By the way, I am trying to prove the following. Suppose $A \subset X$, if we define the boundary by the equation $Bd(A) = \bar{A} \cap \bar{X - A}$
(a)Show that $\bar{A} = Int A \cup Bd(A)$.

Here is my proof so far. It is easy to see that $\bar{A} \subset Int(A) \cup Bd(A)$, so we will do the converse. Let $x \in \bar{A}$. Every neighborhood $U_x \cap A \neq \emptyset$. If $U_x \cap A = U_x$, then $U_x \subset A \implies U_x \subset Int(A) \implies x \in Int(A)$. If $U_x \cap A \neq U_x$, then $U_x = (U_x \cap A) \cup (U_x - A)$. If $x \notin A \implies x \in U_x - A \subset X - A \subset…