Boundary Points and Metric space

Question

Definition:The boundary of a subset of a metric space X is defined to be the set $\partial{E}$ $=$ $\bar{E} \cap \overline{X\setminus E}$

Definition: A subset E of X is closed if it is equal to its closure, $\bar{E}$.

Theorem: Let C be a subset of a metric space X. C is closed iff $C^c$ is open.

Definition: A subset of a metric space X is open if for each point in the space there exists a ball contained within the space

Show that if $E \cap \partial{E}$ $=$ $\emptyset$ then $E$ is open.

Proof:

$E\cap \partial{E}$ being empty means that $ E\subseteq (\bar{E}^c \cup \overline{X\setminus E}^c)$. Since $E \subseteq \bar{E}$ it follows that $E \subseteq \overline{X\setminus E}^c$ which implies that $E \cap \overline{X\setminus E}$ is empty. Since every subset is a subset of its closure, it follows that $X\setminus E$ $=$ $\overline{X\setminus E}$ and so $X\setminus E$ is closed, and therefore $E$ is open.

Is the proof correct? I would really love feedback.

Answer 1

After William Elliot's feedback on your proof and this comment of yours, I don't think there is much that needs to be clarified. Still if you have anything specific regarding your proof to ask me, I welcome you to come here.

In any case, let me try to write a proof that I believe is in line with your attempt.

\begin{align*}E\cap \partial{E}=\emptyset&\implies E\cap(\overline{E}\cap \overline{X\setminus E})=\emptyset\\&\implies (E\cap\overline{E})\cap \overline{X\setminus E}=\emptyset\\&\implies E\cap \overline{X\setminus E}=\emptyset\\&\implies \overline{X\setminus E}\subseteq X\setminus E\\&\implies \overline{X\setminus E}=X\setminus E\end{align*}This shows that $X\setminus E$ is closed and hence $E$ is open.