nLab product topological space

Redirected from "Tychonoff product".
Contents

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

Limits and colimits

Contents

Idea

Given two topological spaces (X 1,τ 1)(X_1,\tau_1) and (X 2,τ 2)(X_2, \tau_2), then the Cartesian product of their underlying sets X 1×X 2X_1 \times X_2 is naturally equipped with a topology τ X 1×X 2\tau_{X_1 \times X_2} itself, generated from the base opens which are themselves Cartesian product U 1×U 2X 1×X 2U_1 \times U_2 \subset X_1 \times X_2, of open subsets of the original spaces: U iX iU_i \subset X_i. The resulting topological space

(X 1×X 2,τ X 1×X 2) (X_1 \times X_2 , \tau_{X_1 \times X_2})

is called the product topological space of the two original spaces.

The formally dual concept is that of disjoint union topological spaces.

More generally, consider any index set II and an II-indexed set {X i,τ i} iI\{X_i, \tau_i\}_{i \in I} of topological spaces. Then the product topology τ prod\tau_{prod} or Tychonoff topology on the Cartesian product iIX i\underset{i \in I}{\prod} X_i of underlying sets is equivalently

  1. the topology generated from the sub-base given by products iIU i\underset{i \in I}{\prod} U_i with U iX iU_i \subset X_i open, but all except one of the factors equal to the corresponding X iX_i,

    hence the topology whose open subsets are precisely those obtained as arbitrary unions of finite intersections of such subsets;

  2. the topology generated from the base given by products iIU i\underset{i \in I}{\prod} U_i with U iX iU_i \subset X_i open, but all except a finite number of factors equal to the corresponding X iX_i,

    hence the topology whose open subsets are precisely those obtained as arbitrary unions of such subsets.

This product topology is singled out by the fact that the resulting product topological space is the category theoretic product of the original space in the category Top of topological spaces:

(iIX i,τ prod)iI(X i,τ i). \left( \underset{i \in I}{\prod} X_i ,\; \tau_{prod} \right) \;\simeq\; \underset{i \in I}{\prod} (X_i, \tau_i) \,.

This means that the product topology enjoys the universal property that for any topological space (Y,τ Y)(Y,\tau_Y) then sets of continuous functions {(Y,τ Y)ϕ i(X i,τ i)} iI\{(Y, \tau_Y) \overset{\phi_i}{\to} (X_i, \tau_i)\}_{i \in I} into the factor spaces are in natural bijection with continuous functions (ϕ i) iI:(Y,τ Y)(iIX i,τ prod)(\phi_i)_{i \in I} \colon (Y, \tau_Y) \to \left(\underset{i \in I}{\prod} X_i, \tau_{prod}\right) into the product topological space.

Beware that (among others) there is also the box topology τ box\tau_{box} on the Cartesian product of underlying sets iIX i\underset{i\in I}{\prod} X_i, whose open subsets are the unions of those of the for iIU i\underset{i \in I}{\prod} U_i with U iX iU_i \subset X_i open and with no further restriction on the factors.

For II a finite set, then these two topologies coincide, but for II not finite then the box topology is a strictly finer topology

τ prodτ box \tau_{prod} \subset \tau_{box}

and hence in this case it does not enjoy the universal property of the product topology above.

examples of universal constructions of topological spaces:

AAAA\phantom{AAAA}limitsAAAA\phantom{AAAA}colimits
\, 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 \,

Definition

Definition

Let {(X i,τ i)} iI\{(X_i, \tau_i)\}_{i \in I} be a set of topological spaces. Then their product topological space has as underlying set the Cartesian product iIX i\underset{i \in I}{\prod} X_i of the underlying sets, and has as topology τ prodP(iIX i)\tau_{prod} \subset P(\underset{i \in I}{\prod} X_i) the coarsest topology such that all the projection maps

p i:iIX iX i p_i \;\colon\; \underset{i \in I}{\prod} X_i \longrightarrow X_i

become continuous functions (called the Tychonoff topology).

This means equivalently that τ prod\tau_{prod} is the topology generated from the sub-base

β prod{p i 1(U i)iIX i|iI,U iX iopen}. \beta_{prod} \;\coloneqq\; \left\{ p_i^{-1}(U_i) \subset \underset{i \in I}{\prod} X_i \;\vert\; i \in I, U_i \subset X_i \, \text{open} \right\} \,.

Notice that

p i 1(U i)=U i×(jI\{i}X j) p_i^{-1}(U_i) = U_i \times \left(\underset{j \in I \backslash \{i\}}{\prod} X_j\right)

and that

p i 1(U i)p j 1(U j)=U i×U j×(kI\{i,j}X k) p_i^{-1}(U_i) \cap p_j^{-1}(U_j) = U_i \times U_j \times \left(\underset{k \in I \backslash \{i,j\}}{\prod} X_k\right)

etc.

Examples

Example

Let XX be any topological space and write Disc({1,2})Disc(\{1,2\}) for the discrete topological space on a set with two elements. Then there is a homeomorphism

X×Disc({0,1})XX X \times Disc(\{0,1\}) \;\simeq\; X \sqcup X

between the product space of XX with Disc({1,2})Disc(\{1,2\}) and the disjoint union space of XX with itself.

Proof

By definition of disjoint union there is a bijection of underlying sets XXX×{1,2}X \sqcup X \simeq X \times \{1,2\}.

By unwinding the definitions

  1. the open subsets of X×Disc({0,1})X \times Disc(\{0,1\}) are unions of those of the form U×SU \times S, where UXU \subset X is an open subset and S{1,2}S \subset \{1,2\} is any subset

  2. the open subsets of XXX \sqcup X are of the form UVU \sqcup V with U,VXU,V \subset X open.

Under the above bijection the we have

UV=(U×{1})(V×{2}). U \sqcup V = \left(U \times \{1\}\right) \cup \left( V \times \{2\}\right) \,.

Conversely, every union of elements of the form U×{1}U' \times \{1\}, V×{2}V' \times \{2\} and W×{1,2}=W×{1}W×{2}W \times \{1,2\} = W \times \{1\} \cup W \times \{2\} is of the form U×{1}V×{2}U \times \{1\} \cup V \times \{2\}.

This shows that the above bijection of underlying sets induces a bijection of the corresponding open subsets, hence is a homeomorphism.

Example

For nn \in \mathbb{N} consider the Cartesian space n\mathbb{R}^n with the metric topology induced from its Euclidean metric structure. Then the product topological spaces satisfy

n 1× n 2 n 1+n 2. \mathbb{R}^{n_1}\times \mathbb{R}^{n_2} \simeq \mathbb{R}^{n_1 + n_2} \,.
Example

Let Disc(S)Disc(S) be a discrete topological space on a set with at least two elements. Then the infinite product space

nDisc(S) \underset{n \in \mathbb{Z}}{\prod} Disc(S)

is itself not a discrete space.

Proof

The open subsets of a discrete space include all the subsets of the underlying set. Hence we need to see that there are subsets of the Cartesian product set nDisc(S)\underset{n \in \mathbb{Z}}{\prod} Disc(S) which are not open subsets in the Tychonoff topology.

But by definition the open subsets in the Tychnoff topology are unions of products of open subsets of the factor spaces such that only a finite number of the factors is not the total space.

Since SS is assumed to contain at least two elements let 1S1 \in S be one of these. Then {1}S\{1\} \subset S is a proper subset. Accordingly the product subset

n𝔹{1}nDisc(S) \underset{n \in \mathbb{B}}{\prod} \{1\} \;\subset\; \underset{n\in \mathbb{N}}{\prod} Disc(S)

is not open. For if it were, it would have to be the union of product subsets that contain the total set SS in at least one entry, which by construction it is not.

Example

(Cantor space)

Write Disc({0,1})Disc(\{0,1\}) for the the discrete topological space with two points. Write nDisc({0,2})\underset{n \in \mathbb{N}}{\prod} Disc(\{0,2\}) for the product topological space of a countable set of copies of this discrete space with itself (i.e. the corresponding Cartesian product of sets n{0,1}\underset{n \in \mathbb{N}}{\prod} \{0,1\} equipped with the Tychonoff topology induced from the discrete topology of {0,1}\{0,1\}).

Then consider the function

n κ [0,1] (a i) i AAAA i=02a i3 n \array{ \underset{n \in \mathbb{N}}{\prod} &\overset{\kappa}{\longrightarrow}& [0,1] \\ (a_i)_{i \in \mathbb{N}} &\overset{\phantom{AAAA}}{\mapsto}& \underoverset{i = 0}{\infty}{\sum} \frac{2 a_i}{ 3^n} }

which sends an element in the product space, hence a sequence of binary digits, to the value of the power series as shown on the right.

One checks that this is a continuous function (from the product topology to the Euclidean metric topology on the closed interval). Moreover with its image κ(n{0,1})[0,1]\kappa\left( \underset{n \in \mathbb{N}}{\prod} \{0,1\}\right) \subset [0,1] equipped with its subspace topology, then this is a homeomorphism onto its image:

nDisc({0,1})AAAAκ(nDisc({0,1}))AAAA[0,1]. \underset{n \in \mathbb{N}}{\prod} Disc(\{0,1\}) \overset{\phantom{AA}\simeq\phantom{AA}}{\longrightarrow} \kappa\left( \underset{n \in \mathbb{N}}{\prod} Disc(\{0,1\}) \right) \overset{\phantom{AAAA}}{\hookrightarrow} [0,1] \,.

This image is the Cantor space as a subspace of the closed interval.

Example

A Cartesian product of countably many copies of the real closed interval [0,1][0,1] (as a topological space) is the Hilbert cube.

Properties

Basic properties

Proposition

(projections are open maps)

For {X i} iI\{X_i\}_{i \in I} a set of topological spaces, the for each jIj \in I the projection out of their product space into the jjth component

p j:iIX iX j p_j \;\colon\; \underset{i \in I}{\prod} X_i \longrightarrow X_j

is an open map.

Proof

Since images preserve unions (this prop.) it is sufficient to check that the image under p jp_j of a base open? is still open. But base opens in the product topology by definition are, in particular, products of open subsets.

Universal property

The product topological space construction from def. is the limit in Top over the discrete diagram consisting of the factor spaces, hence the category theoretic product.

For proof see at Top – Universal constructions.

The Tychonoff theorem

The Tychonoff theorem states that the product space of any set of compact topological spaces (with its Tychonoff topology) is itself compact.

Relation to smash product of pointed spaces

Proposition

(one-point compactification intertwines Cartesian product with smash product)

On the subcategory Top LCHausTop_{LCHaus} of Top on the locally compact Hausdorff spaces with proper maps between them, the functor of one-point compactification (Prop. )

() cpt:Top LCHausTop */ (-)^{cpt} \;\colon\; Top_{LCHaus} \longrightarrow Top^{\ast/}

sends Cartesian products (product topological spaces) to smash products of pointed topological spaces, hence constitutes a strong monoidal functor, in that there is a natural homeomorphism:

(X×Y) cptX cptY cpt. \big( X \times Y \big)^{cpt} \;\simeq\; X^{cpt} \wedge Y^{cpt} \,.

This is briefly mentioned in Bredon 93, p. 199. The argument is spelled out in: MO:a/1645794/, Cutler 20, Prop. 1.6.

Relation to singular (co)homology

The singular homology of product topological spaces is informed by

examples of universal constructions of topological spaces:

AAAA\phantom{AAAA}limitsAAAA\phantom{AAAA}colimits
\, 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 \,

References

The Tychonoff topology is named after A. N. Tychonoff.

On product spaces with Lindelöf topological spaces:

Last revised on January 19, 2024 at 22:40:13. See the history of this page for a list of all contributions to it.