I'm trying to understand the proof that:
The cartesian product $A_1\times \cdots\times A_n$ of open subsets
$A_i\subset M_i$ is an open subset of $M=M_1\times\cdots\times M_n$.
It follows like this:
For each $i=1,\cdots,n$ in the projection $p_i:M\to M_i$ is
continuous, then $p_i^{-1}(A_i)$ is open in $M$. So,
$A_1\times\cdots\times A_n = p_1(A_i)^{-1}\cap\cdots\cap
p_n^{-1}(A_n)$ is a finite intersection of opens, so its open.
Why is $A_1\times\cdots\times A_n = p_1^{-1}(A_i)\cap\cdots\cap p_n^{-1}(A_n)$ true?
And what exactly does $p_i^{-1}(A_i)$ mean?
Best Answer
$p_i^{-1}(A_i)$ is the set of all things mapped by the projection onto $A_i$. This is called the pre image of $A_i$ under the projection onto the $i^{th}$ coordinate and it is precisely $M_1\times \dots M_{i-1} \times A_i \times M_{i+1} \times \dots M_n$. So then what is the intersection $\bigcap_{i=1}^n p^{-1}(A_i)$? Well, its the intersection of all of the $M_1\times \dots M_{i-1} \times A_i \times M_{i+1} \times \dots M_n$ which is just $A_1\times \dots \times A_n$.
Additional comment: The product topology is defined to be the weakest topology in which the projection map onto each each coordinate is continuous. And a continuous function is one under which the pre image of an open set is open. Therefore, if the $A_i$ are open then the pre images $p_i^{-1}(A_i)$ are open by the definitions of product topology and continuity. And since every topology has the property that finite intersections of open sets are open, the finite intersection of these pre images is also open.