Let $(X_j)_{j\in J}$ be a family of nonempty toplogical spaces. Set $X=\prod_{j\in J} X_j$. Now let $A_j\subseteq X_j$ be closed.
It is stated that $A_j\times\prod_{i\neq j} X_i\subseteq X$ is closed.
How can we see this? Or in particular how can we deduce that the complement $(A_j\times\prod_{i\neq j} X_i)^c\subseteq X$ is open?
Because taking the complement of an infinite product gets messy, doesnt it?
Is it: $\bigcup_{j\in J} A_j^c\times\prod_{i\neq j} X_i$ Where this would be indeed an open set.
Oh, I think I finally understand why there is most of the time intersection involved when there is set equality to such closed sets in product topology.
Would this be correct?
Are there other things good to know about product sets?
Best Answer
The best thing to keep in mind about the product topology, imo, is the following “universal” property. “The product topology is the coarsest topology such that all the projections are continuous.”
The “coarsest” part is hard to wrap your head around, but it is certainly true that the projection maps are all continuous. So in particular, if $\pi_j$ is the projection to X_j, it should be true that $\pi_j^{-1}(A_j)$ is closed. But this is exactly $$ A_j \times \prod_{i\neq j} X_i, $$ so sets of this form are closed. Similarly, preimages of open sets $U_j\subseteq X_j$ are of the form $$ U_j \times \prod_{i\neq j} X_i $$ and so things of this form are open. These are often called “open (or closed) cylinders.”
The “coarsest” part of the definition means we want the product topology to have as few open (or closed) sets as possible while keeping the $\pi_j$ all continuous. So we need cylinder sets to be open, and any arbitrary union or finite intersection of them should also be open. But this is all. Consider a product of open sets in each component: $$ \prod_i U_i, \quad U_i \subseteq X_i. $$ Any finite intersection or union of cylinder sets will still contain the full $X_i$ in all but finitely many components. So the product above can only be open if $U_i \subsetneq X_i$ for finitely many $i$, and $U_i = X_i$ otherwise.
Addendum: Complements in the product topology can be messy, but in this case are not: $$(A_j \times \prod_{i \neq j} X_i)^C = A_j^C \times \prod_{i \neq j} X_i,$$ which is clearly open since it is $\pi_j^{-1}(A_j)$. "The complement of an open cylinder is a closed cylinder." We can check it manually: both sides are the set of all tuples $(a_i)_i$ such that $a_j \not\in A_j$.