The Product Topology

Definition 1 A topological space is a pair ${(X,\tau)}$ where ${X}$ is a set and ${\tau}$ is a collection of subsets of ${X}$ such that

(i) the subsets ${\emptyset}$ and ${X}$ belongs to ${\tau}$ ,

(ii) any finite intersection of elements of ${\tau}$ also belongs to ${\tau}$ , and

(iii) any union of elements of ${\tau}$ also belongs to ${\tau}$ .

The elements of ${\tau}$ are called open sets and the complements of those elements of ${\tau}$ are called closed sets.

Often, it is difficult to identify the topology on a set ${X}$ , that is, the whole collection ${\tau}$ of subsets of ${X}$ that satisfy the above requirements. Instead, one usually identifies a smaller collections of subsets of ${X}$ , and then constructs a topology from this smaller collection.

Definition 2 Let ${X}$ be a set. A basis for a topology on ${X}$ is a collection ${\mathcal{B}}$ of subsets of ${X}$ such that

(a) For each ${x \in X}$ , there is a basis element ${B}$ such that ${x \in B}$ .

(b) If ${x}$ is in the intersection of two basis elements ${B_1}$ and ${B_2}$ , then there is a third basis element ${B_3}$ containing ${x}$ such that ${B_3 \subset (B_1 \cap B_2)}$ .

A basis ${\mathcal{B}}$ induces a topology ${\tau_\mathcal{B}}$ on ${X}$ in the following way: a subset ${U}$ of ${X}$ is said to be open in ${X}$ (i.e., ${U \in \tau_\mathcal{B}}$ ), if for each ${x \in U}$ , there is a basis element ${B \in \mathcal{B}}$ such that ${x \in B}$ and ${B \subset U}$ . We note that by the above characterization, each basis element ${B \in \mathcal{B}}$ is open in ${X}$ .

Now, we must check that ${\tau_\mathcal{B}}$ is indeed a topology as described in Definition 1.

Theorem 3 The topology ${\tau_\mathcal{B}}$ induced by the basis ${\mathcal{B}}$ on ${X}$ is a topology on ${X}$ .

Proof: We establish properties (i), (ii), and (iii) of Definition 1.

It is clear that (i) is satisfied.

For (ii), suppose that ${U_1}$ and ${U_2}$ are two elements of ${\tau_\mathcal{B}}$ . Let ${x}$ be an arbitrary element of the intersection ${U_1 \cap U_2}$ . Since ${U_i}$ is an element of ${\tau_\mathcal{B}}$ , there is a basis element ${B_i}$ such that ${x \in B_i}$ . By property (b) of the definition of a basis, there is a third basis element ${B_3}$ such that ${B_3 \subset (B_1 \cap B_2) \subset (U_1 \cap U_2)}$ .

For (iii), let ${ \{U_\alpha \}_{\alpha \in I}}$ be a set of elements in ${\tau_\mathcal{B}}$ , and set ${U = \cup_{\alpha \in I}U_\alpha }$ . If ${x \in U}$ , there is an ${\alpha \in I}$ such that ${x \in U_\alpha}$ . Since ${U_\alpha}$ is open, there is a basis element ${B \in \mathcal{B}}$ such that ${x \in B}$ and ${B \subset U_\alpha \subset U}$ . $\Box$

Lemma 4 Let ${X}$ be a set, and let ${\mathcal{B}}$ be a basis that induces a topology ${\tau_\mathcal{B}}$ on ${X}$ . Then ${\tau_\mathcal{B}}$ is equal to the collection of all unions of sets of ${\mathcal{B}}$ .

Proof: For the backward inclusion, let ${\{B_\alpha\}_{\alpha \in I}}$ be a collection of sets in ${\mathcal{B}}$ . We have seen that each ${B_\alpha}$ is a member of ${\tau_\mathcal{B}}$ . Furthermore, ${\tau_{\mathcal{B}}}$ is a topology, so the union of these elements is also in ${\tau_{\mathcal{B}}}$ .

For the forward inclusion, let ${U}$ be an element of the topology ${\tau_{\mathcal{B}}}$ . Since ${\mathcal{B}}$ is a basis, there is an element ${B_x}$ of ${\mathcal{B}}$ such that ${x \in B}$ and ${B_x \subset U}$ . Set ${V = \cup\{B_x: x \in U\}}$ . Evidently, ${V = U}$ . Therefore, ${U}$ is a union of a collection of elements of ${\mathcal{B}}$ . $\Box$

This Lemma showed that, using a basis, it is easy to construct a topology. However, we might also wonder whether we can find the underlying basis for a topology. That is, if we are given a topology, can we obtain a basis for this topology? The following theorem will answer our question.

Theorem 5 Let ${(X,\tau)}$ be a topological space. Suppose that ${\mathcal{B}}$ is a collection of open sets of ${X}$ such that for each open set ${U}$ of ${X}$ and each ${x}$ in ${U}$ , there is an element of ${B}$ of ${\mathcal{B}}$ such that ${x \in B}$ and ${B \subset U}$ . Then ${\mathcal{B}}$ is a basis for the topology of ${X}$ . Moreover, the topology induced by ${\mathcal{B}}$ , say ${\tau_{\mathcal{B}}}$ , coincides with the topology ${\tau}$ of ${X}$ .

Proof: To show that ${\mathcal{B}}$ is a basis, let ${x}$ be in ${X}$ . Since ${X}$ is open, there is, by assumption, an element ${B}$ of ${\mathcal{B}}$ such that ${x \in B}$ and ${B \subset X}$ . Now, let ${B_1}$ and ${B_2}$ be two elements of ${\mathcal{B}}$ . Suppose that ${p \in B_1 \cap B_2}$ . By assumption ${B_1}$ and ${B_2}$ are open, and thus ${B_1 \cap B_2}$ is open. Therefore, by assumption, there is an element ${B_3 \in \mathcal{B}}$ such that ${p \in B_3}$ and ${B_3 \subset B_1 \cap B_2}$ . Thus, ${\mathcal{B}}$ is a basis.

We now show that ${\tau = \tau_{\mathcal{B}}}$ . Let ${U}$ be an element of ${\tau}$ and let ${x}$ be in ${U}$ . By assumption, there is an element ${B \in \mathcal{B}}$ such that ${x \in B}$ and ${B \subset U}$ , that is, ${U}$ belongs to ${\tau_{\mathcal{B}}}$ . Conversely, suppose that ${V}$ is in ${\tau_{\mathcal{B}}}$ . By the above lemma, ${V}$ is a union of elements of ${\mathcal{B}}$ . Since each element of ${\mathcal{B}}$ is open, each element of ${\mathcal{B}}$ is also in ${\tau}$ . Since ${\tau}$ is a topology, the union ${V}$ must also be in ${\tau}$ . $\Box$

We now work to give a basis for a topology on the real line. However, we are first required to give some preliminary definitions.

Let ${{\mathbb R}}$ be the set of real numbers. Let ${a,b}$ be two real numbers such that ${a<b}$ . The open interval from ${a}$ to ${b}$ , denoted ${(a,b)}$ , is the set of real numbers ${(a,b) := \{x: a<x<b\}}$ . In the case where ${a = b}$ , we define the open interval to be the null set ${\emptyset}$ . The standard topology on ${{\mathbb R}}$ is the set of all unions of open intervals of ${{\mathbb R}}$ .

The careful reader will note that we have set up our standard topology so that each “open interval” is indeed open, that is, each open interval is an element of the topology. It is clear that this collection of subsets satisfies the properties to be a topology on ${{\mathbb R}}$ .

Proposition 6 Let ${\mathcal{C}}$ be the set of all open intervals in the real line, i.e., ${\mathcal{C} = \{(a,b): a,b \in {\mathbb R}\}}$ . Then ${\mathcal{C}}$ is a basis for the standard topology ${\mathcal{S}}$ on ${{\mathbb R}}$ .

Proof: Since ${\mathcal{C}}$ clearly satisfies the requirements for a basis, the claim follows directly from Lemma 4. $\Box$

From now on, whenever we consider the real line ${{\mathbb R}}$ , we suppose it has the standard topology. Therefore, we shall consider open intervals in ${{\mathbb R}}$ to be open sets, i.e., elements of the standard topology. There is a natural extension of this topology to the higher-dimensional space ${{\mathbb R}^2}$ , which we go on to define now. We recall that ${{\mathbb R}^2 = {\mathbb R} \times {\mathbb R}}$ , where ${\times}$ denotes the cartesian product. Recall that the cartesian product of two sets ${X}$ and ${Y}$ is the new set ${X \times Y := \{(x,y): x \in X, y \in Y\}}$ .

Lemma 7 Let ${X}$ and ${Y}$ be topological spaces. Suppose that ${U_1}$ and ${U_2}$ are opens subsets of ${X}$ and that ${V_1}$ and ${V_2}$ are open subsets of ${Y}$ . Then

$\displaystyle (U_1 \times V_1) \cap (U_2 \times V_2) = (U_1 \cap U_2) \times (V_1 \cap V_2).$

Proof: Let us first consider the forward inclusion. Let ${r}$ be in ${(U_1 \times V_1) \cap (U_2 \times V_2)}$ . Then, in particular ${r}$ is in ${U_1 \times V_1}$ , so we can write ${r = (x_1, y_1)}$ where ${x_1}$ is in ${U_1}$ and ${y_1}$ is in ${V_1}$ . Also, ${r}$ belongs to ${U_2 \times V_2}$ so we can write ${r = (x_2, y_2)}$ , where ${x_2}$ is in ${U_2}$ and ${y_2}$ is in ${V_2}$ . Therefore, ${(x_1, y_1) = r = (x_2, y_2) = (x,y)}$ , where ${x \in U_1 \cap U_2}$ and ${y \in V_1 \cap V_2}$ . Thus, we see that ${r}$ belongs to ${(U_1 \cap U_2) \times (V_1 \cap V_2)}$ .

Now, let ${r}$ be ${(U_1 \cap U_2) \times (V_1 \cap V_2)}$ . Write ${r = (x,y)}$ , where ${x \in U_1 \cap U_2}$ and ${y \in V_1 \cap V_2}$ . Then in particular, ${(x,y)}$ belongs to ${U_1 \times V_1}$ . On the other hand, ${(x,y)}$ also belongs to ${U_2 \times V_2}$ . Therefore, ${r = (x,y)}$ belongs to ${(U_1 \times V_1) \cap (U_2 \times V_2)}$ . $\Box$

Proposition 8 Let ${X}$ and ${Y}$ be topological spaces. Let ${\mathcal{B}}$ be the collection of all sets of the form ${U \times V}$ , where ${U}$ is an open subset of ${X}$ and ${V}$ is an open subset of ${Y}$ . Then ${\mathcal{B}}$ is a basis for a topology on ${X \times Y}$ .

Proof: Let ${r}$ be an element of ${X \times Y}$ . Since ${X}$ and ${Y}$ are topological spaces, both ${X}$ and ${Y}$ are open sets. Therefore, their product ${X \times Y}$ is an element of the basis, which contains ${r}$ .

Now, suppose that ${r}$ is in the intersection of two basis elements ${B_1}$ and ${B_2}$ . Write ${B_1 = U_1 \times V_1}$ and ${B_2 = U_2 \times V_2}$ where ${U_1}$ and ${U_2}$ are open subsets of ${X}$ and ${V_1}$ and ${V_2}$ are open subsets of ${Y}$ . The previous lemma gives

$\displaystyle B_1 \cap B_2 = (U_1 \times V_1) \cap (U_2 \times V_2) = (U_1 \cap U_2) \times (V_1 \cap V_2).$

Note that, since ${U_1}$ and ${U_2}$ are open, so is ${U_1 \cap U_2}$ . Similarly for ${V_1 \cap V_2}$ . Let ${U = U_1 \cap U_2}$ and ${V = V_1 \cap V_2}$ . Let ${B_3 = U \times V}$ . Then, we see that ${r}$ belongs to ${B_3}$ and ${B_3 \subset B_1 \times B_2}$ . Thus, ${\mathcal{B}}$ meets the requirements for a basis. $\Box$

If ${X}$ and ${Y}$ are 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 ${Y}$ .

Theorem 9 Let ${\mathcal{B}}$ be a basis for the topology of ${X}$ and let ${\mathcal{C}}$ be a basis for the topology of ${Y}$ . Then the collection of sets ${\mathcal{D} = \{B \times C: B \in \mathcal{B}, C \in \mathcal{C}\}}$ is a basis for the product topology of ${X \times Y}$ .

Proof: We apply Theorem 5. Let ${W}$ be an open set of ${X \times Y}$ and let ${w}$ be in ${W}$ . Write ${w = (x,y)}$ . By definition of the product topology, there is a basis element of the product topology ${U \times V}$ such that ${(x,y) \in U \times V \subset W}$ . Since ${\mathcal{B}}$ is a basis for ${X}$ , there is some basis element ${B}$ of ${\mathcal{B}}$ such that ${x \in B \subset U}$ . Since ${\mathcal{C}}$ is a basis for ${Y}$ , there is some basis element ${C}$ of ${\mathcal{C}}$ such that ${y \in C \subset V}$ . Thus, ${(x,y) \in B \times C \subset W}$ . By Theorem 5, ${\mathcal{D}}$ is a basis for the topology of ${X \times Y}$ . $\Box$

We are now in a position to define a natural topology on ${{\mathbb R}^2}$ . Recall that we defined the standard topology on ${{\mathbb R}}$ to be the collection of unions of all open intervals on the real line. The product of this topology with itself is what we call the standard topology on ${{\mathbb R}^2}$ . It has as a basis the collection of all products ${(a,b) \times (c,d)}$ of open intervals in ${{\mathbb R}}$ . Similarly, one can define the standard topology on ${{\mathbb R}^n}$ to be the that topology induced by the collection of all ${n}$ order cartesian products ${(a_1, b_1) \times \cdots \times (a_n, b_n)}$ of open intervals in ${{\mathbb R}}$ . Although this topology is useful, we will see that it is not the only topology on ${{\mathbb R}^n}$ .