TITLE: Prove A is an open set if and only if $A \cap Bd(A) = \emptyset $
QUESTION [9 upvotes]: 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.

REPLY [6 votes]: 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.