Contents

# 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

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

for $a \lt b \in \mathbb{R}$.

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

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

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

###### Example

A product in $Top$ of totally disconnected spaces is totally disconnected. A subspace of a totally disconnected space is totally disconnected. Hence limits in $Top$ 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 $2^\mathbb{N}$ 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.

## Properties

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

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

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

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

Last revised on June 20, 2021 at 04:52:13. See the history of this page for a list of all contributions to it.