A description/proof of that for finite-product topological space, product of neighborhoods is neighborhood
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of product topology.
- The reader knows a definition of neighborhood of point.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-product topological space, the product of any constituent neighborhoods is a neighborhood.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any finite number of topological spaces, \(T_1, T_2, . . ., T_n\), the product, \(T = T_1 \times T_2 \times . . . \times T_n\), any point, \(p = (p^1, p^2, ..., p^n)\in T\), and any neighborhoods, \(\{N_{p^j} \subseteq T_j \vert j \in \{1, 2, ..., n\}\}\), of \(\{p^j \vert j \in \{1, 2, ..., n\}\}\), \(N_p = N_{p^1} \times N_{p^2} \times . . . \times N_{p^n} \subseteq T\) is a neighborhood of \(p\).
2: Proof
Certainly, \(p \in N_p\).
For each \(j\), there is an open neighborhood, \(U_{p^j} \subseteq T_j\), of \(p^j\) such that \(U_{p^j} \subseteq N_{p^j}\), by the definition of neighborhood of point. \(U_p := U_{p^1} \times U_{p^2} \times ... \times U_{p^n} \subseteq N_p\). \(p \in U_p\), and \(U_p\) is an open neighborhood of \(p\), by the definition of product topology.
So, \(N_p\) is a neighborhood of \(p\).
3: Note
For an infinite-product topological space, the logic does not apply, because \(U_p\) would not be any open subset of \(T\), because \(\times_{\alpha \in A} U_{p^\alpha}\) would not be open unless only each of some finite of \(U_{p^\alpha}\) s is different from \(T_\alpha\).
