[Munkres Topology] 15. The Product Topology on $X\times Y$

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

If $X$ and $Y$ are topological spaces, there is a standard way of defining a topology on the cartesian product $X\times Y$. We consider this topology now and study some of its properties.

Definition. Let $X$ and $Y$ be topological spaces. The product topology on $X\times Y$ is the topology having as basis the collection $\mathcal{B}$ of all sets of the form $U\times V$, where $U$ is an open subset of $X$ and $V$ is an open subset of $Y$.

Let us check that $\mathcal{B}$ is a basis. The first condition is trivial, since $X\times Y$ itself a basis element. The second condition is almost as easy, since the intersection of any two basis elements $U_1 \times V_1$ and $U_2 \times V_2$ is another basis element. For $$(U_1 \times V_1) \cap (U_2 \times V_2)=(U_1 \cap U_2) \times (V_1 \cap V_2),$$ and the latter set is a basis element because $U_1 \cap U_2$ and $V_1 \cap V_2$ are open in $X$ and $Y$, respectively. See Figure 15.1.

Note that the collection $\mathcal{B}$ is not a topology on $X\times Y$. The union of the two rectangles pictured in Figure 15.1, for instance, is not a product of two sets, so it cannot belong to $\mathcal{B}$; however, it is open in $X\times Y$.

Each time we introduce a new concept, we shall try to relate it to the concept that have been previously introduced. In the present case, we ask: What can one say if the topologies on $X$ and $Y$ are given by basis? The answer is as follows:

Theorem 15.1. If  $\mathcal{B}$ is a basis for the topology of $X$ and $\mathcal{C}$ is a basis for the topology of $Y$, then the collection $$\mathcal{D}=\{ B \times C | B \in \mathcal{B} \mbox{ and } C \in \mathcal{C} \}$$ is a basis for the topology of $X\times Y$.

Proof. We apply Lemma 13.2. Given an open set $W$ of $X\times Y$ and a point $x\times y$ of $W$, by definition of the product topology there is a basis element $U\times V$ such that $x \times y \in U \times V \subset W$. Because $\mathcal{B}$ and $\mathcal{C}$ are bases for $X$ and $Y$, respectively, we can choos an element $B$ on $\mathcal{B}$ such that $x\in B \subset U$, and an element $C$ on $\mathcal{C}$ such that $y\in C \subset V$. Then $x\times y \in B\times C \subset W$. Thus the collection $\mathcal{D}$ meets the criterion of Lemma 13.2, so $\mathcal{D}$ is a basis for $X\times Y$.

It is sometimes useful to express the product topology in terms of a subbasis. To do this, we first define certain functions called projections.

Definition. Let $\pi_1: X\times Y \rightarrow X$ be defined by the equation $$\pi_1 (x,y) = x;$$ let $\pi_2: X\times Y \rightarrow Y$ be defined by the equation $$\pi_2(x,y)=y.$$ The maps $\pi_1$ and $\pi_2$ are called the projections of $X\times Y$ onto its first and second factors, respectively.

We use the word 'onto' because $\pi_1$ and $\pi_2$ are surjective (unless one of the spaces $X$ or $Y$ happens to be empty).

If $U$ is an open subset of $X$, then the set $\pi_1^{-1}(U)$ is precisely the set $U\times Y$, which is open in $X\times Y$. Similarly, if $V$ is open in $Y$, then $$\pi_2^{-1}(V)=X\times V,$$ which is also open in $X\times Y$. The intersection of these two sets is the set $U\times V$, as indicated in Figure 15.2. This fact leads to the following theorem:

Theorem 15.2. The collection $$S=\{ \pi_1^{-1}(U) | U \mbox{ open in } X\} \cup \{ \pi_2^{-1}(V)| V \mbox{ open in } Y\}$$ is a subbasis for the product topology on $X \times Y$.

Proof. Let $\mathcal{T}$ denote the product topology on $X\times Y$; let $\mathcal{T}'$ be the topology generated by $\mathcal{S}$. Because every element of $\mathcal{S}$ belongs to $\mathcal{T}$, so do arbitrary unions of finite intersections of elements of $\mathcal{S}$. Thus $\mathcal{T}'\subset \mathcal{T}$. On the other hand, every basis element $U\times V$ for the topology $\mathcal{T}$ is a finite intersection of elements of $\mathcal{S}$, since $$U\times V=\pi_1^{-1} (U) \cap \pi_2^{-1} (V).$$ Therefore, $U\times V$ belongs to $\mathcal{T}'$, so that $\mathcal{T}' \subset \mathcal{T}'$ as well.