Product Topology

The product topology is the natural topology on a Cartesian product of topological spaces. It is defined as the coarsest topology making all projection maps continuous, and it plays a central role throughout topology and analysis.

Finite Products

Definition3.1Product Topology (Finite)

Let $(X_1, \tau_1), \ldots, (X_n, \tau_n)$ be topological spaces. The product topology on $X_1 \times \cdots \times X_n$ is the topology generated by the basis: $\mathcal{B} = \{U_1 \times \cdots \times U_n : U_i \in \tau_i \text{ for each } i\}.$

A set $U_1 \times \cdots \times U_n$ where each $U_i$ is open in $X_i$ is called a basic open set (or open box).

Proof

We verify the basis conditions. (B1): $X_1 \times \cdots \times X_n$ is itself a basic open set. (B2): If $x \in (U_1 \times \cdots \times U_n) \cap (V_1 \times \cdots \times V_n)$ , then $x \in (U_1 \cap V_1) \times \cdots \times (U_n \cap V_n)$ , which is a basic open set contained in the intersection.

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ExampleThe Euclidean Plane as a Product

The product topology on $\mathbb{R} \times \mathbb{R}$ with the standard topology on each factor coincides with the standard (Euclidean) topology on $\mathbb{R}^2$ . The basic open sets $U_1 \times U_2$ (products of open intervals) are open rectangles, which generate the same topology as open disks. This extends to $\mathbb{R}^n = \mathbb{R} \times \cdots \times \mathbb{R}$ .

Arbitrary Products

Definition3.2Product Topology (Arbitrary)

Let $\{(X_\alpha, \tau_\alpha)\}_{\alpha \in A}$ be a family of topological spaces indexed by an arbitrary set $A$ . The product space is: $\prod_{\alpha \in A} X_\alpha = \left\{f: A \to \bigcup_\alpha X_\alpha : f(\alpha) \in X_\alpha \text{ for all } \alpha\right\}.$

For each $\alpha \in A$ , the projection map $\pi_\alpha: \prod_\beta X_\beta \to X_\alpha$ is defined by $\pi_\alpha(f) = f(\alpha)$ .

The product topology on $\prod_\alpha X_\alpha$ is the coarsest topology making all projections $\pi_\alpha$ continuous. It has subbasis: $\mathcal{S} = \{\pi_\alpha^{-1}(U_\alpha) : \alpha \in A, \; U_\alpha \in \tau_\alpha\}.$

The corresponding basis consists of all finite intersections of subbasis elements: $\mathcal{B} = \left\{\bigcap_{i=1}^n \pi_{\alpha_i}^{-1}(U_{\alpha_i}) : n \in \mathbb{N}, \; \alpha_i \in A, \; U_{\alpha_i} \in \tau_{\alpha_i}\right\}.$

A basic open set in the product topology specifies constraints on only finitely many coordinates. Explicitly, a basic open set is: $\prod_{\alpha \in A} V_\alpha, \quad \text{where } V_\alpha = U_\alpha \text{ for } \alpha \in F \text{ (finite), and } V_\alpha = X_\alpha \text{ otherwise.}$

RemarkWhy Not the Box Topology?

The product topology is deliberately not the box topology (which allows all coordinates to be constrained simultaneously). The product topology is chosen because:

It makes projections continuous.
It satisfies the universal property for products in $\mathbf{Top}$ .
Tychonoff's theorem holds for the product topology but fails for the box topology.
It coincides with the box topology for finite products.

Properties of Product Spaces

Theorem3.1Continuity into Products

A map $f: Z \to \prod_\alpha X_\alpha$ is continuous if and only if $\pi_\alpha \circ f: Z \to X_\alpha$ is continuous for every $\alpha \in A$ .

Proof

( $\Rightarrow$ ): Each $\pi_\alpha$ is continuous by definition of the product topology, so $\pi_\alpha \circ f$ is a composition of continuous maps.

( $\Leftarrow$ ): It suffices to check that preimages of subbasis elements are open. We have $f^{-1}(\pi_\alpha^{-1}(U_\alpha)) = (\pi_\alpha \circ f)^{-1}(U_\alpha)$ , which is open since $\pi_\alpha \circ f$ is continuous.

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ExampleDiagonal Map

The diagonal map $\Delta: X \to X \times X$ defined by $\Delta(x) = (x, x)$ is continuous, since $\pi_1 \circ \Delta = \operatorname{id}_X$ and $\pi_2 \circ \Delta = \operatorname{id}_X$ are both continuous.

More generally, if $f: Z \to X$ and $g: Z \to Y$ are continuous, then $(f, g): Z \to X \times Y$ defined by $z \mapsto (f(z), g(z))$ is continuous.

Subspaces and Products

Theorem3.2Product of Subspaces

If $A_\alpha \subseteq X_\alpha$ for each $\alpha$ , then the product topology on $\prod_\alpha A_\alpha$ (where each $A_\alpha$ has the subspace topology from $X_\alpha$ ) coincides with the subspace topology on $\prod_\alpha A_\alpha$ as a subset of $\prod_\alpha X_\alpha$ .

ExampleThe Torus as a Product

The torus $T^2 = S^1 \times S^1$ is naturally a product of two circles. Its product topology coincides with its subspace topology as a subset of $\mathbb{R}^2 \times \mathbb{R}^2 \cong \mathbb{R}^4$ . The torus is compact (product of compact spaces), connected, and Hausdorff.