§ 19. The Product Topology

Definition 19.1 : J-Tuple / Coordinate

Let \(J\) be an index set. Given a set \(X\), we define a \(J\)-tuple of elements of \(X\) to be a function \(\mathbf{x}: J \rightarrow X\). If \(\alpha\) is an element of \(J\), we often denote the value of \(\mathbf{x}\) at \(\alpha\) by \(x_{\alpha}\) rather than \(\mathbf{x}(\alpha)\); we call it the \(\alpha\) th coordinate of \(\mathbf{x}\). And we often denote the function \(\mathbf{x}\) itself by the symbol

\[ \left(x_{\alpha}\right)_{\alpha \in J}, \]

which is as close as we can come to a "tuple notation" for an arbitrary index set \(J\). We denote the set of all \(J\)-tuples of elements of \(X\) by \(X^J\).

Definition 19.2 : Cartesian Product

Let \(\left\{A_{\alpha}\right\}_{\alpha \in J}\) be an indexed family of sets; let \(X=\bigcup_{\alpha \in J} A_{\alpha}\). The cartesian product of this indexed family, denoted by

\[ \prod_{\alpha \in J} A_{\alpha} \]

is defined to be the set of all \(J\)-tuples \(\left(x_{\alpha}\right)_{\alpha \in J}\) of elements of \(X\) such that \(x_{\alpha} \in A_{\alpha}\) for each \(\alpha \in J\). That is, it is the set of all functions

\[ \mathbf{x}: J \rightarrow \bigcup_{\alpha \in J} A_{\alpha} \]

such that \(\mathbf{x}(\alpha) \in A_{\alpha}\) for each \(\alpha \in J\).

Occasionally we denote the product simply by \(\prod A_{\alpha}\), and its general element by \(\left(x_{\alpha}\right)\), if the index set is understood.

Definition 19.3 : Box Topology

Let \(\left\{X_{\alpha}\right\}_{\alpha \in J}\) be an indexed family of topological spaces. Let us take as a basis for a topology on the product space

\[ \prod_{\alpha \in J} X_{\alpha} \]

the collection of all sets of the form

\[ \prod_{\alpha \in J} U_{\alpha} \]

where \(U_{\alpha}\) is open in \(X_{\alpha}\), for each \(\alpha \in J\). The topology generated by this basis is called the box topology.

Definition 19.4 : Projection Mapping

Let

\[ \pi_{\beta}: \prod_{\alpha \in J} X_{\alpha} \rightarrow X_{\beta} \]

be the function assigning to each element of the product space its \(\beta\) th coordinate,

\[ \pi_{\beta}\left(\left(x_{\alpha}\right)_{\alpha \in J}\right)=x_{\beta} \]

it is called the projection mapping associated with the index \(\beta\).

Definition 19.5 : Product Topology / Product Space

Let \(S_{\beta}\) denote the collection

\[ \mathcal{S}_{\beta}=\left\{\pi_{\beta}^{-1}\left(U_{\beta}\right) \mid U_{\beta} \text { open in } X_{\beta}\right\} \]

and let \(S\) denote the union of these collections,

\[ \mathcal{S}=\bigcup_{\beta \in J} s_{\beta} \]

The topology generated by the subbasis \(S\) is called the product topology. In this topology \(\prod_{\alpha \in J} X_{\alpha}\) is called a product space.

Theorem 19.1 : Comparison of the Box and Product Topologies

The box topology on \(\prod X_{\alpha}\) has as basis all sets of the form \(\prod U_{\alpha}\), where \(U_{\alpha}\) is open in \(X_{\alpha}\) for each \(\alpha\). The product topology on \(\prod X_{\alpha}\) has as basis all sets of the form \(\prod U_{\alpha}\), where \(U_{\alpha}\) is open in \(X_{\alpha}\) for each \(\alpha\) and \(U_{\alpha}\) equals \(X_{\alpha}\) except for finitely many values of \(\alpha\).

Proof

We consider the basis \(\mathcal{B}\) that \(S\) generates. The collection \(\mathcal{B}\) consists of all finite intersections of elements of \(\mathcal{S}\). If we intersect elements belonging to the same one of the sets \(S_{\beta}\), we do not get anything new, because

\[ \pi_{\beta}^{-1}\left(U_{\beta}\right) \cap \pi_{\beta}^{-1}\left(V_{\beta}\right)=\pi_{\beta}^{-1}\left(U_{\beta} \cap V_{\beta}\right) ; \]

the intersection of two elements of \(S_{\beta}\), or of finitely many such elements, is again an element of \(S_{\beta}\). We get something new only when we intersect elements from different sets \(S_{\beta}\). The typical element of the basis \(\mathcal{B}\) can thus be described as follows: Let \(\beta_{1}\), \(\ldots, \beta_{n}\) be a finite set of distinct indices from the index set \(J\), and let \(U_{\beta_{i}}\) be an open set in \(X_{\beta_{i}}\) for \(i=1, \ldots, n\). Then

\[ B=\pi_{\beta_{1}}^{-1}\left(U_{\beta_{1}}\right) \cap \pi_{\beta_{2}}^{-1}\left(U_{\beta_{2}}\right) \cap \cdots \cap \pi_{\beta_{n}}^{-1}\left(U_{\beta_{n}}\right) \]

is the typical element of \(\mathcal{B}\).

Now a point \(\mathbf{x}=\left(x_{\alpha}\right)\) is in \(B\) if and only if its \(\beta_{1}\) th coordinate is in \(U_{\beta_{1}}\), its \(\beta_{2}\) th coordinate is in \(U_{\beta_{2}}\), and so on. There is no restriction whatever on the \(\alpha\) th coordinate of \(\mathbf{x}\) if \(\alpha\) is not one of the indices \(\beta_{1}, \ldots, \beta_{n}\). As a result, we can write \(B\) as the product

\[ B=\prod_{\alpha \in J} U_{\alpha} \]

where \(U_{\alpha}\) denotes the entire space \(X_{\alpha}\) if \(\alpha \neq \beta_{1}, \ldots, \beta_{n}\).

Theorem 19.1.1

For finite products \(\prod_{\alpha=1}^{n} X_{\alpha}\) the box topology and product topology are precisely the same.

The box topology is in general finer than the product topology.

Whenever we consider the product \(\prod X_{\alpha}\), we shall assume it is given the product topology unless we specifically state otherwise.

Theorem 19.2

Suppose the topology on each space \(X_{\alpha}\) is given by a basis \(\mathcal{B}_{\alpha}\). The collection of all sets of the form

\[ \prod_{\alpha \in J} B_{\alpha} \]

where \(B_{\alpha} \in \mathcal{B}_{\alpha}\) for each \(\alpha\), will serve as a basis for the box topology on \(\prod_{\alpha \in J} X_{\alpha}\).

The collection of all sets of the same form, where \(B_{\alpha} \in \mathcal{B}_{\alpha}\) for finitely many indices \(\alpha\) and \(B_{\alpha}=X_{\alpha}\) for all the remaining indices, will serve as a basis for the product topology \(\prod_{\alpha \in J} X_{\alpha}\).

Theorem 19.3

Let \(A_{\alpha}\) be a subspace of \(X_{\alpha}\), for each \(\alpha \in J\). Then \(\prod A_{\alpha}\) is a subspace of \(\prod X_{\alpha}\) if both products are given the box topology, or if both products are given the product topology.

Theorem 19.4

If each space \(X_{\alpha}\) is Hausdorff space, then \(\prod X_{\alpha}\) is a Hausdorff space in both the box and product topologies.

Theorem 19.5

Let \(\left\{X_{\alpha}\right\}\) be an indexed family of spaces; let \(A_{\alpha} \subset X_{\alpha}\) for each \(\alpha\). If \(\prod X_{\alpha}\) is given either the product or the box topology, then

\[ \prod \bar{A}_{\alpha}=\overline{\prod A_{\alpha}} . \]

Proof

Let \(\mathbf{x}=\left(x_{\alpha}\right)\) be a point of \(\prod \bar{A}_{\alpha}\); we show that \(\mathbf{x} \in \overline{\prod A_{\alpha}}\). Let \(U=\prod U_{\alpha}\) be a basis element for either the box or product topology that contains \(\mathbf{x}\). Since \(x_{\alpha} \in \bar{A}_{\alpha}\), we can choose a point \(y_{\alpha} \in U_{\alpha} \cap A_{\alpha}\) for each \(\alpha\). Then \(\mathbf{y}=\left(y_{\alpha}\right)\) belongs to both \(U\) and \(\prod A_{\alpha}\). Since \(U\) is arbitrary, it follows that \(\mathbf{x}\) belongs to the closure of \(\prod A_{\alpha}\).

Conversely, suppose \(\mathbf{x}=\left(x_{\alpha}\right)\) lies in the closure of \(\prod A_{\alpha}\), in either topology. We show that for any given index \(\beta\), we have \(x_{\beta} \in \bar{A}_{\beta}\). Let \(V_{\beta}\) be an arbitrary open set of \(X_{\beta}\) containing \(x_{\beta}\). Since \(\pi_{\beta}^{-1}\left(V_{\beta}\right)\) is open in \(\prod X_{\alpha}\) in either topology, it contains a point \(\mathbf{y}=\left(y_{\alpha}\right)\) of \(\prod A_{\alpha}\). Then \(y_{\beta}\) belongs to \(V_{\beta} \cap A_{\beta}\). It follows that \(x_{\beta} \in \bar{A}_{\beta}\).

Theorem 19.6

Let \(f: A \rightarrow \prod_{\alpha \in J} X_{\alpha}\) be given by the equation

\[ f(a)=\left(f_{\alpha}(a)\right)_{\alpha \in J} \]

where \(f_{\alpha}: A \rightarrow X_{\alpha}\) for each \(\alpha\). Let \(\prod X_{\alpha}\) have the product topology. Then the function \(f\) is continuous if and only if each function \(f_{\alpha}\) is continuous.

Proof

Let \(\pi_{\beta}\) be the projection of the product onto its \(\beta\) th factor. The function \(\pi_{\beta}\) is continuous, for if \(U_{\beta}\) is open in \(X_{\beta}\), the set \(\pi_{\beta}^{-1}\left(U_{\beta}\right)\) is a subbasis element for the product topology on \(X_{\alpha}\). Now suppose that \(f: A \rightarrow \prod X_{\alpha}\) is continuous. The function \(f_{\beta}\) equals the composite \(\pi_{\beta} \circ f\); being the composite of two continuous functions, it is continuous.

Conversely, suppose that each coordinate function \(f_{\alpha}\) is continuous. To prove that \(f\) is continuous, it suffices to prove that the inverse image under \(f\) of each subbasis element is open in \(A\); we remarked on this fact when we defined continuous functions. A typical subbasis element for the product topology on \(\prod X_{\alpha}\) is a set of the form \(\pi_{\beta}^{-1}\left(U_{\beta}\right)\), where \(\beta\) is some index and \(U_{\beta}\) is open in \(X_{\beta}\). Now

\[ f^{-1}\left(\pi_{\beta}^{-1}\left(U_{\beta}\right)\right)=f_{\beta}^{-1}\left(U_{\beta}\right) \]

because \(f_{\beta}=\pi_{\beta} \circ f\). Since \(f_{\beta}\) is continuous, this set is open in \(A\), as desired.

Example

Theorem 19.6 fails when applied to box topology.

Consider \(\mathbb{R}^{\omega}\), the countably infinite product of \(\mathbb{R}\) with itself. Recall that

\[ \mathbb{R}^{\omega}=\prod_{n \in \mathbb{Z}_{+}} X_{n} \]

where \(X_{n}=\mathbb{R}\) for each \(n\). Let us define a function \(f: \mathbb{R} \rightarrow \mathbb{R}^{\omega}\) by the equation

\[ f(t)=(t, t, t, \ldots) \]

the \(n\)th coordinate function of \(f\) is the function \(f_{n}(t)=t\). Each of the coordinate functions \(f_{n}: \mathbb{R} \rightarrow \mathbb{R}\) is continuous; therefore, the function \(f\) is continuous if \(\mathbb{R}^{\omega}\) is given the product topology. But \(f\) is not continuous if \(\mathbb{R}^{\omega}\) is given the box topology. Consider, for example, the basis element

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

for the box topology. We assert that \(f^{-1}(B)\) is not open in \(\mathbb{R}\). If \(f^{-1}(B)\) were open in \(\mathbb{R}\), it would contain some interval \((-\delta, \delta)\) about the point 0 . This would mean that \(f((-\delta, \delta)) \subset B\), so that, applying \(\pi_{n}\) to both sides of the inclusion,

\[ f_{n}((-\delta, \delta))=(-\delta, \delta) \subset(-1 / n, 1 / n) \]

for all \(n\), a contradiction.