totally disconnected space



topology (point-set topology)

see also algebraic topology, functional analysis and homotopy theory


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



Discrete spaces are totally disconnected.


(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.


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.


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.


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

See also Stone space.


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


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


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 (