James Munkres Topology: Theorem 19.6

Theorem 19.6 Let \(f: A \rightarrow \prod_{\alpha \in J} X_{\alpha}\) be given by the equation

\[
f(a) = (f_{\alpha}(a))_{\alpha \in J},
\]

where \(f_{\alpha}: A \rightarrow X_{\alpha}\) for each \(\alpha\). Let \(\prod X_{\alpha}\) have the product topology. Then the function \(f\) is continuous if and only if each function \(f_{\alpha}\) is continuous.

Comment: This is an extension of Theorem 18.4, where only two component spaces are involved.

Proof: a) First, we prove the projection map is continuous, which is defined on the Cartesian space constructed from a \(J\)-tuple of component spaces .

For all \(\beta \in J\), let \(\pi_{\beta}: \prod X_{\alpha} \rightarrow X_{\beta}\) be the projection map. For arbitrary open set \(V_{\beta}\) in \(X_{\beta}\), its pre-image under \(\pi_{\beta}\) is \(\pi_{\beta}^{-1}(V_{\beta})\), which is a subbasis element for the product topology on \(\prod X_{\alpha}\). Therefore, \(\pi_{\beta}^{-1}(V_{\beta})\) is open and the projection map \(\pi_{\beta}\) is continuous.

Next, we notice that for all \(\alpha \in J\), the coordinate function \(f_{\alpha}: A \rightarrow X_{\alpha}\) is a composition of the two continuous functions \(f\) and \(\pi_{\alpha}\), i.e. \(f_{\alpha} = \pi_{\alpha} \circ f\). Then according to Theorem 18.2 (c), \(f_{\alpha}\) is continuous.

Remark: Because the box topology is finer than the product topology, the projection map is also continuous when the box topology is adopted for \(\prod X_{\alpha}\). Therefore, this part of the theorem is true for both product topology and box topology.

b) Analysis: To prove the continuity of a function, showing that the pre-image of any subbasis element in the range space is open in the domain space is more efficient than using basis element or raw open set in the range space. In addition, the subbasis element for the product topology on \(\prod X_{\alpha}\) has the form \(\pi_{\beta}^{-1}(U_{\beta})\) with \(U_{\beta}\) being a single coordinate component and open in \(X_{\beta}\). This is the clue of the proof.

For all \(\beta \in J\) and arbitrary open set \(U_{\beta}\) in \(X_{\beta}\), we have \(f_{\beta}^{-1}(U_{\beta}) = f^{-1} \circ \pi_{\beta}^{-1}(U_{\beta})\). Because \(f_{\beta}\) is continuous and \(U_{\beta}\) is open, \(f_{\beta}^{-1}(U_{\beta})\) is open. In addition, \(\pi_{\beta}^{-1}(U_{\beta})\) is an arbitrary subbasis element for \(\prod X_{\alpha}\) with the product topology, whose pre-image under \(f\) is just the open set \(f_{\beta}^{-1}(U_{\beta})\), therefore \(f\) is continuous.

Remark: Part b) of this theorem really depends on the adopted topology for \(\prod X_{\alpha}\), which can be understood as below.

At first, we will show that for all \(\vect{U} = \prod U_{\alpha}\) being a subset of \(\prod X_{\alpha}\), \(f^{-1}(\vect{U}) = \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha})\).

For all \(x \in f^{-1}(\vect{U})\), because \(f(x) \in \vect{U}\), then for all \(\alpha \in J\), \(f_{\alpha}(x) \in U_{\alpha}\), hence \(x \in \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha})\) and \(f^{-1}(\vect{U}) \subset \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha})\).

On the other hand, for all \(x \in \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha})\), we have for all \(\alpha \in J\), \(f_{\alpha}(x) \in U_{\alpha}\). Therefore, \(f(x) \in \vect{U}\) and \(x \in f^{-1}(\vect{U})\). Hence \(\bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha}) \subset f^{-1}(\vect{U})\).

Next, if we assign the product topology to \(\prod X_{\alpha}\), for any \(\vect{U} = \prod U_{\alpha}\) with \(U_{\alpha}\) open in \(X_{\alpha}\) and only a finite number of them not equal to \(X_{\alpha}\), it is a basis element of the product topology. Let the set of all indices with which \(U_{\alpha} \neq X_{\alpha}\) be \(\{\alpha_1, \cdots, \alpha_n\}\) and also notice that when \(U_{\alpha} = X_{\alpha}\), \(f_{\alpha}^{-1}(U_{\alpha}) = A\), we have

\[
f^{-1}(\vect{U}) = \bigcap_{\alpha \in J} f_{\alpha}^{-1}(U_{\alpha}) = \bigcap_{i=1}^n f_{\alpha_i}^{-1}(U_{\alpha_i}),
\tag{*}
\label{eq:intersection}
\]

where those \(f_{\alpha}^{-1}(U_{\alpha})\) with \(\alpha \notin \{\alpha_1, \cdots, \alpha_n\}\) do not contribute to the intersection. This indicates that \(f^{-1}(\vect{U})\) is a finite intersection of open sets which is still open. Hence \(f\) is continuous.

However, if the box topology is adopted for \(\prod X_{\alpha}\), qualitatively speaking, because the topology for the range space becomes finer, according to our previous post, it makes a function to be continuous more difficult. Specifically in this theorem, \(f^{-1}(\vect{U})\) in \eqref{eq:intersection} can be an intersection of infinite number of open sets \(U_{\alpha}\) not equal to \(X_{\alpha}\). Thus \(f^{-1}(\vect{U})\) may not be open anymore.

After understanding this point, it is not difficult to construct a counter example for part b) as below.

Let \(f: \mathbb{R} \rightarrow \mathbb{R}^{\omega}\) be defined as \(f(t) = (t, t, \cdots)\). Select a basis element \(\vect{U}\) in \(\mathbb{R}^{\omega}\) such that the intersection of all its coordinate components is not open. For example, \(\vect{U} = \prod_{n=1}^{\infty} (-\frac{1}{n}, \frac{1}{n})\), which is a neighborhood of \(f(0) = (0, 0, \cdots)\).

For any basis element \((a, b)\) in \(\mathbb{R}\) containing \(0\), with \(a < 0\) and \(b > 0\), by letting \(\delta = \min\{-a, b\}\), we have \((-\delta, \delta) \subset (a, b)\) and \(0 \in (-\delta, \delta)\). The image of \((-\delta, \delta)\) under \(f\) is \(\prod_{n=1}^{\infty} (-\delta, \delta)\). Then there exist an \(n_0 \in \mathbb{Z}_+\) such that \((-\delta, \delta)\) is not contained in \((-\frac{1}{n_0}, \frac{1}{n_0})\). Therefore, \(\pi_{n_0}(f((-\delta, \delta)))\) is not contained in \(\pi_{n_0}(\vect{U})\) and \(\pi_{n_0}(f((a, b)))\) is not contained in \(\pi_{n_0}(\vect{U})\). Hence the image of \((a, b)\) under \(f\) is not contained in \(\vect{U}\). This contradicts Theorem 18.1 (4) and \(f\) is not continuous.