description/proof of that for finite-product topological space, product of constituent bases is basis
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of product topological space.
- The reader knows a definition of basis of topological space.
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 bases is a basis.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{T_j \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_j\): \(= \text{ the product topological space }\)
\(\{B_j \in \{\text{ the bases for } T_j\} \vert j \in J\}\):
\(B\): \(= \{\times_{j \in J} b_j \vert b_j \in B_j\}\)
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Statements:
\(B \in \{\text{ the bases for } \times_{j \in J} T_j\}\)
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2: Note
It needs to be a finite-product, because otherwise, \(\times_{j \in J} b_j\) would not be open in general, because not only some finite of \(b_j\) s would not be \(T_j\) s.
3: Proof
Whole Strategy: Step 1: see that \(B\) satisfies the conditions to be a basis.
Step 1:
Each \(\times_{j \in J} b_j \in B\) is open on \(\times_{j \in J} T_j\), because each \(b_j \subseteq T_j\) is open and only finite of \(b_j\) s are not \(T_j\) s (there are only finite of \(b_j\) s).
Let \(t \in \times_{j \in J} T_j\) be any.
Let \(N_t \subseteq \times_{j \in J} T_j\) be any neighborhood of \(t\).
There is a \(\times_{j \in J} U_j \subseteq \times_{j \in J} T_j\) such that \(t \in \times_{j \in J} U_j \subseteq N_t\), where \(U_j \subseteq T_j\) is an open subset, by the definition of product topology.
For each \(j \in J\), \(t^j \in U_j\), so, there is a \(b_j \in B_j\) such that \(t^j \in b_j \subseteq U_j\), by the definition of basis for topological space.
\(\times_{j \in J} b_j \in B\) and \(t \in \times_{j \in J} b_j \subseteq \times_{j \in J} U_j \subseteq N_t\).
So, \(B\) is a basis for \(\times_{j \in J} T_j\).
