§ 4. Product of topological spaces

4. Product of topological spaces

1. Product spaces

Definition 1

Given a family $(X_i)_{i \in I}$ of topological spaces, the product space of this family is the product set $X = \prod_{i \in I} X_i$ with the topology which is the product of the topologies of the $X_i$ (§ 2, no. 3, Example 3). The spaces $X_i (i \in I)$ are called the factors of X.

By virtue of § 2, no. 3, Proposition 4, the product topology on $X$ has as a base the set $\mathfrak{B}$ of finite intersections of sets of the form $\overline{\mathrm{pr}}_i^{-1}(U_i)$ , where $U_i$ is open in $X_i$ ; these sets are products $\prod_{i \in I} A_i$ , where $A_i$ is open in $X_i$ for each $i \in I$ and $A_i = X_i$ for all but a finite number of indices. These sets will be called elementary sets.

If $\mathfrak{B}_i$ is a base of the topology of $X_i$ (for each $i \in I$ ), it is clear that the elementary sets $\prod_{i \in I} A_i$ such that $A_i \in \mathfrak{B}_i$ for each index $i$ such that $A_i \neq X_i$ form another base of the product topology. The elementary sets of this type which contain a given point $x \in X$ thus form a fundamental system of neighbourhoods of $x$ ( $§ 1$ , no. 3, Proposition 3).

If $I$ is a finite set, the construction of the product topology from the topologies of the factors $X_i$ is simpler: the elementary sets are just products $\prod_{i \in I} A_i$ , where $A_i$ is any open subset of $X_i$ , for each $i \in I$ (cf. Exercise 9).

Example

\* The product $\mathbf{R}^n$ of $n$ spaces identical with the real line $\mathbf{R}$ is called real number space of $n$ dimensions; $\mathbf{R}^2$ is also called the real plane (cf. Chapter VI, § 1). Likewise, starting from the rational line $\mathbf{Q}$ , we define the rational number space of $n$ dimensions $\mathbf{Q}^n$ (rational plane for $n = 2$ ).

The topology of the space $\mathbf{R}^n$ has as a base the set of all products of $n$ open intervals in $\mathbf{R}$ , which are called open boxes of $n$ dimensions. The open boxes which contain a point $x \in \mathbf{R}^n$ form a fundamental system of neighbourhoods of this point. Likewise the products of $n$ closed intervals in $\mathbf{R}$ are called closed boxes of $n$ dimensions. The closed boxes to which $x$ is interior also form a fundamental system of neighbourhoods of $x$ . There are analogous results for $\mathbf{Q}^n$ . \

Proposition 1

Let $f = (f_i)$ be a mapping of a topological space $Y$ into a product space $X = \prod_{i \in I} X_i$ . Then $f$ is continuous at a point $a \in Y$ if and only if $f_i$ is continuous at $a$ for each $i \in I$ .

Since $f_i = \mathrm{pr}_i \circ f$ , this is just a particular case of Proposition 4 of § 2, no. 3.

Corollary 1

Let $(X_i)_{i \in I}, (Y_i)_{i \in I}$ be two families of topological spaces with the same set of indices. For each $i \in I$ , let $f_i$ be a mapping of $X_i$ into $Y_i$ . In order that the product mapping $f : (x_i) \to (f_i(x_i))$ of

\prod_{i \in I} X_i \text{ into } \prod_{i \in I} Y_i

should be continuous at a point

a = (a_i)

it is necessary and sufficient that

f_i

is continuous at

a_i

for each

i \in I

f can be written as $x \to (f_i(\mathrm{pr}_i(x)))$ , so that by Proposition 1 the condition is sufficient. Conversely, for each $x \in I$ let $g_x$ be the mapping of $X_x$ into $\prod_{i \in I} X_i$ such that $\mathrm{pr}_x(g_x(x_x)) = x_x$ and $\mathrm{pr}_i(g_x(x_x)) = a_i$ whenever $i \neq x$ ; then $g_x$ is continuous at the point $a_x$ , by Proposition 1. Since $f_x = \mathrm{pr}_x \circ f \circ g_x$ it follows that if $f$ is continuous at $a$ , then $f_x$ is continuous at $a_x$ .

Corollary 2

Let $X, Y$ be two topological spaces. In order that a mapping $f : X \to Y$ should be continuous it is necessary and sufficient that the mapping $g : x \to (x, f(x))$ is a homeomorphism of $X$ onto the graph $G$ of $f$ (considered as a subspace of the product space $X \times Y$ ).

Since $f = \mathrm{pr}_2 \circ g$ , the condition is sufficient. It is also necessary, for if $f$ is continuous, then $g$ is bijective and continuous (Proposition 1) and the inverse of $g$ is the restriction of $\mathrm{pr}_1$ to $G$ , which is continuous (cf. Set Theory, Chapter IV, § 2, no. 4, criterion CST 17).

Proposition 2 (Associativity of topological products)

Let $(X_i)_{i \in I}$ be a family of topological spaces, $(J_x)_{x \in K}$ a partition of the set $I$ , and for each $x \in K$ let $X'_x = \prod_{i \in J_x} X_i$ be the product of the spaces $X_i$ for $i \in J_K$ . Then the canonical mapping of the product space*

\prod_{i \in I} X_i \text{ onto the product space } \prod_{x \in K} X'_x

*is a homeomorphism.

This is a particular case of transitivity of initial topologies ( $\S$ 2, no. 3, Proposition 5; cf. Set Theory, Chapter IV, § 2, no. 4, criterion CST 13).

Generally we identify the product spaces $\prod_{i \in I} X_i$ and $\prod_{x \in K} X'_x$ by means of the canonical mapping.

Corollary

Let $\sigma$ be a permutation of the set $I$ . Then the mapping $(x_i) \to (x_{\sigma(i)})$ is a homeomorphism of*

\prod_{i \in I} X_i \text{ onto } \prod_{i \in I} X_{\sigma(i)}.

Take $K = I$ and $J_i = \{ \sigma(i) \}$ for each $i \in I$ in Proposition 2.

Proposition 3

Let $X$ be a set, $(Y_i)_{i \in I}$ a family of topological spaces, and for each $i \in I$ let $f_i$ be a mapping of $X$ into $Y_i$ . Let $f$ be the mapping $x \to (f_i(x))$ of $X$ into $Y = \prod_{i \in I} Y_i$ , and let $\mathfrak{T}$ be the coarsest topology on $X$ for which the mappings $f_i$ are continuous. Then $\mathfrak{T}$ is the inverse image under $f$ of the topology induced on $f(X)$ by the product topology on $Y$ .

This is another particular case of transitivity of initial topologies (\S\ 2, no. 3, Proposition 5; cf. Set Theory, Chapter IV, \S\ 2, no. 4, criterion CST 15).

Corollary

For each $i \in I$ let $A_i$ be a subspace of $Y_i$ . Then the topology induced on $A = \prod_{i \in I} A_i$ by the product topology on $\prod_{i \in I} Y_i$ is the product of the topologies of the subspaces $A_i$ .

Let $j_i$ be the canonical injection $A_i \to Y_i$ , and apply Proposition 3 to the mappings $f_i = j_i \circ \mathrm{pr}_i$ (cf. Set Theory, Chapter IV, \S\ 2, no. 4, criterion CST 14).

2. Section of an open set; section of a closed set; projection of an open set. Partial continuity

Proposition 4

Let $X_1, X_2$ be two topological spaces; then for each $a_1 \in X_1$ , the mapping $x_2 \to (a_1, x_2)$ is a homeomorphism of $X_2$ onto the subspace $\{a_1\} \times X_2$ of $X_1 \times X_2$ .

This is a particular case of Corollary 1 of Proposition 1 of no. 1 applied to the constant function $x_2 \to a_1$ .

The mapping $x_2 \to (a_1, x_2)$ is a continuous section (\S\ 3, no. 5) with respect to the equivalence relation $\mathrm{pr}_2 z = \mathrm{pr}_2 z'$ in $X_1 \times X_2$ ; the quotient space $X_1 \times X_2$ by this equivalence relation is therefore homeomorphic to $X_2$ .

Corollary

The section $A(x_1)$ of an open (resp. closed) set $A$ of the product $X_1 \times X_2$ at an arbitrary point $x_1 \in X_1 (*)$ is open (resp. closed) in $X_2$ .

Proposition 5

The projection of an open set $U$ of the product $X_1 \times X_2$ onto either factor is an open set.

For example, we have $\mathrm{pr}_2 U = \bigcup_{x_1 \in X_1} U(x_1)$ , and the proposition follows from the Corollary to Proposition 4 and axiom (O_1).

Remark 1

The projection of a closed subset of $X_1 \times X_2$ onto a factor need not be closed. For example, in the rational plane $\mathbf{Q}^2$ , the hyperbola whose equation is $x_1 x_2 = 1$ is a closed set, but both its projections are equal to the complement of the point 0 in $\mathbf{Q}$ , and this is not a closed set.

(*) By the section $A(x_1)$ of $A$ at $x_1$ is meant the set of all $x_2 \in X_2$ such that $(x_1, x_2) \in A$ (cf. Set Theory R, \S\ 3, no. 7).

Proposition 6

Let $X_1, X_2, Y$ be three topological spaces, $f$ a mapping of the product space $X_1 \times X_2$ into $Y$ . If $f$ is continuous at the point $(a_1, a_2)$ then the partial mapping $x_2 \to f(a_1, x_2)$ of $X_2$ into $Y$ is continuous at the point $a_2$ .

For this mapping is the composition of $f$ and the mapping $x_2 \to (a_1, x_2)$ ; hence the result follows from Proposition 4.

Proposition 6 is often expressed by saying that a continuous function of two variables is continuous with respect to each of them separately.

Remark 2

It is possible for all the partial mappings determined by a map $f : X_1 \times X_2 \to Y$ to be continuous without $f$ being continuous on $X_1 \times X_2$ (cf. Chapter IX, § 5, Exercise 23). \* For example if $f$ is the mapping of the real plane $\mathbf{R}^2$ into $\mathbf{R}$ defined by

f(x, y) = xy/(x^2 + y^2) \quad \text{if} \quad (x, y) \neq (0, 0)

and

f(0, 0) = 0

, then all the partial mappings are continuous; but

f

is not continuous at

(0, 0)

, since

f(x, x) = 1/2

x \neq 0

If $g$ is a mapping of $X_1$ into $Y$ , continuous at a point $a_1$ , then the mapping $(x_1, x_2) \to g(x_1)$ of $X_1 \times X_1 \to Y$ is continuous at all points $(a_1, x_2)$ , for it is the composition of $g$ and the projection onto $X_1$ .

The results of this subsection are easily extended to an arbitrary product $\prod_{i \in I} X_i$ of topological spaces by remarking that this product is homeomorphic to the product $\left( \prod_{i \in J} X_i \right) \times \left( \prod_{i \in K} X_i \right)$ for any partition $(J, K)$ of $I$ (no. 1, Proposition 2).

3. Closure in a product

Proposition 7

In a product space $\prod_{i \in I} X_i$ , the closure of a product of sets $\prod_{i \in I} A_i$ is the same as the product $\prod_{i \in I} \overline{A}_i$ of their closures.

Suppose that $a = (a_i)$ lies in the closure of $\prod_{i \in I} A_i$ ; then for each $x \in I$ , $a_x = \operatorname{pr}_x a$ is in the closure of $A_x$ because of the continuity of $\operatorname{pr}_x$ (§ 2, no. 1, Theorem 1) and therefore $a \in \prod_{i \in I} \overline{A}_i$ . Conversely, let $b = (b_i) \in \prod_{i \in I} \overline{A}_i$ , and let $\prod_{i \in I} V_i$ be any elementary set containing $b$ ; for each $i \in I$ , $V_i$ contains a point $x_i \in A_i$ ; hence $\prod_{i \in I} V_i$ contains the point $(x_i) \in \prod_{i \in I} A_i$ and therefore $b$ lies in the closure of $\prod_{i \in I} A_i$ .

Corollary

A product* $\prod_{i \in I} A_i$ of non-empty sets is closed in the product space $\prod_{i \in I} X_i$ if and only if $A_i$ is closed in $X_i$ for each $i \in I$ .

If $I$ is finite, a product $\prod_{i \in I} A_i$ is open provided that $A_i$ is open in $X_i$ for each $i \in I$ ; but this is no longer so if $I$ is infinite.

Proposition 8

Let* $a = (a_i)$ be any point of a product space $X = \prod_{i \in I} X_i$ ; then the set $D$ of points $x \in X$ such that $\mathrm{pr}_i x = a_i$ except for a finite number of indices $i$ is dense in $X$ .

For each $x \in X$ and each elementary set $V = \prod_{i \in I} U_i$ which contains $x$ , we have $U_i = X_i$ except for indices $i$ belonging to a finite subset $J$ of $I$ ; if we take $y_i = x_i$ for $i \in J$ and $y_i = a_i$ for $i \notin J$ , it is clear that

y = (y_i) \in D \quad \text{and} \quad y \in V;

hence the result.

4. Inverse limits of topological spaces

Let $I$ be a partially ordered (but not necessarily directed) set (*), in which the order relation is written $\alpha \leq \beta$ . For each $\alpha \in I$ , let $X_\alpha$ be a topological space, and for each pair $(\alpha, \beta)$ such that $\alpha \leq \beta$ let $f_{\alpha \beta}$ be a mapping of $X_\beta$ into $X_\alpha$ . We say that $(X_\alpha, f_{\alpha \beta})$ is an inverse system of topological spaces if: 1) $(X_\alpha, f_{\alpha \beta})$ is an inverse system of sets; 2) $f_{\alpha \beta}$ is a continuous mapping whenever $\alpha \leq \beta$ . Let $X$ denote the set $\varprojlim X_\alpha$ , and for each $\alpha \in I$ let $f_\alpha$ be the canonical mapping $X \to X_\alpha$ ; then the coarsest topology on $X$ for which the $f_\alpha$ are continuous is said to be the inverse limit (with respect to the $f_{\alpha \beta}$ ) of the topologies of the $X_\alpha$ , and the set $X$ with this topology is called the inverse limit of the inverse system of topological spaces $(X_\alpha, f_{\alpha \beta})$ . Whenever we speak of $\varprojlim X_\alpha$ as a topological space, it is always to be understood that the topology of this space is the inverse limit of the topologies of the $X_\alpha$ , unless the contrary is expressly stated.

The set $X$ is the subset of the product $\prod_{\alpha \in I} X_\alpha$ consisting of those points $x$ such that

(1)

\mathrm{pr}_\alpha(x) = f_{\alpha \beta}(\mathrm{pr}_\beta(x))

whenever

\alpha \leq \beta

. It follows from Proposition 3 of no. 1 that the inverse limit of the topologies of the

X_\alpha

is the same as the topology induced on

X

by the topology of the product space

\prod_{\alpha \in I} X_\alpha

. If, for each

\alpha \in I

Y_\alpha

() That is, a set endowed with a reflexive and transitive relation (Set Theory* R, § 6, no. 1).

a subspace of $X_\alpha$ such that the $Y_\alpha$ form an inverse system of subsets of the $X_\alpha$ (Set Theory chapter III, , n° 2), then it is clear that the topological space $\varprojlim Y_\alpha$ is a subspace of $\varprojlim X_\alpha$

Let $(X'_\alpha, f'_{\alpha\beta})$ be another inverse system of topological spaces indexed by the same set $I$ , and for each $\alpha \in I$ let $u_\alpha : X_\alpha \to X'_\alpha$ be a continuous mapping such that $(u_\alpha)$ is an inverse system of mappings; then $u = \varprojlim u_\alpha$ is a continuous mapping of $X = \varprojlim X_\alpha$ into $X' = \varprojlim X'_\alpha$ . For if $f'_\alpha$ is the canonical mapping $X' \to X'_\alpha$ , we have $f'_\alpha \circ u = u_\alpha \circ f_\alpha$ , so that $f'_\alpha \circ u$ is continuous for each $\alpha \in I$ , and the assertion follows from Proposition 4 of § 2, no. 3.

Finally, suppose $I$ is a directed set, and let $J$ be a cofinal subset of $I$ ; let $Z$ be the inverse limit of the inverse system of topological spaces $(X_\alpha, f_{\alpha\beta})_{\alpha \in J, \beta \in J}$ . Then the canonical bijection $g : X \to Z$ (Set Theory, Chapter III, § 7, no. 2, proposition 3) is a homeomorphism. For we have $\mathrm{pr}_\lambda(g(x)) = \mathrm{pr}_\lambda(x)$ for each $\lambda \in J$ ; hence $g$ is continuous (no. 1, Proposition 1); and if $h$ is the inverse of $g$ , then for each $\alpha \in I$ there exists $\lambda \in J$ such that $\alpha \leq \lambda$ , and therefore $\mathrm{pr}_\alpha(h(z)) = f_{\alpha\lambda}(\mathrm{pr}_\lambda(z))$ , which shows that $h$ is continuous (no. 1, Proposition 1), since the $f_{\alpha\lambda}$ are continuous.

Proposition 9

Let $I$ be a directed set and $J$ a cofinal subset of $I$ . Let $(X_\alpha, f_{\alpha\beta})$ be an inverse system of topological spaces indexed by $I$ ; let $X = \varprojlim X_\alpha$ and let $f_\alpha : X \to X_\alpha$ be the canonical mapping. Then the family of sets $f^{-1}_\alpha(U_\alpha)$ , where $\alpha$ runs through $J$ and $U_\alpha$ runs through a base $\mathfrak{B}_\alpha$ of the topology of $X_\alpha$ for each $\alpha \in J$ , is a base of the topology of $X$ .

From § 2, no. 3 we know that the finite intersections of sets of the form $f^{-1}_\alpha(U_\alpha)$ ( $\alpha \in I, U_\alpha$ open in $X_\alpha$ ) form a base of the topology of $X$ . If $(\alpha_i)_{1 \leq i \leq n}$ is a finite family of indices of $I$ , then there exists $\gamma \in J$ such that $\alpha_i \leq \gamma$ for $1 \leq i \leq n$ ; hence $f_{\alpha_i} = f_{\alpha_i\gamma} \circ f_\gamma$ ; if we put

V_\gamma = \bigcap_i f^{-1}_{\alpha_i\gamma}(U_{\alpha_i}),

then we have

f^{-1}_\gamma(V_\gamma) = \bigcap_i f^{-1}_{\alpha_i}(U_{\alpha_i});

but $V_\gamma$ is open and is therefore a union of sets belonging to $\mathfrak{B}_\gamma$ . Hence the result.

Corollary

Let $A$ be a subset of $X$ and let $A_\alpha$ denote $f_\alpha(A)$ for each $\alpha \in I$ . Then:

(i) The $A_\alpha$ (resp. the $\overline{A}_\alpha$ ) form an inverse system of subsets of the $X_\alpha$ , and

\overline{A} = \bigcap_\alpha f^{-1}_\alpha(\overline{A}_\alpha) = \varprojlim \overline{A}_\alpha.

(ii) If $A$ is closed in $X$ , $A = \varprojlim A_\alpha = \varprojlim \overline{A}_\alpha$ .

The first assertion of (i) follows from the relations $f_\alpha = f_{\alpha\beta} \circ f_\beta$ for $\alpha \leq \beta$ and from the continuity of the $f_{\alpha\beta}$ (§ 2, no. 1, Theorem 1). Let $A'$ denote

\bigcap_\alpha \overline{f}_\alpha(\overline{A}_\alpha);

then it is clear that

A'

is closed and contains

A

, so that

\overline{A} \subset A'

. Conversely, let

x \in A'

; we have to show that

x

lies in the closure of

A

. By virtue of Proposition 9 it is enough to prove that every neighbourhood of

x

which is of the form

\overline{f}_\alpha(U_\alpha)

, with

\alpha \in I

and

U_\alpha

open in

X_\alpha

, meets

A

. Now, by hypothesis,

f_\alpha(x) \in U_\alpha

, and since

f_\alpha(x) \in \overline{A}

we have

U_\alpha \cap A_\alpha \neq \varnothing

, which means that

A \cap \overline{f}_\alpha(U_\alpha)

is not empty.

To establish (ii) it is enough to remark that, without any restriction on $A$ , we have $A \subset \lim A_\alpha \subset \lim \overline{A}_\alpha$ ; now if $A$ is closed, then from (i)

A = \lim \overline{A}_\alpha

and (ii) follows.

Example

Let $I$ be a directed set and $(X_\alpha)_{\alpha \in I}$ a family of subsets of a set $Y$ , such that $X_\alpha \supset X_\beta$ whenever $\alpha \leq \beta$ . For each $\alpha \in I$ let $\mathfrak{T}_\alpha$ be a topology on $X_\alpha$ such that $\mathfrak{T}_\beta$ is finer than the topology induced on $X_\beta$ by $\mathfrak{T}_\alpha$ whenever $\alpha \leq \beta$ . If we take $f_{\alpha\beta}$ to be the canonical injection $X_\beta \to X_\alpha$ for $\alpha \leq \beta$ , then $\lim X_\alpha$ may be identified canonically with the intersection $X$ of the $X_\alpha$ , with the topology which is the least upper bound (§ 2, no. 3, Example 2) of the topologies induced on $X$ by the $\mathfrak{T}_\alpha$ .

Exercises

See the exercises for § 4.