Source: A First Course in Topology by Robert Conover
3.3 Definition
Let {$(X_\lambda, T_\lambda$):$\lambda \in \Lambda$} be a nonempty collection of nonempty topological spaces. The Product topology on $X_\lambda$ is the topology generated by the the subbasis {$\pi_\lambda^{-1}$($U_\lambda)$:$U_\lambda \in T_\lambda$}
I would like to define a basic open set on the infinite product topology, but I can’t use the following Lemma cause I haven’t proved it .
I look through the site some were a bit to complex. Others I could not use cause I would have to prove 3.8 which came after this theorem.
I need it to prove a theorem.Most of the theorem is done.
The theorem is:
Let {$X_\lambda:$ $\lambda\in\Lambda$ }be a nonempty collection of nonempty sets with |$\Lambda$|$\geq$$\aleph_0$. If $U_\lambda$ is open in $X_\lambda$ then $\Pi U_i $ is open in the product topology on $\Pi X_i$ iff $U_\lambda=X_\lambda$ for all but finitely many
Here is the Lemma
Let {$X_\lambda:$ $\lambda\in\Lambda$ }be a nonempty collection of nonempty sets with |$\Lambda$|$\geq$$\aleph_0$. Then for $\lambda_1,…,\lambda_n\in\Lambda$ If $U_\lambda\subset X_\lambda$ then $\bigcap \pi^{-1}_\lambda$($U_\lambda)$= $\pi_\lambda$ $A_\lambda$= $U_\lambda if \lambda=\lambda_n$ otherwise $X_\lambda $if $\lambda$ $\ne \lambda_n$
I have definitions of sub-basis and projection maps and the definition of infinite product space
I was thinking to use the theorem that if $B_\lambda$ $\in B$ is a basis. for $T_\lambda$ for each $\lambda\in\Lambda$ then the collection{$\pi_{\lambda}^{-1}$($B_\lambda)$:$B_\lambda$ $\in B$ } is a subbasis for the product topology