Components and Local Connectedness

Connected components partition a space into its maximal connected pieces, providing a structural decomposition. Local connectedness ensures that this decomposition behaves well with respect to the topology.

Connected Components

Definition4.8Connected Component

Let $X$ be a topological space and $x \in X$ . The connected component of $x$ , denoted $C(x)$ , is the largest connected subset of $X$ containing $x$ : $C(x) = \bigcup \{A \subseteq X : A \text{ is connected and } x \in A\}.$

This is well-defined since the union of connected sets with a common point is connected.

Theorem4.6Properties of Connected Components

The connected components of $X$ form a partition of $X$ .
Each connected component is closed.
A connected subset of $X$ is contained in a single connected component.
$X$ is connected if and only if it has exactly one connected component.

Proof

(1): Define $x \sim y$ if $x$ and $y$ belong to a common connected subset. This is an equivalence relation (reflexive: $\{x\}$ is connected; symmetric: obvious; transitive: $x, z \in A$ connected, $z, y \in B$ connected with $z \in A \cap B$ , so $A \cup B$ is connected and contains both $x$ and $y$ ). The equivalence classes are exactly the connected components.

(2): If $C$ is a connected component, then $\overline{C}$ is connected (closure of a connected set) and contains $C$ . By maximality, $C = \overline{C}$ , so $C$ is closed.

(3): If $A$ is connected, all points of $A$ are equivalent under $\sim$ , so $A$ lies in one component.

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RemarkComponents Need Not Be Open

Connected components are always closed but not necessarily open. They are open if and only if $X$ is locally connected (see below). For instance, in $\mathbb{Q}$ with the subspace topology, the connected components are singletons $\{q\}$ , which are not open.

Totally Disconnected Spaces

Definition4.9Totally Disconnected Space

A topological space $X$ is totally disconnected if the connected components are all singletons, i.e., the only connected subsets are single points.

ExampleTotally Disconnected Spaces

$\mathbb{Q}$ with the standard topology is totally disconnected: between any two rationals there is an irrational, which "separates" them.
The Cantor set $C \subseteq [0,1]$ is totally disconnected, compact, and uncountable. It is homeomorphic to $\{0, 1\}^{\mathbb{N}}$ with the product topology.
Any discrete space is totally disconnected.
The $p$ -adic integers $\mathbb{Z}_p$ are totally disconnected, compact, and uncountable.

Local Connectedness

Definition4.10Locally Connected Space

A topological space $X$ is locally connected if for every $x \in X$ and every open neighborhood $U$ of $x$ , there exists a connected open neighborhood $V$ of $x$ with $V \subseteq U$ .

Equivalently, $X$ is locally connected if it has a basis of connected open sets.

RemarkLocal vs. Global Connectedness

Local connectedness and connectedness are independent properties:

| | Connected | Not Connected | |---|---|---| | Locally Connected | $\mathbb{R}^n$ , manifolds | $\mathbb{R} \setminus \{0\}$ | | Not Locally Connected | Topologist's sine curve | $\mathbb{Q}$ |

Theorem4.7Components in Locally Connected Spaces

A space $X$ is locally connected if and only if every connected component of every open subspace of $X$ is open.

In particular, if $X$ is locally connected, then the connected components of $X$ are both open and closed.

Proof

( $\Rightarrow$ ): Assume $X$ is locally connected. Let $U$ be open in $X$ and $C$ a connected component of $U$ (in the subspace topology). For any $x \in C$ , local connectedness gives a connected open neighborhood $V$ of $x$ with $V \subseteq U$ . Since $V$ is connected and intersects $C$ (at $x$ ), we have $V \subseteq C$ . So $C$ is open.

( $\Leftarrow$ ): Let $x \in X$ and $U$ an open neighborhood of $x$ . The connected component $C$ of $x$ in $U$ is open (by hypothesis). So $C$ is a connected open neighborhood of $x$ contained in $U$ .

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Quasi-Components

Definition4.11Quasi-Component

The quasi-component of a point $x \in X$ is the intersection of all clopen subsets of $X$ containing $x$ : $Q(x) = \bigcap \{A \subseteq X : A \text{ is clopen and } x \in A\}.$

Theorem4.8Components and Quasi-Components

For every $x \in X$ , the connected component $C(x) \subseteq Q(x)$ . If $X$ is locally connected, then $C(x) = Q(x)$ .

Proof

$C(x)$ is connected and contained in every clopen set containing $x$ (since a connected set meeting a clopen set is contained in it). Thus $C(x) \subseteq Q(x)$ .

If $X$ is locally connected, components are clopen, so $Q(x) \subseteq C(x)$ (since $C(x)$ is a clopen set containing $x$ ).

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ExampleQuasi-Components Can Differ from Components

Let $X = \{(0, 0), (0, 1)\} \cup \{(1/n, y) : n \in \mathbb{Z}^+, \, 0 \leq y \leq 1\}$ with the subspace topology from $\mathbb{R}^2$ . The connected components of $(0, 0)$ and $(0, 1)$ are the singletons $\{(0,0)\}$ and $\{(0,1)\}$ respectively. However, they share the same quasi-component $\{(0,0), (0,1)\}$ , since every clopen set containing $(0,0)$ must also contain $(0,1)$ (the segments $\{1/n\} \times [0,1]$ connect the "levels" $y = 0$ and $y = 1$ , and any clopen set must include them all or none for large $n$ ).