Connected and Disconnected Spaces

Connectedness captures the intuitive notion that a space is "in one piece." It is one of the fundamental topological properties, preserved by continuous maps, and forms the basis for the intermediate value theorem and many existence results.

Definition of Connectedness

Definition4.1Connected Space

A topological space $X$ is connected if it cannot be written as a union of two nonempty disjoint open sets. Equivalently, $X$ is connected if the only subsets of $X$ that are both open and closed (clopen) are $\emptyset$ and $X$ itself.

A space that is not connected is called disconnected.

Definition4.2Separation

A separation (or disconnection) of a topological space $X$ is a pair of nonempty open sets $U, V \subseteq X$ such that $X = U \cup V$ and $U \cap V = \emptyset$ . A space is connected if and only if it admits no separation.

Equivalently, $\{U, V\}$ is a separation of $X$ if and only if $U$ and $V$ are nonempty, $U \cup V = X$ , and neither $U$ nor $V$ contains a limit point of the other.

ExampleBasic Examples

$\mathbb{R}$ is connected (proved below as a theorem).
$\mathbb{Q}$ is disconnected: $\mathbb{Q} = (\mathbb{Q} \cap (-\infty, \sqrt{2})) \cup (\mathbb{Q} \cap (\sqrt{2}, \infty))$ is a separation.
The discrete topology on a set with more than one point is disconnected: every singleton is clopen.
The indiscrete topology on any set is connected (the only clopen sets are $\emptyset$ and $X$ ).
$\{0\} \cup \{1\}$ with the subspace topology from $\mathbb{R}$ is disconnected.

Connected Subsets

Definition4.3Connected Subset

A subset $A$ of a topological space $X$ is connected if $A$ is connected in the subspace topology. Equivalently, there do not exist open sets $U, V$ in $X$ with $A \subseteq U \cup V$ , $A \cap U \neq \emptyset$ , $A \cap V \neq \emptyset$ , and $A \cap U \cap V = \emptyset$ .

Theorem4.1Properties of Connected Sets

Let $X$ be a topological space.

If $C$ is connected and $C \subseteq A \subseteq \overline{C}$ , then $A$ is connected.
If $\{C_\alpha\}$ is a family of connected subsets with $\bigcap_\alpha C_\alpha \neq \emptyset$ , then $\bigcup_\alpha C_\alpha$ is connected.
The continuous image of a connected space is connected.

Proof

(1): Suppose $A = U \cup V$ is a separation. Since $C \subseteq A$ is connected, $C$ lies entirely in $U$ or entirely in $V$ ; say $C \subseteq U$ . Then $\overline{C} \subseteq \overline{U}$ . Since $U$ and $V$ are disjoint and open in $A$ , $\overline{U} \cap V = \emptyset$ in $A$ . Thus $A \subseteq \overline{C} \subseteq \overline{U}$ , giving $V = A \setminus U \subseteq \overline{U} \setminus U$ . But $V$ is open and $\overline{U} \setminus U$ contains no open subset of $A$ , so $V = \emptyset$ , contradicting the assumption.

(2): Fix $p \in \bigcap_\alpha C_\alpha$ . If $\bigcup_\alpha C_\alpha = U \cup V$ is a separation, then $p$ belongs to exactly one, say $p \in U$ . Each $C_\alpha$ is connected and contains $p \in U$ , so $C_\alpha \subseteq U$ . Hence $\bigcup_\alpha C_\alpha \subseteq U$ , giving $V = \emptyset$ , a contradiction.

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Connectedness of Intervals

Theorem4.2Connected Subsets of $\mathbb{R}$

A subset $A \subseteq \mathbb{R}$ is connected if and only if $A$ is an interval (including rays and $\mathbb{R}$ itself).

This is a deep result that relies on the completeness of $\mathbb{R}$ . We prove the forward direction here; the converse is proved in the theorems section.

Proof

( $\Leftarrow$ , sketch): Suppose $A$ is an interval and $A = U \cup V$ is a separation with $a \in U$ and $b \in V$ , $a < b$ . Let $c = \sup(U \cap [a, b])$ . Since $U$ is closed in $A$ (it equals $A \setminus V$ ), $c \in U$ . Since $U$ is open in $A$ , there exists $\epsilon > 0$ with $(c - \epsilon, c + \epsilon) \cap A \subseteq U$ . If $c < b$ , then $c + \epsilon' \in U$ for small $\epsilon'$ , contradicting $c = \sup(U \cap [a,b])$ . If $c = b$ , then $b \in U \cap V = \emptyset$ , a contradiction.

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Operations on Connected Spaces

ExampleProducts of Connected Spaces

If $X$ and $Y$ are connected, then $X \times Y$ is connected. More generally, an arbitrary product $\prod_\alpha X_\alpha$ is connected (in the product topology) if and only if each $X_\alpha$ is connected.

For the finite case: fix $(a, b) \in X \times Y$ . For any $(x, y) \in X \times Y$ , the sets $X \times \{b\}$ and $\{x\} \times Y$ are connected (homeomorphic to $X$ and $Y$ respectively) and share the point $(x, b)$ . So $X \times \{b\} \cup \{x\} \times Y$ is connected. Since all these sets share the point $(a, b)$ , their union $X \times Y$ is connected.

RemarkTopologist's Sine Curve

The topologist's sine curve $S = \{(x, \sin(1/x)) : x > 0\} \cup \{(0, y) : -1 \leq y \leq 1\}$ is connected but not path-connected. It demonstrates that these two notions of connectedness are genuinely different.

$S$ is connected because it is the closure of the graph $\{(x, \sin(1/x)) : x > 0\}$ , which is a continuous image of $(0, \infty)$ (hence connected), and the closure of a connected set is connected.