Box topology definition

Question

Munkres defines the box topology as shown below. I am trying to understand how come an element of basis as defined is even subset of $\prod_{\alpha \in J}X_{\alpha}$. If we get an element of the basis, then it is $\prod_{\alpha \in J} U_{\alpha}$, where $U_{\alpha}$ is open in $X_{\alpha}$. $\prod_{\alpha \in J} U_{\alpha} = \{ f : J \rightarrow \bigcup_{\alpha \in J} U_{\alpha} : f(\alpha) \in U_{\alpha} \}$, which isn't subset of $\prod_{\alpha \in J} X_{\alpha}$ as $\bigcup_{\alpha \in J}U_{\alpha}$ isn't equal to $\bigcup_{\alpha \in J}X_{\alpha}$ in general right ? So, how come it is a subset?

Answer 1

In general, if $A_\alpha\subseteq B_\alpha$, then $\bigcup A_\alpha\subseteq\bigcup B_\alpha$. This is an easy exercise, which I am leaving to you.

And if $f\colon X\to Y$, and $Y\subseteq W$, then $f$ is also a function from $X$ into $W$. Because functions in this context are sets of ordered pairs, not triplets of domain, codomain and "rule". And even if they are, we can replace $\subseteq$ by some relevant notion to match this definition.

Therefore every function in $\prod U_\alpha$ is in particular a function in $\prod X_\alpha$. So it is indeed a subset of the product.

Answer 2

Suppose that $f\in\prod_{\alpha\in J}U_\alpha$; then $f(\alpha)\in U_\alpha$ for each $\alpha\in J$. And for each $\alpha\in J$ we have $U_\alpha\subseteq X_\alpha$, so $f(\alpha)\in X_\alpha$ for each $\alpha\in J$, which by definition means that $f\in\prod_{\alpha\in J}X_\alpha$.