Product topology with infinitely many factors Given a collection of topological spaces $\{X_a\}_{a∈A}$, the product topology Y is given by $Y := \prod_{a\in A}X_a$.
For every $a\in A$ we have the non empty open subset $U_a \subset X_a$.
Show that:
When there are infinitely many indices $a\in A$ such that $U_a \neq X_a$, then $\prod_{a \in A}U_a \subset Y$ is not open.
Q: Does anyone know how to solve this? I've been trying to see what happens when you take the infinite product but I can't figure it out!
Thanks in advance!
 A: I assume this is what you want to prove:

Let $\{X_a\}_{a\in A}$ be a family of topological spaces and $U_a\subset X_a$ proper open subspaces. Prove that $\prod_{a\in A} U_a$ is not open in $\prod_{a\in A} X_a$ with the usual product topology on $\prod_{a\in A} X_a$.

So let $U_a\subset X_a$ be proper (i.e. $U_a\neq X_a$) open subsets and let
$$U=\prod_{a\in A} U_a$$
Assume that $U$ is open, i.e. there is an indexing set $B$ such that
$$U=\bigcup_{b\in B} V_b$$
for a collection $\{V_b\}_{b\in B}$ of base open subsets of $\prod X_a$. So what is a base open subset in the product topology? By the definition for each $b\in B$ we have
$$V_b=\prod_{c\in A} V_{b, c}$$
for some open subsets $V_{b, c}\subseteq X_c$ such that $V_{b, c}=X_c$ for almost all $c\in A$. Now pick a pair of indexes $b'\in B, c'\in A$ such that $V_{b', c'}=X_{c'}$ and take the projection
$$\pi_{c'}:\prod X_i\to X_{c'}$$
Note that $\pi_{c'}(U)=U_{c'}\neq X_{c'}$ by definition of $U$.
On the other hand $\pi_{c'}(V_{b'})=V_{b', c'}=X_{c'}$ (by the choice of $(b', c')$) and since we've assumed that $U=\bigcup V_{b}$ then $\pi_{c'}(U)=X_{c'}$ because image of union is union of images. Contradiction. $\Box$
