One of the implications stated in Theorem 19.6 holds for the box topology. Which one?
Solution
1. In the box topology, each projection \(\pi_\alpha\) is continuous, so \(\boxed{f\text{ continuous}\Rightarrow\text{every }f_\alpha\text{ continuous}}\).
For \(U\) open in \(X_\alpha\), \(\pi_\alpha^{-1}(U)=U\times\prod_{\beta\ne\alpha}X_\beta\) is a product of open sets, hence box-open. A composite of continuous maps is continuous.
2. The map \(f\colon\mathbb R\to\mathbb R^\omega\), \(f(t)=(t,t,t,\ldots)\), has continuous coordinates but is not continuous for the box topology.
Each \(f_n\) is the identity of \(\mathbb R\). The set \(B=\prod_{n\ge1}(-\frac1n,\frac1n)\) is box-open, and \[ f^{-1}(B)=\bigcap_{n\ge1}\left(-\tfrac1n,\tfrac1n\right)=\{0\}, \] which is not open in \(\mathbb R\).
3. Q.E.D.
Step 1 proves the implication that holds, and step 2 shows that the other implication fails.