Show the statement: $ (X\setminus A)\cap \partial A=\emptyset \Leftrightarrow X\setminus A \text{ open}$ Let $ A\subseteq X $. I tried to show$ (X\setminus A)\cap \partial A=\emptyset \Leftrightarrow X\setminus A \text{  open}$
I have no idea how to proof both directions: $\Rightarrow$ and $\Leftarrow$.
For $\Leftarrow$ I only have this: For all $ x\in X\setminus A $ there is an $ \varepsilon>0 $ such that $ B_{\varepsilon}(x)\subseteq X\setminus A $. Then I assume $ (X\setminus A)\cap \partial A\neq \emptyset $. From here I get stuck.
$\Rightarrow$: I want to show: for all $ x\in X\setminus A$ there exists $ \varepsilon > 0 $ such that $ B_{\varepsilon}(x)\subseteq X\setminus A $. I don't know how can I find $ \varepsilon $ if I only know the last information.
 A: Hint: $B_\varepsilon(x) \subseteq X \setminus A$ if and only if $B_\varepsilon(x) \cap A = \varnothing$. Also, recall that the definition of $\partial A$.
A: For "=>" you can do it by contradiction again. Assuming $X \setminus A$ is not open means there's a $x \in X \setminus A$ where $B_{\varepsilon}(x)$ contains a point outside of $X \setminus A$ for all $\varepsilon>0$.
Build a sequence of such points $\{ y_i \in A \}_{i=0}^{\infty}$ - for example, shrink $\varepsilon$ every time by 2 and take $y_i \in B_{2^{-i}}\,(x)$.
Now you have a sequence outside of $X \setminus A$ which converges to a point inside $X \setminus A$. Do you see how this (almost) completes the proof?
A: $\partial A = \partial (X\setminus A)$ and for any set $B$, $B$ is open iff it is disjoint from its boundary:
If $B \cap \partial B = \emptyset$ then all points of $B$ are interior points (or else they'd be boundary points, which cannot happen).
And if $B$ is open then $\partial B = \overline{B}\setminus \operatorname{int}(B) = \overline{B}\setminus B$ is by definition disjoint from $B$.
