TITLE: Continuity of product of fuctions w.r.t. product and box topology
QUESTION [7 upvotes]: 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 $\Pi  X_\alpha$  have the box topology. Show that the implication; "the function $f$ is continuous if each $f_\alpha$ is continuous" is not true for this topology. 
How do I prove this? Can anyone help?
Obviously this is true for the product topology (Munkres, Thm 19.6), but I can't figure out why it is not true for the box.

REPLY [9 votes]: Take an open set $U=\prod_{\alpha\in J}U_\alpha$, with $U_\alpha$ open in $X_\alpha$. Then $f^{-1}(U)=\bigcap_{\alpha\in J}f_\alpha^{-1}(U_\alpha)$. If you allow infinite products, this yields infinite intersections, and these are not guaranteed to be open.

REPLY [8 votes]: Let $J=\mathbb{N}$, let $X_\alpha=\mathbb{R}$ with the usual topology for all $\alpha\in J$, let $A=\mathbb{R}$ with the usual topology, and let $f_\alpha:A\to X$ be the identity map on $\mathbb{R}$ for all $\alpha\in J$. For all $\alpha\in J$, let $U_\alpha=(-\frac{1}{\alpha},\frac{1}{\alpha})$. Then the set
$$\prod_{\alpha\in J}U_\alpha=(-1,1)\times(-\tfrac{1}{2},\tfrac{1}{2})\times(-\tfrac{1}{3},\tfrac{1}{3})\times\cdots\subset\prod_{\alpha\in J}X_\alpha,$$
which is open in the box topology because each $U_\alpha$ is open in $X_\alpha$, has inverse image under the map $f:A\to \prod_{\alpha\in J}X_\alpha$ equal to $\{0\}$, which is not open in $A$. Therefore $f$ is not continuous, even though each $f_\alpha$ is.