J. Munkres. Topology (2013). Chapter 2: Topological Spaces and Continuous Functions. 19: The Product Topology

Definition

Let \(\{X_i\}_{i \in I}\) be a family of topological spaces. The box topology on \(\prod_{i \in I} X_i\) is the topology generated by the basis \(\{\prod_{i \in I} U_i \mid \forall i \in I. U_i \text{ is open in } X_i\}\).

Definition

Let \(\{X_i\}_{i \in I}\) be a family of topological spaces. The product topology on \(\prod_{i \in I} X_i\) is the topology generated by the subbasis \(\{\pi_i^{-1}(U) \mid i \in I, U \text{ is open in } X_i\}\). In this topology, \(\prod_{i \in I} X_i\) is called a product space. If we write just \(\prod_{i \in I} X_i\), it is to be understood that this set is given the product topology.

Theorem 19.1

Let \(\{X_i\}_{i \in I}\) be a family of topological spaces. A basis for the product topology on \(\prod_{i \in I} X_i\) is given by the set of all sets of the form \(\prod_{i \in I} U_i\) where each \(U_i\) is open in \(X_i\), and \(U_i = X_i\) for all but finitely many \(i \in I\).

Theorem 19.2

Let \(\{X_i\}_{i \in I}\) be a family of topological spaces. Let \(\mathcal{B}_i\) be a basis for \(X_i\) for all \(i \in I\). Then \(\{\prod_{i \in I} B_i \mid \forall i \in I. B_i \in \mathcal{B}_i\}\) is a basis for the box topology on \(\prod_{i \in I} X_i\). A basis for the product topology is given by the set of all sets of the form \(\prod_{i \in I} B_i\) where \(B_i \in \mathcal{B}_i\) for finitely many \(i\) and \(B_i = X_i\) for all other \(i\).

Theorem 19.3

Let \(A_i\) be a subspace of \(X_i\) for all \(i\). Then the box topology on \(\prod_{i \in I} A_i\) is the same as the subspace topology as a subspace of \(\prod_{i \in I} X_i\) under the box topology. The product topology on \(\prod_{i \in I} A_i\) is the same as the subspace topology as a subspace of \(\prod_{i \in I} X_i\) under the product topology.

Theorem 19.4

If each \(X_i\) is Hausdorff, then \(\prod_{i \in I} X_i\) is Hausdorff under either the box or the product topologies.

Theorem 19.5

Let \(A_i \subseteq X_i\) for each \(i \in I\). Give \(\prod_{i \in I} X_i\) either the box or the product topology. Then

\[ \prod_{i \in I} \overline{A_i} = \overline{\prod_{i \in I} A_i} \enspace . \]

Theorem 19.6

Let \(\{X_i\}_{i \in I}\) be a family of topological spaces and \(A\) a topological space. Let \(f : A \rightarrow \prod_{i \in I} X_i\). Then \(f\) is continuous if and only if, for all \(i \in I\), \(\pi_i \circ f : A \rightarrow X_i\) is continuous.

Example 2

The theorem does not hold in general if we give \(\prod_{i \in I} X_i\) the box topology. Define \(f : \mathbb{R} \rightarrow \mathbb{R}^\omega\) by \(f(t) = (t,t,\ldots)\). Then \(\pi_i \circ f = \mathrm{id}_\mathbb{R} \) is continuous but \(f\) is not continuous if we give \(\mathbb{R}^\omega\) the box topology; the set \(B = (-1,1) \times (-1/2,1/2) \times (-1/3,1/3) \times \cdots\) is open but \(f^{-1}(B) = \{0\}\) is not.

From the Exercises

Proposition

Let \(\{X_i\}_{i \in I}\) be a family of topological spaces. Let \((x_n)\) be a sequence of points in \(\prod_{i \in I} X_i\) and let \(l \in \prod_{i \in I} X_i\). Then \(x_n \rightarrow l\) as \(n \rightarrow \infty\) if and only if \(\forall i \in I. \pi_i(x_n) \rightarrow \pi_i(l)\) as \(n \rightarrow \infty\).

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