# product and box topology

Let $f\:A\rightarrow\Pi _{\alpha\in J} X_\alpha$ 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 cant figure out why it is not true for the box.

-

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.
What is $f$ here? (I know it's been 4.5 years) but I'm not sure, surely $f_\beta^{-1}(U_\alpha)_{\alpha\in I}$ is $(b_\alpha)_{\alpha\in I}$ where $b_\alpha=X_\alpha$ (for $\alpha\ne\beta$) and $b_\beta=U_\beta$ –  Alec Teal Feb 27 at 19:56
@AlecTeal: I don't understand the second part of your question (is $I$ meant to be $J$?), but the answer to the first part (what $f$ is) is that it's defined by the $f_\alpha$ all being the identity map, so $f$ maps $x\in\mathbb R$ to the sequence $x,x,x,\ldots$. –  joriki Apr 28 at 10:11
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.