subset X-A \implies A \subset \overline{C}$ we get that $x \in [A]$.

