Theorem. If a set $E \subset \mathbb{R}$ is disconnected, then there exist $x, y \in E$, and some $z \in \mathbb{R} \setminus E$ with $x < z < y$.
I want to prove this theorem using the following definition of disconnectedness:
Definition. A subset $E$ of a metric space $X$ is disconnected if there exist $U, V \subset X$ such that:
- $U$ and $V$ are open relative to $X$ (meaning every point of $U / V$ has some neighborhood which is completely contained in $U / V$)
- $E \subset U \cup V$
- $E \cap U \neq \emptyset$ and $E \cap V \neq \emptyset$
- $E \cap U \cap V = \emptyset$
Here's what I have so far:
There must be some $x \in E \cap U$, and some $y \in E \cap V$ (since they are nonempty), and we assume without loss of generality that $x < y$. Now let $z = $???