Basis generated by subbasis of the product topology

Question

Page 115, Topology, Munkres.

The statement is:

For the product topology, I understand that $U_{ \alpha}$ equals $X_{\alpha}$ for finitely many values of $\alpha$, but I don't understand how the following is true:

$\pi_{i_1}^{-1}(U_{i_1})\cap \cdots \cap \pi_{i_n}^{-1}(U_{i_n}) = \prod_{i\in I}U_i,$

where $U_i = \begin{cases} U_{i_k} & \text{if $i=i_k$ for some $k=1,\ldots,n$},\\ X_i & \text{otherwise}. \end{cases}$

I can only make sense of it for only one term, for example,

$\pi_{i_1}^{-1}(U_{i_1}) = X_i \times ... U_{i_1} \times ...$

is indeed $\prod_{i\in I}U_i$

But when I generalize to finite intersections I don't get how these intersections still give the same relation $\prod_{i\in I}U_i$ , if I understand this part then the conclusion that the box topology is finer than the product topology becomes clear to me, any help would be much appreciated.

Answer 1

Consider the same for $2$:

$\pi_{i_1}^{-1}(U_{i_1}) = X_i \times ... U_{i_1} \times ...$ and $\pi_{i_2}^{-1}(U_{i_2}) = X_i \times ... U_{i_2} \times ...$

Then, (assuming $i_1 \neq i_2$, as this case is trivial) \begin{align} \pi_{i_1}^{-1}(U_{i_1})\cap \pi_{i_2}^{-1}(U_{i_2}) &= (X_i \times ... U_{i_1} \times ...) \cap (X_i \times ... U_{i_2} \times ...) \\ &= (X_i\cap X_i) \times ... (U_{i_1}\cap X_{i_1}) \times ... (X_{i_2}\cap U_{i_2}) \times ... \\ &= X_i \times ... U_{i_1} \times ... U_{i_2} \times ... \\ &= \prod_{i\in I}U_i \\ \end{align}

Answer 2

Maybe this can help.

If $A$ denotes the index-set then it is convenient to think of $\prod_{\alpha\in A}X_{\alpha}$ as the set of functions $f:A\to\bigcup_{\alpha\in A}X_{\alpha}$ that have the special property that $f\left(\alpha\right)\in X_{\alpha}$ for every $\alpha\in A$.

So formally denoted: $$\prod_{\alpha\in A}X_{\alpha}=\left\{ f\in\left(\bigcup_{\alpha\in A}X_{\alpha}\right)^{A}\mid\forall\alpha\in A\;f\left(\alpha\right)\in X_{\alpha}\right\} $$

Then for a fixed $\beta\in A$ the projection $\pi_{\beta}:\prod_{\alpha\in A}X_{\alpha}\to X_{\beta}$ is prescribed by $f\mapsto f\left(\beta\right)$

Consequently the preimage of $U_{\beta}\subseteq X_{\beta}$ under $\pi_{\beta}$ is: $$\left\{ f\in\prod_{\alpha\in A}X_{\alpha}\mid f\left(\beta\right)\in U_{\beta}\right\} =\prod_{\alpha\in A}U_{\alpha}$$ where $U_{\alpha}=X_{\alpha}$ if $\alpha\neq\beta$.

So for fixed and distinct $\beta_{1},\dots,\beta_{n}\in A$ we find: $$\bigcap_{i=1}^{n}\pi_{\beta_{i}}^{-1}\left(U_{\beta_{1}}\right)=\left\{ f\in\prod_{\alpha\in A}X_{\alpha}\mid\forall i\in\left\{ 1,\dots,n\right\} \left[f\left(\beta_{i}\right)\in U_{\beta_{i}}\right]\right\} =\prod_{\alpha\in A}U_{\alpha}$$ where $U_{\alpha}=X_{\alpha}$ if $\alpha\notin\left\{ \beta_{1},\dots,\beta_{n}\right\} $