The Cartesian product X₁ × X₂ × · · · × X_n of sets X₁, X₂, . . . , X_n is defined to be the set of all ordered n-tuples (x₁, x₂, . . . , x_n), where x_i ∈ X_i for i =1, 2, . . . , n.

The sets R² and R³ are the Cartesian products R × R and R × R × R respectively.

Definition: Let X₁, X₂, . . . , X_n be topological spaces. A subset U of the Cartesian product X₁ × X₂ × · · · × X_n is said to be open (with respect to the product topology) if, given any point p of U, there exist open sets V_i in X_i

for i = 1, 2, . . . , n such that {p} ⊂ V₁ × V₂ × · · · × V_n ⊂ U.

Lemma 1.1 Let X₁, X₂, . . . , X_n be topological spaces. Then the collection of open sets in X₁ × X₂ × · · · × X_n is a topology on X₁ × X₂ × · · · × X_n.

Proof Let X = X₁ ×X₂ ×· · ·×X_n. The definition of open sets ensures that the empty set and the whole set X are open in X. We must prove that any union or finite intersection of open sets in X is an open set.

Let E be a union of a collection of open sets in X and let p be a point ofE. Then p ∈ D for some open set D in the collection. It follows from this that there exist open sets V_i in X_i for i = 1, 2, . . . , n such that

                               {p} ⊂ V₁ × V₂ × · · · × V_n ⊂ D ⊂ E.

Thus E is open in X.

be a point of U. Then there exist open sets V_ki in X_i for k = 1, 2, . . . , m and i = 1, 2, . . . , n such that {p} ⊂ V_k1 ×V_k2 × · · · ×V_kn ⊂ U_k for k = 1, 2, . . . , m.

Let V_i = V_1i ∩ V_2i ∩ · · · ∩ V_mi for i = 1, 2, . . . , n. Then

 {p} ⊂ V₁ × V₂ × · · · × V_n ⊂ V_k1 × V_k2 × · · · × V_kn ⊂ U_k

for k = 1, 2, . . . , m, and hence {p} ⊂ V₁ × V₂ × · · · × V_n ⊂ U. It follows that U is open in X, as required.

Let X = X1 ×X2 ×· · ·×Xn, where X1, X2, . . . , Xn are topological spaces

and X is given the product topology, and for each i, let p_i: X → Xi denote the projection function which sends (x₁, x₂, . . . , x_n) ∈ X to x_i. It can be shown that a function f:Z → X mapping a topological space Z into X is continuous if and only if p_i ◦ f:Z → X_i is continuous for i = 1, 2, . . . , n.

One can also prove that usual topology on Rⁿ determined by the Euclidean distance function coincides with the product topology on Rⁿ obtained on regarding Rⁿ as the Cartesian product R x R x…..x R of n copies