description/proof of that for finite-product topological space, product of constituent subbases is subbasis
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 subbasis of topological space.
- The reader admits the proposition that for any finite-product topological space, the product of any constituent bases is a basis.
- The reader admits the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets.
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 subbases is a subbasis.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{T_j \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_j\): \(= \text{ the product topological space }\)
\(\{S_j \in \{\text{ the subbases for } T_j\} \vert j \in J\}\):
\(S\): \(= \{\times_{j \in J} s_j \vert s_j \in S_j\}\)
//
Statements:
\(S \in \{\text{ the subbases for } \times_{j \in J} T_j\}\)
//
2: Note
It needs to be a finite-product, because this is based on the proposition that for any finite-product topological space, the product of any constituent bases is a basis.
3: Proof
Whole Strategy: Step 1: see that \(S\) satisfies the conditions to be a subbasis.
Step 1:
For each \(j \in J\), let \(B_j\) be the set of the intersections of the finite subsets of \(S_j\), which is a basis for \(T_j\), by the definition of subbasis for topological space.
\(B := \{\times_{j \in J} b_j \vert b_j \in B_j\}\) is a basis for \(\times_{j \in J} T_j\), by the proposition that for any finite-product topological space, the product of any constituent bases is a basis.
Let us think of the intersection of any finite subset of \(S\), \(\times_{j \in J} s_{1, j} \cap ... \cap \times_{j \in J} s_{n, j}\).
\(\times_{j \in J} s_{1, j} \cap ... \cap \times_{j \in J} s_{n, j} = \times_{j \in J} (s_{1, j} \cap ... \cap s_{n, j})\), by the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets.
For each \(j \in J\), \(s_{1, j} \cap ... \cap s_{n, j}\) is an element of \(B_j\).
So, \(\times_{j \in J} (s_{1, j} \cap ... \cap s_{n, j}) \in B\).
That means that the set of the intersections of the finite subsets of \(S\) is contained in \(B\).
Let \(J = \{1, ..., m\}\) without loss of generality, just for our convenience of expressions.
Each element of \(B\) is \((s_{1, 1} \cap ... \cap s_{n_1, 1}) \times ... \times (s_{1, m} \cap ... \cap s_{n_m, m})\).
Let us take \(n := Max (\{n_1, ..., n_m\})\).
For each \(j \in J\), if \(n_j \lt n\), let \(s_{n_j + 1, j} = ... = s_{n, j} = s_{n_j, j}\).
Then, \((s_{1, 1} \cap ... \cap s_{n_1, 1}) \times ... \times (s_{1, m} \cap ... \cap s_{n_m, m}) = (s_{1, 1} \cap ... \cap s_{n, 1}) \times ... \times (s_{1, m} \cap ... \cap s_{n, m})\).
\(= \times_{j \in J} s_{1, j} \cap ... \cap \times_{j \in J} s_{n, j}\), by the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the sets, which is an element of the set of the intersections of the finite subsets of \(S\).
That means that \(B\) is contained in the set of the intersections of the finite subsets of \(S\).
So, the set of the intersections of the finite subsets of \(S\) is nothing but \(B\), a basis.
So, \(S\) is a subbasis for \(\times_{j \in J} T_j\).
