Suppose that we have $\{X_{\alpha}\}_{\alpha \in J}$, an indexed family of topological spaces. Let $X := \prod_{\alpha \in J}X_{\alpha}$. When we have a map $f_{\alpha} : A \rightarrow X_{\alpha}$ with a topological space $A$. Define $f : A \rightarrow \prod_{\alpha \in J}X_{\alpha}$ by $a \mapsto (f_{\alpha}(a))_{\alpha \in J}$.
Property. We know that $f$ is continuous if and only if each $f_{\alpha}$ is continuous, once we are given product topology on $X$. We further know that this property does not generally hold with box topology on $X$.
After reviewing the definition, it seems to me that this property seems rather trivial (or natural) since product topology only collects finite intersections of inverse images under projections as open sets.
Can someone give more examples outside this property that explains why we prefer product topology? A good list of evident examples will be much preferred than statement of theorems.
Best Answer
Consider $\mathbb{R}^\omega$, the countably infinite product of $\mathbb{R}$ with itself. That is, $$\mathbb{R}^{\omega} = \prod_{n \in \mathbb{Z}_{+}}X_{n}$$ where $X_{n} = \mathbb{R}$ for each $n$. Let us define a function $f: \mathbb{R} \rightarrow \mathbb{R}^{\omega}$ by the equation $$f(t)=(t,t,t,...)$$ the $n^{\text{th}}$ coordinate function of $f$ is the function $f_{n}(t)=t$. Each of the coordinate functions $f_{n}: \mathbb{R} \rightarrow \mathbb{R}$ is continuous; therefore, the function $f$ is continuous if $\mathbb{R}^\omega$ is given the product topology. But $f$ is not continuous if $\mathbb{R}^\omega$ is given the box topology. Consider, for example, the basis element $$B = (-1,1) \times \left(-\frac{1}{2},\frac{1}{2}\right) \times \left(-\frac{1}{3},\frac{1}{3}\right) \times \cdot\cdot\cdot$$ for the box topology. We assert that $f^{-1}(B)$ is not open in $\mathbb{R}$. If $f^{-1}(B)$ were open in $\mathbb{R}$, it would contain some interval $(-\delta,\delta)$ about the point $0$. This would mean that $f((-\delta,\delta)) \subset B$, so that applying $\pi_{n}$ to both sides of the inclusion, $$f_{n}((-\delta,\delta)) = (-\delta,\delta) \subset \left(-\frac{1}{n},\frac{1}{n}\right)$$ for all $n$, which is a contradiction.
Hopefully this is the sort of explicit example you were looking for that compares the product topology and the box topology.