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?

  • $\begingroup$ What is your definition of the product topology? You really can't say that the projection is continuous if you don't know that the set in question is open. $\endgroup$ – rnrstopstraffic May 22 '16 at 4:39

$p_i^{-1}(A_i)$ 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.


In general, if $f\colon M\to N$ is a function, and $S\subset N$, then $f^{-1}(S)$ is the set of all $x\in M$ such that $f(x)\in S$. By definition, we say that $f$ is continuous if $f^{-1}(U)$ is an open set in $M$ whenever $U$ is an open set in $N$.

What is $p_i^{-1}(A_i)$? It is the set of all points $x=(x_1,\dots,x_n)\in M_1\times\dots\times M_n$ such that $p_i(x)\in A_i$. Now $p_i(x)$ is just $x_i$. So the elements $x=(x_1,\dots,x_n)\in p_i^{-1}(A_i)$ are just those elements where $x_i\in A_i$ - the other components are allowed to take any value.

This means that $$ p_i^{-1}(A_i)=M_1\times \dots M_{i-1}\times A_i\times M_{i+1}\times \dots\times M_n $$

Taking the intersection of all these sets gives us $A_1\times\dots A_n$.

Since $A_i$ is open for each $i$, and since $p_i$ is continuous, we know that $p_i^{-1}(A_i)$ is open for each $i$, and so therefore $A_1\times\dots A_n$ may be written as the intersection of fintely many open sets, so it is open.

  • $\begingroup$ I don't think the sentence "It is the set of all points $x=(x_1,\dots,x_n)\in A_1\times\dots\times A_n$ such that $p_i(x)\in A_i$" is correct, it should be "the set of all points $x=(x_1,\dots,x_n)\in M_1\times\dots\times M_n$ such that $p_i(x)\in A_i$" $\endgroup$ – user140776 May 21 '16 at 22:48

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