2.4 Product Topology · Topological Spaces and Continuous Functions

[2.4.1]Definition(Product Topology)#

Suppose $X$ and $Y$ are sets endowed with certain topologies $\Tau_X, \Tau_Y$ . Define $\mathcal{B}$ to be the collection of sets $U \times V$ , where $U$ and $V$ are open subsets of $X$ and $Y$ (with respect to $\Tau_X$ and $\Tau_Y$ , respectively). Then $\mathcal{B}$ generates the product topology on $X \times Y$ .

Proof.

We show $\mathcal{B}$ is, indeed, a basis. Fix $(x, y) \in (X, Y)$ . Since topologies cover their respective set, there exist open subsets $U, V$ of $X, Y$ —thus $(x, y) \in U \times V$ , meaning $\mathcal{B}$ covers $X \times Y$ . Now let $U_1, U_2 \subseteq X$ and $V_1, V_2 \subseteq Y$ be open. Then

(U_1 \times V_1) \cap (U_2 \times V_2) = \underbrace{(U_1 \cap U_2)}_{\in \Tau_X} \times \underbrace{(V_1 \times V_2)}_{\in \Tau_Y}.

Since the intersection of two open sets is open, we have that the intersection of two basis elements is, indeed, another basis element.

[2.4.2]Theorem(Product of Bases Generates Product Topology)#

Suppose $\mathcal{B}_X$ and $\mathcal{B}_X$ are bases for topologies on sets $X$ and $Y$ . Then the collection of sets

\mathcal{B} = \{B_X \times B_Y: B_X \in \mathcal{B}_X, B_y \in \mathcal{B}_Y\}

forms a basis for the product topology of $X \times Y$ .

Proof.

First, we show $\mathcal{B}$ covers $X \times Y$ . Fix $(x, y) \in X \times Y$ . Then note $\mathcal{B}_X$ and $\mathcal{B}_Y$ cover $X$ and $Y$ , so there exists $B_X \in \mathcal{B}_X$ and $B_Y \in \mathcal{B}_Y$ such that $(x, y) \in B_X \times B_Y$ —thus $\mathcal{B}$ covers $X \times Y$ .

Now let $(x, y) \in U \times V \subseteq X \times Y$ , where $U$ and $V$ are open subsets of $X$ and $Y$ , respectively. Of course, $U \times V$ is open with respect to the product topology, so we simply have to locally refine it for the specific point $(x, y)$ . Note that there is some $B_X \in \mathcal{B}_X$ (resp. $B_Y \in \mathcal{B}_Y$ ) such that $x \in B_X \subseteq U$ (resp. $y \in B_Y \subseteq V$ ), and so $(x, y) \in (B_X \times B_Y) \subseteq (U \times V)$ . Thus elements of an open subset of $X \times Y$ can be locally refined by $\mathcal{B}$ . Thus it is a basis, completing the proof.

[2.4.3]Remark(Basis for

\bR^2

)#

Recall the standard topology on $\bR$ is the order topology. Then, in endowing $\bR \times \bR = \bR^2$ with the product topology, we seek a basis for $\bR^2$ . By definition, the basis is simply the product of all open subsets of $\bR$ . But note, with our previous theorem, we may in fact reduce the basis to simply all sets of form $(a, b) \times (c, d)$ , noting that the set of all open intervals of $\bR$ forms a basis for the standard topology on $\bR$ .

[2.4.4]Definition(Projection Maps)#

Suppose $X$ and $Y$ are sets. Define the following projection maps

\pi_X: X \times Y \to X \quad \pi_X(x, y) = x,

\pi_Y: X \times Y \to Y \quad \pi_Y(x, y) = y.

[2.4.5]Theorem(Subbasis for Product Topology)#

The collection

\mathcal{S} = \{\pi_X^{-1}(U): U \text{ open in } X\} \cup \{\pi_Y^{-1}(V): V \text{ open in } Y\}

forms a subbasis for the product topology on $X \times Y$ .

Proof.

Note that $\pi_X^{-1}(U) = U \times X$ and $\pi_Y^{-1}(V) = V \times Y$ . For any two open subsets $U \subseteq X$ and $V \subseteq Y$ , we have that

\pi_X^{-1}(U) \cap \pi_Y^{-1}(V) = (U \times Y) \cap (V \times X) = (U \times V) \cap (X \times Y) = U \times V.

Thus every element of the standard basis for the product topology on $X \times Y$ is generated by the subbasis. Moreover, since $U, V$ are arbitrary, these are the only types of elements produced by the finite intersection of subbasis elements. Thus the intersection of all subbasis elements produces the standard basis for the product topology.

2.4Product Topology