nLab point space




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

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


Basic concepts

Universal constructions

Extra stuff, structure, properties


Basic statements


Analysis Theorems

topological homotopy theory



The point space *\ast is the topological space whose underlying set is the singleton, and equipped with the unique topology that this set carries.



The point space is the terminal object in the category Top of topological spaces.

For XX any topological space, then for every element of its underlying set there is a continuous function from the point space

*X \ast \longrightarrow X

whose image is that point, and every such continuous function arises this way

Relation to irreducible closed subspaces

For the following we write the point space explicitly as

*={{1},τ *={,{1}}} \ast = \left\{ \{1\}, \, \tau_\ast = \left\{ \emptyset, \{1\} \right\} \right\}

For (X,τ)(X,\tau) a topological space, then there is a bijection between the irreducible closed subspaces of (X,τ)(X,\tau) and the frame homomorphisms from τ X\tau_X to τ *\tau_\ast from the frame of opens of XX to that of the point space. Moreover, this is given by

Hom Frame(τ X,τ *) IrrClSub(X) ϕ X\U (ϕ) \array{ Hom_{Frame}(\tau_X, \tau_\ast) &\underoverset{\simeq}{}{\longrightarrow}& IrrClSub(X) \\ \phi &\mapsto& X \backslash U_\emptyset(\phi) }

where U (ϕ)U_\emptyset(\phi) is the union of all elements Uτ xU \in \tau_x such that ϕ(U)=\phi(U) = \emptyset:

U (ϕ)Uτ Xϕ(U)=U. U_{\emptyset}(\phi) \coloneqq \underset{{U \in \tau_X} \atop {\phi(U) = \emptyset} }{\cup} U \,.

See also (Johnstone 82, II 1.3).


First we need to show that the function is well defined in that given a frame homomorphism ϕ:τ Xτ *\phi \colon \tau_X \to \tau_\ast then X\U (ϕ)X \backslash U_\emptyset(\phi) is indeed an irreducible closed subspace.

To that end observe that:

(*)(\ast) If there are two elements U 1,U 2τ XU_1, U_2 \in \tau_X with U 1U 2U (ϕ)U_1 \cap U_2 \subset U_{\emptyset}(\phi) then U 1U (ϕ)U_1 \subset U_{\emptyset}(\phi) or U 2U (ϕ)U_2 \subset U_{\emptyset}(\phi).

This is because

ϕ(U 1)ϕ(U 2) =ϕ(U 1U 2) ϕ(U (ϕ)) =, \begin{aligned} \phi(U_1) \cap \phi(U_2) & = \phi(U_1 \cap U_2) \\ & \subset \phi(U_{\emptyset}(\phi)) \\ & = \emptyset \end{aligned} \,,

where the first equality holds because ϕ\phi preserves finite intersections by def. , the inclusion holds because ϕ\phi respects inclusions by remark , and the second equality holds because ϕ\phi preserves arbitrary unions by def. . But in τ *={,{1}}\tau_\ast = \{\emptyset, \{1\}\} the intersection of two open subsets is empty precisely if at least one of them is empty, hence ϕ(U 1)=\phi(U_1) = \emptyset or ϕ(U 2)=\phi(U_2) = \emptyset. But this means that U 1U (ϕ)U_1 \subset U_{\emptyset}(\phi) or U 2U (ϕ)U_2 \subset U_{\emptyset}(\phi), as claimed.

Now according to prop. the condition (*)(\ast) identifies the complement X\U (ϕ)X \backslash U_{\emptyset}(\phi) as an irreducible closed subspace of (X,τ)(X,\tau).

Conversely, given an irreducible closed subset X\U 0X \backslash U_0, define ϕ\phi by

ϕ:U{ |ifUU 0 {1} |otherwise. \phi \;\colon\; U \mapsto \left\{ \array{ \emptyset & \vert \, \text{if} \, U \subset U_0 \\ \{1\} & \vert \, \text{otherwise} } \right. \,.

This does preserve

  1. arbitrary unions

    because ϕ(iU i)={}\phi(\underset{i}{\cup} U_i) = \{\emptyset\} precisely if iU iU 0\underset{i}{\cup}U_i \subset U_0 which is the case precisely if all U iU 0U_i \subset U_0, which means that all ϕ(U i)=\phi(U_i) = \emptyset and because i=\underset{i}{\cup}\emptyset = \emptyset;

    while ϕ(iU 1)={1}\phi(\underset{i}{\cup}U_1) = \{1\} as soon as one of the U iU_i is not contained in U 0U_0, which means that one of the ϕ(U i)={1}\phi(U_i) = \{1\} which means that iϕ(U i)={1}\underset{i}{\cup} \phi(U_i) = \{1\};

  2. finite intersections

    because if U 1U 2U 0U_1 \cap U_2 \subset U_0, then by (*)(\ast) U 1U 0U_1 \in U_0 or U 2U 0U_2 \in U_0, whence ϕ(U 1)=\phi(U_1) = \emptyset or ϕ(U 2)=\phi(U_2) = \emptyset, whence with ϕ(U 1U 2)=\phi(U_1 \cap U_2) = \emptyset also ϕ(U 1)ϕ(U 2)=\phi(U_1) \cap \phi(U_2) = \emptyset;

    while if U 1U 2U_1 \cap U_2 is not contained in U 0U_0 then neither U 1U_1 nor U 2U_2 is contained in U 0U_0 and hence with ϕ(U 1U 2)={1}\phi(U_1 \cap U_2) = \{1\} also ϕ(U 1)ϕ(U 2)={1}{1}={1}\phi(U_1) \cap \phi(U_2) = \{1\} \cap \{1\} = \{1\}.

Hence this is indeed a frame homomorphism τ Xτ *\tau_X \to \tau_\ast.

Finally, it is clear that these two operations are inverse to each other.

examples of universal constructions of topological spaces:

\, point space\,\, empty space \,
\, product topological space \,\, disjoint union topological space \,
\, topological subspace \,\, quotient topological space \,
\, fiber space \,\, space attachment \,
\, mapping cocylinder, mapping cocone \,\, mapping cylinder, mapping cone, mapping telescope \,
\, cell complex, CW-complex \,

Last revised on May 9, 2017 at 07:52:52. See the history of this page for a list of all contributions to it.