Prove A is an open set if and only if $A \cap Bd(A) = \emptyset $ Prove A  is an open set if and only if $A \cap Bd(A) = \emptyset $
Here is my start:
Suppose A is an open set.  We know $X-A$ is closed. Need to show $A \cap Bd(A) = \emptyset$
Let $ x \in A$. 
Going the other direction, Suppose $ A \cap Bd(A) = \emptyset$. Need to show $A$ is an open set.
Let $x \in A$.  since $x \in A$ $x \notin Bd(A)$
Am I headed in the right direction? I feel more confident about where I have started for the second part then the 1st. 
 A: You can knock out the first direction with a little set algebra. 
$(\implies)$ Suppose $A$ is open, meaning $A = A^\circ$. Then we have $$\begin{align} A \cap \text{Bd}(A) = A \cap (\overline{A}\setminus A^\circ) \\ =  A \cap (\overline{A}\setminus A) \\ = A \cap (\overline{A}\cap A^c) \\ = (A\cap A^c)\cap \overline{A} \\ = \emptyset \cap \overline{A} \\ = \emptyset\end{align}$$
$(\impliedby)$ Now suppose $A \cap \text{Bd}(A) = \emptyset $ but for the sake of contradiction we will additionally assume that $A$ is not open. Then there is an element $x \in A$ such that no open set containing $x$ is a subset of $A$. But $A^\circ$ is always open and $A^\circ \subset A$ so $x\notin A^\circ$. This means $x \in \overline{A}\setminus A^\circ$, or that $x$ is a boundary point of $A$. Since $x$ is an element of $A$ and an element of the boundary, then $x\in A\cap\text{Bd}(A) \neq \emptyset$, a contradiction.
A: Suppose $A$ is open. Then let $x \in A$. Then $A$ is an open set containing $x$ that does not intersect with $X \setminus A$. So $x$ is not in the boundary.
Suppose the intersection of the boundary of $A$ and $A$ is empty. Then all points of $A$ are not in the boundary. So for every $x \in A$, there exists an open set $U_x$ containing x that does not intersect with $A$ or $X \setminus A$.  Obviously $U_x$ must intersect with $A$ as $x \in A$. So it cannot intersect with $X \setminus A$. That means $U_x \subseteq A$. (why?) Now every point $i$ in $A$ is contained in an open set $U_i$, contained in $A$. What is the union of all these open sets? What does that imply about $A$?
A: for the first direction we can say that BdA is the intersection of closure(A) and closure(X\A). If x belongs to both A and BdA then x belongs to both A and closure(X\A) but X\A is closed (since A is open) so closure(X\A) is X\A. So x belongs to both A and X\A which can't be true.
For the other direction if x belongs to A then it doesn't belong to BdA. So x belongs to either the int(X\A) or intA. The first case is not true so x is in the interior of A which means A is open.
A: By the symmetry of the definition, it is clear that
$$
\operatorname{Bd}(A)=\operatorname{Bd}(X\setminus A)
$$
Since $A\cap B=\emptyset$ is equivalent to $B\subseteq X\setminus A$, for any subsets $A$ and $B$ of $X$, the statement is equivalent to

$C$ is closed if and only if $\operatorname{Bd}(C)\subseteq C$

It also follows from the definition that any point in the boundary of a set $B$ belongs to the closure of $B$.
So, if $C$ is closed we have $\operatorname{Bd}(C)\subseteq C$.
Conversely, suppose $\operatorname{Bd}(C)\subseteq C$ and let $x$ belong to the closure of $C$. If a neighborhood $U$ of $x$ doesn't intersect $X\setminus C$, then $U\subseteq C$, so $x\in C$. If all neighborhoods of $x$ intersect $X\setminus C$, then $x\in\operatorname{Bd}(C)$, so $x\in C$ as well.

Your direction is good, however.
Suppose $A\cap\operatorname{Bd}(A)=\emptyset$ and let $x\in A$. Then $x\notin\operatorname{Bd}(A)$ and so a neighborhood $U$ of $x$ doesn't intersect $X\setminus A$, which means $U\subseteq A$. Then…
Suppose $A$ is open and let $x\in A$. Then there exists a neighborhood $U$ of $x$ such that $U\subseteq A$. Therefore…
