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 
 A: 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.
A: It is important to note that the box topology and the product topology are equivalent on any product of finitely many topological spaces.  Therefore, to make this fail, we need to look at infinite products for counterexamples.  Let's investigate precisely where problems arise as we pass from finite to infinite products.
First, let's think about what the preimage of a subset $\displaystyle \prod_{i \in I} T_i \subset \prod_{i \in I} X_i$ looks like.  If $f_j^{-1}(T_j) = S_j$ for some $j$, notice that $\displaystyle f^{-1} \left( \prod_{i \in I} T_i \right)$ cannot contain points outside of $S_j$.  As this will be true for all $j \in I$, we'll have $\displaystyle f^{-1} \left( \prod_{i \in I} T_i \right) = \bigcap_{i \in I} S_i$.  
Next, recall that the box topology has a basis consisting of sets of the form $\displaystyle \prod_{i \in I} U_i$, where each $U_i$ is open in $X_i$.  Using this fact, one can show $f$ is continuous $\iff$ the preimage  $\displaystyle \bigcap_{i \in I} f_i^{-1}(U_i)$ of each basis element is open in $A$.  Since finite intersections of open sets are open, $f$ is continuous if $I$ is finite.  On the other hand, infinite intersections of open sets are not necessarily open, so $f$ is not necessarily continuous if $I$ is infinite. For something concrete, this example takes advantage of $\displaystyle \bigcap_{n \in \mathbb{N}} (-1/n, \ 1/n) = \{0\}$ not being open in $\mathbb{R}$ under the usual topology.
However, if we endowed the product space instead with the product topology, then notice only finitely many of the preimages $f_i^{-1}(U_i)$ would be proper subsets of $A$, even if $I$ is infinite.  So $\displaystyle \bigcap_{i \in I} f_i^{-1}(U_i)$ would be an intersection of finitely many open sets and thus would be open in $A$.  This is to say: $f$ would be continuous.
A: Let $X,Z$ be topological spaces and $Y$ a set. 
Suppose that you have functions $f:X \to Y$ and $g:Y \to Z$. What is the best topology to put on $Y$ to make both maps continuous? The theorem loosely says that the product topology is the 'right' one to do this (when you replace $g$ by the appropriate family of functions). Observe (using the definition of continuity) that if $Y$ has a very fine topology is 'hard' to make $f$ continuous but easy to make $g$ continuous. 
Now onto your question. The box topology is finer than the product topology, the above paragraph suggests that the extra open sets in the box topology make it possible to have $f$ discontinuous and each $f_\alpha$ continuous. 
I'll leave the rest for you to work out, but I suggest letting $f_\alpha = \pi_\alpha$ where $\pi_\alpha$ is the projection map (which is continuous with respect to the product and therefore box topology).
A: Let $J=\mathbb N$ and, for each $n\in\mathbb N,$ let $X_n=\{0,1\}$ with the discrete topology. Let $A=\prod_{n\in\mathbb N}X_n$ with the Tychonoff product topology, and let $B=\prod_{n\in\mathbb N}X_n$ with the box topology. I think you will agree that the spaces $A$ and $B$ are not homeomorphic; $A$ is compact (homeomorphic to the Cantor set) and $B$ is discrete. The identity function $f:A\to B,\ f(x)=x$ is an open bijection, it is not continuous, although the projection functions $f_n:A\to X_n$ are continuous.
