nLab
totally disconnected space

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

Contents

Definition

Definition

A (topological) space whose only connected subspaces are singletons is called totally disconnected.

Examples

Example

Discrete spaces are totally disconnected.

Example

(the rational numbers are totally disconnected)

The rational numbers \mathbb{Q} \subset \mathbb{R} equipped with their subspace topology inherited from the Euclidean metric topology on the real numbers, form a totally disconnected space.

Proof

By construction, a base for the topology is given by the open subsets which are restrictions of open intervals of real numbers to the rational numbers

(a,b) (a,b) (a,b)_{\mathbb{Q}} \coloneqq (a,b) \cap \mathbb{Q}

for a<ba \lt b \in \mathbb{R}.

Now for any such a<ba \lt b there exists an irrational number r\r \in \mathbb{R}\backslash \mathbb{Q} with a<r<ba \lt r \lt b. This being irrational implies that (a,r) (a,r)_{\mathbb{Q}} \subset \mathbb{Q} and (r,b) (r,b)_{\mathbb{Q}} \subset \mathbb{Q} are disjoint subsets. Therefore every basic open subset is the disjoint union of (at least) two open subsets:

(a,b) =(a,r) (r,b) . (a,b)_{\mathbb{Q}} = (a,r)_{\mathbb{Q}} \cup (r,b)_{\mathbb{Q}} \,.

Hence no inhabited open subspace of the rational numbers is connected.

Example

A product in TopTop of totally disconnected spaces is totally disconnected. A subspace of a totally disconnected space is totally disconnected. Hence limits in TopTop of diagrams of totally disconnected spaces are totally disconnected.

For example, the Baire space of irrational numbers is homeomorphic to a countable product space \mathbb{N}^\mathbb{N} (via continued fractions), so it is totally disconnected. Similarly, Cantor space 2 2^\mathbb{N} is totally disconnected. Another notable special case of the preceding class of examples is the following.

Example

Every profinite group is totally disconnected and in particular the set of p-adic numbers is totally disconnected.

See also Stone space.

Properties

The general class of examples in Example 3 may be seen in the following light.

Theorem

The category of totally disconnected spaces and continuous maps is a reflective subcategory of Top.

Proof

The reflector takes a space XX to the space of connected components, i.e., the quotient space X/X/\sim of XX where \sim-equivalence classes are precisely the connected components of XX. We check that connected components CC of X/X/\sim are singletons. Let q:XX/q: X \to X/\sim be the quotient map; it suffices to check that q 1(C)Xq^{-1}(C) \subseteq X is connected (because then q 1(C)q^{-1}(C) is contained in a single \sim-equivalence class, making C=qq 1(C)C = q q^{-1}(C) a single point). So, suppose the (closed) set q 1(C)q^{-1}(C) is a disjoint union K 1+K 2K_1 + K_2 of closed sets K 1,K 2K_1, K_2; we are required to show one or the other is empty. For each cCc \in C, the inverse image q 1({c})q^{-1}(\{c\}) is connected, hence we must have q 1({c})K 1q^{-1}(\{c\}) \subseteq K_1 or q 1({c})K 2q^{-1}(\{c\}) \subseteq K_2. Thus we can partition CC into sets

C 1{cC:q 1({c})K 1},C 2{cC:q 1({c})K 2}C_1 \coloneqq \{c \in C: q^{-1}(\{c\}) \subseteq K_1\}, \qquad C_2 \coloneqq \{c \in C: q^{-1}(\{c\}) \subseteq K_2\}

and we observe q 1(C 1)=K 1q^{-1}(C_1) = K_1 and q 1(C 2)=K 2q^{-1}(C_2) = K_2. By definition of quotient topology, since K 1,K 2K_1, K_2 are closed we infer C 1,C 2C_1, C_2 are closed. They are also disjoint and C=C 1+C 2C = C_1 + C_2, so by connectedness of CC either C 1=C_1 = \emptyset or C 2=C_2 = \emptyset, and therefore K 1=q 1(C 1)K_1 = q^{-1}(C_1) or K 2=q 1(C 2)K_2 = q^{-1}(C_2) is empty, as required.

Finally, given a continuous map f:XYf: X \to Y with YY totally disconected, each connected component CC of XX is mapped to a connected set f(C)f(C) of YY which is a singleton by total disconnectedness of YY, and so we get a (unique) factoring through a map X/YX/\sim \to Y, continuous of course by virtue of the quotient topology. This completes the proof.

Revised on May 22, 2017 09:08:42 by Urs Schreiber (92.218.150.85)