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 disconected, 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 April 29, 2019 at 13:59:27. See the history of this page for a list of all contributions to it.