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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.

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2 Answers 2

up vote 7 down vote accepted

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.

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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.

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I am grateful to both of you. –  smanoos Oct 15 '11 at 7:28
1  
@smanoos: You're welcome! –  joriki Oct 15 '11 at 7:33

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