Separated, Disconnected, Connected Sets

Separated, Disconnected

$A, B$ are separated if both:

\overset{―}{A} \cap B = \emptyset, A \cap \overset{―}{B} = \emptyset

If $E \subseteq R$ can be written as $E = A \cup B$ where $A, B \neq \emptyset$ and $A, B$ are separated, then $E$ is disconnected.

This is an implication definition: Suppose

E = A \cup B

where

A, B

nomempty and separated implies that

E

is disconnected.

The idea here is that the limit points of $A$ aren't in $B$ , and vice versa.

connected

A set is connected if it is not disconnected. Namely, $E$ is connected for any $A, B \neq \emptyset$ and $A, B$ are not separated ( $\overset{―}{A} \cap B \neq \emptyset$ or $A \cap \overset{―}{B} \neq \emptyset$ ) then $E \neq A \cup B$ .

Examples

$A = (0, 1), B (1, 2)$ then these two sets are separated (the closure of $A$ throws on $0, 1 \notin B$ and the closure of $B$ with the limit points $1, 2$ are not in $A$ neither). Thus $E = A \cup B = (0, 1) \cup (1, 2)$ must be a disconnected set (by construction).

Note that

A, B

being disjoint is not equivalent. For example

A = (0, 1), B = [1, 2)

are not separated since the limit point

1

from

A

is in

B

. Namely:

$\overset{―}{A} \cap B = {1} \neq \emptyset$

Let $E = Q$ . Is $E$ disconnected or connected? Since $Q$ is dense in $R$ , then it is disconnected via the following construction. Let $A = {x \in Q | x < \sqrt{2}}$ and $B = {x \in Q : x > \sqrt{2}}$ . Here $A, B \neq \emptyset$ and $A, B$ are indeed separated. However, $\overset{―}{A} = (- \infty, \sqrt{2}]$ and $\overset{―}{B} = [\sqrt{2}, \infty)$ but $A \cap \overset{―}{B} = \overset{―}{A} \cap B = \emptyset$ showing that $E = A \cup B$ must be disconnected.