Let $E$ be a subset of a metric space $X$, and say $E$ is connected if and only if it is not disconnected. I have the following two definitions of disconnected:
I) $E$ is disconnected if there exist two nonempty subsets $A$, $B$ of $X$ such that:
- $E = A \cup B$
- $A \cap \overline{B} = \emptyset$ and $\overline{A} \cap B = \emptyset$
II) $E$ is disconnected if there exist two open (relative to the metric space $X$) subsets $A$, $B$ of $X$ such that:
- $E \subset A \cup B$
- $E \cap A \neq \emptyset$ and $E \cap B \neq \emptyset$
- $E \cap A \cap B = \emptyset$
Is there a concise way to prove that these definitions are equivalent?
Here are the definitions I'm using: A neighborhood of radius $r > 0$ around a point $p \in X$ is the set $N_r(p) = \{q \in X \; | \; d(p, q) < r\}$. A limit point of a subset $E$ of $X$ is a point $p$ such that every neighborhood of $p$ contains at least one other point of $E$. A subset $E$ of $X$ is closed if it contains all of its limit points. The closure of $E$, written $\overline{E}$, is the set $E$ together with its limit points. A subset $E$ of $X$ is open if, for every point $p \in E$, there is some neighborhood of $p$ which is contained in $E$.
Assume we have already proved that every closure is closed, and that a set is closed if and only if its complement is open.