KindiakMath

Multivariable Topology

Let (K_1, \mathcal T_1), (K_2, \mathcal T_2) be topological spaces. What would be a suitable topology on K_1 \times K_2? For convenience, we will denote x_1 \times x_2 \equiv (x_1,x_2) \in K_1 \times K_2.

Theorem 1. Let \mathcal B_i be a basis for \mathcal T_i. Then

\mathcal B_1 \times \mathcal B_2 := \{B_1 \times B_2 : B_i \in \mathcal B_i\} \subseteq \mathcal P(K_1 \times K_2)

forms a topological basis that generates a topology \mathcal T on K_1 \times K_2. Furthermore, \mathcal T_1 \times \mathcal T_2 \subseteq \mathcal T.

Proof. To check that \mathcal B_1 \times \mathcal B_2 forms a topological basis, we need to verify that it satisfies the two properties of a topological basis. Fix x_1 \times x_2 \in K_1 \times K_2. For each i, there exists B_i \in \mathcal B_i such that x_i \in B_i. Therefore, x_1 \times x_2 \in B_1 \times B_2, as required for our first property.

For the second property, fix x_1 \times x_2 \in K_1 \times K_2. Suppose there exist B_1 \times B_2, B_1' \times B_2' \in \mathcal B_1 \times \mathcal B_2 that contain x_1 \times x_2. For each i, there exists B_i'' \in \mathcal B_i such that x_i \in B_i'' \subseteq B_i \cap B_i'. Therefore, B_1'' \times B_2'' \in \mathcal B_1 \times \mathcal B_2 is the desired set such that

x_1 \times x_2 \in B_1'' \times B_2'' \subseteq (B_1 \times B_2) \cap (B_1' \times B_2').

Finally, for the subset relation, fix V_1 \times V_2 \in \mathcal T_1 \times \mathcal T_2. Then for any x_i \in V_i, there exists B_i \in \mathcal B_i such that

x_1 \times x_2 \in B_1 \times B_2 \subseteq V_1 \times V_2,

so that V_1 \times V_2 \in \mathcal T, since B_1 \times B_2 \in \mathcal B_1 \times \mathcal B_2.

In fact, we can generalise this result to any index set I. We leave its proof as an exercise.

Definition 1. We define the product of the sets K_\alpha, where \alpha \in I, by

\displaystyle \prod_{\alpha \in I} K_\alpha := \left\{\mathbf x \in \mathcal F\left(I, \bigcup_{\alpha \in I} K_\alpha\right) : \mathbf x(\alpha) \in K_\alpha \right\}

We denote \mathbf x_\alpha = \mathbf x(\alpha) for brevity. When I = \{1,\dots, n\}, we recover the usual finite cartesian product by the identification

\mathbf x = (\mathbf x_1,\dots, \mathbf x_n) \equiv \mathbf x_1 \times \cdots \times \mathbf x_n.

In the event K_\alpha = K for all \alpha, we denote K^I \equiv \prod_{\alpha \in I} K_\alpha. In particular, we denote K^\omega \equiv K^{\mathbb N}.

Lemma 1. For any \alpha \in I, let (K_\alpha, \mathcal T_\alpha) be a topological space and \mathcal B_{\alpha} a basis for \mathcal T_{\alpha}. Then

\displaystyle \prod_{\alpha \in I} \mathcal B_\alpha := \left\{ \prod_{\alpha \in I} B_\alpha : B_\alpha \in \mathcal B_\alpha \right\} \subseteq \mathcal P \left( \prod_{\alpha \in I} K_\alpha \right)

forms a topological basis that generates a topology \mathcal T on \prod_{\alpha \in I} K_\alpha, called the box topology. Then for any V_\alpha \subseteq K_\alpha,

\displaystyle \prod_{\alpha \in I} V_\alpha \in \mathcal T \quad \iff \quad \prod_{\alpha \in I} K_\alpha \in \prod_{\alpha \in I} {\mathcal T}_\alpha.

Example 1. Let \mathbb R be equipped with the standard topology (i.e. generated by open intervals). Then \mathbb R^I is a topological space when equipped with the box topology.

Now, for topological spaces (K, \mathcal S), (L, \mathcal T), recall that for any function (or map) f : K \to L, f is continuous if and only if f^{-1}(V) \in \mathcal S for any V \in \mathcal T.

Lemma 2. For each \alpha \in I, the projection map \pi_\alpha : \prod_{\alpha \in I} K_\alpha \to K_\alpha is continuous. Here, we equipped \prod_{\alpha \in I} K_\alpha with the box topology \mathcal T.

Proof. Fix \alpha \in I and V_\alpha \in \mathcal T_\alpha. Then \pi_\alpha^{-1}(V_\alpha) = \prod_{\beta \in I} U_\beta \in \mathcal T, where

U_\beta := \begin{cases} K_\beta, & \alpha \neq \beta,\\ V_\beta, & \alpha = \beta. \end{cases}

Lemma 3. Let (K, \mathcal S) be a topological space. For any map f : K \to \prod_{\alpha \in I} K_\alpha and any \alpha \in I, define the coordinate map f_\alpha := \pi_\alpha \circ f : K \to K_\alpha.

  • If f is continuous, then f_\alpha is continuous for each \alpha \in I.
  • Furthermore, if I = \{1,\dots, n\} and f_i is continuous for each i \in I, then f is continuous.

Proof. The first point is obvious since composing continuous functions yields yet another continuous function. For the second point, it suffices to prove the case n =2, then extend the result by induction. Furthermore, it suffices to check the continuity property for basis elements. Fix any basis element B_1 \times B_2 in K_1 \times K_2. Then

\begin{aligned} f^{-1}(B_1 \times B_2) &= f^{-1}(\pi_1^{-1}(B_1) \cap \pi_2^{-1}(B_2)) \\ &= f^{-1}(\pi_1^{-1}(B_1)) \cap f^{-1}(\pi_2^{-1}(B_2)) \\ &= (\pi_1 \circ f)^{-1}(B_1) \cap (\pi_2 \circ f)^{-1}(B_2) \\ &= f_1^{-1}(B_1) \cap f_2^{-1}(B_2). \end{aligned}

Conversely, coordinate maps f_\alpha induce a map f defined by (f( \cdot ))(\alpha) := f_\alpha( \cdot ). Remarkably, if I is infinite, we don’t immediately get continuity. This is where counterexamples play a crucial role in adding nuance in our topological adventure.

Example 2. The map f : \mathbb R \to \mathbb R^\omega defined by the continuous coordinate maps f_n = \mathrm{id}_{\mathbb R} : \mathbb R \to \mathbb R is not continuous.

Proof. For each n,

\begin{aligned} (-1/n, 1/n) &= f_n^{-1}((-1/n, 1/n))\end{aligned}

is open in \mathbb R. Therefore, \prod_{n \in \mathbb N} (-1/n, 1/n) is open in \mathbb R^\omega when equipped with the box topology. However,

\displaystyle f^{-1} \left( \prod_{n \in \mathbb N} (-1/n, 1/n) \right) = \bigcap_{n \in \mathbb N} (-1/n, 1/n) = \{0\}

does not contain any interval (-\delta, \delta) and thus is not open. Thus, f is not continuous. The key idea here is that only finite intersections of open sets are open, and infinite intersections may not guarantee that effect.

This raises a question: if the box topology causes f to become not continuous, what would be the topology that allows f to be continuous? Given the continuous maps f_\alpha : K \to K_\alpha, we can define the map f : K \to \prod_{\alpha \in I} K_\alpha via (f(\cdot))(\alpha) := f_\alpha(\cdot).

Now firstly, suppose \mathcal T is a topology such that all \pi_\alpha : \prod_{\alpha \in I} K_\alpha \to K are continuous. Then for any V_\alpha \in \mathcal T_\alpha,

\displaystyle \prod_{\beta \in I} U_\beta = \pi_{\alpha}^{-1}(V_\alpha) \in \mathcal T,

as per the notation in Lemma 2. The converse turns out to be true as well.

Lemma 4. Define the product topology \mathcal T on \prod_{\alpha \in I} K_\alpha as the coarsest topology such that each \pi_\alpha : \prod_{\alpha \in I} K_\alpha \to K_\alpha is continuous. Then \mathcal T is generated by the basis

\displaystyle \mathcal U := \left\{ \prod_{\alpha \in I} U_\alpha : U_\alpha \in \mathcal T_\alpha, U_\alpha \neq K_\alpha\ \text{for finitely many}\ \alpha\right\}.

Furthermore, \mathcal T is contained in the box topology.

Proof. We leave it as an exercise to verify that \mathcal U is indeed a topological basis for \prod_{\alpha \in I} K_\alpha, and generates a topology \mathcal T_{\mathcal U} on \prod_{\alpha \in I} K_\alpha. We have proven that \mathcal T \subseteq \mathcal T_{\mathcal U}.

For the direction (\supseteq), fix the open set U := \prod_{\alpha \in I} U_\alpha and suppose there exist indices i = 1,\dots,n such that the open set U_{\alpha_i} \neq K_{\alpha_i}. Then

\displaystyle U = \prod_{\alpha \in I} U_\alpha = \bigcap_{i=1}^n \pi_{\alpha_i}^{-1}(U_{\alpha_i}).

Since each \pi_{\alpha} : \prod_{\alpha} K_\alpha \to K_\alpha is continuous, we have U \in \mathcal T. Therefore, \mathcal U \subseteq \mathcal T so that \mathcal T_{\mathcal U} \subseteq \mathcal T.

Theorem 2. If \prod_{\alpha \in I} K_\alpha is equipped with the product topology, then f is continuous if and only if each f_\alpha is continuous.

Proof. We have proven (\Rightarrow) in Lemma 3. For the direction (\Leftarrow), suppose each f_\alpha = \pi_\alpha \circ f is continuous. Fix any basis element V \in \mathcal T. Then there exist finitely many indices \alpha_1,\dots,\alpha_n such that the open sets U_{\alpha_i} \neq K_{\alpha_i}, and

\displaystyle V = \bigcap_{i=1}^n \pi_{\alpha_i}^{-1}(U_{\alpha_i}).

Therefore,

\displaystyle \begin{aligned} f^{-1}(V) &= f^{-1} \left( \bigcap_{i=1}^n \pi_{\alpha_i}^{-1}(U_{\alpha_i}) \right) \\ &= \bigcap_{i=1}^n f^{-1}(\pi_{\alpha_i}^{-1}(U_{\alpha_i})) \\ &= \bigcap_{i=1}^n (\pi_{\alpha_i} \circ f)^{-1}(U_{\alpha_i}) \\ &= \bigcap_{i=1}^n f_{\alpha_i}^{-1}( U_{\alpha_i} ) \in \mathcal S.\end{aligned}

It turns out that we don’t see the apparent differences because the product topology and the box topology coincide when the index set is finite, say I = \{1, \dots, n\}. All the results we derived remain consistent.

—Joel Kindiak, 27 Mar 25, 2025H

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