# Set Theory (Proof)

1. Mar 14, 2010

### CornMuffin

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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: Mar 14, 2010
2. Mar 14, 2010

### union68

I've seen this question before. Are you working out of the metric space chapter in Gamelin and Greene's "Intro to Topology?" That's the only place I've ever seen "adherent points."

You're trying to prove

$$\partial E \subset E \Rightarrow E=\overline{E}$$?

I got this by looking at the contrapositive. We know that if the closure is the set of all adherent points, then a set is always a subset of its own closure, correct? Then the contrapositive would look like

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

Try taking it from there. Perhaps there is a direct method of proof, but the contrapositive was the first route that popped into my head.