Set Theory (Proof)

Homework Statement

The boundary $$\partial E$$ of a set E if defined to be the set f points adherent to both E and the complement of E,
$$\partial E=\overline{E}\bigcap \overline{(X\backslash E)}$$

Show that E is open if and only if $$E \bigcap \partial E$$ is empty. Show that E is closed if and only if $$\partial E \subseteq E$$

I did the first part, but I need help with the second part.

The Attempt at a Solution

Assume E is closed, then $$E = \overline{E}$$ and its complement is open
so, $$(X\backslash E) \subset \overline{(X\backslash E)}$$ and $$\overline{(X\backslash E)}$$
(contains points in X but not in X\E)
So, $$E\bigcap \overline{(X\backslash E)}=\overline{E} \bigcap \overline{(X\backslash E)}$$ is non empty and every point in $$\overline{E} \bigcap \overline{(X\backslash E)}$$ is in E since $$E=\overline{E}$$
So, $$\partial E \subseteq E$$

But I am having trouble going in the other direction

Last edited:

$$\partial E \subset E \Rightarrow E=\overline{E}$$?
$$\overline{E} \not\subset E \Rightarrow \partial E \not\subset E$$.