description/proof of that finite product of 2nd-countable topological spaces is 2nd-countable
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
- 4: Note
Starting Context
- The reader knows a definition of 2nd-countable topological space.
- The reader knows a definition of product topology.
Target Context
- The reader will have a description and a proof of the proposition that the product of any finite number of 2nd-countable topological spaces is 2nd-countable.
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:
\(\{T_1, ..., T_n\}\): \(T_j \in \{\text{ the 2nd-countable topological spaces }\}\)
\(T\): \(= T_1 \times . . . \times T_n\), \(= \text{ the product topological space }\)
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Statements:
\(T \in \{\text{ the 2nd-countable topological spaces }\}\)
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2: Natural Language Description
For any 2nd-countable topological spaces, \(\{T_1, ..., T_n\}\), and the product topological space, \(T := T_1 \times . . . \times T_n\), \(T\) is a 2nd-countable topological space.
3: Proof
Whole Strategy: Step 1: construct countable \(B := \{U_{1, k_1} \times ... \times U_{n, k_n} \vert k_j \in J_j\}\) where \(B_j = \{U_{j, k_j} \vert k_j \in J_j\}\) is a countable basis for \(T_j\); Step 2: see that \(B\) is a basis for \(T\).
Step 1:
Each \(T_j\) has a countable basis, \(B_j = \{U_{j, k_j} \vert k_j \in J_j\}\), where \(J_j\) is a countable index set. \(B := \{U_{1, k_1} \times ... \times U_{n, k_n} \vert k_j \in J_j\}\) is countable.
Step 2:
Let us prove that \(B\) is a basis for \(T\). Each \(U_{1, k_1} \times ... \times U_{n, k_n}\) is open on \(T\), by the definition of product topology. For any neighborhood of any point, \(p = (p^1, ..., p^n) \in T\), \(N_p \subseteq T\), there is an open neighborhood of \(p\), \(U_p \subseteq T\), such that \(U_p \subseteq N_p\), by the definition of neighborhood of point; \(U_p = \cup_{\alpha \in A} U_{1, \alpha} \times ... \times U_{n, \alpha}\) where \(A\) is a possibly uncountable index set and \(U_{j, \alpha} \subseteq T_j\) is an open subset, by Note of the article on the definition of product topology; \(p \in U_{1, \alpha} \times ... \times U_{n, \alpha}\) for an \(\alpha\); there is an element, \(U_{j, k_j}\), of \(B_j\) such that \(p^j \in U_{j, k_j} \subseteq U_{j, \alpha}\); \(U_{1, k_1} \times ... \times U_{n, k_n}\) is an element of \(B\), and \(p \in U_{1, k_1} \times ... \times U_{n, k_n} \subseteq U_{1, \alpha} \times ... \times U_{n, \alpha} \subseteq U_p \subseteq N_p\).
4: Note
The product has to be finite for this proposition, while the product of Hausdorff topological spaces does not need to be finite for the product to be Hausdorff.
