2.8Box and Product Topology

Updated 29 Jun 2026Chapter (PDF)

We have a notion of Cartesian product for two sets $X, Y$ . Namely, $X \times Y = \{(x, y): x \in X, y \in Y\}$ . But what about the Cartesian product of three sets? Four? Five? Countably many? To tackle this, we redefine the Cartesian product in terms of an arbitrary indexing set.

[2.8.1]Definition(Cartesian Tuple, Cartesian Product)#

Let $\Lambda$ be an indexing set. Given a set $X$ , we define a $\Lambda$ -tuple of $X$ to be some function $\vb{x}: \Lambda \to X$ . We define $x(\lambda)$ , more commonly denoted as $\vb{x}_\lambda$ , as the $\lambda$ -th component of $\vb{x}$ . We often write $(x_\lambda)_{\lambda \in \Lambda}$ in lieu of $\vb{x}$ . We denote the set of all $\Lambda$ -tuples on $X$ by $X^{\Lambda}$ , the set of all functions from $\Lambda$ to $X$ .

Now let $\{A_\lambda\}_{\lambda \in \Lambda}$ be an indexed family of set. We define the Cartesian product of this family of sets as the set of all $\Lambda$ -tuples $(x_\lambda)_{\lambda \in \Lambda}$ such that each $x_\lambda \in A_\lambda$ (that is, the $\lambda$ -th coordinate of $\vb{x}$ lies in $A_\lambda$ ). That is,

\text{Cartesian product of } \{A_\lambda\}_{\lambda \in \Lambda} = \prod_{\lambda \in \Lambda} A_\lambda = \left\{x: \Lambda \to \bigcup_{\lambda \in \Lambda} A_\lambda: x_\lambda \in A_\lambda \, \forall \lambda \in \Lambda \right\}.

From here, we can extend the product topology of two sets in different ways. Recall how the basis and subbasis were defined for the product topology on $X \times Y$ . We extend these notions to general indexing sets.

[2.8.2]Definition(Box Topology, Product Topology)#

Suppose $\{X_\lambda\}_{\lambda \in \Lambda}$ is an indexed family of sets.

We define the box topology to be the topology generated by the basis

\mathcal{B} = \left\{\prod_{\lambda \in \Lambda} U_\lambda: U_\lambda \text{ is open in } X_\lambda \right\}.

We define the product topology to be the topology generated by the subbasis

\mathcal{S}_\lambda = \{\pi_\lambda^{-1}(U_\lambda): U_\lambda \text{ is open in } X_\lambda \} \qquad \mathcal{S} = \bigcup_{\lambda \in \Lambda} \mathcal{S}_\lambda.

Of course, if $\Lambda = \{1, 2\}$ , then these are identical to our previous notion of the product topology on the Cartesian product of two sets. In fact, as we will realize soon, the box and product topology coincide for Cartesian products that come from finite indexing sets.

[2.8.3]Remark(Comparison Between Box and Product Topology)#

We will compare these topologies by noticing the basis generated by the subbasis for the product topology, then comparing this basis with the basis for the box topology. From the previous definition, let $\{X_\lambda\}_{\lambda \in \Lambda}$ be a family of indexed sets, and define $\mathcal{S}_\lambda$ and $\mathcal{S}$ accordingly.

Recall that a basis is formed by finite intersection of elements of a subbasis. Let's first discover what the finite intersection of elements of $\mathcal{S}_\lambda$ do. Let $O_1, \dots, O_n$ be open subsets of $X_\lambda$ , then note

\bigcap_{k = 1}^n \pi_\lambda^{-1}(O_k) = \pi_\lambda^{-1}\underbrace{\Bigg(\bigcap_{k = 1}^n O_k\Bigg)}_{\text{open in } X_\lambda},

which is another element of $S_\lambda$ . Thus, a general basis element $B \in \mathcal{B}$ can be written as

B = \bigcap_{k = 1}^n \pi_{\lambda_k}^{-1}(O_{\lambda_k}),

where $\pi_{\lambda_k}: \prod_{\lambda \in \Lambda} X_\lambda \to X_{\lambda_k}$ is the projection map on $X_{\lambda_k}$ , and $O_{\lambda_k}$ is an arbitrary open subset of $X_{\lambda_k}$ . This implies an element $\vb{x} = (x_\lambda)_{\lambda \in \Lambda} \in \prod_{\lambda \in \Lambda} X_\lambda$ belongs to some basis $B$ if $x_{\lambda_k} \in O_{\lambda_k}$ for each $k \in \{1, 2, \dots, n\}$ . Thus, we write $B$ as a product

B = \prod_{\lambda \in \Lambda} U_\lambda,

where $U_\lambda \subseteq X_\lambda$ is open for $\lambda \in \{1, 2, \dots, n\}$ , and $U_\lambda = X_\lambda$ otherwise. Recalling that a basic open set in the box topology is the product of any open subset of $X_\lambda$ for each $\lambda \in \Lambda$ , not just finitely many of them, we conclude that the box topology is strictly finer than the product topology for an infinite indexing set, and they are otherwise equal for a finite one.

The implication of this is that, for infinite products, an element of a product belongs to a basic open set in the product topology if and only if finitely many of its coordinates lie in the finitely many open sets that generate the basis element. On the contrary, a basic open set in the box topology is determined by $|\Lambda|$ many open sets, not just finitely many.

The standard topology for the Cartesian product is the product topology.

[2.8.4]Theorem#

Suppose $\{X_\lambda\}_{\lambda \in \Lambda}$ is an indexed family of sets, and let $X = \prod_{\lambda \in \Lambda} X_\lambda$ . Then let $\{\mathcal{B}_\lambda\}_{\lambda \in \Lambda}$ be an indexed family of sets such that $\mathcal{B}_{\lambda}$ is a basis for each $X_\lambda$ . Then

\mathcal{C} = \prod_{\lambda \in \Lambda} B_\lambda,

where $B_\lambda \in \mathcal{B}_\lambda$ for each $\lambda \in \Lambda$ , forms a basis for the box topology on $X$ . Alternatively, if $B_\lambda \in \mathcal{B}_\lambda$ for finitely many values of $\lambda \in \Lambda$ , and otherwise $B_\lambda = X_\lambda$ , then the product forms a basis for the product topology.

Proof.

Let $X := \prod_{\lambda \in \Lambda} X_\lambda$ be endowed with the box topology, and let $O \subseteq X$ be open. We seek to show that, for each $x \in O$ , there is some $C \in \mathcal{C}$ such that $x \in C \subseteq O$ . If $x \in O$ , then by the definition of the box topology, there exists a standard basic open set such that $x \in \prod_{\lambda \in \Lambda} O_\lambda \subseteq O$ , where $O_\lambda \subseteq X_\lambda$ is open for each $\lambda \in \Lambda$ .

But by definition, because $\mathcal{B}_\lambda$ is a basis for $X_\lambda$ and the coordinate $x_\lambda \in O_\lambda$ , we have that for each $\lambda \in \Lambda$ , there exists a basis element $B_\lambda \in \mathcal{B}_\lambda$ such that

x_\lambda \in B_\lambda \subseteq O_\lambda.

But this implies there is a specific choice of sets to form $C = \prod_{\lambda \in \Lambda} B_\lambda \in \mathcal{C}$ such that

x \in \prod_{\lambda \in \Lambda} B_\lambda \subseteq \prod_{\lambda \in \Lambda} O_\lambda \subseteq O,

completing the proof that $\mathcal{C}$ is a basis for the box topology. An entirely analogous proof follows for the product topology, with the added condition that $O_\lambda = X_\lambda$ (and thus we can choose $B_\lambda = X_\lambda$ ) for all but finitely many values of $\lambda$ .

[2.8.5]Theorem(Component-Wise Continuity)#

Suppose $\{X_\lambda\}_{\lambda \in \Lambda}$ is a family of indexed sets, where $X = \prod_{\lambda \in \Lambda} X_\lambda$ . Suppose $f: A \to X$ is such that $f(a) = (f_\lambda(a))_{\lambda \in \Lambda}$ for every $a \in A$ . Then $f$ is continuous if and only if each $f_\lambda$ is continuous.

Proof.

$(\Longrightarrow)$ : Suppose $f$ is continuous. For any $\lambda \in \Lambda$ , note $f_\lambda = \pi_\lambda \circ f$ . Note that the product topology forces $\pi_\lambda$ to be continuous, $f$ is continuous by hypothesis, and the composition of continuous functions is continuous, so $f_\lambda$ is continuous.

$(\Longleftarrow)$ : Suppose each $f_\lambda$ is continuous. To show that $f$ is continuous, it suffices to show that the preimage of every subbasis element of $X$ is open in $A$ . Let $S$ be a standard subbasis element of the product topology on $X$ . Then $S = \pi_\beta^{-1}(U_\beta)$ for some index $\beta \in \Lambda$ and some open set $U_\beta \subseteq X_\beta$ . We examine the preimage of $S$ under $f$ :

f^{-1}(S) = f^{-1}(\pi_\beta^{-1}(U_\beta)) = (\pi_\beta \circ f)^{-1}(U_\beta) = f_\beta^{-1}(U_\beta).

Because $f_\beta$ is continuous by hypothesis and $U_\beta$ is open in $X_\beta$ , the set $f_\beta^{-1}(U_\beta)$ is open in $A$ . Thus, $f^{-1}(S)$ is open in $A$ , meaning $f$ is continuous.

[2.8.6]Remark(Component-Wise Continuity Fails for Box Topology)#

Define $\bR^\omega$ to be the countably infinite product of $\bR$ with itself. That is, $\bR^\omega = \prod_{k \in \bZ_+} \bR$ . Define the function $f: \bR \to \bR^\omega$ as the “identity” function $f(t) = (t, t, t, \dots)$ , where $\bR$ takes the standard topology and $\bR^\omega$ takes the box topology. Note then that $f(t) = (f_n(t))_{n \in \bZ_+}$ , where each $f_n(t) = t$ . Clearly, each $f_n$ is continuous. But is $f$ continuous? Consider the basis element

B = (-1, 1) \times \left(-\frac{1}{2}, \frac{1}{2}\right) \times \left(-\frac{1}{3}, \frac{1}{3}\right) \times \cdots,

which is open in the box topology. If $f$ is continuous, then $f^{-1}(B)$ must be open. But suppose $x \ne 0$ were in $f^{-1}(B)$ —then $f(x) \in B$ , meaning $f(x) = (x, x, x, \dots) \in B$ , but there is some $n \in \bZ_+$ such that $x \notin (-1/n, 1/n)$ , and so $f(x) \notin B$ . $0$ , however, does lie in $f^{-1}(B)$ , so this concludes $f^{-1}(B) = \{0\}$ , which is not open in $\bR$ . Thus $f$ is not continuous, emphasizing the previous result only holds for the product topology and not the box topology.

2.8.1Exercises#

[2.8.7]Problem#

Let $\{X_\lambda\}$ be an indexed product of sets, and let $X = \prod_{\lambda \in \Lambda} X_\lambda$ . Let $A_\lambda \subseteq X_\lambda$ be a subspace for each $\lambda \in \Lambda$ , and let $A = \prod_{\lambda \in \Lambda} A_\lambda$ . Show $A$ is a subspace of $X$ if both products are given the box topology or product topology.

Solution.

To show $A \subseteq X$ is a subspace for $X$ , we show that the basis for the box topology on $A$ is the same as the subspace topology. Let $B$ be an arbitrary basis element for the box topology on $A$ , meaning $B = \prod_{\lambda \in \Lambda} O_\lambda$ where $O_\lambda \subseteq A_\lambda$ is open for each $\lambda \in \Lambda$ . Since $A_\lambda$ takes on the subspace topology, it follows that $O_\lambda = U_\lambda \cap A_\lambda$ for some open subset $U_\lambda \subseteq X_\lambda$ . Then

B = \prod_{\lambda \in \Lambda} (U_\lambda \cap A_\lambda) = \left(\prod_{\lambda \in \Lambda} U_\lambda\right) \cap \left(\prod_{\lambda \in \Lambda} A_\lambda\right) = O \cap A,

where $O \subseteq X$ is open in the topology of $X$ . Thus $O \cap A$ is open in the subspace topology of $A$ . Conversely, we show that an arbitrary basis element–say, B—for the subspace topology on $A$ is open in the box topology. Write $B = O \cap A$ , where $O$ is a basic open subset of $X$ . Then $O = \prod_{\lambda \in \Lambda} O_\lambda$ , where each $O_\lambda \subseteq X_\lambda$ is open, meaning we may write

B = \left(\prod_{\lambda \in \Lambda} O_\lambda\right) \cap \left(\prod_{\lambda \in \Lambda} A_\lambda\right) = \prod_{\lambda \in \Lambda} \underbrace{(O_\lambda \cap A_\lambda)}_{\text{open in } A_\lambda},

which is open in the box topology of $A$ (by virtue of being a basis element). Thus the subspace and box topologies are equivalent for $A$ , making $A$ a subspace of $X$ .

The proof for the product topology is entirely analogous.