If $\{X_i\}_i\in I$ are topological spaces, $I$ is infinite, $U_i \subset X_i$ are proper open subsets, then $\prod U_i$ is not open in $\prod X_i$
This is a question from my H.W. assignment. I’ve been staring at this question for 3-4 hours but I couldn’t even find a way to initiate the question. I would appreciate any help or hint to solve the question.
Thanks.
Let $T_i$ be the topology on $X_i.$ The product topology $T$ on $X=\prod_{i\in I}X_i$ is defined as the weakest topology such that each projection $p_j:X\to X_j$ is continuous. So when $c_j\in T_j$ then $p_j^{-1}c_j=\prod_{i\in I}A_i\in T$, where $A_j=c_j,$ and $A_i=X_i$ for each $i\ne j.$
So we have $S=\bigcup_{j\in I}\{p_j^{-1}c_j:c_j\in T_j\}\subseteq T.$
Now $S$ is a sub-base for a topology $T'$ on $X,$ and $$T'\subseteq T$$ because $S\subseteq T.$
Now each $p_j$ is continuous with respect to the topology $T'$ so by the $\subseteq$-minimality ("weakest") definition of $T,$ we have $$T\subseteq T'.$$ Therefore $T'=T.$
Consider any $U\in T$. Since $S$ is a sub-base for $T$, we have $U\supseteq\bigcap_{i\in E}\;p_i^{-1}c_i$ for some finite $E\subset I$ with each $c_i\in T_i.$ So $U\supseteq\prod_{i\in I}A_i$ where $A_i=c_i$ if $i\in E,$ and $A_i=X_i$ if $i\not\in E.$ Now $E$ is finite so if $I$ is infinite then $\{i\in I: A_i=X_i\}\ne\emptyset.$