Open Subset of Infinite Product of Spaces Let $U$ be any basic open subset of $\prod_{a \in A} X_a$ where $X_a$ are arbitrary topological spaces, and $A$ is infinite. Is it true that there always exist a $b \in A$ s.t. $\pi_b(U) = X_b$, Why?
 A: Yes, and in fact, given an open set $U$ of the product $X$, there are only finitely many $a\in A$ with $\pi_a(U)\neq X_a$.
How you show this depends on the definition of the product topology. For example, one may define the product topology to have as a basis the sets of the form $\prod_{a\in A} U_a$ where $U_a$ is open in $X_a$ for all $a$ and $U_a\neq X_a$ for only finitely many $a\in A$. In this case the answer to your question is clearly “yes”. 
However, Wikipedia defines the product topology differently, so one needs to check that the definitions are equivalent. Wikipedia defines the product topology to be the coarsest topology for which the projections are continuous. Clearly the topology above has this property, so we need only show that every topology with the projections being continuous contains the topology above. If $\tau$ is a topology on $X$ with that property, then let $U=\prod_{a\in A} U_a$ be a basis element of the topology above (so $U_a\neq X_a$ for only finitely many $a\in A$). Suppose that $a_1,\dots,a_n$ are the indices for which $U_{a_i}\neq X_{a_i}$. Then $U=\bigcap_{i=1}^n \pi_{a_i}^{-1} (U_{a_i})$ is open in $\tau$, as desired.
