Question regarding box topology

In James.E.Munkres there is a theorem that says

Let $f:A \rightarrow \sqcap_{\alpha\in J} X_{\alpha}$ be given by the equation $f (a)=(f_{\alpha}(a))_{\alpha\in J}$. Where $f_{\alpha}:A \rightarrow X_{\alpha}$. For each $\alpha$. Let $\sqcap _{\alpha\in J} X_{\alpha}$ have the product topology. Then f is continuous iff each function $f_{\alpha}$ is continuous. I can understand the proof given for product topology. What I can't understand why the theorem is not valid if we apply box topology. I mean I can see the example given. But don't how to prove it is not true for box topology.

Any help would be appriciated. Thanks

Answer

For a counterexample, consider $X = \mathbb{R}$ (with the usual topology), $f_n(x) = nx$, and $f\colon X \to \Pi_{n\in \mathbb{N}}\; X$ defined by $f(x) = (f_n(x))_n$. Then pick $U = \Pi_{n\in\mathbb{N}}(-1, 1)$, which is open in the box topology. We have $f^{-1}(U) = \{0\}$, which is not open in X. Hence $f$ is not continuous. See also box topology (failure continuity).

The problem is that the intersection of an arbitrary family of open sets is not open in general.