Connected Sets

# Connected Sets

Two sets $$A$$ and $$B$$ in a space $$X$$ are separated if both $$A \cap \overline{B}$$ and $$\overline{A} \cap B$$ are empty.

A set $$E$$ is connected if $$E$$ is not the union of two separated sets. Or, equivalently, a set $$E$$ is connected if $$E$$ is not the union of two non-trivial relatively open sets in $$E$$. Or, equivalently, a set $$E$$ is connected if $$E$$ is not the union of two non-trivial relatively closed sets in $$E$$.

Two separated sets are called a separation.

For example,

1. In $$\mathbb{R}^2$$, $$E = \{(x, y) : x, y \in \mathbb{Q}\}$$. One separation for $$E$$ is $$A = \{(x, y) : x, y \in \mathbb{Q}, x < \pi\}$$ and $$B = \{(x, y) : x, y \in \mathbb{Q}, x \geq \pi\}$$.

Theorem. Any closed interval $$[a, b]$$ is connected.

Proof (by contradiction). If not, then there exists a separation $$A$$ and $$B$$ with $$a \in A$$. Let $$s = \sup A$$. Then, $$s \in \overline{A}$$. Hence, $$s \notin B$$. So, $$s \in A$$. Thus, $$s \notin \overline{B}$$. Then, there exists $$(s - \epsilon, s + \epsilon)$$ containing no point of $$B$$. Hence, $$(s - \epsilon, s + \epsilon) \subset A$$, which contradicts the fact that $$s$$ is the supremum of $$A$$. QED.