[Munkres Topology] 19. The Product Topology

 This article is one of the posts in the Textbook Commentary Project.

We now return, for the remainder of the chapter, to the consideration of various methods for imposing topologies on sets.

Definition. Let $J$ be an index set. (Not need to be $\mathbb{Z}_+$) 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 $$(x_\alpha)_{\alpha \in J},$$ which is as close as we can come to a 'tuple notation' for an arbitrary index set $J$. We denote the set of all $J$-tuple of elements of $X$ by $X^J$.

Definition. Let $\{A_\alpha\}_{\alpha \in J}$ 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 $(x_\alpha)_{\alpha\in J}$ 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 $(x_\alpha)$, if the index set if understood.

If all the sets $A_\alpha$ are equal to one set $X$, then the cartesian product $\prod_{\alpha \in J} A_\alpha$ is just the set $X^J$ of all $J$-tuple of elements of $X$. We sometimes use 'tuple notation' for the elements of $X^J$, and sometimes we use functional notation, depending on which is more convenient.

Definition. Let $\{X_\alpha \}_{\alpha \in J}$ 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.

This collection satisfies the first condition for basis because $\prod X_a$ is itself a basis element; and it satisfies the second condition because the intersection of any two basis elements is another basis element: $$(\prod_{\alpha \in J} U_\alpha)\cap (\prod_{\alpha\in J} V_\alpha)=\prod_{\alpha\in J} (U_\alpha \cap V_\alpha).$$

Now we generalize the subbasis formulation of the definition. 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, $$\pi_\beta((x_\alpha)_{\alpha\in J})=x_\beta;$$ it is called the projection mapping associated with the index $\beta$.

Definition. Let $\mathcal{S}_\beta$ denote the collection $$\mathcal{S}_\beta = \{\pi_\beta^{-1}(U_\beta)|U_\beta \mbox{ open in } X_\beta\},$$ and let $\mathcal{S}$ denote the union of these collections, $$\mathcal{S}=\bigcup_{\beta\in J} \mathcal{S}_\beta.$$ The topology generated by the subbasis $\mathcal{S}$ is called the product topology. In this topology $\prod_{\alpha \in J} X_\alpha$ is called a product space.

To compare these topologies, we consider the basis $\mathcal{B}$ that $\mathcal{S}$ 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 $\mathcal{S}_\beta$, we do not get anything new, because $$\pi_\beta^{-1}(U_\beta)\cap \pi_\beta^{-1}(V_\beta)=\pi_\beta^{-1}(U_\beta \cap V_\beta);$$ the intersection of two elements of $\mathcal{S}_\beta$, or of finitely many such elements, is again and element of $\mathcal{S}_\beta$. We get something new only when we intersect elements from different sets $\mathcal{S}_\beta$. The typical element of the basis $\mathcal{B}$ can thus be described as follows: Let $\beta_1,\cdots,\beta_n$ 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 $i=1,\cdots,n$. Then $$B=\pi_{\beta_1}^{-1}(U_{\beta_1})\cap \pi_{\beta_2}^{-1}(U_{\beta_2})\cap \cdots\cap \pi_{\beta_n}^{-1}(U_{\beta_n})$$ is the typical element of $\mathcal{B}$.

Now a point $\mathbf{x}=(x_a)$ is in $B$ if and only if its $\beta_1$the 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 $\beta_1,\cdots,\beta_n$. As a result, we write $B$ as the product $$B=\prod_{\alpha\in J} U_\alpha,$$ where $U_\alpha$ denotes the entire space $X_\alpha$ is $\alpha\ne \beta_1,\cdots,\beta_n$.

All this is summarized in the following theorem:

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 $\alpha$.

Two things are immediately clear. First, for finite products $\prod^n_{\alpha =1}X_\alpha$ the two topologies are precisely the same. Second, 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 (Basis for the box topology). 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 (Subspace). 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 it both products are given the product topology.

Theorem 19.4  (Hausdorff). 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 (Closure). Let $\{X_\alpha\}$ be an indexed family of spaces; let $A_\alpha \subset X_\alpha$ for each $\alpha$. If $\prod X_\alpha$ is given either the product of the box topology, then $$\prod \bar{A}_\alpha = \bar{\prod A_\alpha}.$$

Proof. 

[$\subset$] Let $\mathbf{x}=(x_\alpha)$ be a point of $\prod \bar{A}_\alpha$; we show that $\mathbf{x}\in \bar{\prod A_\alpha}$. 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 $\mathbf{y}=(y_\alpha)$ belong to both $U$ and $\prod A_\alpha$. Since $U$ is arbitrary, if follows that $\mathbf{x}$ belongs to the closure of $\prod A_\alpha$. 

[$\supset$] Conversely, suppose $\mathbf{x}=(x_\alpha)$ 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 $\pi_\beta^{-1}(V_\beta)$ is open in $\prod X_\alpha$ in either topology, it contains a point $\mathbf{y}=(y_\alpha)$ 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 (Continuity on Product Topology). Let $f:A\rightarrow \prod_{\alpha \in J}X_\alpha$ be given by the equation $$f(a)=(f_\alpha (a))_{\alpha \in J},$$ 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. 

[$\Rightarrow$] 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 $\pi_\beta^{-1} (U_\beta)$ 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.

[$\Leftarrow$] Conversely, suppose that each coordinate function $f_\alpha$ is continuous. To prove that $f$ is continuous, it suffices to porve 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 $\pi_\beta^{-1}(U_\beta)$, where $\beta$ is some index and $U_\beta$ is open in $X_\beta$. Now $$f^{-1}(\pi_\beta^{-1}(U_\beta))=f_\beta^{-1}(U_\beta),$$ because $f_\beta=\pi_\beta\circ f$. Since $f_\beta$ is continuous, this set is open in $A$, as desired.

This theorem fail if we use the box topology.

Example 2. 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 $$f(t)=(t,t,t,\cdots);$$ 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 $$B=(-1,1)\times (-\frac{1}{2},\frac{1}{2})\times (-\frac{1}{3},\frac{1}{3})\times \cdots$$ 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.